Applied Differential Equations Lecture Notes

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Applied Di↵erential Equations I Brandon Sweeting

August 19, 2022

LATEX Brandon Sweeting

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Applied Di↵erential Equations I

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Overview

1

Chapter 1 - Introduction Section 1.2 - Solutions & Initial Value Problems

LATEX Brandon Sweeting

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Explicit Solutions The general form for an n-th order equation with dependent variable y and independent variable x is: ✓ ◆ dy d ny F x, y , , . . . , n = 0; (1) dx dx which, in many cases, is equivalently expressed as ✓ ◆ d ny dy d n 1y = f x, y , , . . . , n 1 dx n dx dx

(2)

Definition A function (x) that when substituted for y in either equation (1) [or (2)] satisfies the equation for all x in the interval I is called an explicit solution on I .

LATEX Brandon Sweeting

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Explicit Solutions cont. Example Show that (x) = x 2

x

1

is an explicit solution to the linear equation d 2y dx 2

but

2 y = 0, x2

(3)

(x) = x 3 is not.

Example Show that for any choice of the constants c1 and c2 , the function (x) = c1 e

x

+ c2 e 2x

is an explicit solution to the linear equation y 00 Brandon Sweeting

y0

2y = 0.

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(4)

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Implicit Functions Sometimes, methods used to a solve di↵erential equation may fail to yield an explicit solution, i.e. some function of the form (x) = y . We may instead have to settle for a solution that is defined implicitly.

Definition We say that the dependent variable, y , is an implicit function of the independent variable, x, if y satisfies some implicit equation with respect to x, i.e. if G (x, y ) = 0 for some relation G .

Example The function y + x = 2 gives y as an implicit function of x. (in this case G (x, y ) = y + x 2). As x, the independent variable, changes we can see that the value of y so too will change (so as to satisfy the relation). Clearly, y can also be expressed as an explicit function of x, namely y = f (x) where f (x) = 2 x. Note that it may not always be possible to express an implicit function as an explicit one, e.g. y 2 + x 2 = 4. LATEX Brandon Sweeting

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Implicit Di↵erentiation Recall that to di↵erentiate the variable y when given as an implicit function x, we simply di↵erentiate both sides of the relation G (x, y ) = 0 with respect to x. The result will be an equation giving dy dx implicitly as a function of x and y .

Example Use implicit di↵erentiation to find y2

dy dx ,

where y is given by the relation

x 3 + 8 = 0.

Example Use implicit di↵erentiation to find x 2y 3

dy dx ,

where y is given by the relation

xy = 10

LATEX Brandon Sweeting

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Implicit Solutions Definition A relation G (x, y ) = 0 is said to be an implicit solution to a di↵erential equation on the interval I if it defines one or more explicit solutions on I .

Example Show that y 2

x 3 + 8 = 0 is an implicit solution to the nonlinear equation dy 3x 2 = . dx 2y

(5)

Example Show that x + y + e xy = 0 is an implicit solution to the nonlinear equation (1 + xe xy ) Brandon Sweeting

dy + 1 + ye xy = 0. dx

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Family of Solutions As we saw with the example (x) = c1 e x + c2 e 2x , which was a solution for each value of c1 and c2 to the equation y 00 y 0 2y = 0, a single di↵erential equation can have infinitely many solutions:

Example Verify that for every constant C the relation 4x 2 solution to the nonlinear equation y

dy dx

y 2 = C is an implicit

4x = 0.

(7)

Graph the solution curves for C = 0, ±1, ±4. (We call the collection of all such solutions a one-parameter family of solutions.

LATEX Brandon Sweeting

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Constants We’ll see later that solving n-th order di↵erential equations will produce n-arbitrary constants of integration. These constants of integration will result in an n-parameter family of solutions. To specify a unique solution to a di↵erential equation, we must impose initial values on the dependent variable and its first n 1 derivatives.

Second-order Free-fall Observation: An object’s mass times its acceleration equals the force acting on it (Newton’s Second Law). Let h(t) be height of a vertically falling object released from a stationary position at time t. d 2h Di↵erential Equation: m 2 = mg dt To solve this equation, we divide by m and integrate twice: gt 2 + c1 t + c2 . 2 We see we have a two parameter family of solutions. h(t) =

Brandon Sweeting

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Initial Value Problem Definition By an initial value problem for an n-th order di↵erential equation ✓ ◆ dy d ny F x, y , , . . . , n = 0, dx dx we mean: find a solution to the di↵erential equation on an interval I that satisfies at x0 the n initial conditions y (x0 ) = y0 dy (x0 ) = y1 dx .. . d n 1y = yn dx n where x0 2 I and y0 , y1 , . . . , yn Brandon Sweeting

1

1

LATEX

are given constants.

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Initial Value Problem cont. Example Show that (x) = sin(x)

cos(x) is a solution to the initial value problem

d 2y + y = 0; dx 2

y (0) =

1,

dy (0) = 1. dx

Example As shown previously, the function (x) = c1 e d 2y dx 2

dy dx

x

+ c2 e 2x is a solution to

2y = 0

for any choice of the constants c1 and c2 . Determine c1 and c2 so that satisfies the initial conditions y (0) = 2 Brandon Sweeting

and

dy (0) = dx

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Applied Di↵erential Equations I Brandon Sweeting

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LATEX Brandon Sweeting

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Overview

1

Chapter 1 - Introduction Section 1.3 Direction Fields

LATEX Brandon Sweeting

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Direction Fields The existence and uniqueness theorem of Section 1.2 gives us conditions we can check to verify whether a solution to an initial value problem exists and whether this solution is unique, but doesn’t tell us anything about the behavior of solutions. One technique that allows us to visually analyze the behavior of solutions to di↵erential equations and initial value problems is direction fields. Direction fields model solution curves of a di↵erential equation by using the fact that solutions are approximated locally by their tangent lines. Specifically, if the di↵erential equation can be written in the form dy = f (x, y ), dx then we can easily compute the exact value for the slope of the tangent line to the graph of a solution to the di↵erential equation at any (x, y ). All we must do is evaluate f (x, y ). A

LTEX

Brandon Sweeting

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Tangent Lines

Figure: (Wikipedia) A tangent line locally approximating the graph of a function.

Knowing the line tangent to the graph of a function at a point gives us information about the function’s behavior near that point of tangency.

LATEX Brandon Sweeting

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Direction Fields cont When we populate the plane with short line segments at points (x, y ) whose slopes equal f (x, y ) (i.e. create a field of directed line segments), we can start to approximate how solutions to the di↵erential equation dy dx = f (x, y ) behave based on how they “flow” through this field. By tangency, the solutions will “graze” the segments in this field.

LATEX Brandon Sweeting

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Examples

What conclusions can we draw about the solutions to these di↵erential equations based on the given direction fields? Also, can we say anything about a solution to a particular initial value problem (i.e. a solution that LATEX must pass through a specified (x0 , y0 ))? Brandon Sweeting

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Examples The logistic equation for the population p (in thousands) at time t of a certain species is given by dp p). dt = p(2

From the direction field sketched in Figure 1.10, answer the following: 1 If the initial population is 3000 what can you say about lim t!1 p(t)? 2 Can a population of 1000 ever increase to 3000? LATEX 3 Can a population of 1000 ever decline to 500? Brandon Sweeting

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Method of Isoclines One method for generating a direction field for a di↵erential equation dy = f (x, y ) dx if via the use of isoclines. An isocline is a curve in the x, y plane along which the solutions of the di↵erential equation have the same slope; i.e. an isocline is the level set of the function f (x, y ). Rather than haphazardly sampling points in the plane and computing f (x, y ) at each of these points (which may be di↵erent for each point). We can find the level set for one value of f (x, y ), say C , and then populate as many points as we’d like along this curve with tangent segments of slope C .

LATEX Brandon Sweeting

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Method of Isoclines cont. Method of Isoclines 1

Fix some real number C .

2

Draw the curve expressed by f (x, y ) = C in plane.

3

At points along the curve from step 2, draw small ticks with slope C .

Figure: Method of Isoclines to find direction field for Brandon Sweeting

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Applied Di↵erential Equations I

dy dy

=x +y

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Applied Di↵erential Equations I Brandon Sweeting

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LATEX Brandon Sweeting

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Overview

1

Chapter 1 - Introduction Section 1.4 The Approximation Method of Euler

LATEX Brandon Sweeting

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Solution Approximations In Section 1.3, given a first order di↵erential equation of the form dy = f (x, y ), dx we were able to visualize and examine the behavior of its solutions via the use of direction fields. The guiding principle was that knowing the tangent lines of solutions at many points allows us to model the behavior of solutions collectively. However, when it cames to analyzing the solution of a particular initial value problem (IVP): dy = f (x, y ); dx

y (x0 ) = y0 ,

i.e. finding the exact solution curve that passes through the point (x0 , y0 ), this method had it’s shortcomings. LATEX Brandon Sweeting

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Direction Field Shortcomings

In the direction field above, we can see that if we were to try to trace the solution curve passing through a particular point some ambiguity arises. In particular, it’s not clear how the solution would behave once it hits the x-axis. The solution could pass right over the axis and continue on its way, or travel horizontally for some distance. Both possibilities abide by the LATEX direction field but both may not be solutions. Brandon Sweeting

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Three Initial Techniques Euler’s approximation method (a.k.a Euler’s method), which also relies on the principle of tangent line approximation, addresses this issue as it allows us to numerically estimate the solution to a particular initial value problem. With this method, we will have concluded three techniques for analyzing solutions to di↵erential equations and initial value problems: 1

Existence & Uniqueness Theorem: See if an IVP solution exists

2

Direction Fields - Visualize the collective behaviour of solutions

3

Euler’s Method - Numerically approximate particular IVP solution

We will later be dealing with techniques that will allow us to solve for solutions to di↵erential equations and IVPs exactly.

LATEX Brandon Sweeting

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Euler’s Approximation Method Suppose we want to know the behavior of the solution to the IVP dy = f (x, y ); dx

y (x0 ) = y0 .

(1)

Without any further information, we at least know two things: 1

The graph of the solution must pass through the point (x0 , y0 )

2

The solution is locally approximated by its tangent line at any (x, y )

Therefore, if (x) is the solution to (1), then at the point (x0 , y0 ) we have: 1

(x0 ) = y0

2

(x) ⇡ y0 +

0 (x

0 )(x

x0 ), for x ⇡ x0

LATEX Brandon Sweeting

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Euler’s Approximation Method cont. Since (x) is a solution to the IVP, we have 0 (x) = f (x, (x)) and we can quantify x being close to x0 by saying x = x0 + h for some small h > 0. Therefore, these two conditions on the solution (x) at (x0 , y0 ) become: 1

(x0 ) = y0

2

(x) ⇡ y0 + hf (x0 , y0 ), for h > 0 small

We can compute y0 + hf (x0 , y0 ) explicitly since f (x, y ) is given as part of the IVP. Therefore, if we fix some small h > 0, we then have a concrete approximation of our solution by some linear function near the initial value, i.e. for those x in the interval (x0 , x0 + h). The smaller h is, the better our linear approximation will be; however, the interval on which our approximation holds, i.e. (x0 , x0 + h), will be smaller.

LATEX Brandon Sweeting

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Euler’s Approximation Method cont. We can take this idea further and repeat this procedure at the endpoint of this interval, i.e. at x0 + h since (x0 + h) ⇡ y0 + hf (x0 , y0 ) and (x) is approximated by a tangent line at every point on its graph. Let x1 = x0 + h and y1 = y0 + hf (x0 , y0 ). Then (x) ⇡ y1 + f (x1 , y1 )(x

x1 ), for x 2 (x1 , x1 + h).

We can set x2 = x1 + h, y2 = y1 + hf (x1 , y1 ) and repeat this procedure yet again at (x2 , y2 ) to approximate (x) on the interval (x2 , x2 + h). All together, after iterating this procedure n times, this will produce an approximation of (x) by some polygonal-line on the interval (x0 , xn ) where xn = x0 + nh. This procedure is called Euler’s Method. The number h that we fix for this procedure is called the step size. A

LTEX

Brandon Sweeting

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Euler’s Approximation Method cont.

LATEX Brandon Sweeting

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Euler’s Method Formula

When wanting to find the values that the approximation gives only at the nodes of the iterations, i.e. at the points x0 , x1 = x0 + h, ..., xn = x0 + nh (as is the case for your homework problems) we can use the following recursive formulas:

Euler’s Approximation Method xn+1 = xn + h, yn+1 = yn + hf (xn , yn ),

(2) n = 0, 1, 2, ....

(3)

Note that yn is the value of the approximation for the solution (x) at xn .

LATEX Brandon Sweeting

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Examples Example Use Euler’s method with step size h = 0.1 to approximate the solution to the initial value problem p y0 = x y,

y (1) = 4

at the points x = 1.1, 1.2, 1.3, 1.4, and 1.5. Since the step size is 0.1 and x0 = 1, we may simply use the recursive formula. This is because all points of interest for the approximation are equal to x0 plus some multiple of the step size. More specifically, if we iterate the formula 5 times, then we’ll obtain the value the approximation gives at the points x1 = x0 + 0.1 = 1.1, x2 = x1 + 0.1 = 1.2 and so on.

LATEX Brandon Sweeting

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Example cont. p For this equation, we have x0 = 1, y0 = 4 and f (x, y ) = x y . Therefore, recalling the recursive formula xn+1 = xn + h, yn+1 = yn + hf (xn , yn ),

(4) n = 0, 1, 2, ....

(5)

we have n 0 1 2 3 4 5

xn 1 1.1 1.2 1.3 1.4 1.5

Euler’s Method (i.e. yn ) 4 p 4 + 0.1( p 4) = 4.2 4.2 + 0.1(p4.2) = 4.42543 4.42543 + 0.1(p4.42543) = 4.67787 4.67787 + 0.1(p4.67787) = 4.95904 4.95904 + 0.1( 4.95904) = 5.27081

Exact Value 4 4.21276 4.45210 4.71976 5.01760 5.34766

LATEX Brandon Sweeting

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Examples cont.

Example Use Euler’s method with step size h = 0.1 to approximate the solution to the initial value problem y0 = 1

sin(y ),

y (0) = 0

at the points x = 1.1, 1.2, 1.3, 1.4, and 1.5. For the exact same reason, we may apply the recursive formula as we had in the previous problem to obtain an approximation via Euler’s method at these values.

LATEX Brandon Sweeting

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Example cont. For this equation, we have x0 = 0, y0 = 0 and f (x, y ) = 1 Therefore, recalling the recursive formula

sin(y ).

xn+1 = xn + h, yn+1 = yn + hf (xn , yn ),

(6) n = 0, 1, 2, ....

(7)

we have n 0 1 2 3 4 5

xn 0 0.1 0.2 0.3 0.4 0.5

0.1 0.19002 0.27113 0.34435

Euler’s Method (i.e. yn ) 0 0 + 0.1(1-sin(0)) = 0.1 + 0.1(1-sin(0.1)) = 0.19002 + 0.1(1-sin(0.19002)) = 0.27113 + 0.1(1-sin(0.27113)) = 0.34435 + 0.1(1-sin(0.34435)) = 0.41059

Exact Value 0 0.09517 0.18132 0.25941 0.33029 0.39479

LATEX Brandon Sweeting

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Examples cont. Example Use Euler’s method to approximate the solution to the initial value problem y0 = 1

sin(y ),

y (0) = 0

at x = ⇡, taking n = 1, 2, 4, and 8 steps. To solve this problem, we just find the required step size for each value of n. The step size must be chosen such that xn = ⇡, i.e. after n iterations we’ll end up at the x value we want the approximation at. For the first case, we require that x1 = x0 + h so that h = pi. In the second case, we require that x2 = ⇡; since xn = x0 + nh, we have require that h = ⇡2 . Once we’ve identified the required step size, we proceed as we had in the previous problems; i.e. for each value of n, we use the step size noted above in the recursive formula, which we iterate n items (yn will be the LATEX desired value for our approximation in n steps). Brandon Sweeting

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Applied Di↵erential Equations I Brandon Sweeting

September 2, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 2 - First-Order Di↵erential Equations Section 2.2 Separable Equations

LATEX Brandon Sweeting

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Separable Equations A simple class of first-order equations that can be solved using integration is the class of separable equations. These are equations that can be rewritten so as to isolate the variables x and y (along with their di↵erentials dx and dy ) on opposite sides of the equation. That is to say, the di↵erential equation dy = f (x, y ), dx is separable if we can rearrange the equation to look like h(y )dy = g (x)dx for some function h of y only and g of x only. To rearrange the equation, we can manipulate the di↵erentials dx and dy algebraically as usual.

