Boundary Value Problems Part 2

7/29/2019 Boundary Value Problems Part 2

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ACM 100bBoundary value problems and the Sturm-Liouville ODE - part 2

Dan Meiron

Caltech

February 6, 2013

D. Meiron (Caltech) ACM 100b - Methods of Applied Mathematics February 6, 2013 1 / 25
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Recap - boundary value problems

The general solution to the heat problem was

(x, t) =

n=1

Bnexp(n22t) sin(nx).

which 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)

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Recap - boundary value problems

So we would have a solution that satisfies all the conditions if wecould 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 seriesconverges to 0(x) as n?

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

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The Sturm-Liouville ODE

We will answer all these questions shortly.

But what we want to emphasize right now is that the type of ODEproblem we just solved is actually quite common.

It turns out that the ODE we solved above in the x-direction is a

special case of a second order ODE boundary value problem

called the Sturm-Liouville problem.

The Sturm-Liouville ODE is given by

d

dx

p(x)

dy

dx

q(x)y(x) + r(x)y(x) = 0, a< x< b,

You will also see this ODE written as

d

dx

p(x)

dy

dx

+ q(x)y(x) = r(x)y(x), a< x< b,

This is called the positive definite form.



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Boundary conditions for the Sturm-Liouville ODE

The boundary conditions are homogeneous

c1y(a) + c2y(a) = 0,

d1y(b) + d2y(b) = 0

For the boundary conditions

c1y(a) + c2y(a) = 0,

d1y(b) + d2y(b) = 0

the coefficients c1, c2, d1, d2, are assumed to be real constants.The boundary conditions above are said to be separated because

they provide conditions on only one end point at a time.

Later on we will relax this a bit.

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The Sturm-Liouville ODE

We will assume in the ODE

d

dx

p(x)

dy

dx

q(x)y(x) + r(x)y= 0, a< x< b,

that the coefficient functions p(x), q(x) and r(x) are all continuousin the interval a x b.

We also assume that p(x) is also continuous in this interval.

Most importantly, we will assume that p(x) and r(x) are strictlypositive over the interval a x b.

And, as usual, there is no loss of generality if we restrict our

attention to a specific interval so we will assume in what follows

that a= 0 and b= 1.

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Regular vs. singular Sturm-Liouville ODE

Note we have already insisted in the S-L ODE

ddx

p(x)dy

dx

q(x)y(x) + r(x)y= 0, a< x< b,

that p(x) and r(x) be strictly positiveThere are a few more restrictions that we will impose and then

relax later.

First well insist the boundary conditions be of the separable form

c1y(a) + c2y(a) = 0,

d1y(b) + d2y(b) = 0

Second, we insist that the domain a x b be finiteYou can see that if p(x) vanishes our ODE becomes singularBut the ODE will also be singular if the domain is made infinite

A S-L problem on a finite domain with separable boundary

conditions and p(x) > 0 and w(x) > 0 is called a regular

Sturm-Liouville problemD. Meiron (Caltech) ACM 100b - Methods of Applied Mathematics February 6, 2013 9 / 25
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Another example of the S-L ODE - the Bessel equation

We will examine a Sturm-Liouville ODE that does not have simple

solutions like sines and cosines

This is the Bessel equation

It comes up when we solve the heat equation in cylindrical

coordinates as we will show later

The Bessel ODE is given by

d2y(x)

dx2+

1

x

dy(x)

dx+

2

m2

x2

y(x) = 0 x1 x x2.

Note as written its not in the typical S-L form which is

d

dx

p(x)

d

dxy(x)

+ q(x)y(x) = r(x)y(x)

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Boundary conditions for the Bessel equation

Because

p(x) = x q(x) =m2

x r(x) = x

we see that p(x) > 0 and r(x) > 0 exceptwhere x= 0 which is asingular point

So for now well take our domain x1 x x2 to be 1 x 2

This is safely away from x= 0 so the ODE is nonsingularFor boundary conditions well take

y(1) = 0 y(2) = 0.

which are of separable type.

