General Theory of Differential Equations - ACU Blogs€¦ · Solve the differential equation dy dx = 2 y x : Solution: This equation is separable and so we proceed as follows: dy

A B I L E N E C H R I S T I A N U N I V E R S I T Y

Department of Mathematics

General Theory of Differential EquationsSections 2.8, 3.1-3.2, 4.1

Dr. John EhrkeDepartment of Mathematics

Fall 2012

A B I L E N E C H R I S T I A N U N I V E R S I T Y D E P A R T M E N T O F M A T H E M A T I C S

Questions Of Existence and UniquenessPart of the difficulty in working with differential equations is understanding whatconstitutes a solution and knowing when a solution even exists. In this lecture, wewill try to address questions of existence and uniqueness as they relate to solutionsof linear differential equations. This will allow us to build up a general theorysupporting our study of differential equations throughout the semester. We willbegin with a small example to illustrate what can go wrong.

Example

Solve the differential equationdydx

= 2(y

x

).

Solution: This equation is separable and so we proceed as follows:

dyy

=2x

dx =⇒ ln y = 2 ln x + c

=⇒ eln y = e2 ln x+c

=⇒ y = ec ·(

eln x)2

=⇒ y = A · x2

Slide 2/27 — Dr. John Ehrke — Lecture 3 — Fall 2012


Specifying a SolutionThe solutions of this equation are parabolas of the form y = A · x2, but we have acouple of questions to still be resolved.

• Are there some initial conditions for which no solutions exist?

• If a solution does exist, over what interval(s) are they uniquely defined?

In answering the second question, consider what would happen if I asked you tosolve the IVP corresponding to y(0) = 0. This is bad for two reasons:

• There is no rate of change defined at the origin, since f (x, y) is undefined there.

• A solution could come in on one branch and leave on a completely differentbranch, i.e. a solution of the IVP corresponding to y(1) = 2 is the parabolay = 2x2, but this parabola passes through the origin and so cannot be extendeduniquely for negative x. The solution is unique only from [0,∞).

Even for the first question above, no solution exists for any initial value along they-axis other than the origin, because all the parabolas must pass through the origin.This example highlights just a few of the things that can go wrong when solvingdifferential equations.



Existence Uniqueness TheoremAt this point one might wonder if there is any way to predict the behavior ofsolutions of a given differential equation without solving the ODE. The answer tothis question is yes, and is summarized by the existence uniqueness theorem (EUT).

First Order Existence Uniqueness TheoremSuppose that both the function f (x, y) and its partial derivative ∂f/∂y are continuouson some rectangle R in the xy-plane containing the point (a, b) in its interior. Then,for some open interval I ⊆ R containing the point a, the initial value problem

dydx

= f (x, y), y(a) = b (1)

has one and only one solution that is defined on the interval I.

We will not do a complete proof of this result, but will introduce the machinerybehind the proof and a make a few remarks. Our text does not have a completeproof of this result but many undergraduate texts include such a proof in theappendices. If you are interested in reading through the complete proof let me knowand I would be happy to provide the proof for you.



Method of Successive ApproximationsThe approach we employ to establish the existence of solutions is called themethod of successive approximations and utilizes a series of approximationscalled Picard iterates. This method is based on the fact that a function y(x)satisfies the initial value problem, (1) on the open interval I containing x = aif and only if it satisfies the integral equation

y(x) = b +∫ x

af (t, y(t)) dt (2)

for all x ∈ I. In particular, if y(x) satisfies (2), then clearly y(a) = b, anddifferentiating both sides gives y′(x) = f (x, y(x)) as desired. It is importantto note that we traded solving a differential equation for an integralequation. This is almost always the case when one studies differentialequations, as we will see.