LATEX Brandon Sweeting

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Separable Equations cont. Equivalently, a di↵erential equation is separable if it can be written as dy = g (x)p(y ), dx in which case p(y ) = 1/h(y ) for the function h of the previous slide.

Example The di↵erential equation

is separable since

dy 2x + xy = 2 dx y +1 2x + xy 2+y =x 2 = g (x)p(y ). 2 y +1 y +1

However, the following di↵erential equation is not dy = 1 + xy . dx Brandon Sweeting

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Identifying Separable Equations Example Determine whether the given di↵erential equations are separable dy 1 sin(x + y ) = 0 dx 2

dy = 4y 2 dx

3

ds = t ln(s 2t ) + 8t 2 dt

4

dy ye x+y = 2 dx x +2

5

s2 +

3y + 1

ds s +1 = dt st

LATEX Brandon Sweeting

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Solving Separable Equations To solve the equation

dy = g (x)p(y ) dx multiply by dx and by h(y ) := 1/p(y ) to obtain

(1)

h(y )dy = g (x)dx. Then integrate both sides (wrt their respective variables): Z Z h(y ) dy = g (x) dx, to obtain H(y ) = G (x) + C ,

(2)

where both constants of integration are absorbed into C . The last equation gives an implicit solution to the di↵erential equation. Brandon Sweeting

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Solving Separable Equations cont.

Example Solve the nonlinear equation dy x 5 = dx y2

Example Solve the initial value problem dy y 1 = , dx x +3

y ( 1) = 0.

LATEX Brandon Sweeting

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Solving Separable Equations cont.

Example Solve the nonlinear equation dy 6x 5 2x + 1 = . dx cos(y ) + e y

Example Solve the initial value problem 1 dy y sin(✓) = 2 , ✓ d✓ y +1

y (⇡) = 1.

LATEX Brandon Sweeting

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Solving Separable Equations cont.

Example Solve the nonlinear equation x

dv 1 4v 2 = . dx 2v

Example Solve the initial value problem dy = 8x 3 e dx

2y

,

y (1) = 0.

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Pitfalls of Method When performing algebraic manipulations on equalities, we may lose track of information when dividing by values that may be 0. For example, x(x

2) = 4(x

2)

has two solutions: x = 2 and x = 4, but if we rewrote this equation as x = 4 by dividing by (x 2) we lose track of the root x = 2. The same issue can arise when solving separable equations by our method. We may end up disregarding constant solutions when dividing by p(y ). Take the example equation from before: dy y 1 = . dx x +3 Then y = 1 is a solution to this equation but it doesn’t show up if the family of solutions we obtained. To address this issue, we must check for the roots of p(y ) beforehand as these will define our constant solutions.LATEX Brandon Sweeting

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Mathematical Justification We re-frame our method of solving separable equations as an application of the chain rule. If a first-order di↵erential equation is separable, then it may be written in the form h(y )

dy = g (x). dx

(3)

If H(y ) and G (x) denote the primitives of h(y ) and g (x), respectively, i.e. H 0 (y ) = h(y ),

G 0 (x) = g (x),

we can rewrite (3) as dy = G 0 (x). dx We recall H(y ) = H(y (x)), i.e. H is a function of x through composition. H 0 (y )

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Mathematical Justification cont. Therefore, by the chain rule, we have d dy H(y (x)) = H 0 (y (x)) , dx dx thus we can rewrite (3) as d d H(y (x)) = G (x). dx dx So H(y (x)) and G (x) are two functions of x with the same derivative. Therefore, they must di↵er by only a constant, i.e. H(y (x)) = G (x) + C .

(4)

Equation (4) is equivalent to (3) (i.e. if y satisfies one then it satisfies the other), and gives the solution function y implicitly up to knowledge of H and G , which may be explicitly computed via integration. LATEX Brandon Sweeting

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Applied Di↵erential Equations I Brandon Sweeting

September 6, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 2 - First-Order Di↵erential Equations Section 2.3 Linear Equations

LATEX Brandon Sweeting

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Linear Equations Recall from Section 1.1 that a linear first-order equation is an equation that can be expressed in the form: a1 (x)

dy + a0 (x)y = b(x), dx

(1)

where a1 (x), a0 (x) and b(x) are functions of the variable x only.

Example The equation x 2 sin(x)

(cos x)y = (sin x)

dy dx

is linear; whereas the equation y

dy + (sin x)y 3 = e x + 1 dx

is not. Brandon Sweeting

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Simple Solutions There are two situations in which the solution of a linear equation a1 (x)

dy + a0 (x)y = b(x) dx

can be readily found. The first is when a0 (x) = 0, in which case we have a1 (x)

dy = b(x). dx

Therefore, provided a1 (x) 6= 0, we may solve by integration: Z b(x) y (x) = dx + C . a1 (x)

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Simple Solutions The second situations arises when product rule, we have a1 (x)

d dx a1 (x)

= a0 (x). In this case, by the

dy d + a0 (x)y = (a1 (x)y ) = b(x). dx dx

Integrating the above equation gives, Z a1 (x)y = b(x) dx + C . Therefore, provided a1 (x) 6= 0, we have Z 1 y (x) = b(x) dx + C . a1 (x)

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General Solution

It’s rarely the case that a0 (x) = 0 for a given first-order linear equation. However, it is possible to transform any first-order linear equation by multiplication with a well-chosen function µ(x) such that this new equation has the form d (a1 (x)y ) = b(x). dx We call such a function µ(x) an integrating factor. An integrating factor, in most general terms, is a function that is chose/construct to facilitate the solving of a given di↵erential equation.

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Standard Form

How do we find this function µ(x)? Let’s first transform our equation a1 (x)

dy + a0 (x)y = b(x) dx

by dividing through by a1 (x) to obtain the equivalent equation dy + P(x)y = Q(x), dx where we have P(x) = a0 (x)/a1 (x) and Q(x) = b(x)/a1 (x). We call this the standard form for first-order linear di↵erential equations.

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Special Function Given a first-order linear di↵erential equation in standard form dy + P(x)y = Q(x), dx we want µ(x) such that the left-hand side of the multiplied equation µ(x)

dy + µ(x)P(x)y = µ(x)Q(x) dx

is just the derivative of the product µ(x)y , i.e. µ(x)

dy d dy + µ(x)P(x)y = [µ(x)y ] = µ(x) + µ0 (x)y . dx dx dx

For this happen, we clearly need that µ0 (x) = µ(x)P(x).

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Special Function cont.

Therefore, we require that µ(x) satisfy the separable di↵erential equation y 0 = yP(x). Using our method for solving separating equations, we have µ(x) = e

R

P(x) dx

.

We can disregard the constant of integration (as µ(x) isn’t unique).

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Finishing up With this choice of µ(x), our equation µ(x)

dy + µ(x)P(x)y = µ(x)Q(x) dx

becomes

d [µ(x)y ] = µ(x)Q(x), dx which, after integrating and dividing by µ(x), we see has the solution Z 1 y (x) = µ(x)Q(x) dx + C . µ(x) We call such a solution (which involves the constant of integration in its expression) a general solution to the di↵erential equation.

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Solving Linear Equations Summary 1

Write the equation in the standard form dy + P(x)y = Q(x) dx

2

Calculate the integrating factor µ(x) by the formula µ(x) = e

3

P(x) dx

Multiply the equation in standard form by µ(x) and, recalling the d LHS is just dx [µ(x)y ] since µ(x) was specially chosen, obtain µ(x)

4

R

dy d + µ(x)P(x) = [µ(x)y ] = µ(x)Q(x). dx dx

Integrate the last equation and solve for y to obtain solution Z 1 y (x) = µ(x)Q(x) dx + C . µ(x) Brandon Sweeting

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LATEX 11 / 15

Solving Linear Equations

Example Find the general solutions to 1 dy x dx

2y = x cos(x), x2

x > 0.

Example Find the general solutions to y

dx + 2x = 5y 3 . dy

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Solving Linear Equations cont.

Example Solve the initial value problem dy dx

y = xe x , x

y (1) = e

1.

Example Solve the initial value problem dy + 4y dx

e

x

= 0,

4 y (0) = . 3

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Solving Linear Equations cont.

Example Find the general solutions to (x 2 + 1)

dy + xy dx

x = 0.

Example Solve the initial value problem t2

dx + 3tx = t 4 ln(t) + 1, dt

x(1) = 0.

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Identifying Equations Example Determine whether the given equation is separable, linear, neither or both. dy 1 x2 + sin(x) y = 0 dx dx 2 + xt = e x dt dy 3 (t 2 + 1) = yt y dt dy 4 3t = e t + y ln(t) dt dx 5 x + t 2 x = sin(t) dt dr 6 3r = ✓3 d✓

LATEX

Brandon Sweeting

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Applied Di↵erential Equations I Brandon Sweeting

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LATEX Brandon Sweeting

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Overview

1

Chapter 2 - First-Order Di↵erential Equations Section 2.6 Substitutions and Transformations

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Substitutions and Transformations

As we’ve seen, many equations fail to be separable or linear, e.g. dy y =2 dx x

x 2y 2.

However, when it comes to solving such equations all hope is not lost. In certain situations, we can transform an equation, via a substitution of variables or other transformation, so that our new equation reduces to a simpler form we know how to solve (i.e. a linear or separable equation). This is quite similar to the use of u-substitution for simplifying complicated integrals into one’s we know how to compute.

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Bernoulli Equations

Bernoulli Equation A first-order equation that can be written in the form dy + P(x)y = Q(x)y n , dx

(1)

where P(x) and Q(x) are continuous on an interval (a, b) and n is a real number, is called a Bernoulli equation. Notice that when n = 0 or n = 1, equation (1) is also a linear equation and can be solved by the method discussed in Section 2.3. However, this is certainly not the case for other values of n.

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Bernoulli Equations cont. For all other values of n, we can perform the substitution v = y1

n

to transform our Bernoulli equation into a linear equation we can solve. Indeed, suppose we’re given a Bernoulli equation dy + P(x)y = Q(x)y n . dx Then dividing both sides of this equation by y n gives y

n dy

dx

+ P(x)y 1

n

= Q(x).

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Bernoulli Equations cont. Taking v = y 1

n,

we find via the chain rule that dv = (1 dx

n dy

n)y

dx

,

so that our transformed equation y now becomes

n dy

dx 1

1

+ P(x)y 1

n

= Q(x),

dv + P(x)v = Q(x), n dx

which is indeed linear in the variable v (since the value 1 1 n is constant). Note that both P(x) and Q(x) remain unchanged after the transformation.

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Bernoulli Equations Examples Example Use the method discussed under “Bernoulli Equations” to solve the following di↵erential equations dy 5 3 1 5y = xy dx 2 2

dy y + = x 2y 2 dx x

3

dy dx

4

dy 2y = dx x

5

dx x + tx 3 + = 0 dt t

y = e 2x y 3

Brandon Sweeting

x 2y 2

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LATEX Brandon Sweeting

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Overview

1

Chapter 3: Mathematical Models and Numerical Methods Involving First-Order Equations Section 3.2 - Compartmental Analysis

LATEX Brandon Sweeting

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Mathematical Modeling

The process of mimicking reality by using the language of mathematics is called mathematical modeling. Formulating problems in mathematical terms allows us to place complicated real-world problems into a concrete and unbiased context.

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Compartmental Analysis Many complicated processes can be broken down into distinct stages and the entire system modeled by describing the interactions between the various stages. Such systems are called compartmental and are graphically depicted by block diagrams. The basic one-compartment system consists of a function x(t) which gives the amount of a substance in the compartment at time t, an input rate at which the substance enters the compartment, and an output rate at which the substance leaves the compartment.

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Compartmental Analysis cont.

Since the derivative of x with respect to t represents the rate of change in the substance in the compartment with respect to time, we have the following relationship: dx = input rate dt

output rate

In the models that we’ll build using compartmental analysis, we’ll construct this di↵erential equation by determining those functions of the independent variable which comprise the input and output quantities of the substance into our compartment.

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Mixing Problems

A problem for which the one-compartment system provides a useful representation is the mixing of fluids in a tank. More specifically, we seek to model the amount of some dissolved substance contained in a tank of fluid (e.g. salt) as fluid enters the tank (of a certain concentration), mixes with the fluid in the tank, then leaves the tank. In what follows, we will develop a general process for constructing a model that will give the amount of substance within the tank at any time t. Here, when using compartmental analysis, our compartment is an actual compartment, i.e. the tank itself, and the function we are interested in modeling, i.e. x(t), is the amount of the desired substance that is dissolved in the fluid within the tank at time t.

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Mixing Problems cont.

To setup the di↵erential equation that represents our mixing problem, we must determine the input rate and output rate of the substance: dx = input rate dt

output rate

The input rate of the substance can be derived from the rate of fluid entering the tank (volume/time) and concentration (amount/volume). Multiplying these together yields the substance input rate (amount/time).

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Mixing Problems cont.

The output rate requires an additional assumption that the substance is dissolved uniformly in the fluid within the tank. This is an example of an idealization (a simplification of reality to make it easier to find a model). With this, we have that the concentration of the fluid leaving the tank at any time t is given by dividing x(t) by the volume (amount/volume). We multiply this by the output rate of fluid leaving the tank (volume/time) to get the output rate of the substance (amount/time).

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Mixing Problems Example Example Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to flow at a constant rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a rate of 6 L/min. If the concentration of the salt in the brine entering the tank is 0.1 kg/L, determine when the concentration of salt in the tank will reach 0.05 kg/L.

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Mixing Problems Example cont. To solve this problem we’ll find the function x(t) which gives the amount of salt dissolved in the brine solution of the tank (in kilograms) at any time t (in minutes). We’ll then find t such that x(t) = 0.05. We have dx = input rate dt

output rate

The units of this equation, as suggested by the units of the problem, will be kilograms for the amount of substance and minutes for time. The input rate, as mentioned, is given by finding the amount of substance entering the tank in kilograms/minute. We know the concentration of salt in the brine solution entering the tank: 0.1 kg/L. We also know the flow rate of the brine solution entering the tank: 6 L/min. Multiplying these quantities (and their units) together gives the input rate of salt entering the tank: (6 L/min)(0.1 kg/L) = 0.6 kg/min.

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Mixing Problems Example cont. With this information, keeping in mind our units, our equation becomes dx = 0.6 dt

output rate

The output rate, as mentioned, is found by assuming uniform concentration of the solution in the tank. The output rate is then given by multiplying this quantity by the flow rate of the solution leaving the tank. Since the amount of substance in the tank at time t is given by x(t) and the volume of solution in the tank is constant in time (since the input and output rates of liquid are the same 6 L/min) the concentration of solution in the tank at time t is x(t)/1000 kg/L. Multiplying this by the flow rate of solution leaving the tank, we have that the total amount of salt leaving the tank is ✓ ◆ x(t) 3x(t) (6 L/min) kg/L = kg/min. 1000 500 LATEX Brandon Sweeting

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Mixing Problems Example cont. In all, this gives the di↵erential equation, dx = 0.6 dt

3x , 500

which is both separable and linear. To find an explicit solution, we must add some initial condition. Since the liquid in the tank is pure water at the start, we have that x(0) = 0. So to find our desired function x(t), we must solve the initial value problem dx = 0.6 dt

3x , 500

x(0) = 0.

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Mixing Problems Example cont. The IVP we derived from our compartmental analysis dx = 0.6 dt

3x , 500

x(0) = 0,

has as its solution x(t) = 100(1

e

3t/500

).

Therefore, we seek the t such that x(t) = 0.1(1 1000

e

3t/500

) = 0.05;

hence,

500 ln(2) ⇡ 115.52 min. 3 That is to say, after 115.52 minutes the concentration of salt in our tank will reach 0.05 kg/L. LATEX t=

Brandon Sweeting

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Mixing Problems Examples Example Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to flow at a constant rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a rate of 5 L/min. If the concentration of the salt in the brine entering the tank is 0.1 kg/L, determine the amount of salt in the tank after t minutes.

