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4.1 Boundary value problems

Note: 2 lectures, similar to §3.8 in [EP], §10.1 and §11.1 in [BD]

4.1.1 Boundary value problems

Before we tackle the Fourier series, we need to study the so-called boundary value problems (or endpoint problems). For example, suppose we have

 ′′ x + λx = 0, x(a) = 0, x(b) = 0,

for some constant λ , where x(t) is defined for t in the interval [a, b] . Unlike before, when we specified the value of the solution and its derivative at a single point, we now specify the value of the solution at two different points. Note that x = 0 is a solution to this equation, so existence of solutions is not a problem here. Uniqueness of solutions is another issue. The general solution to  ′′ x + λx = 0 has two arbitrary constants present. It is, therefore, natural (but wrong) to believe that requiring two conditions guarantees a unique solution.

Example 4.1.1: Take λ = 1 , a = 0 , b = π . That is,

 ′′ x + x = 0, x (0 ) = 0, x (π) = 0.

Then x = sint is another solution (besides x = 0 ) satisfying both boundary conditions. There are more. Write down the general solution of the differential equation, which is x = Aco st + B sin t . The condition x(0) = 0 forces A = 0 . Letting x (π ) = 0 does not give us any more information as x = B sin t already satisfies both boundary conditions. Hence, there are infinitely many solutions of the form x = B sin t , where B is an arbitrary constant.

Example 4.1.2: On the other hand, consider λ = 2 . That is,

 ′′ x + 2x = 0, x(0) = 0, x(π) = 0.

Then the general solution is  √ -- √ -- x = A cos( 2t) + B sin( 2t) . Letting x(0) = 0 still forces A = 0 . We apply the second condition to find  √ -- 0 = x(π) = B sin( 2π) . As  √ -- sin( 2π) ⇔ 0 we obtain B = 0 . Therefore x = 0 is the unique solution to this problem.

What is going on? We will be interested in finding which constants λ allow a nonzero solution, and we will be interested in finding those solutions. This problem is an analogue of finding eigenvalues and eigenvectors of matrices.

4.1.2 Eigenvalue problems

For basic Fourier series theory we will need the following three eigenvalue problems. We will consider more general equations, but we will postpone this until chapter 5.

x′′ + λx = 0, x(a) = 0, x(b) = 0,
(4.1)

x′′ + λx = 0, x ′(a) = 0, x′(b) = 0,
(4.2)

and

x′′ + λx = 0, x(a) = x(b), x ′(a) = x′(b).
(4.3)

A number λ is called an eigenvalue of (4.1) (resp. (4.2) or (4.3)) if and only if there exists a nonzero (not identically zero) solution to (4.1) (resp. (4.2) or (4.3)) given that specific λ . A nonzero solution is called a corresponding eigenfunction.

Note the similarity to eigenvalues and eigenvectors of matrices. The similarity is not just coincidental. If we think of the equations as differential operators, then we are doing the same exact thing. For example, let L = − d2 dt2 . We are looking for nonzero functions f satisfying certain endpoint conditions that solve (L − λ)f = 0 . A lot of the formalism from linear algebra can still apply here, though we will not pursue this line of reasoning too far.

Example 4.1.3: Let us find the eigenvalues and eigenfunctions of

 ′′ x + λx = 0, x(0) = 0, x(π) = 0.

We have to handle the cases λ > 0 , λ = 0 , λ < 0 separately. First suppose that λ > 0 . Then the general solution to x′′ + λx = 0 is

 √ -- √ -- x = A cos( λt) + Bsin( λt).

The condition x(0 ) = 0 implies immediately A = 0 . Next

 -- 0 = x(π) = B sin(√ λπ).

If B is zero, then x is not a nonzero solution. So to get a nonzero solution we must have that  √ -- sin( λπ) = 0 . Hence, √-- λ π must be an integer multiple of π . In other words, √-- λ = k for a positive integer k . Hence the positive eigenvalues are k2 for all integers k ≥ 1 . Corresponding eigenfunctions can be taken as x = sin(kt) . Just like for eigenvectors, we get all the multiples of an eigenfunction, so we only need to pick one.

