Introduction to Partial Differential Equations
Exercise Sheet 14

Ross Ogilvie      2nd December, 2024
______________________________________________________________________________________________________________________________

42.
Method of Descent

In this exercise we will apply the method of descent to solve the wave equation on 2 for a particular set of initial conditions. The idea is to help you understand the key ideas and notation of the method. It is a combination of results from Sections 5.1–5.

Consider the wave equation on 2 with initial conditions

t2u Δu = 0 on (x,t) 2 × (0,), u(x,0) = g(x) = χ[0,)(x1),tu(x,0) = h(x) = 0.
(a)
Suppose u is a solution of the wave equation for n = 2. Why does (x1,x2,x3,t)u(x1,x2,t) solve the wave equation on 3? (note, the Laplacians are different in different dimensions). (1 point)
(b)
Conversely, prove that a solution ū to the 3-dimensional wave equation that does not depend on x3 gives a solution to the 2-dimensional wave equation. (1 point)
(c)
By (a) and (b), we now must solve a wave equation on 3. The key to solving the 3-dimensional wave equation is to consider the (spatial-)spherical means
U(x,t,r) = 1 4πr2∂𝐵(x,r)ū(z,t)𝑑𝜎(z),

and likewise let G and H be the spherical means of and h¯ respectively. Explain why (x1,x2,x3) = χ[0,)(x1) and h¯ = 0 (or give the definition of bar). Show that

G(x,r) = { 0 for x1 r 1 2 x1+r r for |x1| r 1 for  r x 1 and H(x,r) = 0.

You may use the following geometric fact: for R < a < b < R, the surface area of the part of the sphere ∂𝐵(0,R) with a < x1 < b is 2𝜋𝑅(b a). (4 points)

(d)
We know by Lemma 5.2 that U obeys the Euler-Poisson-Darboux equation. Let Ũ(x,t,r) := 𝑟𝑈(x,t,r). Show that Ũ obeys the following PDE t2Ũ r2Ũ = 0 on (t,r) [0,) × [0,), Ũ(x,0,r) = 𝑟𝐺(x,r),tŨ(x,0,r) = 𝑟𝐻(x,r).

Note that there are no x-derivatives in this PDE, so we can think of it as a family of PDEs in the variables r,t parametrised by x. (2 points)

(e)
Thus we see that Ũ obeys the 1-dimensional wave equation on the half-line r [0,). This is solved by a trick using reflection, and the formula is at the end of Section 5.2. We only need the solution for small r, so it is enough to consider the case 0 r t. In this case, show
Ũ(x,t,r) = { 0 for x1 (t + r) 1 4(x1 + t + r) for  (t + r) x1 (t r) 1 2r for |x1| t r 1 4(x1 t + 3r)for t r x1 t + r r for  x 1 t + r.

(4 points)

(f)
Recover ū from Ũ using a certain property of spherical means. (3 points)

Observe that ū does not depend on x3. So by part (b) we have a solution to the 2-dimensional wave equation:

u(x1,x2,t) = { 0 for x1 < t 0.25for x1 = t 0.5 for  t < x1 < t 0.75for x1 = t 1 for x1 > t.

This solution has jump discontinuities, but this is unsurprising since the initial conditions also had them.

43.
Wave energy modes

Let us work with n = 1 for simplicity. If we apply the Fourier transform to the wave equation, we arrive at

2û t2 + 4π2k2û = 0.

This leads us to define the spectral energy density

2E(k,t) := |∂û ∂𝑡 |2 + 4π2k2 |û|2.
(a)
Through calculation, show that E is constant in t. Note that û is a complex valued function and it is important to respect complex conjugation: |g|2 = 𝑔ḡ. (1 point)
(b)
Consider a complex valued function g and its conjugate h = . Prove that ĥ(k) = ĝ(k)¯.
(1 point)
(c)
Using an equation on page 73 for the script, establish Plancherel’s theorem (1 point)
|h|2𝑑𝑥 =|ĥ|2𝑑𝑘.
(d)
Use this to show that the energy of a wave is constant. (2 points)