Example A nitric acid solution flows at a constant rate of 6 L/min into a large tank that initially held 200L of a 0.5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 8 L/min. If the solution entering the tank is 20% nitric acid, determine the volume of nitric acid in the tank after t minutes. When will the percentage of nitric acid in the tank reach 10%?

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Population Models We can apply our compartmental analysis to model the changing of some population of organisms (e.g. bacteria, humans, etc.). The compartment in this case would be either a petri dish, an island, a country, etc., the substance changing would be the number of organisms (i.e. population); lastly, the input would be those factors influencing reproduction and the output would be those factors influencing the death of these organisms. It’s reasonable to assume that these latter two quantities are proportional to the size of the population at any given time t, i.e. if p(t) denotes the population at time t and p0 the initial population at time t = 0, we have, for some proportionality constants k1 , k2 > 0, the following IVP:

where k := k1

dp = k1 p k2 p = kp, p(0) = p0 dt k2 and k2 is the death rate proportionality constant.

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Malthusian for Population Growth

Solving this initial value problem, we obtain the function p(t) = p0 e kt . We call this the Malthusian or exponential law of population growth.

Example In 1790 the population of the United States was 3.93 million, and in 1890 it was 62.98 million. Using the Malthusian model, estimate the U.S. population as a function of time.

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Population Model Comparison The Malthusian model holds for some time but eventually overshoots.

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Logistic Model

To better our model, we can improve our estimate of the death rate by considering deaths caused by two-party interactions (e.g. deaths caused by violent crimes, communicable diseases, resource inequality etc.). For a population of p individuals there are p(p 1)/2 possible two-party interactions. Therefore, for some third proportionality constant k3 > 0, we have the new and improved model given by the IVP dp = k1 p dt

k3

p(p

1) 2

=

Ap(p

p1 ),

p(0) = p0

where A = k3 /2 > 0 and p1 = (2k1 /k3 ) + 1 > 0.

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Logistic Model cont. The equation we just derived dp = dt

Ap(p

p1 ),

p(0) = p0

where A, p1 > 0, is separable and has as its solution p(t) =

p0 p1 p0 + (p1 p0 )e

Ap1 t

.

Such functions are called logistic functions. Below are its possible graphs

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Logistic Model Example

Example Taking the 1790 population of 3.93 million as the initial population and given the 1840 and 1890 populations of 17.07 and 62.98 million, respectively, use the logistic model to estimate the population at time t. To solve this problem, we simply take the general logistic model we just derived for population growth and determine the auxiliary constants A and p1 based on the data given (i.e. we’re fitting our model to the given data). When fitting our data, we’ll end up with two nonlinear equations in the variables A and p1 that we can then solve via numerical techniques.

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Overview

1

Chapter 3: Mathematical Models and Numerical Methods Involving First-Order Equations Section 3.3 - Heating and Cooling of Buildings

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Problem Setup In this section, we’re interested in modeling the heat inside a building as a function of the outside temperature, the heat generated inside the building and the furnace heating or air conditioner cooling within the building. From this model, we’d like to answer the following three questions: How long does it take to change the building temperature substantially? How does the building temperature vary during spring and fall when there is no furnace heating or air condition cooling? How does the building temperature vary in the summer when there is air conditioning or in the winter when there is furnace heating?

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Compartmental Analysis This model is perfectly suited to the compartmental analysis that we established in Section 3.2. The quantity we’re interested in measuring is the heat within the building at any given time t, say T (t), as measured by the temperature. Our compartment is the building itself and among the factors we’re considering that influence the temperature in the building, we have: U(t) - The heating/cooling provided by the furnace or air conditioner. H(t) - The heating provided by people/machines inside the building. M(t) - The temperature of the building’s surroundings. How do these components combine to give us our di↵erential equation: dT = input rate dt

output rate.

expressing the change in temperature with respect to time? Brandon Sweeting

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Compartmental Analysis cont. We have that the heating produced by the people/machines in the building, H(t), and the heating/cooling provided by the furnace or air conditioner, U(t), are themselves given as rates of change in temperature (BTUs x heat capacity). So they factor directly into the equation for dT dt . As for the outside temperature, this will a↵ect dT dt via Newton’s law of cooling. This law states that the change in temperature of a body is proportional to the di↵erence between it and its surroundings. In all, dT = K [M(t) T (t)] + H(t) + U(t). dt where K > 0 is some constant of proportionality (dependant on things like the building’s materials, sun exposure, outside humidity etc.). Here, U(t) will either be positive or negative depending on whether its the furnace or AC that’s operating.

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Visual Depiction of Situation

Figure: A depiction of the factors influencing temperature of building at timeAt. LT

EX

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Solving for T (t)

The di↵erential equation we just derived dT = K [M(t) dt

T (t)] + H(t) + U(t).

is first-order linear and can therefore be solved by using an integrating factor. This will yield the following general temperature model: ⇢Z Kt T (t) = e e Kt [KM(t) + H(t) + U(t)] dt + C , whose components we may simply tweak for each problem we consider.

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Example Example Suppose at the end of the day (at time t0 ), when people leave the building, the outside temperature stays constant M0 , the additional heating rate H inside the building is zero, and the furnace/air conditioner rate U is zero. Determine T (t), given the initial condition T (t0 ) = T0 . In this situation, we have H(t) = 0, U(t) = 0 and M(t) = M0 ; thus ⇢Z Kt T (t) = e e Kt KM0 dt + C = M0 + Ce

Kt

By the initial condition, we have T0 = M0 + Ce T (t) = M0 + (T0

M0 )e

Kt0

so that

K (t t0 )

.

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Further Analysis We now address one of the questions we posed earlier: How long does it take to change the the building temperature substantially? If M0 < T0 , the temperature function T (t) = M0 + (T0 M0 )e K (t t0 ) will decay exponentially fast to the outside temperature M0 as t ! 1. Notice that, regardless of the exact values of M0 and T0 , equilibrium is reached when t = K1 . We call this quantity the time constant for the building (without heating or air conditioning). This value is often easier to observe directly through experimentation rather than estimating K . Therefore, to provide an explicit answer to this question, the building temperature changes exponentially with a time constant of 1/K ; that is to say, the building temperature will match that of its surroundings after an elapsed time of 1/K .

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Problem 1

Example On a hot Saturday morning while people are working inside, the air conditioner keeps the temperature inside the building at 24°C. At noon the air conditioner is turned o↵, and the employees go home. The temperature outside is a constant 35°C the rest of the afternoon. If the time constant for the building is 4 hr, what will be the temperature inside the building at 2:00 p.m.? At 6:00 p.m.? When will the temperature inside the building reach 27°C?

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Problems cont.

Example It was noon on a cold December day in Tampa: 16°C. Detective Taylor arrived at the crime scene to find the sergeant leaning over the body. The sergeant said there were several suspects. If they knew the exact time of death, then they could narrow the list. The Detective took a thermometer and measured the temperature of the body: 34.5°C. He then left for lunch. Upon returning at 1:00 p.m., he measured the body temperature to be 33.7°C. When did the murder occur? [Hint: Normal body temp. is 37°C.]

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Overview

1

Chapter 4: Linear Second-Order Equations Section 4.2 - Homogeneous Linear Equations: The General Solution

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Linear Second-Order Equations

We will begin our study of linear second-order di↵erential equations, a(t)y 00 + b(t)y 0 + c(t)y = f (t),

a(t) 6= 0,

with the special case where the coefficients are constant and f (t) = 0, i.e. ay 00 + by 0 + cy = 0,

a 6= 0.

Such equations arise in the study of mass-spring oscillators that vibrate freely and are called homogeneous (since f (t) = 0).

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Towards a Solution Second-order homogeneous linear equations with constant coefficients ay 00 + by 0 + cy = 0,

a 6= 0

(1)

imply that the second derivative of the solution y is expressible as a linear combination of its first two derivatives (zeroth and first). This suggests a solution of the form y = e rt for some constant r . Indeed, plugging this function into equation (1) we have e rt (ar 2 + br + c) = 0. Dividing this equation by e rt , which is always non-zero, gives: ar 2 + br + c = 0.

(2)

Consequently, y = e rt is a solution provided r solves equation (2), which we call the auxiliary equation (or characteristic equation) of the ODE. LATEX Brandon Sweeting

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Characteristic Equation The characteristic equation ar 2 + br + c = 0 is a quadratic equation whose roots are given by p b + b 2 4ac b r1 = and r2 = 2a

p

b2 2a

4ac

.

Therefore, y = e r1 t and y = e r2 t are the two solutions of our given form. When the discriminant, b 2 4ac, is positive, the roots r1 and r2 are real and distinct. If the discriminant is zero then the roots are real and identical. Otherwise, the roots are complex (Section 4.3).

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Existence and Uniqueness How do we know the solutions we derived by our initial guess are the only possible solutions to the di↵erential equation? There exists an analog of the Existence and Uniqueness theorem from Section 1.2:

Existence and Uniqueness: Homogeneous Case For any real numbers a (6= 0), b, c, t0 , Y0 and Y1 , there exists a unique solution to the initial value problem: ay 00 + by 0 + cy = 0;

y (t0 ) = Y0 ,

y 0 (t0 ) = Y1 .

So whatever solution we find to an IVP will be unique. We will now consider problems dependent on the value of the discriminant of the underlying auxiliary equation (Section 4.3 gives the complex case).

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Positive Discriminant, b2

4ac > 0

In such cases, the general solution is given by a linear combination of the two solutions we derived, i.e. y = C1 e r1 t + C2 e r2 t .

Example Find a pair of solutions to y 00 + 5y 0 + 6y = 0.

Example Solve the initial value problem y 00 + 2y 0

y = 0;

y (0) = 0,

y 0 (0) =

1.

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Zero Discriminant, b2

4ac = 0

In such cases, the general solution is given by a linear combination of e rt and te rt , which is also a solution, i.e. y = C1 e rt + C2 te rt . Why is y = te rt a solution? Plugging this function into the equation gives: a [r (1 + rt) + r ] e rt + b [1 + rt] e rt + cte rt = 0. Dividing this equation by e rt and rearranging gives ⇥ ⇤ [2ar + b] + ar 2 + br + c = 0.

Recall that r = 2ab is a root of the auxiliary equation ar 2 + br + c = 0. Therefore, the above equation and thus y = te rt is a solution.

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Zero Discriminant cont.

Example Find a pair of solutions to y 00 + 6y 0 + 9y = 0.

Example Solve the initial value problem y 00 + 4y 0 + 4y = 0;

y (0) = 1,

y 0 (0) = 3.

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Further Problems Example Find a general solution to the given di↵erential equations 1

2y 00 + 7y 0

2

y 00

3

6y 00

+

4y = 0

8y 0

+ 16y = 0

y0

2y = 0

+

Example Solve the given initial value problem 1

y 00 + 2y 0

2

y 00

3

y 00

+

y0

+

2y 0

8y = 0;

= 0;

y (0) = 3,

y (0) = 2,

+ y = 0;

y 0 (0)

y (0) = 1,

y 0 (0) =

12

=1 y 0 (0) =

3

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Overview

1

Chapter 4: Linear Second-Order Equations Section 4.3 - Auxiliary Equations with Complex Roots

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Recap In Section 4.2, we discussed second-order, homogeneous linear equations with constant coefficients ay 00 + by 0 + cy = 0.

(1)

We conjectured a solution of the form y = e rt , which we found was a solution to (1) provided r was a root to of the auxiliary equation ar 2 + br + c = 0. We considered the case when this equation has two distinct roots and a repeated real root and derived the general solution in each case. We will now consider what happens when the roots of this equation are complex, i.e. when the discriminant of the equation, b 2 4ac, is negative.

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Complex Case If b 2

4ac < 0, then ar 2 + br + c = 0 has two complex solutions: r1 = ↵ + i ,

where ↵ and

and

r2 = ↵

i

are the real numbers ↵=

b , 2a

and

=

p

4ac b 2 . 2a

We can still try to make sense of the corresponding functions y (t) = e r1 t ,

and

y (t) = e r2 t .

To do so, we’ll need to establish a few facts about complex functions.

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Complex Numbers Complex numbers are an algebraic extension of the real numbers (just like the real numbers are an algebraic extension of the integers). p Each number is of the form ↵ + i , where i = 1 and ↵, are real. Complex numbers can be thought of as 2-dimensional vectors with an additional multiplication operation: if ↵ + i and a + ib are complex numbers, then we define (↵ + i )(a + ib) := (↵a

b) + i(↵b + a).

We can also define a norm on complex numbers and do analysis as usual (i.e. define complex functions, complex limits, complex derivatives, etc.). As the real numbers are embedded in the complex numbers in a nice way (r = r + i0), the conclusions obtained in the setting of complex numbers can be applied to problems involving real functions (as we’ll soon see).

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Complex Exponential For what follows, we will assume that the Maclaurin series for e z holds even when z is a complex number. Observing that i 2 = 1, we have (i✓)2 (i✓)n + ··· + + ··· 2! n! ✓2 i✓3 i✓4 i✓5 = 1 + i✓ + + + ··· 2! 3! ◆4! ✓ 5! ✓ ◆ ✓2 ✓4 ✓3 ✓5 = 1 + + ··· + i ✓ + + ··· 2! 4! 3! 5! = cos(✓) + i sin(✓)

e i✓ = 1 + (i✓) +

thus we have derived Euler’s formula for the complex exponential e i✓ = cos(✓) + i sin(✓).

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Complex Exponential cont. Furthermore, from the Maclaurin series for e z , we can also derive the familiar law of exponents, i.e. if z, w are complex numbers then e z+w = e z e w . In all, this allows us to make sense of the function e r when r is complex, i.e. r = ↵ + i . In particular, this value can be expressed in terms of the real-valued functions sine and cosine as follows: e ↵+i = e ↵ e i = e ↵ (cos( ) + i sin( )) Lastly, from the Maclaurin series for e z , we also find that d (↵+i e dt

)t

= (↵ + i )e (↵+i

)t

.

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General Solution Therefore, considering our initial situation when the auxiliary equation ar 2 + br + c = 0 has complex roots r1 = ↵ + i ,

and

r2 = ↵

i ,

we can (and shall) consider the complex functions y = e r1 t , y = e r2 t as solutions to our equation. Hence, we have the general solution: y (t) = C1 e r1 t + C2 e r2 t .

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Example

Example Solve the initial value problem y 00 + 2y 0 + 2y = 0;

y (0) = 0,

y 0 (0) = 2.

The above example illustrates that, despite having a general solution that’s constructed from complex functions, we may still end up with a solution that’s real-valued. In fact, this is always the case for such equations!

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Complex Di↵erentiation Real Solutions Derived from Complex Solutions Lemma. Let z(t) = u(t) + iv (t) be a solution to ay 00 + by 0 + cy = 0, where a, b, c are real numbers. Then the real part u(t) and the imaginary part v (t) are real-valued solutions of this di↵erential equation. This lemma shows that the general complex solution we derived y (t) = C1 e (↵+i

)t

is in fact real. Taking the solution e (↵+i e (↵+i

)t

+ C2 e (↵ )t ,

i )t

we have

= e ↵t cos( t) + ie ↵t sin( t).

So, by the above lemma, e ↵t cos( t) and e ↵t sin( t) are real solutions.

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General Solution part 2 Complex Conjugate Roots If the auxiliary equation has complex conjugate roots ↵ ± i , then the two linearly independent solutions to the equation ay 00 + by 0 + cy = 0 are e ↵t cos( t)

and

e ↵t sin( t)

and a general solution is y (t) = C1 e ↵t cos( t) + C2 e ↵t sin( t), where C1 and C2 are arbitrary constants. We first lifted the original problem into the realm of complex numbers. We then defined a solution in this case and brought this solution back into the realm of real numbers. This technique works with many other problems.

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Examples

Example Solve the initial value problem y 00 + 36y = 0;

y 0 (0) = 5.

y (0) = 3,

Example Solve the initial value problem y 00 + 2y 0 + 5y = 0;

y (0) =

1,

y 0 (0) = 1.

