CHAPTER 18

FOUR SAMPLE MIDTERM EXAMINATIONS

These midterm examinations are actual exams held during the academic term, and each lasted 50 minutes.

18.1 MIDTERM EXAM 1

Exercise 18.1.

Find the values of λ² for which the boundary value problem

has nontrivial solutions.

Solution.

Case (i): λ = 0. In this case, the general solution to d²u/dx² = 0 is given by u(x) = Ax + B, and u(0) = 0 implies that B = 0, so that u(x) = Ax. The condition implies that

which implies that A = 0, and the boundary value problem has only the trivial solution in this case.

Case (ii): λ ≠ 0. In this case, the general solution to d²u/dx² + λ²u = 0 is given by

and u(0) = 0 implies that A = 0 so that u(x) = B sin λx. Now , implies that

so either B = 0 or cos(λπ/2) = 1. Therefore, a nontrivial solution exists if and only if we have cos(λπ/2) = 1; that is, λπ/2 = 2πn, where n ≠ 0 is an integer. The values of λ² for which the boundary value problem has nontrivial solutions are

for n = 1, 2, 3, ….

Exercise 18.2.

Let f(x) = cos² x, 0 ≤ x ≤ π, and f(x + 2π) = f(x) otherwise.

(a) Find the Fourier sine series for f on the interval [0, π].

Hint: For n ≥ 1,

(b) Find the Fourier cosine series for f on the interval [0, π].

(c) For which values of x in [0, π] do the series in (a) and (b) converge to f(x)?

Solution.

(a) Writing f(x) = cos² x ~ b_n sin nx, the coefficients b_n in the Fourier sine series are computed as follows:

if n ≠ 2 is even, while if n = 2, since sin 2x = 2 sin x cos x, we have

Therefore, b_n = 0 for all even n ≥ 2. If n is odd,

The Fourier sine series for f is therefore

for 0 < x < π

(b) Using the double-angle formula, we have

which is the Fourier cosine series for f. If you integrate cos² x cos nx, you will find

(c) From Dirichlet’s theorem, the Fourier sine series in part (a) converges to cos² x for all x ∈ (0, π) and converges to 0 for x = 0 and x = π. The Fourier cosine series in part (b) converges to cos² x for all x ∈ [0, π] since the series is actually finite.

Exercise 18.3.

Let v(x) be the steady-state solution to the boundary value-initial value problem

where r is a constant. Find and solve the boundary value problem for the steady-state solution v(x).

Solution. The steady-state solution v(x) satisfies the boundary value problem

and the general solution to the differential equation is

and

Therefore,

while

so that

The steady-state solution is therefore

for 0 ≤ x ≤ a.

18.2 MIDTERM EXAM 2

Exercise 18.4.

The neutron flux u in a sphere of uranium obeys the differential equation

for 0 < r < a, where λ is the effective distance traveled by a neutron between collisions, A is called the absorption cross section, and k is the number of neutrons produced by a collision during fission. In addition, the neutron flux at the boundary of the sphere is 0.

(a) Make the substitution

and show that v(r) satisfies d²v/dr² + μv = 0, 0 < r < a,

(b) Find the general solution to the differential equation in part (a) and then find u(r) that satisfies the boundary condition and boundedness condition:

(c) Find the critical radius, that is, the smallest radius a for which the solution is not identically 0.

Solution.

(a) Letting u = v/r, then

so that

and the differential equation for v(r) is

for 0 < r < a.

(b) The general solution to the differential equation in part (a) is

for 0 < r < a, and the solution to the neutron flux equation is

for 0 < r < a. Applying the boundedness condition, since

doesn’t exist, we need c₁ = 0, and the solution is

for 0 < r < a.

(c) Applying the boundedness condition

clearly, there is a nontrivial solution if and only if μa = nπ for some positive integer n. The critical radius is a = π/μ.

Exercise 18.5.

Show that

for –∞ < x < ∞.

Hint: Using the identity , find the Fourier cosine series of the function f(x) = sin x, for 0 ≤ x ≤ π, and then use Dirichlet’s theorem.

Solution. For 0 ≤ x ≤ π, the Fourier cosine series is

For n = 0,

for n = 1,

and for n > 1,

The Fourier cosine series for sin x on the interval 0 ≤ x ≤ n is therefore

Since cos nx is an even function, this is also the Fourier series for the even extension of sin x, that is, | sin x|, on the interval –π ≤ x ≤ π, so that

and since this is a continuous piecewise smooth function on –π ≤ x ≤ π such that | sin(–π)| = |sin π| = 0, then by Dirichlet’s theorem the Fourier series converges to |sin x| for all x ∈[–π, π], and

for all x ∈ [–π, π]. However, |sin x| is a continuous periodic function on the interval –∞ < x < ∞, so the Fourier series converges to | sin x| for all x, – ∞ < x < ∞.

Exercise 18.6.

Solve Laplace’s equation in the square 0 < x, y < π with the boundary conditions

Solution. We use separation of variables and assume a solution to Laplace’s equation

of the form u(x, y) = X(x) · Y(y). Separating variables, we have

and we obtain the two ordinary differential equations

Solving the complete boundary value problem for X, the eigenvalues and eigenfunctions are given by

for n > 1. The corresponding problem for Y is

with solutions

for n ≥ 1.

From the superposition principle, we write

and setting y = π, we need

and from the orthogonality of the eigenfunctions,

so that

for 0 < x < π, 0 < y < π.

18.3 MIDTERM EXAM 3

Exercise 18.7.

Consider the following eigenvalue problem on the interval (0,1):

(a) Explain the meaning of eigenvalue problem.

(b) Show that this eigenvalue problem is not of Sturm-Liouville type.

(c) Multiply the equation above by e^2x to obtain a Sturm-Liouville problem. Identify p(x), q(x), and σ(x).