Now suppose that λ = 0 . In this case the equation is x ′′ = 0 , and its general solution is x = At + B . The condition x(0 ) = 0 implies that B = 0 , and x(π ) = 0 implies that A = 0 . This means that λ = 0 is not an eigenvalue.

Finally, suppose that λ < 0 . In this case we have the general solution

 √ --- √ --- x = A cosh( − λt) + Bsinh( −λt).

Letting x (0) = 0 implies that A = 0 (recall cosh0 = 1 and sinh 0 = 0 ). So our solution must be x = B sin h(√ −λt) and satisfy x(π) = 0 . This is only possible if B is zero. Why? Because sin hξ is only zero when ξ = 0 . You should plot sinh to see this fact. We can also see this from the definition of sinh. We get  eξ−e−ξ 0 = sin hξ = --2-- . Hence  ξ −ξ e = e , which implies ξ = − ξ and that is only true if ξ = 0 . So there are no negative eigenvalues.

In summary, the eigenvalues and corresponding eigenfunctions are

λk = k2 with an eigenfunction xk = sin(kt) for all integers k ≥ 1.

Example 4.1.4: Let us compute the eigenvalues and eigenfunctions of

x′′ + λx = 0, x ′(0) = 0, x′(π ) = 0.

Again we have to handle the cases λ > 0 , λ = 0 , λ < 0 separately. First suppose that λ > 0 . The general solution to  ′′ x + λx = 0 is  √-- √-- x = A cos( λt) + B sin( λt) . So

 ′ √-- √-- √ -- √-- x = − A λ sin( λt) + B λcos( λt).

The condition x′(0) = 0 implies immediately B = 0 . Next

 ′ √-- √-- 0 = x (π) = − A λ sin( λπ ).

Again A cannot be zero if λ is to be an eigenvalue, and  √ -- sin( λπ) is only zero if √-- λ = k for a positive integer k . Hence the positive eigenvalues are again k2 for all integers k ≥ 1 . And the corresponding eigenfunctions can be taken as x = cos(kt) .

Now suppose that λ = 0 . In this case the equation is x ′′ = 0 and the general solution is x = At + B so  ′ x = A . The condition  ′ x (0) = 0 implies that A = 0 . The condition  ′ x (π) = 0 also implies A = 0 . Hence B could be anything (let us take it to be 1). So λ = 0 is an eigenvalue and x = 1 is a corresponding eigenfunction.

Finally, let λ < 0 . In this case the general solution is  √--- √--- x = A cosh( − λt) + B sin h( − λt) and hence

 √--- √ --- √ --- √ --- x ′ = A − λsinh( −λt) + B −λ cosh( −λt).

We have already seen (with roles of A and B switched) that for this expression to be zero at t = 0 and t = π , we must have A = B = 0 . Hence there are no negative eigenvalues.

In summary, the eigenvalues and corresponding eigenfunctions are

 2 λk = k with an eigenfunction xk = cos(kt) for all integers k ≥ 1,

and there is another eigenvalue

λ0 = 0 with an eigenfunction x0 = 1.

The following problem is the one that leads to the general Fourier series.

Example 4.1.5: Let us compute the eigenvalues and eigenfunctions of

x′′ + λx = 0, x(− π) = x(π), x ′(−π ) = x′(π).

We have not specified the values or the derivatives at the endpoints, but rather that they are the same at the beginning and at the end of the interval.

Let us skip λ < 0 . The computations are the same as before, and again we find that there are no negative eigenvalues.

For λ = 0 , the general solution is x = At + B . The condition x(−π) = x(π) implies that A = 0 (Aπ + B = − Aπ + B implies A = 0 ). The second condition x′(− π) = x′(π) says nothing about B and hence λ = 0 is an eigenvalue with a corresponding eigenfunction x = 1 .