Picard Iterates (Demonstrating Existence)To attempt to solve (2), we begin with the initial function y0(x) ≡ b, andthen define iteratively a sequence y1, y2, y3, . . . of functions that we hopewill converge to the solution. Specifically, we let

y1(x) = b +∫ x

af (t, y0(t)) dt

y2(x) = b +∫ x

af (t, y1(t)) dt

...

yn+1(x) = b +∫ x

af (t, yn(t)) dt

Suppose we know that each of these functions {yn(x)}∞0 is defined on someopen interval (the same for each n) containing x = a, and that the limity(x) = lim

n→∞yn(x) exists at each point of this interval.



Continuing...Continuing from the previous slide, it follows that

y(x) = limn→∞

yn+1(x)

= limn→∞

[b +

∫ x

af (t, yn(t)) dt

]= b + lim

n→∞

∫ x

af (t, yn(t)) dt

= b +∫ x

af(

t, limn→∞

yn(t))

dt

= b +∫ x

af (t, y(t)) dt

This solution method works provided that:

• we can validate the interchange of limit operations above.• we can show there exists only this one solution.



Demonstrating UniquenessIf we assume at this point that we have established the existence of a solution viasuccessive approximations (there still is some work to be done to make thatargument solid), then we only need to show that the hypotheses of the theorem leadto a unique solution. When proving uniqueness you generally assume there existsanother solution to the IVP different from y(x), say φ(x) and look at the difference|y(x)− φ(x)| and try to reach a contradiction by showing this difference must bezero. In this case, that is done by proving that

|y(x)− φ(x)| ≤ A∫ x

0|y(t)− φ(t)| dt.

To do this requires the use of some heavy duty analysis and the determination of aLipschitz condition. A Lipschitz condition for two functions exists if there is apositive constant A such that

|f (x, y1)− f (x, y2)| ≤ A|y1 − y2|.

Arguments of this type are generally considered in a first semester ODE course ingraduate school. We will not pursue this problem beyond this point.



Applying the Existence Uniqueness Theorem

ExampleConsider the first order ODE, xy′ = y− 1. Identify for what values of (a, b), theinitial value problem xy′ = y− 1, y(a) = b has a unique solution, no solution, ormore than one solution.

Solution: Consider the point(s), (0, b) for all b 6= 1. At these points, x = 0, and sof (x, y) = (y− 1)/x fails to be continuous at x = 0 and so by the EUT there can be nosolution to the IVP,

dydx

=y− 1

x, y(0) = b.

Consider the point, (0, 1). When we consider the point (0, 1), f (x, y) = 0/0 which isindeterminate. So the EUT says nothing about this point in regards to existence. Wemay have solutions, or we may not have solutions. In the case, a solution existsmight it be unique? We will check the partial derivative,

∂

∂y

(y− 1

x

)=

1x

which is discontinuous at x = 0. This means, that solutions to the IVP y(0) = a whenthey exist are not guaranteed to be unique and in particular we see this for (0, 1).



Working With Complex NumbersTo achieve our goals this semester we will need to learn how to handlecomplex numbers and operations involving complex variables. A complexnumber is of the form z = a + bi where a, b ∈ R. The complex conjugate of z,denoted z is given by

z = a− bi.Remember, that multiplying a number by its complex conjugate turns thenumber to a real number. In particular,

z · z = a2 + b2.

The absolute value of z, denoted | z |, involves conjugation since

| z |=√

z · z =√

a2 + b2.

Moreover, using z and z we obtain

a = <(z) = z + z2

and b = =(z) = z− z2i

.

Recall that when working with complex numbers you will often need tomultiply or divide by i. Recall that i =

√−1 and so i2 = −1, i3 = −i and

i4 = 1.Slide 10/27 — Dr. John Ehrke — Lecture 3 — Fall 2012


Polar Representation of a Complex Number

z = a + bi = r cos θ + i(r sin θ)= r (cos θ + i sin θ)

= reiθ

Euler’s Formula: eiθ = cos θ + i sin θ

The form z = reiθ is called the polar form of the complex number a + bi.Under this description, a is called the real part, b the imaginary part when z iswritten in rectangular form. In polar form, r is called the modulus of z(denoted | z |), and θ is called the argument of z (denoted arg(z)).