This problem is now a regular S-L ODE problem

Well take m= 0 for simplicity

d

dx

x

dy

dx

+ 2xy(x) = 0. 1 x 2 y(1) = 0 y(2) = 0

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The solutions of the Bessel equation

The solutions of this ODE are new functions called Bessel

functionsThey are not simple functions likes sines and cosines but we will

see shortly that they share some important properties with sines

and cosines.

The ODE is second order so has two linearly independent

solutionsIt has a regular singular point at x= 0.

The Frobenius theory tells us there is one singular solution that

blows up at x= 0 and one regular solution that does not blow up

at the originThe regular solution is labeled J0(x)

The singular solution is labeled Y0(x)

Note that for our purposes since our domain is 1 x 2 both of

these linearly independent solutions are smooth over this interval.

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Solution of the Bessel S-L problem

For our purposes all we need to know is that J0 and Y0 are linearly

independent

The general solution to

d

dx

x

dy

dx

+ 2xy(x) = 0.

is

y(x) = c1J0(x) + c2Y0(x).

Now in order to satisfy the boundary conditions we must have

C1J0(x1) + C2Y0(x1) = 0,

C1J0(x2) + C2Y0(x2) = 0.or using that x1 = 1 and x2 = 2 we have

C1J0() + C2Y0() = 0,

C1J0(2) + C2Y0(2) = 0.



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Homogeneous solutions of the Bessel S-L problem

Note this system of equations is homogeneous

So in general there is only the trivial solution C1 = C2 = 0But if for some special values of we could get the determinant

J0() Y0()J0(2) Y0(2)

to vanish, we could get nontrivial special solutions.

But because the system is homogeneous they would not be

unique.

So the question is - does this determinant

() =

J0() Y0()J0(2) Y0(2) = J0()Y0(2) Y0()J0(2)

vanish for some values of ?

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The determinant for our S-L problem

It is not easy to see where this determinant

() = J0()Y0(2) Y0()J0(2)

might vanish

The Bessel functions are not simple things like sines and cosines.

So we resort to computing it numerically and evaluating the

determinant for various values of .



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Plotting the determinant

The determinant () is plotted below.

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Locations of the roots of the determinant

As can be seen, the determinant oscillates and crosses zero at

various values of .

At these values of , we expect we can get nontrivial solutions that

satisfy the boundary conditions.

While these are not simple values, like = n, n= 1,2, 3 . . ., wesee that there is some structure that can be seen in the locations

where the determinant vanishes.Below we calculate the location of the first 20 zeros of the

determinant:



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The zeros take on a simple pattern as n

If we calculate the crossings of adjacent zeroes and see how far

apart they are, we can see that the spacing of adjacent zeroes

approaches a constant.

That constant seems to be getting close to .

In fact the numbers themselves seem to be approaching = nfor n large

This is similar to the values of we calculated when we solved the

heat equation.

It will turn out this is not an accident.

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Th l ti f th l t l f


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The solution for the lowest value of

Figure: The solution corresponding to the first value of for which thedeterminant vanishes


Th l ti f th 6th t
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The solution for the 6th root

Figure: The solution corresponding to the sixth value of for which thedeterminant vanishes


Th l ti f th 11th t
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The solution for the 11th root

Figure: The solution corresponding to the 11th value of for which thedeterminant vanishes

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Overview of the solutions


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Overview of the solutions

We can see that these solutions look increasingly sinusoidalLater in a quantitative sense which we will make precise, they do

get increasingly close to sin mx (times an envelope function

which modulates the amplitude)

Here m describes the m+ 1th value of for which thedeterminant vanishes.

This too is not an accident

Its a general feature of the S-L problem which we will

demonstrate.

In order to see why this ODE is so special, we first have to derivean important identity which allows us to understand the properties

of the solutions of this boundary value problem.

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Boundary Value Problems Part 2