(d) Use the Rayleigh quotient to show that the leading eigenvalue is positive; that is, λ₁ > 0.

(e) Find an upper bound for the leading eigenvalue.

Solution.

(a) The eigenvalue problem consists of finding the values of λ (eigenvalues) for which there are nontrivial solutions (eigenfunctions) satisfying both the differential equation and the boundary conditions.

(b) If the eigenvalue problem

were of Sturm-Liouville form, we would have

that is, p(x) = 1 and p′(x) = 2, which is impossible.

(c) If we multiply the differential equation by e^2x, we have

that is,

which is of Sturm-Liouville type with

for 0 < x < 1.

(d) The eigenvalue λ and corresponding eigenfunction u are related by the Rayleigh quotient:

and

since u′(0) = 0 and u′(1) = 0, so the Rayleigh quotient becomes

and all the eigenvalues of the boundary value problem are nonnegative.

To show that λ = 0 is not an eigenvalue, we can see immediately that since u and u′ are continuous on the interval [0,1], then

and this implies that u(x) = u′(x) = 0 for all 0 < x 1. Hence, there is no nontrivial eigenfunction corresponding to λ = 0. Alternatively, the equation u″ + 2u′ – u = 0, 0 < x < 1, has general solution

and from the boundary conditions u′(0) = 0 and u′(1) = 0, the solution is u(x) = 0 for 0 < x < 1. Therefore, the leading eigenvalue λ₁ > 0.

(e) To get an upper bound on λ₁, we try a quadratic test function v which satisfies the boundary conditions v′(0) = 0 and v(l) = 0, say

then the boundary conditions imply that a = b = 0, so that v(x) = c for 0 ≤ x ≤ 1. The Rayleigh quotient for this test function is

and since λ₁ is the minimum of R(u) as u runs over all twice continuously differentiable functions that satisfy the boundary conditions,

Exercise 18.8.

Consider Laplace’s equation for the steady-state temperature distribution in a square plate of side length 1.

Obviously, the solution is u(x, y) = 1 for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Show that this is the case by using separation of variables.

Solution.

Using separation of variables, assume that u(x,y) = X(x) · Y(y), and obtain the two boundary value problems

Note that we cannot meet the boundary conditions u(0,y) = u(l,y) = 1 for all 0 < y < 1 by requiring that X(0) = 1 and X(l) = 1.

We solve the complete boundary value problem for Y first since it has homogeneous boundary conditions. The eigenvalues are λ_n = n²π² and the corresponding eigenfunctions are

for n = 0,1,2, The corresponding solutions to the problem for X are

for n ≥ 1.

From the superposition principle, we have

for 0 < x < 1 and 0 < y < 1. Applying the boundary condition u(0,y) = 1, we get

and from the orthogonality of the eigenfunctions on the interval [0,1], this implies that

The solution is now

for 0 < x < 1 and 0 < y < 1. Applying the boundary condition u(l, y) = 1, we get

that is,

for 0 < y < 1. Again, from the orthogonality of the eigenfunctions on the interval [0,1], this implies that

Therefore, the solution is u(x, y) = 1 for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, as promised.

18.4 MIDTERM EXAM 4

Exercise 18.9.

Solve the normalized wave equation

Solution. Since the partial differential equation and the boundary conditions are linear and homogeneous, we can use separation of variables, and assuming a solution of the form u(x, t) = ϕ(x) · G(t), we get two ordinary differential equations:

where λ is the separation constant. We solve the spatial problem first since it has a complete set of boundary conditions. These are homogeneous Dirichlet conditions, so the eigenvalues and eigenfunctions are given by

for n ≥ 1, and the corresponding solutions to the temporal equation are

Using the superposition principle, we write the solution as an “infinite” linear combination of {ϕ_n · G_n}_n≥1; that is,

where the constants a_n and b_n are determined from the initial conditions

and

From the orthogonality of the eigenfunctions, we find

and the solution is

for 0 < x < π, and t > 0.

Exercise 18.10.

Consider the regular Sturm-Liouville problem

where h > 0. Show that there is a single negative eigenvalue λ₀ if and only if h < 1. Find λ₀ and the corresponding eigenfunction φ₀(x).

Hint: Assume that λ = –μ² for some real number μ≠ 0.

Solution. Following the hint, the differential equation becomes

with general solution

for 0 < x < 1. Applying the first boundary condition, we have

so that

Applying the second boundary condition, we have

so that

and we have a nontrivial solution if and only if

for some μ ≠ 0. However, the graphs of y = tanhμ and y = hμ intersect only at μ = 0 if h ≥ 1, and they intersect at exactly one positive value μ₀ if 0 < h < 1. Therefore, there is exactly one negative eigenvalue for this Sturm-Liouville problem if and only if 0 < h < 1, and the eigenvalue is

where μ₀ is the positive root of the equation tanhμ = hμ. The corresponding eigenfunction is

for 0 ≤ x ≤ 1.

Exercise 18.11.

Consider Laplace’s equation

in a semicircular disk of radius a centered at the origin with boundary conditions

Solve this problem using separation of variables.

Solution. Assuming a solution of the form u(r,0) = R(r) · φ(0), and separating variables, we have the following problems for R and φ:

We solve the θ-problem first. The eigenvalues and corresponding eigenfunctions are

for n ≥ 1. The corresponding r-equation is the Cauchy-Euler equation

with general solution

for n ≥ 1. The boundedness condition |R(0)| < ∞ requires that B_n = 0, so that

for n ≥ 1. Using the superposition principle, we write

for 0 r a, 0 < θ < λ, and from the boundary condition we want

From the orthogonality of the eigenfunctions on the interval [0,π], we have

so the solution is

for 0 < r < a, 0 < θ < π λ.