The Complex ExponentialTypically, when we call something an exponential in mathematics, we havein mind two very specific properties that it must satisfy.

• An exponential should satisfy the exponential rule. That is, if e is anexponential, then ex · ey = ex+y.

• An exponential should satisfy the differential equation

dydt

= ay.

This says that an exponential has the property that it is self-replicatingunder differentiation (and integration).

Question?Does our definition of the complex exponential, eiθ as described by Euler’sformula define an exponential as stated above? eiθ = cos θ+ i sin θ is called acomplex-valued function of a real variable



Power Series RepresentationsThe final motivation for our description of cos θ + i sin θ as being thecomplex exponential is provided by our knowledge of the Taylor series forthe exponential, which is given by

ex = 1 +x1!

+x2

2!+

x3

3!+

x4

4!+ · · · .

Inserting x = iθ we obtain

eiθ = 1 +iθ1!

+(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!+ · · ·

=

(1− θ2

2!+θ4

4!+ · · ·

)+ i(θ − θ3

3!+ · · ·

).

This points to the fact that the real part of eiθ should be cos θ and theimaginary part is the Taylor series for sin θ.



The Art of ComplexifyingTo see how truly amazing and wondrous operations with complex variablescan be we will demonstrate two examples which take advantage of the artof complexifying (yes that’s a word).

ExampleObtain a solution to the integral

∫e−x cos x dx by complexifying the integral.

Solution: Looking at the integrand, we can rewrite it as the real part of acomplex variable,

e−x cos(x) = <(e−x · eix) .

By substituting this in for the real integrand we complexify the integral,∫e−x cos(x) dx = <

(∫e(−1+i)x dx

)= <

(e(−1+i)x

−1 + i

).

Taking the real part of this expression gives

<(

1−1 + i

e(−1+i)x)

= <(−1− i

2e−x(cos x + i sin x)

)=

e−x

2(sin x− cos x) .



Constant Coefficient EquationsIn this example, we will introduce a technique that we will lean on later inthe semester called exponential response.

ExampleObtain a general solution to the first order linear equation with constantcoefficients given by

x′(t) + 2x(t) = cos(t).

Solution: Complexifying the ODE leads to z′(t) + 2z(t) = eit. We change toz(t) to represent the fact that we expect the solution to such an equation tobe complex. We guess that the solution to such an equation is of the formz(t) = Aeit (i.e. matches the exponential response term up to a constant)where A is complex valued. Substituting our guess into the ODE givesA = 1/(2 + i) so

z(t) =1

2 + ieit ⇐⇒ <(z(t)) = x(t) =

25

cos t +15

sin t.



The Exponential Response TheoremThis method is so useful that we will establish the general case, that is, solvex′ + kx = Aert. The results of this are summarized in the theorem below:

Exponential Response TheoremThe exponential response to the first order linear constant coefficient ODE

x′ + kx = Aert

is given by

x(t) =A

r + kert (3)

as long as r + k 6= 0.

This formula is valid for complex values of r, but depending on the specificnature of the original oscillation you might have to obtain the real orimaginary part of (3).



Second Order Linear EquationsThe general form of a second order differential equation is given byy′′ = f (t, y, y′). In this lecture, we will consider the second order, linearhomogeneous ODE with constant coefficients, given by

ay′′ + by′ + cy = 0 (4)

We are going to make some assumptions about solutions to (4) which wewill come back and verify later.

1 A general solution (4) is of the form y(x) = c1y1 + c2y2.2 y1 and y2 are linearly independent solutions of (4). (y1 6= cy2, for some

constant c 6= 0)3 Initial conditions involve both y and y′ which help to determine c1 and

c2 above.4 The solutions, y1 and y2 are called the fundamental set of solutions for the

equation (4).



