Duhamel's Principle for the wave equation

Question

attempt

a) To prove Duhamel's principle, we must show that (3) satisfies (1) and (2). One has

$$ u_t = U(x,0,t)$$ and so $u_{tt} = U_t(x,0,t) = f(x,t) $. Also, is $u_x = 0$? Can someone clarify this? I believe we have to differentiate under the integral, there is some trick that I forgot.

for b) then we see that I think we have to let $v(x,t) = u (x,t) - h(x) $ and observe that $v(x,0) = 0 $ and $v_t(x,0) = u_t(x,0)=g(x)$ so this choice doesnt work. How do we make this work?

Answer 1

To prove Duhamel's principle all you have to do is differentiate. Remember that if

$$F(t) = \int _{a(t)} ^{b(t)} f(x, t) \ \mathrm d x$$

then

$$F'(t) = \int _{a(t)} ^{b(t)} \frac {\partial f} {\partial t} (x, t) \ \mathrm d x + f(b(t), t) \ b'(t) - f(a(t), t) \ a'(t) \ .$$

In your case, this means

$$u_t (x,t) = \int _0 ^t U_t (x, t-s, s) \ \mathrm d s + \underbrace{ U(x, 0, t) } _{=0} - 0 = \int _0 ^t U_t (x, t-s, s) \ \mathrm d s$$

so that, when differentiating once more,

$$u_{tt} (x,t) = \int _0 ^t U_{tt} (x, t-s, s) \ \mathrm d s + U_t (x, 0, t) - 0 = c^2 \int _0 ^t U_{xx} (x, t-s, s) \ \mathrm d s + f(x,t) = \\ = c^2 \left( \frac \partial {\partial x} \right) ^2 \int _0 ^t U (x, t-s, s) \ \mathrm d s + f(x,t) = c^2 u_{xx} (x,t) + f(x,t)$$

which is exactly your equation.

To check the initial conditions,

$$u(x,0) = \int _0 ^0 U (x, 0-s, s) \ \mathrm d s = 0$$

and (using the calculation of $u_t$ performed above)

$$u_t (x,0) = \int _0 ^0 U_t (x, 0-s, s) \ \mathrm d s = 0 \ .$$

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To solve the problem

$$\left\{ \begin{eqnarray} & u_{tt} - c^2 u_{xx} &= f(x,t) \\ & u(x,0) &= g(x) \\ & u_t (x,0) &= h(x) \end{eqnarray} \right.$$

consider the two auxiliary problems

$$\left\{ \begin{eqnarray} & v_{tt} - c^2 v_{xx} &= 0 \\ & v(x,0) &= g(x) \\ & v_t (x,0) &= h(x) \end{eqnarray} \right.$$

and

$$\left\{ \begin{eqnarray} & w_{tt} - c^2 w_{xx} &= f(x,t) \\ & w(x,0) &= 0 \\ & w_t (x,0) &= 0 \ . \end{eqnarray} \right.$$

The first one is homogeneous, so it can be solved with the classical approach. The second one is non-homogeneous, so you reduce it with Duhamel's principle to the homogeneous problem

$$\left\{ \begin{eqnarray} & W_{tt} - c^2 W_{xx} &= 0 \\ & W(x,0,s) &= 0 \\ & W_t (x,0,s) &= f(x,s) \end{eqnarray} \right.$$

that you solve with the same classical method as above, and then

$$w(t,x) = \int _0 ^t W(x, t-s, s) \ \mathrm d s \ .$$

Check for yourself (it is easy) that if $v$ and $w$ solve the problems above, then their sum $v + w$ solves the problem given initially.

Finally, you show that this solution is unique by using some uniqueness theorem given to you in the lecture notes.