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Overview

1

Chapter 4: Linear Second-Order Equations Section 4.4 - Nonhomogeneous Equations - the Method of Undetermined Coefficients

LATEX Brandon Sweeting

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Nonhomogenous Equations In this section, we’ll be considering second-order linear equations with constant coefficients that are not homogeneous, i.e. those of the form ay 00 + by 0 + cy = f (t),

f (t) 6= 0.

We’ll approach these equations much as we had in the homogeneous case. When f (t) = 0, we remarked that the second derivative of any solution must be expressible as a linear combination of the first two derivatives. This led us to conjecture solutions of the form y (t) = e rt , which possess this desired behavior. From here, we were able to obtain general solutions. Our strategy for the non-homogenous case will mirror this. For various choices of f , we’ll identify the conditions imposed on the relationships of the derivatives of our solutions. We’ll then use this to conjecture solutions. We call this technique the method of undetermined coefficients.

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Linear Case We begin with a simple example

Example Find a particular (i.e. not necessarily general) solution to y 00 + 3y 0 + 2y = 3t.

(1)

We can see here that, for a function satisfying this equation, combining the function with its first and second derivatives will yield a linear function. The simplest functions with this property are linear functions themselves, i.e. functions of the form y (t) = At + B. Indeed, for such a function y 0 (t) = A

and

y 00 (t) = 0.

When we substitute this function into the LHS of (1) we obtain y 00 + 3y 0 + 2y = 3A + 2At + 2B. Brandon Sweeting

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Linear Case cont. Therefore, our di↵erential equation is satisfied provided for all t: 3A + 2At + 2B = 3t; hence, we require that A and B satisfy the system of equations: 3A + 2B = 0 and

2A = 3.

The solution to this system is A = 32 and B = 94 . Therefore, the linear solution we conjectured (thus our particular solution) is: 3 yp (t) = t 2

9 . 4

The term method of undetermined coefficients comes from this process of conjecturing a solution of a given form then determining associated coefficients. A

LTEX

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Extension of Simple Case A similar technique can be developed to solve equations of the form ay 00 + by 0 + cy = Ct m ,

m = 0, 1, 2, . . . ;

namely, we guess a solution of the form yp (t) = Am t m + · · · + A1 t + A0 , with undetermined coefficients Aj , and match the corresponding powers of t in the expression ay 00 + by 0 + cy with Ct m (as in the previous example). This will require solving a system of m + 1 linear equations in m + 1 unknowns. The result (if it exists) is a particular solution to our ODE.

Example Find a particular solution to the equation y 00 + 3y 0 + 2y = 3t 2 . Brandon Sweeting

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Exponential Case We now consider the following example

Example Find a particular solution to y 00 + 3y 0 + 2y = 10e 3t . In this case, for a function satisfying this equation, combining the function with its first and second derivatives will yield the function 10e 3t . The simplest functions with this property are multiples of e 3t , i.e. functions of the form y (t) = Ae 3t . Indeed, for such a function y 0 (t) = 3Ae 3t

and

y 00 (t) = 9Ae 3t .

When we substitute this function into the LHS of (1) we obtain y 00 + 3y 0 + 2y = 9Ae 3t + 9Ae 3t + 2Ae 3t = 20Ae 3t . Brandon Sweeting

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Exponential Case cont. Therefore, for y (t) = Ae 3t to be a solution we require that 20Ae 3t = 10e 3t , which implies that A = 12 . Therefore, a particular solution is 1 yp (t) = e 3t . 2 the general extension of this procedure for arbitrary f (t) = Ce rt is clear. We’ll next cover the case when f is a trigonometric function (either sine or cosine) and will work towards handling functions that are combinations of these three classes of functions. A final procedure will be given at the end for the most general case we’ll consider.

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Trigonometric Case We consider the next example

Example Find a particular solution to y 00 + 3y 0 + 2y = sin(t). In this case, for a function satisfying this equation, combining the function with its first and second derivatives will yield the trig function sin(t). The simplest functions with this property are combinations of sin and cos, i.e. functions of the form y (t) = A sin(t) + B cos(t). We need the cosine to account for the derivatives (which introduce cosine terms). We have, y 0 (t) = A cos(t)

B sin(t)

and

y 00 (t) =

A sin(t)

B cos(t).

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Trigonometric Case cont. When we substitute y (t) = A sin(t) + B cos(t) into our equation we get y 00 + 3y 0 + 2y =

A sin(t)

B cos(t) + 3A cos(t)

3B sin(t)

+ 2A sin(t) + 2B cos(t) = (A

3B) sin(t) + (B + 3A) cos(t).

Therefore, our di↵erential equation is satisfied provided for all t: (A

3B) sin(t) + (B + 3A) cos(t) = sin(t);

hence, we require that A and B satisfy the system of equations: A

3B = 1 and

B + 3A = 0.

The solution to this system is A = 0.1 and B =

0.3.

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Trigonometric Case cont. Therefore, the solution we conjectured (thus our particular solution) is: yp (t) = 0.1 sin(t)

0.3 cos(t).

More generally, for an equation of the form ay 00 + by 0 + by = C sin( t)

(or C cos( t)),

(2)

the method of undetermined coefficients suggests that we guess yp (t) = A cos( t) + B sin( t) and solve (2) for the unknowns A and B.

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Combined Case We now consider an example which combines those we’ve seen previously

Example Find a particular solution to y 00 + 4y = 5t 2 e t . Our previous experience suggests we guess a solution of the form (At 2 + Bt + C )e t to match the form of f (t). For such a function, we have y 0 = (2At + B)e t + (At 2 + Bt + C )e t y 00 = 2Ae t + 2(2At + B)e t + (At 2 + Bt + C )e t y 00 + 4y = e t (2A + 2B + C + 4C ) + te t (4A + B + 4B) + t 2 e t (A + 4A)

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Combined Case cont. Therefore, our di↵erential equation is satisfied provided for all t: e t (2A + 2B + C + 4C ) + te t (4A + B + 4B) + t 2 e t (A + 4A) = 5t 2 e t ; hence, we require that A, B and C satisfy the system of equations: 2A + 2B + 5C = 0 4A + 5B = 0 5A = 5 2 The solution to this system is A = 1, B = 45 and C = 25 . Therefore, the solution we conjectured (thus our particular solution) is: ✓ ◆ 4 2 2 yp (t) = t t et . 5 25

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Examples

Example Find a particular solution to the di↵erential equations 1

y 00 + 3y =

2

y 00 + y = 2t

3

y 00

y 0 + 9y = 3 sin(t)

4

y 00

5y 0 + 6y = te t

5

y 00 + 4y = 8 sin(2t)

9

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Possible Pitfall Issues may arise when using the method of undetermined coefficients.

Example Find a particular solution to y 00 + y 0 = 5. Our method would suggest looking for a solution of the form y (t) = A. However, this doesn’t work; since y 0 = 0 and y 00 = 0 we have y 00 + y 0 = 0 6= 5. In di↵erentiating, we’ve “lost” our undetermined coefficient. This is because the function y (t) = A solves the homogeneous equation. Brandon Sweeting

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A Simple Fix We can correct this by appending a “t” to our conjectured function y (t) = At. This way, our undetermined coefficient will “survive” the di↵erentiation: y 0 (t) = A y 00 (t) = 0. With this, we have y 00 + y 0 = A and thus a particular solution is given by yp (t) = 5t.

LATEX Brandon Sweeting

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Another Pitfall The same issue arises when considering f (t) = Ce rt .

Example Find a particular solution to y 00

6y 0 + 9y = e 3t .

This time, our method would suggest looking for a solution of the form y (t) = Ae 3t . However, this would not work. Since e 3t solves the homogeneous equation y 00

6y 0 + 9y = 0,

our undetermined coefficient won’t “survive” the expression on the left.

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A Similar Fix Furthermore, since the characteristic equation for this ODE r2

6r + 9 = 0,

has 3 as a repeated root, we’ll also have that the function te 3t also solves the homogeneous ODE (by the results of Section 4.2). Therefore, the undetermined coefficient in Ate 3t will also get “lost”. We must take things a step further. The simplest way to do so is to append another factor of t to the function te 3t and to conjecture a solution of the form y (t) = At 2 e 3t . This time, when evaluating y 00 6y 0 + 9y for our conjectured solution, we’ll obtain some non-zero expression containing A. From here, we can proceed to solve for A to satisfy the equation as before.

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A Similar Fix cont. For the function y (t) = At 2 e 3t , we have y 0 (t) = 2Ate 3t + 3At 2 e 3t y 00 (t) = 2Ae 3t + 12Ate 3t + 9At 2 e 3t . So for this new conjectured solution, we have y 00

6y 0 + 9y = (2Ae 3t + 12Ate 3t + 9At 2 e 3t ) 6(2Ate 3t + 3At 2 e 3t ) + 9(At 2 e 3t ) = 2Ae 3t .

Therefore, we require that 2Ae 3t = e 3t if our conjectured function is to be a solution; so A = 1/2, thus we obtain the particular solution 1 yp (t) = t 2 e 3t . 2 Brandon Sweeting

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Generalization To generalize this fix we’ve described to all equations of the form ay 00 + by 0 + cy = Ct m e rt , we start with our initial conjectured solution for such equations y (t) = (Am t m + · · · + A1 t + A0 )e rt and we simply do one of three things depending on whether or not r is a root of the corresponding characteristic equation ar 2 + br + c = 0 1 2 3

If r is not a root, we keep our conjectured solution as is. If r is a root, we append t to our conjectured solution. If r is a double root, we append t 2 .

These rules are summarized in the following theorem

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Method of Undetermined Coefficients Part I Method of Undetermined Coefficients To find a particular solution to the di↵erential equation ay 00 + by 0 + cy = Ct m e rt where m is a nonnegative integer, use the form yp (t) = t s (Am t m + · · · + A1 t + A0 )e rt , with 1

s = 0 if r is not a root of the associated auxiliary equation.

2

s = 1 if r is a simple root of the associated auxiliary equation.

3

s = 2 if r is a double root of the associated auxiliary equation.

LATEX Brandon Sweeting

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Method of Undetermined Coefficients Part II Method of Undetermined Coefficients To find a particular solution to the di↵erential equation ay 00 + by 0 + cy = Ct m e ↵t cos( t) (or Ct m e ↵t sin( t)) for

6= 0, use the form yp (t) = t s (Am t m + · · · + A1 t + A0 )e ↵t cos( t)

t s (Bm t m + · · · + B1 t + B0 )e ↵t sin( t)

with 1

s = 0 if ↵ + i

is not a root of the associated auxiliary equation.

2

s = 1 if ↵ + i

is a root of the associated auxiliary equation.

LATEX Brandon Sweeting

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Examples Example Find the form for a particular solution to y 00 + 2y 0

3y = f (t),

where f (t) equals (a)7 cos(3t) (b)2te t sin(t) (c)t 2 cos(⇡t) (d)5e

3t

(e)2te t (f)t 2 e t

Example Find the form of a particular solution to y 00 + 2y 0

3y = f (t),

for the same set of nonhomogeneities f (t) as in the previous example.

LATEX Brandon Sweeting

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Example cont. Example Find the form of a particular solution to y 00

2y 0 + 2y = 5te t cos(t).

Example Find a particular solution to the di↵erential equation y 00

5y 0 + 6y = te t .

Example Find a particular solution to the di↵erential equation y 00

y 0 + 9y = 3 sin(3t).

LATEX Brandon Sweeting

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Applied Di↵erential Equations I Brandon Sweeting

October 4, 2022

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Overview

1

Chapter 4: Linear Second-Order Equations Section 4.5 - The Superposition Principle and Undetermined Coefficients Revisited

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Superposition Principle We now discuss a very important concept that will allow us to greatly extend the set of solutions we’ve derived for nonhomogeneous equations.

Superposition Principle Theorem 3. If y1 is a solution to the di↵erential equation ay 00 + by 0 + cy = f1 (t), and y2 is a solution to ay 00 + by 0 + cy = f2 (t), then for any constants k1 and k2 , the function k1 y1 + k2 y2 is a solution to the di↵erential equation ay 00 + by 0 + cy = k1 f1 (t) + k2 f2 (t).

LATEX Brandon Sweeting

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Superposition Principle cont. Proof. This result is a direct consequence of our equation’s linearity. Indeed, by substituting and rearranging we have, a(k1 y1 + k2 y2 )00 + b(k1 y1 + k2 y2 )0 + c(k1 y1 + k2 y2 ) = k1 (ay100 + by10 + cy1 ) + k2 (ay200 + by20 + cy2 ) = k1 f1 (t) + k2 f2 .

Example Find a particular solution to the following equations 1

y 00 + 3y 0 + 2y =

2

y 00 + 3y 0 + 2y = 3t + 10e 3t

9t + 20e 3t

LATEX Brandon Sweeting

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Superposition Principle Consequences Recall that for any real numbers a, b, and c, the di↵erential equation ay 00 + by 0 + cy = 0, is solvable and has general solution y = C1 y1 + C2 y2 , where the particular form of y1 and y2 depends on the roots of the related auxiliary equation. Consequently, if we’ve found a particular solution, say yp , to the ODE ay 00 + by 0 + cy = f (t), we have for any choice of C1 and C2 , by the superposition principle, the function y = yp + C1 y1 + C2 y2 will also be a solution to this equation. Moreover, the coefficients C1 and C2 can be chosen so that the IVP ay 00 + by 0 + cy = f (t);

y (t0 ) = Y0 ,

y 0 (t0 ) = Y1

has a solution. Brandon Sweeting

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Existence and Uniqueness: Nonhomogeneous Case

The solution derived this way will be unique since if yI also satisfies ay 00 + by 0 + cy = f (t); then the function yII (t) := yI ay 00 + by 0 + cy = 0;

y (t0 ) = Y0 ,

y 0 (t0 ) = Y1 ,

(yp + C1 y1 + C2 y2 ) will satisfy y (t0 ) = 0,

y 0 (t0 ) = 0.

Since the above initial value problem also admits the zero solution, by the existence and uniqueness theorem of the homogeneous case, we must have that yII = 0, i.e. yI = yp + C1 y1 + C2 y2 .

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Existence and Uniqueness: Nonhomogeneous Case cont. To summarize, if we can find a particular solution, say yp , to ay 00 + by 0 + cy = f (t),

(1)

then we can solve any corresponding IVP by taking the function y (t) = yp + C1 y1 + C2 y2

(2)

where y1 and y2 are the linearly independent solutions to the homogeneous equation and C1 and C2 chosen accordingly. Furthermore, this solution will be unique. As a direct consequence of this, all solutions to (1) will be of the form (2). That is to say, the general solution to (1) is yg (t) = yp + C1 y1 + C2 y2 .

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Examples

Example Given that yp (t) = t 2 is a particular solution to y 00

y =2

t 2,

find a general solution and a solution satisfying y (0) = 1, y 0 (0) = 0.

Example Find a particular solution to y 00

y = 8te t + 2e t

LATEX Brandon Sweeting

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Streamlined Method of Undetermined Coefficients Part I Method of Undetermined Coefficients To find a particular solution to the di↵erential equation ay 00 + by 0 + cy = Pm (t)e rt where Pm (t) is a polynomial of degree m, use the form yp (t) = t s (Am t m + · · · + A1 t + A0 )e rt , with 1

s = 0 if r is not a root of the associated auxiliary equation.

2

s = 1 if r is a simple root of the associated auxiliary equation.

3

s = 2 if r is a double root of the associated auxiliary equation.

LATEX Brandon Sweeting

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Streamlined Method of Undetermined Coefficients Part II Method of Undetermined Coefficients To find a particular solution to the di↵erential equation ay 00 + by 0 + cy = Pm (t)e ↵t cos( t) + Qn (t)e ↵t sin( t) where

6= 0 and Pm (t), Qn (t) are polynomials of degrees m and n, yp (t) = t s (Ak t k + · · · + A1 t + A0 )e ↵t cos( t)

t s (Bk t k + · · · + B1 t + B0 )e ↵t sin( t)

where k is the larger of m and n and 1

s = 0 if ↵ + i

is not a root of the associated auxiliary equation.

2

s = 1 if ↵ + i

is a root of the associated auxiliary equation.

LATEX Brandon Sweeting

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Examples Example Write down the form of a particular solution to the equation y 00 + 2y 0 + 2y = 5e

t

sin(t) + 5t 3 cos(t).

