Boundary Value Problems Part 1

7/31/2019 Boundary Value Problems Part 1

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ACM 100cBoundary value problems and the Sturm-Liouville ODE - part 1

Dan Meiron

Caltech

April 4, 2012

D. Meiron (Caltech) ACM 100c - Methods of Applied Mathematics April 4, 2012 1 / 20
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Boundary value problems

So far we have explored linear (and nonlinear) ODEs but we have

focused on the initial value problems.That is we looked at equations of the form

dx

dz= A(z)x

where A is an n n matrix

In order to consider some unique solution we applied nconditions

at some initial point typically taken to be z= 0:

x(z= 0) = x0.

Under these conditions, we showed the existence and uniqueness

of solutions provided A(z) obeyed various mild smoothnesscriteria.

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In many applications, we are asked to solve the same ODE

dx

dz= A(z)x,

But we want the solution over some fixed domain z0 < z< z1rather than some neighborhood of z0

Most importantly not all n conditions are given at z= z0.Instead, some are applied at z= z0 and some at z= z1.

Of course, for any hope of uniqueness in the linear case we must

have a total of nsuch conditions.

Such problems are called boundary value problems in contrast tothe initial value problems we have studied up till now.

In the cases we will study we will write the system as an nth order

ODE where n is typically 2

But the theory we will present holds for the system as well.

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The theory of existence and uniqueness for boundary value

problems is considerably more complicated.This is to be expected because many of the guarantees we had

for initial value problems are not present for boundary value

problems.

Consider a simple example of a second order homogeneous ODE

y + p(z)y + q(z)y= 0 z0 z z1

Suppose we know the general solution:

y(z) = c1y1(z) + c2y2(z).

Well assume that the coefficient functions are nice and smooth for

any z

So we can be assured the functions y1(z) and y2(z) are alsosimilarly nice and smooth in the region z0 z z1.

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Suppose we ask for a solution subject to the following conditions:

y(z= z0) = a, y(z= z1) = b.

This is different from what we have done previously

We are now asking that the solution satisfy two conditions as

before.

But they both involve the value of the solution at the two boundary

end points.

Plugging these conditions in, we get a 2 2 system to solve for c1and c2:

c1y1(z0) + c2y2(z0) = a,

c1y1(z1) + c2y2(z1) = b.

Whether this system has a solution depends on the values the

solutions take on at the boundary.

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Now consider solving this 2 2 linear system:

c1y1(z0) + c2y2(z0) = a,

c1y1(z1) + c2y2(z1) = b.

For example, suppose neither aor b are zero.

In that case we will have a solution as long asy1(z0) y2(z0)y1(z1) y2(z1) = 0,

In turn this clearly depends on what happens at the boundary and

the values the solutions take on there.Suppose, on the other hand we had a= b= 0.

Then, in general, we would expect the trivial solution y(z) = 0

This is because if the above determinant did not vanish, we would

have to take c1 = c2 = 0 which is just the trivial solution

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On the other hand, it might be that in some cases we did get that

y1(z0) y2(z0)y1(z1) y2(z1) = 0.

This might happen depending on the equation and the locations of

the boundaryIn that case we would get nontrivial solutions but they would not

be unique.

We see then that such problems are harder to analyze.

They seem to depend on matrices such as

y1(z0) y2(z0)y1(z1) y2(z1).

which are really about global information as regards the solution.



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In contrast, linear initial value problems depend on the Wronskian

determinant.For example for a second order ODE initial value problem

y + p(z)y + q(z)y= 0 z z0

the Wronskian is given by

W(z) =

y1(z0) y2(z0)y1(z0) y2(z0) = 0.

Abels theorem guarantees that this never vanishes as long as the

matrix coefficients of a linear system are smooth.

Boundary value problems (BVP) turn up in many applications andwe will explore quite a few of these in ACM 100c.

One particular class of BVP the Sturm-Liouville ODE and its

associated boundary value problems is very important.

We will motivate this next.D. Meiron (Caltech) ACM 100c - Methods of Applied Mathematics April 4, 2012 8 / 20


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Example of a BVP - the heat equation

To motivate the importance of BVPs and also give a preview of apartial differential equation (PDE) we consider the solution of the

heat equation in one space dimension.

The heat equation is a partial differential equation that describes

the evolution of heat in a uniform medium.

The equation is given by

(x, t)

t= D

2(x, t)

x2

We want to solve it in the domain

0 x 1, t > 0.

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We wont worry for now about where this equation came from -

well derive it later.

Here x describes the length along some ideal rod of material that

extends from x= 0 to x= 1.

(x, t) is the temperature of the rod

D is called the diffusivity, a measure of how well heat diffusesthrough the rod.

Given some initial distribution of heat, we want to see how it

evolves for t> 0.

So at t= 0 we have

(x, t= 0) = 0(x).

This is the initial condition for the problem.

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We also need to say what happens at the edges of the rod during

the time we are interested in getting the evolution of the

temperature.