For λ > 0 we get that  √-- √ -- x = Aco s( λt) + B sin( λt) . Now

 √ -- √ -- √ -- √-- Aco s(− λπ) + B sin(− λπ) = A cos( λπ) + B sin( λ π). ◟--------------------------------◝ ◜--------------------------------◞ ◟----------------------------◝ ◜----------------------------◞ x(−π) x(π)

We remember that cos(−𝜃) = cos(𝜃) and sin(−𝜃) = − sin(𝜃) . Therefore,

 √ -- √ -- √-- √-- A cos( λπ) − B sin( λπ) = Aco s( λπ ) + B sin( λπ ).

Hence either B = 0 or sin(√ λπ) = 0 . Similarly (exercise) if we differentiate x and plug in the second condition we find that A = 0 or  √-- sin( λπ ) = 0 . Therefore, unless we want A and B to both be zero (which we do not) we must have  √ -- sin( λπ) = 0 . Hence, √ -- λ is an integer and the eigenvalues are yet again λ = k2 for an integer k ≥ 1 . In this case, however, x = A cos(kt) + Bsin(kt) is an eigenfunction for any A and any B . So we have two linearly independent eigenfunctions sin(kt) and cos(kt) . Remember that for a matrix we can also have two eigenvectors corresponding to a single eigenvalue if the eigenvalue is repeated.

In summary, the eigenvalues and corresponding eigenfunctions are

 2 λk = k with eigenfunctions co s(kt) and sin (kt) for all integers k ≥ 1, λ 0 = 0 with an eigenfunction x0 = 1.

4.1.3 Orthogonality of eigenfunctions

Something that will be very useful in the next section is the orthogonality property of the eigenfunctions. This is an analogue of the following fact about eigenvectors of a matrix. A matrix is called symmetric if  T A = A . 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 (4.1), (4.2) or (4.3) for two different eigenvalues λ1 and λ 2 . Then they are orthogonal in the sense that

∫ b x1(t)x2(t) dt = 0. a

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.

 ′′ ′′ x 1 + λ1x1 = 0 and x 2 + λ2x2 = 0.

Multiply the first by x2 and the second by x1 and subtract to get

(λ1 − λ2)x 1x 2 = x′2′x 1 − x2x′1′.

Now integrate both sides of the equation:

 ∫ b ∫ b ′′ ′′ (λ1 − λ2) a x 1x 2 dt = a x2 x1 − x2x1 dt ∫ b = d-(x′x1 − x2x′) dt a dt 2 1 [ ′ ′]b = x 2x 1 − x2x1 t=a = 0.

The last equality holds because of the boundary conditions. For example, if we consider (4.1) we have x1(a) = x1(b ) = x2(a) = x 2(b) = 0 and so x′2x1 − x2x′1 is zero at both a and b . As λ1 ⇔ λ2 , the theorem follows.

Exercise 4.1.1 (easy): Finish the proof of the theorem (check the last equality in the proof) for the cases (4.2) and (4.3).

The function sin(nt) is an eigenfunction for the problem x′′ + λx = 0 , x(0) = 0 , x(π) = 0 . Hence for positive integers n and m we have the integrals

∫ π sin(mt)sin(nt) dt = 0, when m ⇔ n. 0

Similarly

∫ π ∫ π 0 cos(mt)cos(nt) dt = 0, when m ⇔ n, and 0 cos(nt) dt = 0.

And finally we also get

∫ π ∫ π sin(mt)sin(nt) dt = 0, when m ⇔ n, and sin(nt) dt = 0, −π −π

∫ ∫ π π cos(mt)cos(nt) dt = 0, when m ⇔ n, and cos(nt) dt = 0, −π −π

and

∫ π cos(mt)sin(nt) dt = 0 (even if m = n). −π

4.1.4 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 chapter 5.

Theorem 4.1.2 (Fredholm alternative1). Exactly one of the following statements holds. Either

 ′′ x + λx = 0, x (a) = 0, x(b) = 0
(4.4)

has a nonzero solution, or

x′′ + 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 if λ is not an eigenvalue, the nonhomogeneous equation (4.5) has a unique solution for every right hand side. On the other hand if λ is an eigenvalue, then (4.5) 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. Let A be an n × n matrix. The Fredholm alternative then states that either (A − λI)⃗x = ⃗0 has a nontrivial solution, or (A − λI)⃗x = ⃗b has 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.