Example Find a general solution to the di↵erential equations 1

y 00

2y 0

2

y 00

3y 0

3y = 3t 2 + 2y =

ex

5

sin(t)

Example Find the solution to the initial value problems 1

y 00 + y = 2e

2

y 00

t;

y = sin(t)

Brandon Sweeting

y (0) = 0, e 2t ;

y 0 (0) = 0

y (0) = 1,

y 0 (0) =

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LATEX October 4, 2022

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Applied Di↵erential Equations I Brandon Sweeting

October 6, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 4: Linear Second-Order Equations Section 4.6 - Variation of Parameters

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Method of Undetermined Coefficients In Sections 4.4 and 4.5, we were concerned with solving linear equations: ay 00 + by 0 + cy = f (t), where a, b, c were real numbers and f (t) had a special form, i.e. f (t) = Pn (t)e ↵t cos( t) + Qm (t)e ↵t sin( t). where Pn was some polynomial of degree n, Qm some polynomial of degree m and ↵ and were real numbers. We obtained a particular solution by guessing a function of the form yp (t) = t s [ Ak t k + · · · + A1 t + A0 e ↵t cos( t) h i + t s Bk t k + · · · + B1 t + B0 e ↵t sin( t)

and choosing the Ai ’s and Bi ’s so that the equation would be satisfied. A

LTEX

Brandon Sweeting

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Variation of Parameters We now seek to handle more general functions f (t) through a technique called Variation of Parameters. From Section 4.2, we know that the corresponding homogeneous equation ay 00 + by 0 + cy = 0,

(1)

has two linearly independent solutions, y1 (t) and y2 (t), through which the general solution is expressed as an arbitrary linear combination: yg (t) = c1 y1 (t) + c2 y2 (t). We seek to modify this general solution to the homogeneous equation by taking the coefficients c1 and c2 to now be functions of t. The hope is that by slightly perturbing the homogeneous solutions y1 and y2 by functions of t, we can obtain a particular solution to the nonhomogeneous equation.

LATEX Brandon Sweeting

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Variation of Parameters cont. Therefore, we seek a particular solution of the form yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t). As was the case for the method of undetermined coefficients, to determine v1 and v2 we must see what conditions these functions must satisfy by plugging yp into the di↵erential equation. Therefore, we compute yp0 (t) = (v10 y1 + v20 y2 ) + (v1 y10 + v2 y20 ). To simplify our analysis, we impose the additional condition v10 y1 + v20 y2 = 0. So, in all, we have yp0 (t) = v1 y10 + v2 y20 yp00 (t) = v10 y10 + v1 y100 + v20 y20 + v2 y200 Brandon Sweeting

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Variation of Parameters cont. We now plug yp into the equation ay 00 + by 0 + cy = f (t). Since y1 and y2 solve the homogeneous equation, we obtain f = ayp00 + byp0 + cyp = a(v10 y10 + v1 y100 + v20 y20 + v2 y200 ) + b(v1 y10 + v2 y20 ) + c(v1 y1 + v2 y2 ) = a(v10 y10 + v20 y20 ) + v1 (ay100 + by10 + cy1 ) + v2 (ay200 + by20 + cy2 ) = a(v10 y10 + v20 y20 ) + 0 + 0. We can see that things simplify considerably when working with our homogeneous solutions. So we have the two conditions on v1 and v2 : v10 y1 + v20 y2 = 0; Brandon Sweeting

v10 y10 + v20 y20 =

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f . a

(2)

LAT

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Variation of Parameters cont. Solving these equations for v10 and v20 , we obtain v10 (t) =

f (t)y2 (t) f (t)y1 (t) ; v20 (t) = 0 y1 (t)y2 (t)] a [y1 (t)y20 (t) y10 (t)y2 (t)]

a [y1 (t)y20 (t)

Therefore, we can just integrate to find that Z f (t)y2 (t) v1 (t) = dt, 0 a [y1 (t)y2 (t) y10 (t)y2 (t)] v2 (t) =

Z

f (t)y1 (t) dt. a [y1 (t)y20 (t) y10 (t)y2 (t)]

Thus, for v1 and v2 above, we obtain the particular solution yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t).

LATEX Brandon Sweeting

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Variation of Parameters Recap Variation of Parameters To determine a particular solution to ay 00 + by 0 + cy = f : 1

Find two linearly independent solutions {y1 (t), y2 (t)} to the corresponding homogeneous equation and take yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t).

2

3

Determine v1 (t) and v2 (t) by solving the system in (2) for v10 (t) and v20 (t) and integrating. Substitute v1 (t) and v2 (t) into the expression for yp (t) to obtain a particular solution.

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Examples Example Find a general solution on ( ⇡/2, ⇡/2) to d 2y + y = tan(t) dx 2

Example Find a particular solution on ( ⇡/2, ⇡/2) to d 2y + y = tan(t) + 3t dx 2

1

Note that neither of the above equations could have been solved by the method of undetermined coefficients (as tan(t) is not of the form we addressed). We’ll next see how variation of parameters can be used to solve some non-constant coefficient equations. LATEX Brandon Sweeting

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Examples Example Find a particular solution to the variable coefficient equation t 2 y 00 + 4ty 0 + 6y = 4t 3 ,

t > 0,

given that y1 (t) = t 2 and y2 (t) = t 3 solve the homogeneous equation.

Example Find a general solution to the given ODEs using variation of parameters. 1

y 00 + y = sec(t)

2

y 00

3

y 00 + 4y 0 + 4y = e

2y 0 + y = t

1e t 2t

ln(t)

LATEX Brandon Sweeting

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Applied Di↵erential Equations I

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Applied Di↵erential Equations I Brandon Sweeting

October 18, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 4: Linear Second-Order Equations Section 4.9 - A Closer Look at Free Mechanical Vibrations

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Governing Equation In Section 4.1, we considered the physical system consisting of a weight (i.e. mass) attached to a spring that was fixed at one end. With damping and possibly external forces present, assuming the spring satisfied Hooke’s law, we derived the following di↵erential equation for the function y (t) that gave the displacement of the weight from equilibrium at time t: d 2y dy + [damping] + [sti↵ness]y 2 dt dt = my 00 + by 0 + ky

Fext = [inertia]

This equation was derived using Newton’s second law of motion. We will now continue our analysis of the various motions of the weight when it’s displaced from equilibrium.

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Simplest Case When b = 0 and Fext = 0, i.e. the so-called undamped, free case, we were left with the following equation: my 00 + ky = 0; If we let ! =

p k/m, we may rewrite this equation as y 00 + ! 2 y = 0.

By the techniques of Section 4.3, we found the general solution y (t) = C1 cos(!t) + C2 sin(!t) and noticed that such functions had oscillatory behavior with frequency !. We then looked at a simple example by considering the initial conditions y (0) = y0 and y 0 (0) = 0, which had the solution y (t) = y0 cos(!t). We’ll now analyze this behavior further in the general case. LAT X

E

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General Undamped, Free Case We begin by noting that we may rewrite our solution y (t) = C1 cos(!t) + C2 sin(!t) in the form A sin(!t + ) with A 0. Indeed, if we can find A and that C1 = A sin( ) and C2 = A cos( ), then by the sum to product formulas we’d have

such

A sin(!t + ) = A cos(!t) sin( ) + A sin(!t) cos( ) = C1 cos(!t) + C2 sin(!t). Solving for A and

we find q A = C12 + C22

and

tan( ) =

C1 . C2

With this representation in hand, we continue our analysis. Brandon Sweeting

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General Undamped, Free Case cont. Recall the previously derived form for our solution, y (t) = A sin(!t + ). This function describes simple harmonic motion, i.e. the oscillation with angular frequency ! between the values A and A (i.e. with amplitude A). The period is the amount of time needed for an oscillation from A to A and back and is thus 2⇡/!. The natural frequency is the number of full oscillations per time unit and is thus !/2⇡.

LATEX Brandon Sweeting

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Undamped, Free Case Example

Example A 1/8-kg mass is attached to a spring with sti↵ness k = 16 N/m. The mass is displaced 1/2 m to the right p of the equilibrium point and given an outward velocity (to the right) of 2 m/sec. Neglecting any damping or external forces that may be present, determine the equation of motion of the mass along with its amplitude, period, and natural frequency. How long after release does the mass pass through the equilibrium position?

LATEX Brandon Sweeting

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Damped, Free Case We now consider the case that there is some damping present from various factors like friction and air resistance, i.e. b > 0, but the system is still free, i.e. Fext = 0. In this case, we have the following equation: my 00 + by 0 + ky = 0. The corresponding auxiliary equation mr 2 + br + k = 0, has roots,

b 1 p 2 ± b 2m 2m

4mk.

We consider each case separately based on the discriminant, b 2

4mk.

LATEX Brandon Sweeting

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Underdamped or Oscilltory Motion (b 2 < 4mk) In this case, the discriminant, b 2 4mk, is negative; consequently, our auxiliary equation has two complex complex roots ↵ ± i , where: ↵ :=

b ; 2m

thus we have the general solution

:=

1 p 4mk 2m

b2

y (t) = e ↵t (C1 cos( t) + C2 sin( t)) = Ae ↵t sin( t + ). We call the quantity Ae ↵t the damping factor of the motion. In this situation, the damping isn’t great enough to completely overpower the oscillatory behavior (hence why we call this system underdamped). Our weight will still oscillate (a motion we refer to quasioscillation in this context), but now with quasiperiod 2⇡/ and quasifrequency 1/P. A

LTEX

Brandon Sweeting

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Underdamped or Oscilltory Motion (b 2 < 4mk) cont.

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Overdamped Motion (b 2 > 4mk) In this case, the discriminant, b 2 4mk, is positive; consequently, our auxiliary equation has two real roots p p b + b 2 4mk b b 2 4mk r1 = and r2 = , 2m 2m thus we have the general solution y (t) = C1 e r1 t + C2 e r2 t . Clearly r2 < 0. Furthermore, since b 2 > b 2 4mk, it also follows that r1 < 0. In this situation, the damping is so great that it overpowers all oscillatory behavior (hence why we call this system overdamped).

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Overdamped Motion (b 2 > 4mk) cont. Looking at the derivative of this solution (which has at most one root), y 0 (t) = C1 r1 e r1 t + C2 r2 e r2 t = e r1 t (C1 r1 + C2 r2 e (r2

r1 )t

),

we see that there are three possibilities for the motion (depicted below).

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Critically Damped Motion (b 2 = 4mk) In this case, the discriminant, b 2 4mk, is zero; consequently, our auxiliary equation has one repeated real root r = b/2m. Therefore, we have the geneal solution y (t) = C1 e rt + C2 te rt . Clearly r < 0. Also, since e at decays slower that t (by l’Hˆopital’s rule), both terms will decay to 0 as t gets large. Therefore, the damping is again so great that it overpowers all oscillatory behavior. This situation, which we call a critically damped system, is very similar to the overdamped case and we again have three possible (non-trivial) behaviors given by the figures on the last slide.

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Overdamped Motion (b 2 > 4mk) Example Example A 1/4-kg mass is attached to a spring with a sti↵ness 4 N/m as shown in Figure 4.31(a). The damping constant b for the system is 1 N-sec/m. If the mass is displaced 1/2 m to the left and given an initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum displacement that the mass will attain?

LATEX Brandon Sweeting

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Applied Di↵erential Equations I Brandon Sweeting

October 25, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 5: Introduction to Systems and Phase Analysis Section 5.1 - Interconnected Fluid Tanks Section 5.2 - Di↵erential Operators and the Elimination Method for Systems

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Motivating Example Imagine we had two tanks, each holding 24 liters of a brine solution, that are interconnected by pipes (as in the figure below). Fresh water flows into tank A at a rate of 6L/min, and is drained out of tank B at the same rate; also 8 L/min of fluid are pumped from tank A to tank B, and 2 L/min from tank B to tank A. The liquids inside each tank are kept well stirred so that each mixture is homogeneous. If, initially, the brine solution in tank A contains x0 kg of salt and the brine solution in tank B initially contains y0 kg of salt, determine the mass of salt in each tank at time t > 0.

LATEX Brandon Sweeting

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Compartmental Analysis We can treat each tank separately, using compartmental analysis to find the rate of change of salt in each tank. Starting with tank A, we want to determine the input and output rate for the equation: dx = input rate output rate. dt For the input rate, as pure water is being pumped in, the only salt entering tank A is that from tank B. The rate is 2 L/min and the concentration is y /24 kg/L (since the volume in both tanks is a constant 24L). For the output rate, the only salt leaving tank A is that being pumped into tank B. The rate is 8 L/min and the concentration is x/24 kg/L. Therefore, our final equation above becomes: dx 2 = y dt 24

8 1 x= y 24 12

1 x. 3

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Compartmental Analysis cont. We then apply compartmental analysis to tank B; we want to now determine the input and output rate for the equation: dy = input rate output rate. dt For the input rate, the only salt entering tank B is that from tank B. The rate is 8 L/min and the concentration is x/24 kg/L. For the output rate, the situation is a bit more complicated. We lose salt to tank A and we also lose salt from solution being pumped out of the two tank system. The former is at a rate of 2 L/min the latter is at a rate of 6 L/min and both are at a concentration of x/24 kg/L. Consequently, our equation above becomes: dy 8 = x dt 24

2 y 24

6 1 y= x 24 3

1 y. 3

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System of Equations Therefore, we have the following system of di↵erential equations for the unknown functions x(t) and y (t): 1 1 1 1 x + y, y0 = x y. 3 12 3 3 Since each unknown function is the solution of di↵erential equation in which the other appears, we say the equations (and accordingly the solutions) are coupled. x0 =

To find a solution to this system, we might be inclined to apply familiar algebraic techniques like substitution and elimination. This will turn out to be the correct approach, though care must be taken to ensure this process makes sense for all applications of these techniques.

LATEX Brandon Sweeting

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System of Equations cont. Recall our system of di↵erential equations, x0 =

1 1 x + y, 3 12

1 y0 = x 3

1 y. 3

Solving the second equation for x, x = 3y 0 + y , and substituting this into the first equation to eliminate x, we obtain (3y 0 + y 0 ) = 3y 00 + y 0 =

1 1 (3y 0 + y ) + y , 3 12 1 1 y0 y + y. 3 12

LATEX Brandon Sweeting

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System of Equations cont. The equation we derived, 3y 00 + y 0 =

y0

1 1 y + y, 3 12

can be rewritten as

1 3y 00 + 2y 0 + y = 0, 4 which is a 2nd order linear constant coefficient homogeneous di↵erential equation (of the kind we studied in chapter 4). Using our established techniques for solving such equations, we find: y (t) = c1 e

t/2

+ c2 e

t/6

and plugging this back into the second equation gives x(t) = Brandon Sweeting

1 c1 e 2

t/2

1 + c2 e 2

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t/6

.

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System of Equations cont. Lastly, adding our initial conditions, x(0) = x0 and y (0) = y0 , gives

x(t) = y (t) =

✓

✓

y0

◆ y0 + 2x0 e + e t/6 , 4 4 ◆ ✓ ◆ 2x0 y0 + 2x0 t/2 e + e t/6 . 4 4

y0

2x0

◆

t/2

✓

This final solution indicates that the salt concentration within each tank will eventually reach 0. This is in line with our expectations, as only pure water is being pumped into the system. The next section will formalize this process of solving systems of di↵erential equations using the familiar techniques of substitution and elimination (as care must be taken to make sense of these operations in this context).