Making a guess...A natural question to ask at this point is, “What is the nature of thesolutions y1 and y2?” This question is best answered by making a guess (aswe saw in the last lecture) that such solutions are exponential in nature. So,to try and solve the equation

ay′′ + by′ + cy = 0, (5)

we guess that y(t) = ert is a solution. Substituting into (2) gives

ar2ert + brert + cert = 0 =⇒ (r2 + pr + q)ert = 0

where p = b/a and q = c/a and a 6= 0. This implies that y(t) = ert is asolution if and only if

r2 + pr + q = 0.



The Characteristic EquationSo we have determined that y = ert if and only if r2 + pr + q = 0. Thisequation is called the characteristic equation for the ODE, (2). There are threecases to consider, each of which leads to different solution behaviors.

1 distinct real roots r1, r2 ⇐⇒ y1 = er1t, y2 = er2t

2 complex conjugate pair r = a± bi⇐⇒ y1 = e(a+bi)t, y2 = e(a−bi)t

3 repeated real root, r gives you y1 = ert, but how do we find y2?

We will explore the physical interpretations of these cases, as well assolution techniques in the next lecture, but for the remainder of this lecture,we will discuss the general theory which underpins these techniques. Thequestions we need to answer are:

1 Why is c1y1 + c2y2 a solution to the equation, (2)?

2 Why are all solutions of this form?



Superposition Principle

Superposition PrincipleIf y1 and y2 are solutions to a linear homogeneous ODE, then y = c1y1 + c2y2is also a solution.

To make our lives easier in manipulating differential equations we oftenwant to think about differential equations as linear operators acting on y.We define the linear operator L corresponding to the ODE y′′ + py′ + qy = 0by

y′′ + py′ + qy = 0 =⇒ D2y + pDy + qy = 0

=⇒ (D2 + pD + q)y = 0=⇒ Ly = 0

We denote by D, the differential operator, D = d/dy. L = D2 + pD + q is alinear operator if it satisfies the following properties:• L(u1 + u2) = L(u1) + L(u2)

• L(αu) = αL(u), for some constant α.Slide 20/27 — Dr. John Ehrke — Lecture 3 — Fall 2012


Linear OperatorsUnder the definition of a linear operator on the previous slide, we show thatL = D2 + pD + q is linear.

L(u1 + u2) = (D2 + pD + q)(u1 + u2) = D2(u1 + u2) + pD(u1 + u2) + q(u1 + u2)

= u′′1 + u′′

2 + p(u′1 + u′

2) + qu1 + qu2

= (u′′1 + pu′

1 + qu1) + (u′′2 + pu′

2 + qu2)

= Lu1 + Lu2

We claim it is obvious that L(αu) = αL(u) by properties of the derivative. We arenow prepared to answer our first question: “If we suppose that y1 and y2 aresolutions to y′′ + py′ + qy = 0, does that automatically mean c1y1 + c2y2 is asolution?”

L(c1y1 + c2y2) = L(c1y1) + L(c2y2)

= c1L(y1) + c2L(y2)

= 0 (since y1 solves the ODE, then L(y1) = 0)

Thus, c1y1 + c2y2 solves y′′ + py′ + qy = 0 if y1 and y2 do.



Two solutions are better than one...In answering the second of our questions for this lecture, “Why do all solutions looklike c1y1 + c2y2?” we will show that this solution form is necessary to satisfy IVPs.

TheoremThe set of solutions {c1y1 + c2y2} is enough to satisfy any initial values imposed onthe equation y′′ + py′ + qy = 0. This means there does not exist some initialcondition y(x0) = a, y′(x0) = b such that a solution not of the form c1y1 + c2y2 is theonly solution that satisfies this IVP.

Proof: Let y(x0) = a, y′(x0) = b. Then applying these initial conditions to the solutionc1y2 + c2y2 we obtain the system of linear equations with unknowns c1, c2 given by

c1y1(x0) + c2y2(x0) = ac2y′

1(x0) + c2y′2(x0) = b

This system of equations is solvable if and only if the determinant,∣∣∣∣y1 y2

y′1 y′

2

∣∣∣∣ 6= 0.