4.1.5 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 density ρ , for example in kg∕m . Let us put this problem into the xy -plane and both x and y are in meters. The x axis represents the position on the string. The string rotates at angular velocity ω , in radians∕s2 . Imagine that the whole xy -plane rotates at angular velocity ω . This way, the string stays in this xy -plane and y measures its deflection from the equilibrium position, y = 0 , on the x axis. Hence the graph of y gives 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 constant T in Newtons. Assuming that the deflection is small, we can use Newton’s second law (let us skip the derivation) to get the equation

 ′′ 2 Ty + ρω y = 0.

To check the units notice that the units of y′′ are m∕m2 , as the derivative is in terms of x .

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 and y(L) = 0 . See Figure 4.1.


PIC

Figure 4.1: Whirling string.


We rewrite the equation as  ρω2 y′′ +-T-y = 0 . The setup is similar to Example 4.1.3, except for the interval length being L instead of π . We are looking for eigenvalues of y′′ + λy = 0, y(0) = 0,y(L) = 0 where λ = ρω2- T . As before there are no nonpositive eigenvalues. With λ > 0 , the general solution to the equation is  √-- √ -- y = Acos( λx) + B sin( λx) . The condition y(0 ) = 0 implies that A = 0 as before. The condition y(L ) = 0 implies that  √ -- sin( λL ) = 0 and hence √ -- λL = kπ for some integer k > 0 , so

ρω 2 k2π2 ----= λ = --2-. T L

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 ρω2= k2π22 T L , then the string will “pop out” some distance B . We cannot compute B with the information we have.

Let us assume that ρ and T are fixed and we are changing ω . For most values of ω the string is in the equilibrium state. When the angular velocity ω hits a value  √-- ω = kπ√T L ρ , then the string pops out and has the shape of a sin wave crossing the x axis k − 1 times between the end points. When ω changes again, the string returns to the equilibrium position. The higher the angular velocity, the more times it crosses the x axis when it is popped out.

For another example, if you have a spinning jump rope (then k = 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”.

4.1.6 Exercises

Hint for the following exercises: Note that when λ > 0 , then  (√ -- ) cos λ(t − a) and  (√ -- ) sin λ(t − a) are also solutions of the homogeneous equation.

Exercise 4.1.2: Compute all eigenvalues and eigenfunctions of x′′ + λx = 0, x(a) = 0, x(b) = 0 (assume a < b ).

Exercise 4.1.3: Compute all eigenvalues and eigenfunctions of  ′′ ′ ′ x + λx = 0, x (a) = 0, x (b) = 0 (assume a < b ).

Exercise 4.1.4: Compute all eigenvalues and eigenfunctions of x′′ + λx = 0, x′(a) = 0, x(b) = 0 (assume a < b ).

Exercise 4.1.5: Compute all eigenvalues and eigenfunctions of x′′ + λx = 0, x(a) = x(b), x′(a) = x′(b) (assume a < b ).

Exercise 4.1.6: We skipped the case of λ < 0 for the boundary value problem x′′ + λx = 0, x(−π) = x(π), x′(−π) = x′(π) . Finish the calculation and show that there are no negative eigenvalues.

Exercise 4.1.101: Consider a spinning string of length 2 and linear density 0.1 and tension 3. Find smallest angular velocity when the string pops out.

Exercise 4.1.102: Suppose  ′′ x + λx = 0 and x(0) = 1 , x(1) = 1 . Find all λ for which there is more than one solution. Also find the corresponding solutions (only for the eigenvalues).

Exercise 4.1.103: Suppose x′′ + x = 0 and x(0) = 0 , x′(π) = 1 . Find all the solution(s) if any exist.

Exercise 4.1.104: Consider  ′ x + 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?

Exercise 4.1.105 (challenging): Suppose x ′′′ + λx = 0 and x(0) = 0 , x′(0 ) = 0 , x(1) = 0 . Suppose that λ > 0 . Find an equation that all such eigenvalues must satisfy. Hint: Note that  √ -- − 3 λ is a root of r3 + λ = 0 .

1Named after the Swedish mathematician Erik Ivar Fredholm (1866–1927).