LATEX Brandon Sweeting

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Di↵erentiation as an Operator To begin, we need to reformulate our interpretation di↵erentiation. Specifically, we want to think of the derivative as an operation that takes functions in and spits out their derivatives (i.e. the derivative is a function whose inputs and outputs are other functions). We’ll call this operation D so that: dy = y 0. dt where we use brackets to avoid ambiguity. We can take this interpretation further and define the product DD = D 2 to be the composition of D with itself. Therefore, D[y ] :=

D 2 [y ] := D[D[y ]] = D[y 0 ] = y 00 . We can also define how D interacts with constants as follows. If a is some fixed real number (e.g. a = 4), then we write: aD[y ] := a(D[y ]) = ay 0 . Brandon Sweeting

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Operator Notation Lastly, it’s clear how we can add such operations together (D + D 2 )[y ] := D[y ] + D 2 [y ] = y 0 + y 00 . With this new notation, we have that the di↵erential equation y 00 + 4y 0 + 3y = 0, becomes (D 2 + 4D + 3)[y ] = 0. To solve systems of equations, we’ll treat our derivatives like “coefficients” of our variables (i.e. the unknown functions x and y ). From here, we can apply our familiar techniques for solving systems of equations. We can do this since we’ve established an algebra for these objects (i.e. we can add and multiply them). Note that we implicitly associated to each constant an operation corresponding to multiplication. LAT X

E

Brandon Sweeting

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Solving Systems - Elimination Method To begin the process of solving a system of di↵erential equations using the elimination method, we must first rewrite the equations using our operator notation, i.e. we write the original system a1 x 0 (t) + a2 x(t) + a3 y 0 (t) + a4 y (t) = f1 (t), a5 x 0 (t) + a6 x(t) + a7 y 0 (t) + a8 y (t) = f2 (t), where a1 , a2 , . . . , a8 are constants and x(t), y (t) our unknown solutions, as (a1 D + a2 )[x] + (a3 D + a4 )[y ] = f1 , (a5 D + a6 )[x] + (a7 D + a8 )[y ] = f2 . We can then add or subtract multiples of one equation from the other to eliminate a variable (i.e. either x or y ) to “decouple” the equations and solve for one of the unknown variables. Once found, we can plug this solution back into either equation to solve for the other unknown. LATEX Brandon Sweeting

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Examples Example Solve the system x 0 (t) + 2y (t) = 0, x 0 (t)

y 0 (t) = 0.

Example Solve the system x 0 (t) = 3x(t) 0

y (t) = 4x(t)

4y (t) + 1, 7y (t) + 10t.

LATEX Brandon Sweeting

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Special Cases If one of the equations is missing a derivative of the unknown variables, as we had seen in the introductory example where we had the system: x0 =

1 1 + y, 3 12

y0 =

1 3

1 y, 3

then we can immediately proceed to find the solution via substitution. Another special case arises when it so happens that (a1 D + a2 )(a7 D + a8 )

(a3 D + a4 )(a5 D + a6 ) = 0.

If the above holds, we say that our system is degenerate. In such cases, further analysis is needed to determine if we have any solutions. This is analogous to the problem of solving for the points of intersection of two parallel or coincident lines (there may be no solutions or, if there are solutions, they may involve any number of arbitrary constants).

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Elimination Procedure for 2x2 Systems Elimination Procedure for 2x2 Systems To find a general solution for the system L1 [x] + L2 [y ] = f1 , L3 [x] + L4 [y ] = f2 , where L1 , L2 , L3 , and L4 are polynomials in D = d/dt: 1

Make sure that the system is written in operator form.

2

Eliminate one variable (say, y ) and solve the resulting equation for x. If the system is degenerate, stop! Need further analysis for solutions.

3

(Shortcut) If an equation is missing a derivative of one of the unknown functions, then we can solve directly via substitution.

4

Otherwise, eliminate x from the system and solve for y (t).

5

Remove additional constants by substitution. Brandon Sweeting

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Examples Example Find a general solution to x 0 (t) + y 0 (t) 0

x(t) = 5,

0

x (t) + y (t) + y (1) = 1

Example Find a general solution to x 0 (t) + y 0 (t) + 2x(t) = 0, x 0 (t) + y 0 (t)

x(t)

y (t) = sin(t).

LATEX Brandon Sweeting

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Applied Di↵erential Equations I Brandon Sweeting

October 30, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 7: Laplace Transforms Section 7.2 - Definition of the Laplace Transform

LATEX Brandon Sweeting

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The Laplace Transform We now begin our study of the Laplace transform. Just like the derivative, the Laplace transform is an operator, i.e. it’s a function whose input and output are both functions. The Laplace transform is defined as follows:

Laplace Transform Definition 1. Let f (t) be a function on [0, 1). The Laplace Transform of f is the function F defined by the integral Z 1 F (s) = e st f (t) dt. 0

The domain of F (s) is all the values of s for which the integral above exists. The Laplace transform of f is denoted both by F and L{f }. Note that the integral above is an improper one and must be formally computed as the limit of integrals over intervals [0, N] as N ! 1. Brandon Sweeting

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Purpose of the Laplace Transform Recall our study of Bernoulli equations, where we made a substitution to transform a nonlinear equation into a linear one. This simplified our equation allowing us to find a solution using our basic techniques. The Laplace transform will serve a similar purpose. Applying the Laplace transform to a certain class of di↵erential equations will transform these equations into algebraic ones (in the independent variable s)! We’ll then be able to resolve our solution using basic techniques of algebra and then convert this solution back into the initial independent variable x using the inverse Laplace transform (more on this later). The difficulty of this technique will no longer lie in solving the equation but rather in computing the Laplace transform and the inverse Laplace transform (which is also defined as an integral).

LATEX Brandon Sweeting

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Comparison of Solution Methods

Before embarking on solving equations using the Laplace transform, we’ll first look at computing the Laplace transform of some common functions and we’ll study some of its properties. LATEX Brandon Sweeting

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Computing L: Constant Functions Example 1. Compute L{f } for the function f (t) ⌘ 1: F (s) =

Z

1

e

st

N!1 0

0

= lim

N!1

· 1 dt = lim

Z

e s

When s > 0 we have limN!1 e

st t=N

= lim t=0 sN

N!1

N

e 

1 s

st

dt e

sN

s

= 0, thus

1 F (s) = , s

s>0

R1 When s  0, we have 0 e st clearly diverges. So F (s) = 1/s on (0, 1). By a slight extension of this argument, we can see that if a is a constant and f (t) ⌘ a, then F (s) = a/s with domain (0, 1).

LATEX

Brandon Sweeting

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Computing L: Exponential Functions Example 2. Compute L{f } for the function f (t) = e at (a constant). Z 1 Z 1 F (s) = e st e at dt = e (s a)t dt 0

0

Z

= lim

N

e

N!1 0

= lim

N!1

= If s  a, then

R1 0

Brandon Sweeting

e

"

1 s

a

(s a)t

(s a)

N!1

1 s

e a

for

t dt = lim (s a)N

s

a

#

e s

(s a)t

a

N

0

s > a.

dt diverges hence F (s) = 1/(s

a) on (a, 1).

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Computing L{f }: Trigonometric Functions Example 3. Compute L{f } for the function f (t) = sin(bt) (b constant). To solve the following integral, we’ll use a pre-comptued table integral. F (s) =

Z

1 0

= lim

e "

N!1

= lim

N!1

= Since limN!1 e

s2



st

N

N!1 0

e

st

s2 + b

( s sin(bt) 2

b s 2 + b2

b + b2

sN (s

sin(bt) dt = lim

Z

for

e

st

sin(bt) dt # N

b cos(bt))

0

sN

e (s sin(bN) + b cos(bN)) s 2 + b2 s > 0.

sin(bN) + b cos(bN)) = 0.

LATEX Brandon Sweeting

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Computing L{f }: Piecewise Functions Example 4. Compute L{f } for the (discontinuous) piecewise function 8 > 0 < t < 5,

: 4t e , 10 < t. Using the linearity of the integral operator, we have Z 1 F (s) = e st f (t) dt 0 Z 5 Z 10 Z 1 st st = e · 2 dt + e · 0 dt + e 0 5 10 Z 5 Z 1 st =2 e dt + lim e (s 4)t dt 0

st 4t

e

dt

N!1 10

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Computing L{f }: Piecewise Functions

F (s) = 2 =

2 s

=

2 s

Z

5

e

st

0 5s

2e

+ lim

s s

1

s

N!1

5s

2e

Z

e (s 4)t dt N!1 10 " # e 10(s 4) e (s 4)N

dt + lim

+

e

4

s

4

10(s 4)

s

4

for

s > 4.

Notice that the jump discontinuities at t = 5 and t = 10 are reflected in the exponential terms e 5s and e 10s . Furthermore, it’s quite remarkable that our discontinuous function in t becomes continuous in the new variable s after applying the Laplace transform.

LATEX Brandon Sweeting

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Linearity of Laplace Transform An important property of the Laplace transform is the following

Linearity of the Laplace Transform Theorem 1. Let f , f1 and f2 be functions whose Laplace transforms exist for s > ↵ and let c be a constant. Then for s > ↵ we have the following L{f1 + f2 } = L{f1 } + L{f2 }

(1)

L{cf } = cL{f }.

(2)

These properties immediately follow from linearity of the integral operator. With this in hand, we’ll be able to compute the Laplace transform of a wider class of functions more easily.

LATEX Brandon Sweeting

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Linearity Example Example 5. Determine L{11 + 5e 4t

6 sin(2t)}.

By the linearity of Laplace transform, we have L{11 + 5e 4t

6 sin(2t)} = 11L{1} + 5L{e 4t }

6L{sin(2t)}.

By our previous calculations, we have 1 L{1}(s) = ; s

L{e 4t }(s) =

1 s

4

;

L{sin(2t)}(s) =

s2

2 . + 22

Therefore, it follows that L{11 + 5e 4t

6 sin(2t)} =

11 5 + s s 4

12 . +4

s2

As each of these transforms is defined for s > 4, the domain is s > 4.

LATEX Brandon Sweeting

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Table of Laplace Transforms Below is a list of pre-computed Laplace transforms. We may use the values from tables like these (along with linearity) to build up the functions we’re able to take the Laplace transform of.

LATEX Brandon Sweeting

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Table Example Example 6. Determine L{5t 2 e

3t

e 12t cos(8t)}.

By linearity of the Laplace transform, we have L{5t 2 e

3t

e 12t cos(8t)} = 5L{t 2 e

3t

}

L{e 12t cos(8t)}.

Using the table of pre-computed transforms, we have L{t 2 e

3t

}(s) =

2 ; (s + 3)3

L{e 12t cos(8t)}(s) =

(s

s 12 . 12)2 + 64

Therefore, it follows that L{5t 2 e

3t

e 12t cos(8t)} =

10 (s + 3)3

s 12 . (s 12)2 + 64

As each of these transforms is defined for s > 12, the domain is s > 12.

LATEX

Brandon Sweeting

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Existence of the Transform Now that we know more about the Laplace transform’s properties and the transform of common functions, it would help to know when the Laplace transform is defined in general. For this we’ll need two definitions:

Piecewise Continuity Definition 2. A function on a finite interval [a, b] is said to be piecewise continuous if it is continuous on [a, b] except (possibly) a finite number of points, where there may be jump discontinuities. A function on [0, 1) is piecewise continuous if it’s piecewise continuous on [0, N] for all N > 0.

Exponential Order Definition 3. A function f (t) is said to be of exponential order ↵ if there exists positive constants T and M such that |f (t)|  Me ↵t , Brandon Sweeting

for all t

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T

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Existence of the Transform cont.

Conditions for Existence of the Transform Theorem 2. If f (t) is piecewise continuous on [0, 1) and of exponential order ↵, then L{f }(s) exists for s ↵. The above is proven using the comparison text for improper integrals. The above gives firm criteria to determine when the Laplace transform of a given function exists. Essentially all the functions we’ve been working with in this class satisfy these conditions. 2

A simple example of a function that doesn’t, is the function e t .

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Examples Example Determine the Laplace transform of the given functions 1

t2

2

te 3t

Example Compute the Laplace transform of the function ( 0, 0 < t < 2, f (t) = t, 2 < t.

Example Use the Laplace transform table and linearity of the transform to find 1 2

L{t 3

te 2t + 6t 2 }

L{e 3t sin(6t) Brandon Sweeting

t3 + et }

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November 3, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 7: Laplace Transforms Section 7.3 - Properties of the Laplace Transform

LATEX Brandon Sweeting

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Further Properties In this section, we’ll look at some additional properties of the Laplace transform. In particular, we’ll study how the transform interacts with di↵erentiation and multiplication by certain functions. These properties will help to establish the Laplace transform’s ability to modify IVP’s in the t variable into algebraic equations in the s variable. We begin with the following property

Translation in s Theorem 3. If the Laplace transform L{f } exists for s > ↵, then L{e at f (t)}(s) = F (s

a)

for

s >↵+a

This property follows directly from the definition of the Laplace transform. Summarily, this theorem says that multiplying a function by e at shifts the graph of its transform by |a| to the left (if a < 0) or to the right (if a > 0).

LATEX

Brandon Sweeting

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Translation in s

Example 1. Determine the Laplace transform of e at sin(bt). In Section 7.2, we found using an integral table that L{sin(bt)}(s) =

s2

b . + b2

Therefore, by Theorem 3 we have that L{e at sin(bt)}(s) = L{sin(bt)}(s

a) =

(s

b . a)2 + b 2

LATEX Brandon Sweeting

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Laplace Transform of the Derivative The following theorem is of particular importance as it tells us how the Laplace transform interacts with di↵erentiation:

Laplace Transform of the Derivative Theorem 4. Let f (t) be continuous on [0, 1) and f 0 (t) be piecewise continuous on [0, 1), with both of exponential order ↵. Then for s > ↵, L{f 0 }(s) = sL{f }(s)

f (0).

Summarily, this theorem says that di↵erentiation of a function in the t variable is equivalent to multiplication of it’s Laplace transform by s. This is quite remarkable, as di↵erentiation is a much more difficult operation to work with (computationally) than multiplication. This theorem hints at the Laplace transform’s ability to transform di↵erential equations in t to algebraic equations in s. Brandon Sweeting

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Laplace Transform of the Derivative Proof st

Proof. Using integration by parts (u = e

Z N e st f 0 (t) dt = lim e N!1 0 0 " Z sZ N N st = lim e f (t) + s e

L{f 0 }(s) =

Z

and dv = f 0 (t)dt) we have

1

N!1

= lim e

sN

= lim e

sN

N!1 N!1

0

0

0

st 0

f (t) dt

st

Z

f (t) dt N

f (N)

f (0) + s lim

f (N)

f (0) + sL{f }(s).

N!1 0

e

st

# f (t) dt

Recall that f is of exponential order ↵; hence, for some ↵ and M lim |e

N!1

sN

f (N)|  lim e N!1

sN

Me ↵N = lim Me N!1

(s ↵)N

.

LATEX Brandon Sweeting

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Laplace Transform of the Derivative Proof cont. Consequently, if s > ↵ then limN!1 e

sN f (N)

L{f 0 }(s) = sL{f }(s)

= 0 so that

f (0).

By induction, we can obtain the following for higher-order derivatives

Laplace Transform of Higher-Order Derivatives Theorem 5. Let f (t), f 0 (t), . . . , f (n 1) (t) be continuous on [0, 1) and let f (n) (t) be piecewise continuous on [0, 1), with all these functions of exponential order ↵. Then for s > ↵ L{f (n) }(s) = s n L{f }

sn

1

f (0)

sn

2 0

f (0)

···

f (n

1)

(0).

LATEX Brandon Sweeting

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Laplace Transform of Derivatives Example Example 2. Determine L{cos(bt)} using Theorem 4 and the fact that L{sin(bt)}(s) =

s2

b . + b2

let f (t) := sin(bt). Then f (0) = 0 and f 0 (t) = b cos(bt). Consequently, L{f 0 }(s) = sL{f }(s)

f (0),

L{b cos(bt)}(s) = sL{sin(bt)}(s) sb bL{cos(bt)}(s) = 2 . s + b2

0,

Dividing both sides of this equation by b gives L{cos(bt)}(s) =

s . s 2 + b2

LATEX Brandon Sweeting

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Derivatives of the Laplace Transform Another question arises concerning the Laplace transform. If F (s) is the Laplace transform of f (t), is F 0 (s) itself the Laplace transform of some function of t? The following theorem gives an affirmative answer.