The PDE wont have a unique solution unless we do this.

Intuitively this makes sense since the ends could be insulating or

perhaps connected to some heat bath that maintains a constanttemperature.

To specify what happens at the ends we specify boundary

conditions at x= 0 and x= 1.

We will assume that the temperature of the rod is fixed at = 0by some mechanism:

(x= 0, t) = 0, (x= 1, t) = 0, t> 0.

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Solving the PDE

One very useful approach to such problems is called the method

of separation of variables.We will motivate this approach in great detail later on.

But for now we will just assume that

(x, t) = T(t)X(x),

or that the solution can be written in separable formIf we substitute this into the PDE (well omit the details right now) it

turns out to be possible to get such a solution if T(t) and X(x)satisfy the following ODEs:

d2

X(x)dx2

CX = 0,

dT(t)

dtDCT = 0

where C is a constant called the separationconstant.



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Solving the PDE

Now lets try to solve the ODEs

d2X(x)

dx2 CX = 0,

dT(t)

dtDCT = 0

There are many types of solutions to these equations as we willdiscuss later.

But were interested in the ones that make physical sense.

The diffusivity D is always positive (D> 0).

We can solve the ODE for the time variable t to give

T(t) = aexp(DCt),

where a is some arbitrary constant and C is again the separation

constant.



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Solving the PDE

Now, physically, we dont expect that the temperature will grow

exponentially without somehow externally heating the rod (whichwere not doing).

So we try to obtain solutions in which C< 0 meaning our

temperature decays in time which seems reasonable for this type

of problem.

So we set

C= 2,

Here is hopefully real or at least 2 has a positive real part.

Making this substitution in the ODE for X(x) we get the followingboundary value problem:

d2X(x)

dx2+ 2X = 0 X(0) = X(1) = 0.



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Solving the boundary value problem

We get the following boundary value problem:

d2X(x)dx2

+ 2X = 0 X(0) = X(1) = 0.

Now it looks like were stuck

We have a homogeneous ODE and homogeneous boundary

conditions.

It seems the only solution is X(x) = 0.

This is, in fact, true for almost any value of

But recall from our previous discussion that one can find nontrivial

solutions to even a homogeneous boundary value problem

This is because in the system we typically solve

c1y1(z0) + c2y2(z0) = 0,

c1y1(z1) + c2y2(z1) = 0,

it might happen that the matrix has zero determinant



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In that case we could get nontrivial solutions although they wont

be unique.

So we ask whether its possible to play with so that solutions

exist.

Up till now, we have not said anything about .

We can solve the ODE,

d2X(x)dx2

+ 2X = 0 X(0) = X(1) = 0.

since its a linear constant coefficient ODE to get the general

solution:

X(x) = c1 sin(x) + c2 cos(x)We want X(x) to satisfy the boundary conditions.

At x= 0 in order to have X(0) vanish, we must have c2 = 0.

Using this and trying to satisfy the other boundary condition at

x= 1 gives us the equation c1 sin() = 0.



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Now normally, the only solution to this equation

c1 sin() = 0

is c1 = 0

We only get the trivial solution X(x) = 0.

But we note that if we set so that the sin() vanishes then c1

could be arbitrary.We dont yet know what this means but at least we get some kind

of solution.

From the properties of the sine function we know that

sin(n) = 0 where n = n, n= 1,2,3, .We have a set of solutions of the following type:

n(x, t) = Bnexp(n22Dt) sin(nx) n= 1,2, 3, .

where the Bn are arbitrary constants.



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The general solution

We see that we went from no solutions to a countable infinity of

solutions.

The heat equation is a linear homogeneous PDE because each

term (like the time and space derivatives) appear linearly and

there is no inhomogeneous term

So it seems to be like a linear homogeneous ODE for which weknow we can use the principle of superposition of solutions

Because the whole problem is linear we can see that a more

general solution of this heat equation is a superposition of the

solutions we just found:

(x, t) =

n=1

Bnexp(n22t) sin(nx).



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Clearly, the sum

(x, t) =

n=1

Bnexp(n22t) sin(nx).

satisfies the boundary conditions, because the sines vanish at

x= 0,1

But there is also an initial condition to satisfy.At t= 0 we have some starting distribution of heat in the rod:

(x,0) = 0(x).

In order to satisfy this condition we substitute t= 0 into

(x, t) =

n=1

Bn exp(n22t) sin(nx) to get

0(x) =

n=1

Bn sin(nx)


Th l l i
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So we would have a solution that satisfies all the conditions if we

could figure out the coefficients Bn in the expression

0(x) =

n=1

Bn sin(nx)

As promising as this looks, there are some unanswered questions:

How does one determine Bn?

If you can determine Bn is there only one choice that works?

Even if there is a unique choice of Bn can you show the series

converges to 0(x) as n?

If it converges at t= 0 does it converge for t> 0?

The answers to these questions will take up the next few lectures.


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Boundary Value Problems Part 1