The WronskianGiven y1, y2, the Wronskian of y1, y2, denoted W(y1, y2) is given by

W(y1, y2) =

∣∣∣∣ y1 y2y′1 y′2

∣∣∣∣ = y1y′2 − y2y′1

The Wronskian explains why linearly independent solutions are requiredsince if y2 = cy1, then

W(y1, y2) =

∣∣∣∣ y1 y2y′1 y′2

∣∣∣∣ = ∣∣∣∣ y1 cy1y′1 cy′1

∣∣∣∣ = cy1y′1 − cy1y′1 = 0

In this case, because W(y1, y2) = 0, then the resulting system of equationsfrom the initial conditions is not solvable. This leads to a very importanttheorem,

Abel’s TheoremIf y1, y2 are solutions to the ODE y′′ + py′ + qy = 0, then either

1 W(y1, y2) ≡ 0 for all x

2 or W(y1, y2) 6= 0 for all x



Abel’s TheoremProof: To prove Abel’s theorem we note that y1 and y2 satisfy

y′′1 + p(t)y′

1 + q(t)y1 = 0 (6)

y′′2 + p(t)y′

2 + q(t)y2 = 0 (7)

If we multiply the first equation by −y2, the second by y1 and add the results, weobtain

(y1y′′2 − y′′

1 y2) + p(t)(y1y′2 − y′

1y2) = 0. (8)

Next, we let W(t) = W(y1, y2)(t) and observe that W′(t) = y1y′′2 − y′′

1 y2. Then we canwrite (3) in the form

W′ + p(t)W = 0 (9)

which is a first order linear equation in W, and has solution

W(t) = c · exp(−∫

p(t) dt)

(10)

where c is a constant. The value of c depends on the fundamental solutions, y1 andy2. However, since the exponential function is never 0, W(t) is not 0 unless c = 0, inwhich case W(t) ≡ 0 for all t, which completes the proof.



Normalized SolutionsNot all solutions y1 and y2 are created equal. There are in fact, “bestsolution”, called normalized solutions which we will introduce below.

Normalized SolutionsTwo solutions Y1 and Y2 of the second order linear ODE, y′′ + py′ + qy = 0are called normalized if they satisfy the initial conditions,

Y1(0) = 1,Y′1(0) = 0Y2(0) = 0,Y′2(0) = 1

ExampleUsing the above definition, find the normalized solutions for y′′ + y = 0 andy′′ − y = 0. Compare your results. Remember to use the characteristicequation to determine your fundamental set of solutions.



Why do we care about normalized solutions?If Y1 and Y2 are normalized at x = 0, then the solution to the IVP

y′′ + py′ + qy = 0

with initial conditions y(0) = a, y′(0) = b is

y(x) = aY1 + bY2.

In example, we can simply read the solutions off straight from the ODE.This is highly desired by those in the engineering fields, and is generally anicer version of linearly independent solutions than what we have beenworking with thus far. The general solution

y(x) = c1Y1 + c2Y2

is called the normalized general solution.



Wrapping up the discussion...We are now ready to finish answering the second of our two questions, “Why allsolutions look like c1y1 + c2y2?” First, we state the second order existenceuniqueness theorem.

Second Order Existence Uniqueness TheoremGiven the second order linear differential equation

y′′ + p(x)y′ + q(x)y = 0, (11)

where p and q are continuous for all x in a rectangle about the initial point x0, there isexactly one solution satisfying the initial conditions, y(x0) = a and y′(x0) = b.

How does this help answer our question? Well, given a solution u(x) satisfyingu(0) = a, u′(0) = b, then by the existence uniqueness theorem above, there can onlybe one such solution, and we know that aY1 + bY2 is a solution, where Y1 and Y2 arenormalized solutions of (11). This says that u(x) is of the form c1y1 + c2y2 for somefundamental set of solutions, y1, y2, which is what we needed to show.


Documents

General Theory of Differential Equations - ACU Blogs€¦ · Solve the differential equation dy dx = 2 y x : Solution: This equation is separable and so we proceed as follows: dy