Derivatives of the Laplace Transform Theorem 6. Let F (s) = L{f }(s) and assume f (t) is piecewise continuous on [0, 1) and of exponential order ↵. Then for s > ↵, L{t n f (t)}(s) = ( 1)n

d nF (s). ds n

The above follows from Leibniz’s rule and induction using the identity Z 1 dF d (s) = e st f (t) dt. ds ds 0 Summarily, this theorem says that di↵erentiation of the Laplace transform in the s variable is equivalent to multiplication of the function by t. LATEX Brandon Sweeting

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Derivatives of the Laplace Transform cont. Example 4. Determine L{t sin(bt)}. We already know that L{sin(bt)}(s) = F (s) =

s2

b . + b2

Di↵erentiating F (s), we obtain dF 2bs (s) = 2 . ds (s + b 2 )2 Hence, using Theorem 6, we have L{t sin(bt)}(s) =

dF 2bs (s) = 2 . ds (s + b 2 )2

LATEX Brandon Sweeting

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Laplace Transform Properties Summary The following table summarizes the properties we’ve established

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Examples

Example Determine the Laplace transform of the following functions 1

t 2 + e t sin(2t)

2

e

3

2t 2 e

4

e

t

cos(3t) + e 6t

tt

t

1

t + cos(4t)

sin(2t)

LATEX Brandon Sweeting

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November 10, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 7: Laplace Transforms Section 7.4 - Inverse Laplace Transform

LATEX Brandon Sweeting

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Inverse Laplace Transform In Section 7.2, the Laplace transform was defined as an integral operator that maps a function f (t) into a function F (s). We will now consider the problem of finding the function f (t) when we are given its transform F (s). That is to say, we seek an inverse mapping for the Laplace transform. In establishing how this inverse is computed, we will have discovered all necessary elements for using the Laplace transform to solve di↵erential equations. To illustrate this we consider the following IVP: y 00

y=

t;

y 0 (0) = 1.

y (0) = 0,

Taking the transform of both sides of the equation above gives L{y 00 }

Y (s) =

1 , s2

where Y (s) := L{y }(s) is the Laplace transform of y (t). Brandon Sweeting

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Motivating Example We recall the e↵ect that the Laplace transform has on derivatives in t, i.e. L{y 00 }(s) = s 2 Y (s)

sy (0)

y 0 (0) = s 2 Y (s)

1.

Where in the last equality we incorporated the initial conditions, y (0) = 0 and y 0 (0) = 1. Substituting this expression back into our equation gives s 2 Y (s)

1

Y (s) =

1 . s2

We may now solve the algebraic equation for Y (s) to obtain 1 Y (s) = 2 s

1 s2

1

=

s2 1 1 = 2. s 2 (s 2 1) s

LATEX Brandon Sweeting

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Motivating Example cont. Lastly, we recall that L{t}(s) = 1/s 2 so that L{t} =

1 = Y (s) = L{y }. s2

Consequently, y (t) = t is our IVP solution (which we can easily verify). It should be noted that a pivotal step in this process was the logical step from the observation that L{y } = L{t} to our conclusion that y = t. Indeed, we have that the discontinuous function ( t , t 6= 6 g (t) := 0 , t=6 also shares the same transform as the two functions above. This is due to the fact that the integral in the transform disregards a function’s values at isolated points. So how are we able to make the deduction above? A

LTEX

Brandon Sweeting

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Motivating Example cont. Since we seek the solution of a di↵erential equation, it’s reasonable to impose that the solution function y we seek be continuous everywhere. Furthermore, although there are infinitely many such functions which share the same Laplace transform as y (e.g. g and other similar functions) it can be proven that there is only one such function that’s continuous. Therefore, it is this unique, continuous function that we’ll take when computing the inverse transform. This motivates the following definition.

Inverse Laplace Transform Definition 4. Given a function F (s), if there is a function f (t) that is continuous on [0, 1) and satisfies L{f } = F , then we say that f (t) is the inverse Laplace transform of F (s) and employ the notation f = L 1 {F }. LATEX Brandon Sweeting

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Inverse Laplace Transform: Starting Point While there is a formula expressing the inverse Laplace transform as an integral operator, this formula is far beyond the technical scope for this class. For those of you who are curious, it’s defined as follows: L

1

{F (s)}(t) =

1 lim 2⇡i T !1

Z

+iT

e st F (s) ds,

iT

where the integral above is taken along the vertical line Re(s) = in the complex plane such that is greater than the real part of all singularities of F (s) and F (s) is bounded on that line. Instead of this formula, we’ll resort to ad hoc methods for computing the inverse Laplace transform. As a starting point we can use the Laplace transforms we’ve computed and the ones we’ve found using tables.

LATEX Brandon Sweeting

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Inverse Transform Examples Example 1. Determine L (a) F (s) =

2 , s3

1 {F },

where

(b) F (s) =

s2

3 , +9

(c) F (s) =

s s2

1 2s + 5

Using the Laplace transforms we’ve previously computed, we have ⇢ ⇢ 2 2! 1 1 (a) L (t) = L (t) = t 2 . 3 s s3 ⇢ ⇢ 3 3 1 1 (b) L (t) = L (t) = sin(3t). 2 2 s +9 s + 32 ⇢ ⇢ s 1 s 1 1 1 (c) L (t) = L (t) = e t cos(2t). 2 s 2s + 5 (s 1)2 + 22

LATEX Brandon Sweeting

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Properties of the Inverse Transform In practice, we don’t always encounter a transform F (s) that exactly corresponds to an entry in a Laplace transform table. To handle more complicated functions F (s), we must exploit various properties of L 1 . The first of which is linearity (which is inherited from L)

Linearity of the Inverse Transform Theorem 7. Assume that L 1 {F }, L 1 {F1 }, and L 1 {F2 } exist and are continuous on [0, 1) and let c be any constant. Then L

1

{F1 + F2 } = L L

1

{cF } = cL

1

{F1 } + L 1

{F }.

1

{F2 }

(1) (2)

LATEX Brandon Sweeting

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Linearity Example I Example 2. Determine L

1

⇢

5 s

6

6s 3 + 2 + 9 2s + 8s + 10

s2

By the linearity of the inverse transform, we have ⇢ 5 6s 3 1 L + 2 (t) 2 s 6 s + 9 2s + 8s + 10 ⇢ ⇢ 1 s 3 = 5L 1 (t) 6L 1 (t) + L 2 2 s 6 s +3 2 ⇢ 3 1 = 5e 6t 6 cos(3t) + L 1 (t) 2 (s + 2)2 + 1 3 = 5e 6t 6 cos(3t) + e 2t sin(t), 2

1

⇢

s2

1 + 4s + 5

(t)

which gives the final answer.

LATEX Brandon Sweeting

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Linearity Example II Example 2. Determine L

1

⇢

5 (s + 2)4

The (s + 4)4 term in the denominator would suggest the formula ⇢ n! 1 L (t) = e at t n . (s a)n+1 If we took a =

2 and n = 3, then we’d have ⇢ 3! L 1 (t) = e (s + 2)4

2t 3

t ,

which is almost what we need. Using linearity of L 1 , we have ⇢ ⇢ 5 5 3! 5 1 1 L (t) = L (t) = e 2t t 3 . 4 4 (s + 2) 3! (s + 2) 6 Brandon Sweeting

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Linearity Example III Example 3. Determine L

1

⇢

s2

3s + 2 + 2s + 10

By completing the square, the quadratic in the denominator becomes s 2 + 2s + 10 = s 2 + 2s + 1 + 9 = (s + 1)2 + 32 . Substituting this into the expression gives ⇢ ⇢ 3s + 2 3s + 2 1 1 L =L 2 s + 2s + 10 (s + 1)2 + 32 Thus, given our prior experience, we should look to the transforms ⇢ ⇢ s a b 1 at 1 L = e cos(t); L = e at sin(t). 2 2 (s a) + b (s a)2 + b 2

LATEX Brandon Sweeting

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Linearity Example III cont. Choosing a =

1 and b = 3, we seek constants A and B such that

3s + 2 s +1 3 =A +B . 2 2 2 2 (s + 1) + 3 (s + 1) + 3 (s + 1)2 + 32 Multiplying both sides of this equation by (s + 1)2 + 32 gives, 3s + 2 = A(s + 1) + 3B = As + (A + 3B). Equating terms above and solving the resulting system of equations gives A = 3 and B = 1/3; thus, by linearity of the inverse transform, we have ⇢ ⇢ ⇢ 3s + 2 s +1 1 1 3 1 1 L = 3L L 2 2 2 2 (s + 1) + 3 (s + 1) + 3 3 (s + 1)2 + 32 1 t = 3e t cos(3t) e sin(3t). 3

LATEX Brandon Sweeting

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Method of Partial Fractions The previous example hints at a technique that we’ll use to compute the inverse transform of rational functions (i.e. quotients of polynomials in s). Namely, we seek to simplify rational functions by rewriting them in simpler terms and then using linearity of the inverse transform, e.g. 7s 2 + 10s 1 2 1 4 = + + . 3 2 s + 3s s 3 s 1 s +1 s +3 Clearly the quotients on the RHS are easier to find the inverse transform of. We’ll use the method of partial fractions (i.e. partial fraction decomposition) to accomplish this simplification. To apply this technique, we require that the denominator polynomial have larger degree than the numerator; that is, if P(s)/Q(s) is our rational function, then deg(P(s)) < deg(Q(s)). There will be three cases to consider (dependent on the factors of the denominator).

LATEX Brandon Sweeting

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Nonrepeated Linear Factors

If Q(s) can be factored into a product of distinct linear factors Q(s) = (s

r1 )(s

r2 ) · · · (s

rn ),

where the ri ’s are distinct real numbers, then we have the expansion P(s) A1 A2 An = + + ··· + , Q(s) s r1 s r2 s rn where the Ai ’s are real numbers.

LATEX Brandon Sweeting

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Nonrepeated Linear Factors Example Example 5. Determine L

1 {F },

F (s) =

where

7s 1 (s + 1)(s + 2)(s

3)

.

We begin by finding the partial fraction expansion of F . Since the denominator consists of three distinct linear factors, we must have 7s 1 (s + 1)(s + 2)(s

3)

=

A B C + + . s +1 s +2 s 3

for some constants A, B and C to be determined. We proceed to solve for these constants by multiplying this entire equation by (s + 1)(s + 2)(s + 3), 7s

1 = A(s + 2)(s

3) + B(s + 1)(s

3) + C (s + 1)(s + 2).

LATEX Brandon Sweeting

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Nonrepeated Linear Factors Example cont. From the equation we derived, 7s

1 = A(s + 2)(s

3) + B(s + 1)(s

3) + C (s + 1)(s + 2),

we may proceed in one of two ways. The first is to rearrange the RHS of the equation by grouping the coefficients of the various powers of s, e.g. 7s

1 = (A + B + C )s 2 + ( A

2B + 3C )s + ( 6A

3B + 2C ).

Equating the coefficients, we obtain a system of equations for A, B and C A+B +C =0 A 6A

2B + 3C = 7 3B + 2C =

1

This is similar to what we had done in the method of undetermined coefficients. Solving this system gives A = 2, B = 3 and C = 1. Brandon Sweeting

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Nonrepeated Linear Factors Example cont. The second way to find the coefficients from the previous equation: 7s

1 = A(s + 2)(s

3) + B(s + 1)(s

3) + C (s + 1)(s + 2)

is to choose three values for s and substitute them into the equation. The values of s should be chosen so as to eliminate as many coefficients from the equation as possible. So, starting with s = 1, we have 7

1 = A(1)( 4) + B(0) + C ()),

thus A = 2. Next, setting s = 14 thus B =

2 gives

1 = A(0) + B( 1)( 5) + C (0),

3. Lastly, setting s = 3 we’ll find that C = 1.

LATEX Brandon Sweeting

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Nonrepeated Linear Factors Example cont. Therefore, having found A = 2, B = F (s) =

7s 1 (s + 1)(s + 2)(s

3)

3 and C = 1, we have =

2 s +1

3 1 + . s +2 s 3

Therefore, by linearity of the inverse transform, we compute ⇢ 3 1 2 1 1 L {F } (t) = L + (t) s +1 s +2 s 3 ⇢ ⇢ ⇢ 1 1 1 1 1 1 = 2L (t) 3L (t) + L s +1 s +2 s 3 = 2e

t

3e

2t

(t)

+ e 3t .

LATEX Brandon Sweeting

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Repeated Linear Factors If s r is a factor of Q(s) and (s r )m is the highest power of s r that divides Q(s), then the portion of the partial fraction expansion of the quotient P(s)/Q(s) that corresponds to the term (s r )m is A1 s

r

+

A2 Am + ··· + . 2 (s r ) (s r )m

The remaining (non-repeated) factors will split as before, e.g.

(s

s 2 + 2s + 1 A B C D = + + + . 1)(s 2)(s 3)2 s 1 s 2 s 3 (s 3)2

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Repeated Linear Factors Example Example 6. Determine L

1 {F },

F (s) =

where s 2 + 9s + 2 . (s 1)2 (s + 3)

Since s 1 is a repeated linear factor with multiplicity two and s + 3 is a nonrepeated linear factor, the partial fraction expansion of F has the form s 2 + 9s + 2 A B C = + + . 2 2 (s 1) (s + 3) s 1 (s 1) s +3 We again multiply this entire equation by the denominator of F to obtain s 2 + 9s + 2 = A(s

1)(s + 3) + B(s + 3) + C (s

1)2 .

We proceed to solve for A, B and C by means of the second method discussed in the previous example (i.e. by inputting values of s). Brandon Sweeting

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Repeated Linear Factors Example cont. Recalling our equation, s 2 + 9s + 2 = A(s

1)(s + 3) + B(s + 3) + C (s

1)2 ,

we see that setting s = 1 gives 1 + 9 + 2 = A(0) + B(4) + C (0), thus B = 3. Next, setting s = 9

3 gives

27 + 2 = A(0) + B(0) + C (16),

thus C = 1. Lastly, no values of s can be chosen to eliminate B and C while leaving A; thus we choose a value of s which makes computation easy and use the values we found for B and C .

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Repeated Linear Factors Example cont. Setting s = 0, recalling that B = 3 and C = 2 = A( 3) + B(3) + C =

1, we have 3A + 9

1

thus A = 2. Putting things together, we have F (s) =

s 2 + 9s + 2 2 3 = + 2 (s 1) (s + 3) s 1 (s 1)2

1 . s +3

Therefore, by linearity of the inverse transform, we compute ⇢ 2 3 1 1 1 L {F } (t) = L + (t) 2 s 1 (s 1) s +3 ⇢ ⇢ 2 3 =L 1 (t) + L 1 (t) L s 1 (s 1)2 = 2e t + 3te t

e

3t

.

1

⇢

1 s +3

(t)

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Quadratic Factors If (s ↵)2 + 2 is a quadratic factor of Q(s) that’s irreducible (i.e. no real roots) and m is the highest power of (s ↵)2 + 2 that divides Q(s), then the part of the partial fraction expansion corresponding to (s ↵)2 + 2 is, C1 s + D1 (s ↵)2 +

2

+

C2 s + D2 [(s

↵)2

+

2 ]2

+ ··· +

Cm s + Dm , [(s ↵)2 + 2 ]m

which we may rewrite as A1 (s ↵) + B1 A2 (s ↵) + B2 Am (s ↵) + Bm + + ··· + . 2 2 2 (s ↵) + [(s ↵)2 + 2 ]m [(s ↵)2 + 2 ] This is to combine the solution of the constants for the partial fraction expansion with the solution of constants for the resulting cos and sin inverse expansions (given by the denominator) as was illustrated in Example 3. Brandon Sweeting

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Quadratic Factors Example Example 7. Determine L

1 {F },

F (s) =

where

(s 2

2s 2 + 10s . 2s + 5)(s + 1)

We begin by observing that the quadratic factor, s 2 2s + 5, is irreducible (its discriminant, i.e. ( 2)2 4(5) = 16, is negative). We must then rewrite this term in the form (s ↵)2 + 2 by completing the square: s2

2s + 5 = s 2

2s + 1 + 4 = (s

1)2 + 22 .

Since all denominator terms are unrepeated, we have the expansion form (s 2

2s 2 + 10s A(s 1) + 2B C = + . 2 2 2s + 5)(s + 1) (s 1) + 2 s +1

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Quadratic Factors Example cont. Multiplying both sides of this equation by the denominator of F , we have 2s 2 + 10s = [A(s Setting s =

2s + 5).

