# BOUNDARY VALUE PROBLEMS

## Fourier series and PDEs

### Orthogonality of eigenfunctions

Something that will be very useful in the next section is theorthogonalityproperty of the eigen- functions. This is an analogue of the following fact about eigenvectors of a matrix. A matrix is calledsymmetricifA =AT. Eigenvectors for two distinct eigenvalues of a symmetric matrix are orthogonal.The differential operators we are dealing with act much like a symmetric matrix. We, therefore, get the following theorem.

Theorem 4.1.1. Suppose that x1(t)and x2(t)are two eigenfunctions of the problem( ),( )or

( )for two different eigenvaluesλ1andλ2. Then they areorthogonalin the sense that

Z b

a

x1(t)x2(t)dt =0.

The terminology comes from the fact that the integral is a type of inner product. We will expand on this in the next section. The theorem has a very short, elegant, and illuminating proof so let us give it here. First, we have the following two equations.

x001 +λ1x1 =0 and x002 +λ2x2 =0.

Multiply the first byx2 and the second byx1and subtract to get

(λ1−λ2)x1x2= x002x1−x2x001.

Now integrate both sides of the equation: (λ1−λ2) Z b a x1x2dt = Z b a x002x1− x2x001 dt =Z b a d dt x 0 2x1− x2x 0 1 dt =h x20x1−x2x01 ib t=a= 0.

The last equality holds because of the boundary conditions. For example, if we consider ( ) we havex1(a)=x1(b)=x2(a)= x2(b)=0 and so x02x1−x2x01is zero at bothaandb. Asλ1 , λ2, the

theorem follows.

Exercise4.1.1(easy): Finish the proof of the theorem (check the last equality in the proof) for the

cases( )and( ).

The function sin(nt) is an eigenfunction for the problem x00+λx=0, x(0)= 0,x(π)=0. Hence for positive integersnandmwe have the integrals

Z π

0

Similarly Z π

0

cos(mt) cos(nt)dt =0, whenm,n, and

Z π

0

cos(nt)dt = 0. And finally we also get

Z π

−π

sin(mt) sin(nt)dt= 0, whenm, n, and

Z π

−π

sin(nt)dt =0, Z π

−π

cos(mt) cos(nt)dt =0, whenm,n, and

Z π −π cos(nt)dt = 0, and Z π −π

cos(mt) sin(nt)dt =0 (even ifm= n).

### Fredholm alternative

We now touch on a very useful theorem in the theory of differential equations. The theorem holds in a more general setting than we are going to state it, but for our purposes the following statement is sufficient. We will give a slightly more general version in .

Theorem 4.1.2(Fredholm alternative ). Exactly one of the following statements holds. Either x00+λx= 0, x(a)=0, x(b)= 0 (4.4) has a nonzero solution, or

x00+λx= f(t), x(a)=0, x(b)=0 (4.5) has a unique solution for every function f continuous on[a,b].

The theorem is also true for the other types of boundary conditions we considered. The theorem means that ifλis not an eigenvalue, the nonhomogeneous equation ( ) has a unique solution for every right hand side. On the other hand ifλis an eigenvalue, then ( ) need not have a solution for every f, and furthermore, even if it happens to have a solution, the solution is not unique.

We also want to reinforce the idea here that linear differential operators have much in common with matrices. So it is no surprise that there is a finite dimensional version of Fredholm alternative for matrices as well. LetA be ann×nmatrix. The Fredholm alternative then states that either (A−λI)~x=~0 has a nontrivial solution, or (A−λI)~x=~bhas a unique solution for every~b.

A lot of intuition from linear algebra can be applied to linear differential operators, but one must be careful of course. For example, one difference we have already seen is that in general a differential operator will have infinitely many eigenvalues, while a matrix has only finitely many.

