Show that the wave equation does not in general satisfy a

Question

Problem 1. Show that the wave equation does not, in general, satisfy a maximum principle.

Problem 2. For a solution u(x, t) of the wave equation

u_tt - u_xx = 0,

the energy density is defined as e = (u_t² + u_x² )/2 and the momentum density is p = u_tu_x.

(a) Show that e_t = p_x and p_t = e_x.

(b) Show that both e and p also satisfy the wave equation.

Problem 3. Consider a traveling wave u(x, t) = f(x - at), where f is a given function of one variable.

(a) Show that if u is a solution of the wave equation u_tt - c²u_xx = 0, then a = ±c (unless f is a linear function).

(b) Show that if u is a solution of the diffusion equation u_t - ku_xx = 0 then 

u(x, t) = C₁exp (-a/k(x - at)) + C₂

where C₁ and C₂ are arbitrary constants, and a ∈ R is arbitrary.

This exercise shows that the speed of propagation of travelling wave solutions of the wave equation is c, while for the heat equation, any arbitrary speed a ∈ R will do (i.e., we have infinite speed of propagation).

Problem 4. Here, we find a direct relationship between the heat and wave equations. Let u(x, t) solve the wave equation on the whole line, and suppose the second derivatives of u are bounded. Let

v(x, t) = c/√(4πkt). _-∞∫^∞e^-s2c2/4kt u(x, s) ds.

(a) Show that v(x, t) solves the diffusion equation u_t - ku_xx = 0.

(b) Show that lim_t→0 v(x, t) = u(x, 0).

Problem 5. Find a formula for the solution of

u_t - ku_xx = 0 for x > 0 and t > 0

u = 0 for x > 0 and t = 0

u = 1 for x = 0 and t > 0.

(1)

Problem 6. Suppose u solves the heat equation

u_t - ku_xx = 0

for x ∈ R and t > 0

u = ? for x ∈ R and t = 0.

(2)

Show that if ? is odd, then for each t > 0 the function x → u(x, t) is odd. Show that the analogous result is true when ? is even.

Problem 7. Consider the diffusion equation on the half-line with Robin boundary condition:

u_t - ku_xx = 0 for x > 0 and t > 0

u = ? for x > 0 and t = 0

ux - hu = 0 for x = 0 and t > 0.

(3)

(a) Motivated by the method of odd and even extensions, we might guess that the solution u(x, t) is of the form

u(x, t) = 1/√(4πkt). _-∞∫^∞e^-(x-y)2/4kt f(y)dy,

where

f(y) = ?(y), if y > 0

         g(-y), if y ≤ 0, (5)

and g is some (yet to be determined) function satisfying g(0) = ?(0). Recall that for the method of odd extension, g = -?, and for the method of even extension g = ?. Show that u given by (4) is the solution of (3) provided the function f - hf is odd.

(b) Show that f'- hf is odd if and only if

g'(y) + hg(y) = ?'(y) - h?(y) for all y ≥ 0.

(c) Suppose that ?(y) = y² . Find g by solving the ODE from part (b) with initial condition g(0) = ?(0) assuming h = 0. What happens when h = 0?