1, we obtain 2

thus C =

1) + 2B](s + 1) + C (s 2

10 = [A( 2) + 2B](0) + C (8),

1. Next, setting s = 1 (noting that C =

1) we have

2 + 10 = [A(0) + 2B](2) + C (4) = 4B

4

thus B = 4. Finally, setting s = 0 (noting our values for B and C ) we have 0 = A( 1) + 2B + C (5) =

A+8

5

thus A = 3. Brandon Sweeting

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Quadratic Factors Example cont. Recall that A = 3, B = 4 and C = F (s) =

(s 2

1. Putting things together gives

2s 2 + 10s 3(s = 2s + 5)(s + 1) (s

1) + 2(4) 1)2 + 22

1 . s +1

Therefore, by linearity of the inverse transform, we compute ⇢ 1) + 2(4) 1 1 1 3(s L {F } (t) = L (t) (s 1)2 + 22 s +1 ⇢ ⇢ 1) + 2(4) 1 1 3(s 1 =L (t) L (t) 2 2 (s 1) + 2 s +1 ⇢ ⇢ 3(s 1) 2 1 1 =L (t) + 4L (s 1)2 + 22 (s 1)2 + 22 = 3e t cos(2t) + 4e t sin(2t) e t .

(t)

e

t

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Further Examples Example Determine L

1 {F }

for the following functions

1

F (s) =

6s 2 13s + 2 s(s 1)(s 6)

2

F (s) =

s + 11 (s 1)(s + 3)

3

F (s) =

5s 2 + 34s + 53 (s + 3)2 (s + 1)

4

F (s) =

7s 2 + 23s + 30 (s 2)(s 2 + 2s + 5)

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Applied Di↵erential Equations I Brandon Sweeting

November 15, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 7: Laplace Transforms Section 7.5 - Solving Initial Value Problems

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Solving Initial Value Problems Having introduced the Laplace transform and having studied its essential properties, we’re now ready to use Laplace transforms to solve initial value problems for linear di↵erential equations. Though we’ve already discussed techniques for solving such problems in Chapter 4, the method we developed was quite cumbersome. Recall that to solve such IVPs, we needed to first find a general solution, after which we incorporated our initial conditions. When using the Laplace transform to solve IVPs, we’ll no longer need to first find a general solution. Another advantage of the Laplace transform is its use in solving equations in which there are discontinuous forcing terms (e.g. Fext from Section 7.1) and for equations with variable coefficients, systems of equations and even partial di↵erential equations.

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Method of Laplace Transforms We’ve already seen Laplace transforms in action to solve IVPs (see example from Section 7.4). Below is a general outline of the steps we followed:

Method of Laplace Transforms To solve an initial value problem: 1

Take the Laplace transform of both sides of the equation.

2

Use the properties of the Laplace transform and the initial conditions to obtain an equation for the Laplace transform of the solution and then solve this equation for the transform.

3

Determine the inverse Laplace transform of the solution by looking it up in a table or by using a suitable method (such as partial fractions) in combination with the table.

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Example 1 Example 1. Solve the initial value problem y 00

2y 0 + 5y =

8e

t

;

y 0 (0) = 12

y (0) = 2,

Since the above inequality is an equality of functions in t, taking the Laplace transform of both sides should yield an equality in s: L{y 00

2y 0 + 5y }(s) = L{ 8e

t

}(s)

Using linearity of L and our computed transform for exponentials gives L{y 00 }(s)

2L{y 0 }(s) + 5L{y }(s) =

8 . s +1

From here on, we’ll denote Y (s) := L{y }(s).

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Example 1 cont. Recalling the formula for the Laplace transform of higher-order derivatives along with our initial conditions y (0) = 2 and y 0 (0) = 12 we have L{y 0 }(s) = sY (s) 00

2

y (0) = sY (s)

L{y }(s) = s Y (s)

sy (0)

2,

0

y (0) = s 2 Y (s)

2s

12.

Substituting these back into our equation and solving for Y (s) yields, [s 2 Y (s)

2s

12]

2[sY (s) (s 2 (s 2

2] + 5Y (s) =

8 s +1

8 s +1 2s 2 + 10s 2s + 5)Y (s) = s +1 2s 2 + 10s Y (s) = 2 . (s 2s + 5)(s + 1) A 2s + 5)Y (s) = 2s + 8

LTEX

Brandon Sweeting

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Example 1 cont.

Having solved for Y (s), we finish by taking its inverse transform, i.e. ⇢ 2s 2 + 10s 1 1 y (t) = L {Y (s)} = L . (s 2 2s + 5)(s + 1) As we’ve already solved the inverse transform of the rational function above (this was the final example of section 7.4, where we used the method of partial fractions for irredicble terms) we have y (t) = 3e t cos(2t) + 4e t sin(2t)

e

t

.

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Example 2 Example 2. Solve the initial value problem y 00 + 4y 0

5y = te t ;

y 0 (0) = 0

y (0) = 1,

Again, with Y (s) := L{y }(s), we take the Laplace transform both sides L{y 00 }(s) + 4L{y 0 }(s)

5Y (s) =

1 (s

1)2

.

Recalling our formulas for the Laplace transform of higher derivatives and the values for our initial conditions we have L{y 0 }(s) = sY (s) 00

2

L{y }(s) = s Y (s)

y (0) = sY (s) sy (0)

0

1,

y (0) = s 2 Y (s)

s.

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Example 2 cont. Substituting these back into our equation and solving for Y (s) gives [s 2 Y (s)

s] + 4[sY (s)

1]

(s 2 + 4s (s + 5)(s

5Y (s) =

1 (s

1)2

5)Y (s) = s + 4 +

1

1)2 s 3 + 2s 2 7s + 5 1)Y (s) = (s 1)2 s 3 + 2s 2 7s + 5 Y (s) = . (s + 5)(s 1)3 (s

Having solved for Y (s), we finish by taking its inverse transform, i.e. ⇢ 3 2 7s + 5 1 1 s + 2s y (t) = L {Y (s)} = L . (s + 5)(s 1)3 Brandon Sweeting

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Example 2 cont. The partial fraction expansion has the form s 3 + 2s 2 7s + 5 A B C D = + + + . 3 2 (s + 5)(s 1) s + 5 s 1 (s 1) (s 1)3 Solving for these coefficients (as we had done in section 7.4) gives A=

35 , 216

B=

181 , 216

1 , 36

C=

1 D= . 6

Therefore, 35 Y (s) = 216

✓

◆ ✓ ◆ 1 181 1 + s +5 216 s 1

1 36

✓

1 (s

1)2

◆

1 + 12

✓

2 (s

1)3

◆

.

So by linearity of the inverse transform, we have y (t) = Brandon Sweeting

35 e 216

5t

+

181 t e 216

1 t 1 te + t 2 e t . 36 12

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Example 3 Example 3. Solve the initial value problem w 00

2w 0 + 5w =

8e ⇡

t

;

w (⇡) = 2,

w 0 (⇡) = 12.

To solve this equation using Laplace transforms, we must first move the initial conditions to t = 0. This is accomplished by means of the substitution y (t) := w (t + ⇡). By the chain rule, we have y 0 (t) = w 0 (t + ⇡),

y 00 (t) = w 00 (t + ⇡).

Therefore, replacing t by t + ⇡ in our initial equation gives the IVP y 00

2y 0 + 5y =

8e

t

;

y (0) = 2,

y 0 (0) = 12

which is the IVP we solved in Example 1.

LATEX Brandon Sweeting

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Example 3 cont.

Therefore, we have that y (t) = 3e t cos(2t) + 4e t sin(2t) Since y (t) = w (t + ⇡), we have w (t) = y (t w (t) = 3e t

⇡

cos(2(t

⇡)) + 4e t

= 3e t

⇡

cos(2t) + 4e t

⇡

⇡

e

t

.

⇡). Therefore, sin(2(t

sin(2t)

e⇡

⇡)) t

e

(t ⇡)

.

LATEX Brandon Sweeting

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Example 4 Example 4. Solve the initial value problem y 00 + 2ty 0

4y = 1;

y (0) = 0,

y 0 (0) = 0.

Here we see our first application of Laplace transforms to an equation with non-constant coefficients (which we’ve, as of yet, not been able to solve). As before, we set Y (s) := L{y }(s), and take the transform of both sides L{y 00 }(s) + 2L{ty 0 }(s)

1 4Y (s) = . s

By properties of the Laplace transform and our initial conditions, we have L{y 00 }(s) = s 2 Y (s) sy (0) y 0 (0) = s 2 Y (s) d d L{ty 0 }(s) = L{y 0 }(s) = [sY (s) y (0)] = ds ds

sY 0 (s)

Y (s).

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Example 4 cont. Substituting these expressions into our equation gives s 2 Y (s) + 2[ sY 0 (s)

Y (s)]

2sY 0 (s) + (s 2 ✓ 3 s 0 Y (s) + s 2

1 s 1 6)Y (s) = s ◆ 4Y (s) =

Y (s) =

1 . 2s 2

We no longer can solve for Y (s) algebraically, as our equation now involves the derivative of Y (s). However, what we have is a di↵erential equation in s for which Y (s) is the solution. Furthermore, this equation is first order linear, thus we may solve for Y (s) using the integrating factor µ(s) = e

R

(3/s s/2) ds

= e ln s

3

s 2 /4

= s 3e

s 2 /4

.

LATEX Brandon Sweeting

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Example 4 cont. Multiplying this equation by our integrating factor gives d d h 3 [µ(s)Y (s)] = s e ds ds

s 2 /4

i Y (s) =

Integrating both sides of this equation gives Z s s 2 /4 2 s 3 e s /4 Y (s) = e ds = e 2

s e 2

s 2 /4

s 2 /4

.

+ C.

Finally, solving for Y (s) we have 2

1 e s /4 Y (s) = 3 + C 3 . s s It remains to find the inverse transform of Y (s).

LATEX Brandon Sweeting

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Example 4 cont. We must first make the following observation: if f is of exponential order ↵, then for all s ↵ we have the following estimate for its transform: Z 1 M |L{f }(s)|  M e (↵ s)t dt = ML{e ↵t }(s) = . s ↵ 0 Therefore, by the squeeze theorem, it follows that lim L{f }(s) = 0.

s!1

Consequently, we must have that C = 0 in the formula for Y (s) (since for any other value of C the above property will be violated); hence, ⇢ 1 t2 1 1 y (t) = L {Y (s)}(t) = L (t) = . s3 2

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Applied Di↵erential Equations I Brandon Sweeting

November 15, 2022

LATEX Brandon Sweeting

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Overview

1

Chapter 7: Laplace Transforms Section 7.6 - Transforms of Discontinuous Functions

LATEX Brandon Sweeting

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Discontinuous Functions In Section 7.2, when we were using the formula to compute Laplace transforms of function in t directly, we enountered the following discontinuous function: 8 > 0 < t < 5,

: 4t e , 10 < t. which we found had as its Laplace transform: F (s) =

2 s

5s

2e s

+

e

10(s 4)

s

4

for

s > 4.

While it’s a straightforward matter to compute such transforms, going the opposite direction (i.e. computing the inverse transform of F (s) where L 1 {F (s)} happens to be discontinuous) is much more difficult. Our techniques from Section 7.4 don’t apply to such F as given above. LATEX Brandon Sweeting

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Motivating Example The following problem depicts a situation in which we’ll need to invert a function whose inverse transform will most likely be continuous.

A Mixing Problem (Section 7.1) Figure 7.1 depicts a mixing problem with valved input feeders. At time t = 0, valve A is opened, delivering 6 L/min of a brine solution containing 0.04 kg of salt per liter. At t = 10 min, valve A is closed and valve B is opened, delivering 6 L/min of brine at a concentration of 0.02 kg/L. Initially, 30 kg of salt are dissolved in 1000L of water in the tank. The exit valve C, which empties the tank at 6 L/min, maintains the contents of the tank at constant volume. Assuming the solution is kept wells stirred, determine the amount of salt in the tank at all times t > 0. Naturally, the solution y (t) will be discontinuous given the dynamics of the problem. The sudden valve change will cause an abrupt change in the di↵erential equation and thus the solution. LATEX Brandon Sweeting

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Motivating Example Depiction

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Step Functions Before we begin about detailing a method of Oliver Heaviside for finding inverse transforms of such functions, we require two definitions:

Unit Step Function Definition 5. The unit step function u(t) is defined by ( 0, t < 0, u(t) := 1, 0 < t. We can manipulate the unit step function in to move this discontinuity; to move the discontinuity from t = 0 to t = a, we simply shift the argument by a, i.e. we take the function u(s a). By multiplying our unit step function by a constant, we change change its height.

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Window Functions The next type of function we consider is essential; such functions will allow us to isolate a behavior of a given function on a specific interval (a, b).

Rectangular Window Function Definition 6. The rectangular window function ⇧a,b (t) is defined by

⇧a,b (t) := u(t

a)

u(t

8 >

: 0,

t < a, a < t < b, b < t.

When we multiply any given function, say f (t), by the rectangular window function for an interval, say (a, b), this produces a function that’s zero everywhere except the interval (a, b), on which it’s equal to f (t). The key result is that any piecewise continuous function can be expressed in terms of window and step functions, as the following example shows.A

LTEX

Brandon Sweeting

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Example 1. Example 1. Express u on [0, 1) in terms of window & step functions: 8 > 3, t < 2, > > >

t, 5 < t < 8, > > > :t 2 /10, 8 < t.

The function u has di↵erent behavior on the intervals (0, 2), (2, 5), (5, 8) and (8, 1). For the first three, we can use window functions and for the last interval we can use a step function. Therefore, u(t) = 3⇧0,2 (t) + 1⇧2,5 (t) + t⇧5,8 (t) + (t 2 /10)u(t

8).

With this representation in hand, we now see how to compute Laplace transforms and inverse transforms of functions involving discontinuities.

LATEX

Brandon Sweeting

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Transform of Step/Window Functions

Translation in t Theorem 8. Let F (s) = L{f }(s) exist for s > ↵ L{f (t

a)u(t

a)}(s) = e

as

0. Then for a > 0, F (s),

and, conversely, an inverse Laplace transform of e L

1

{e

as

F (s)}(t) = f (t

a)u(t

as F (s)

is given by

a).

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Transform of Step/Window Functions Proof. By the definition of the Laplace transform and a change of variable, Z 1 L{f (t a)u(t a)}(s) = e st f (t a)u(t a) dt Z0 1 Z 1 = e st f (t a) dt = e as e sv f (v ) dv a 0 Z 1 as sv =e e f (v ) dv = e as F (s) 0

In practice, it’s more common to encounter the problem of computing a transform of a function like g (t)u(t a) rather than f (t a)u(t a). To compute L{g (t)u(t a)}, we simply identify g (t) with f (t a) so that f (t) = g (t + a). Then the equation we derived previously gives: L{g (t)u(t Brandon Sweeting

a)}(s) = e

as

L{g (t + a)}(s).

AMS Open Math Notes: Works in Progress; Reference # OMN:202212.111348; Last Revised: 2022-12-28 21:03:13

Applied Di↵erential Equations I

November 15, 2022

LATEX 10 / 15

Example 2. Example 2 Determine the Laplace transform of t 2 u(t

1)

To apply the equation we derived previously, we take g (t) = t 2 and a = 1. Then g (t + a) = g (t + 1) = (t + 1)2 = t 2 + 2t + 1. Now the Laplace transform of g (t + a) is L{g (t + a)}(s) = L{t 2 + 2t + 1}(s) =

2 2 1 + + . s3 s2 s

Therefore, we have that 2

L{t u(t

1)}(s) = e

s



2 2 1 + + . s3 s2 s

LATEX Brandon Sweeting

AMS Open Math Notes: Works in Progress; Reference # OMN:202212.111348; Last Revised: 2022-12-28 21:03:13

Applied Di↵erential Equations I

November 15, 2022

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Example 3. Example 3. Determine L

1

⇢

2s

e s

To use the translation property, we first write e 2s /s 2 as e as F (s). For this, we can take e as = e 2s and F (s) = 1/s 2 ; thus, a = 2 and ⇢ 1 1 f (t) = L (t) = t. s2 It now follows from the translation property that ⇢ 2s 1 e L (t) = f (t 2)u(t 2) = (t s2

2)u(t

2).

LATEX Brandon Sweeting

AMS Open Math Notes: Works in Progress; Reference # OMN:202212.111348; Last Revised: 2022-12-28 21:03:13

Applied Di↵erential Equations I

November 15, 2022

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Example 4. Example 4. Solve the IVP: ( 0, t < 4, 00 y +y = 2, 4