4.1. BOUNDARY VALUE PROBLEMS 173

### Application

Let us consider a physical application of an endpoint problem. Suppose we have a tightly stretched quickly spinning elastic string or rope of uniform linear densityρ, for example inkg/m. Let us put

this problem into the xy-plane and both xandyare in meters. The xaxis represents the position on the string. The string rotates at angular velocityω, inradians/s. Imagine that the wholexy-plane

rotates at angular velocityω. This way, the string stays in thisxy-plane andymeasures its deflection from the equilibrium position,y = 0, on thexaxis. Hence the graph ofygives the shape of the string. We consider an ideal string with no volume, just a mathematical curve. We suppose the tension on the string is a constantT in Newtons. Assuming that the deflection is small, we can use Newton’s second law (let us skip the derivation) to get the equation

T y00+ρω2y=0.

To check the units notice that the units ofy00 arem/m2, as the derivative is in terms ofx.

Let L be the length of the string (in meters) and the string is fixed at the beginning and end points. Hence,y(0)=0 andy(L)=0. See .

L x

y y

0

Figure 4.1: Whirling string.

We rewrite the equation asy00+ρω2

T y=0. The setup is similar to on page 168,

except for the interval length beingLinstead ofπ. We are looking for eigenvalues of y00+λy= 0,y(0) =0,y(L)=0 whereλ= ρωT2. As before there are no nonpositive eigenvalues. Withλ >0, the general solution to the equation isy=Acos(√λx)+Bsin(√λx). The conditiony(0)=0 implies thatA=0 as before. The conditiony(L)=0 implies that sin(

√ λL)=0 and hence √ λL=kπfor some integerk >0, so ρω2 T = λ= k2π2 L2 .

What does this say about the shape of the string? It says that for all parameters ρ,ω,T not satisfying the above equation, the string is in the equilibrium position,y=0. When ρωT2 = k2Lπ22, then

the string will “pop out” some distanceB. We cannot computeBwith the information we have. Let us assume thatρandT are fixed and we are changingω. For most values ofωthe string is in the equilibrium state. When the angular velocityωhits a valueω = kπ

√ T

L√ρ, then the string pops

changes again, the string returns to the equilibrium position. The higher the angular velocity, the more times it crosses the xaxis when it is popped out.

For another example, if you have a spinning jump rope (thenk =1 as it is completely “popped out”) and you pull on the ends to increase the tension, then the velocity also increases for the rope to stay “popped out”.

### Exercises

Hint for the following exercises: Note that whenλ >0, then cos √λ(t−a)

and sin √λ(t−a) are also solutions of the homogeneous equation.

Exercise4.1.2: Compute all eigenvalues and eigenfunctions of x00+λx =0, x(a) =0, x(b) =0

(assume a<b).

Exercise4.1.3: Compute all eigenvalues and eigenfunctions of x00+λx=0, x0(a)=0, x0(b)=0

(assume a<b).

Exercise4.1.4: Compute all eigenvalues and eigenfunctions of x00+λx=0, x0(a)=0, x(b)=0

(assume a<b).

Exercise4.1.5: Compute all eigenvalues and eigenfunctions of x00+λx= 0, x(a)= x(b), x0(a)=

x0(b)(assume a< b).

Exercise4.1.6: We skipped the case ofλ <0for the boundary value problem x00+λx=0, x(−π)=

x(π), x0(−π)= x0(π). Finish the calculation and show that there are no negative eigenvalues.

Exercise4.1.101: Consider a spinning string of length 2 and linear density 0.1 and tension 3. Find

smallest angular velocity when the string pops out.

Exercise4.1.102: Suppose x00+λx=0and x(0)=1, x(1)=1. Find allλfor which there is more

than one solution. Also find the corresponding solutions (only for the eigenvalues).

Exercise4.1.103: Suppose x00+x=0and x(0)=0, x0(π)=1. Find all the solution(s) if any exist.

Exercise 4.1.104: Consider x0 +λx = 0 and x(0) = 0, x(1) = 0. Why does it not have any

eigenvalues? Why does any first order equation with two endpoint conditions such as above have no eigenvalues?

Exercise4.1.105(challenging): Suppose x000+λx= 0and x(0)=0, x0(0)=0, x(1)= 0. Suppose

thatλ >0. Find an equation that all such eigenvalues must satisfy. Hint: Note that−√3λis a root of

Outline

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