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In PDE Evans, 2nd edition, page 80 ...

2.4.2 Nonhomogenous problem

We next investigate the initial value problem for the nonhomogeneous wave equation \begin{cases} u_{tt} - \Delta u = f & \text{in } \mathbb{R}^n \times (0,\infty) \\ u=0, u_t=0 & \text{on } \mathbb{R}^n \times \{t=0\} \end{cases} Motivated by Duhamel's principle, we define $u=u(x,t;s)$ to be the solution of \begin{cases} u_{tt}(\cdot;s) - \Delta u(\cdot;s) = 0 & \text{in } \mathbb{R}^n \times (s,\infty) \\ u(\cdot;s)=0, u_t(\cdot;s)=f(\cdot;s) & \text{on } \mathbb{R}^n \times \{t=s\} \end{cases} Now set $$ u(x,t) := \int_0^t u(x,t;s) \, ds \, \, \, (x \in \mathbb{R}^n, t \ge 0)$$ Duhamel's principle asserts this is a solution of the nonhomogeneous wave equation\begin{cases} u_{tt} - \Delta u = f & \text{in } \mathbb{R}^n \times (0,\infty) \\ u=0, u_t=0 & \text{on } \mathbb{R}^n \times \{t=0\} \end{cases}

Now I am asked to prove that $u_{tt} - \Delta u = f$ in $\mathbb{R}^n \times (0,\infty)$. I try to compute for \begin{align} u_{t}(x,t) &=\underbrace{\require{cancel}{\cancelto{0}{u(x,t;t)}}+\int_0^t u_t(x,t;s) \, ds}_{\text{Differentiation under the integral sign}} = \int_0^t u_t(x,t;s) \, ds \\ u_{tt}(x,t)&=\underbrace{\require{cancel}{\cancelto{f(x,t)}{u_t(x,t;t)}}+\int_0^t u_{tt}(x,t;s) \, ds}_{\text{Differentiation under the integral sign}} = f(x,t)+\int_0^t u_{tt}(x,t;s) \, ds \end{align} Also, as $u_{tt}(\cdot,s)=\Delta u(\cdot;s)$, \begin{align} \Delta u(x,t) = \int_0^t \Delta u(x,t;s) \, ds = \int_0^t u_{tt}(x,t;s) \, ds \end{align} Thus, $$u_{tt}(x,t)-\Delta u(x,t) = f(x,t) \, \, \, (x \in \mathbb{R}^n, t > 0)$$ and clearly $u(x,0)=u_t(x,0)=0$ for $x \in \mathbb{R}^n$.

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  • $\begingroup$ What does the pink arrow mean? $\endgroup$ – mvw Jun 4 '14 at 19:04
  • $\begingroup$ @mvw I think it means: "reduces to". $\endgroup$ – Hakim Jun 4 '14 at 19:06
  • $\begingroup$ Yeah, what @حكيم الفيلسوف الضائع said. Also, I wonder why one person downvoted this question .... $\endgroup$ – Cookie Jun 4 '14 at 19:07
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The solution was $$ u(x,t) = \int\limits_0^t u(x,t;s) \, ds \quad x \in \mathbb{R}^n, t \ge 0 $$ Partial differentiation for $t$ of the integral function according to Leibniz $$ \frac{\partial}{\partial t} \int\limits_0^{b(t)} u(x, t; s) \, ds = u(x,t; b(t)) \, b'(t) + \int\limits_0^{b(t)} \frac{\partial}{\partial t} u(x, t; s) \, ds \quad (*) $$

gives \begin{align} \frac{\partial}{\partial t} u(x,t) &= \frac{\partial}{\partial t} \int\limits_0^t u(x,t;s) \, ds \\ &= \underbrace{\left. u(x,t;s) \right|_{s=t}}_0 \frac{dt}{dt} + \int\limits_0^t \frac{\partial}{\partial t} u(x,t;s) \, ds \\ &= \int\limits_0^t u_t(x,t;s) \, ds \\ \end{align}

Applying Leibniz's rule again: \begin{align} \frac{\partial^2}{\partial t^2} u(x,t) &=\frac{\partial}{\partial t} \int\limits_0^t u_t(x,t;s) \, ds \\ &= \underbrace{\left. u_t(x,t;s) \right|_{s=t}}_{f(x,t)} \frac{dt}{dt} + \int\limits_0^t \frac{\partial}{\partial t} u(x,t;s) \, ds \\ &=f(x,t) + \int\limits_0^t u_{tt}(x,t;s) \, ds \end{align}

Derivation of equation $(*)$:

Define $$ \varphi(x,t) := \int\limits_0^{b(t)} u(x,t;s)\; ds $$

then using some integral mean value theorem one gets

\begin{align} \Delta \varphi(x, t) &= \varphi(x, t + \Delta t) - \varphi(x, t) \\ &= \int\limits_0^{b+\Delta b} u(x,t + \Delta t;s) \, ds - \int\limits_0^b u(x,t;s) \, ds \\ &= \int\limits_0^{b} u(x,t + \Delta t;s) \, ds + \int\limits_b^{b+\Delta b} u(x,t + \Delta t;s) \, ds - \int\limits_0^b u(x,t;s) \, ds \\ &=\int\limits_b^{b+\Delta b} u(x,t + \Delta t;s) \, ds + \int\limits_0^{b} \left[ u(x,t + \Delta t;s) - u(x,t;s) \right] \, ds \\ &=u(x, t + \Delta t;\sigma) \, \Delta b + \int\limits_0^{b} \frac{u(x,t + \Delta t;s) - u(x,t;s)}{\Delta t} \, ds \, \Delta t \end{align}

with $\sigma \in [b, b + \Delta b]$ and therefore $$ \frac{\Delta \varphi(x, t)}{\Delta t} = u(x, t + \Delta t;\sigma) \, \frac{\Delta b}{\Delta t} + \int\limits_0^{b} \frac{u(x,t + \Delta t;s) - u(x,t;s)}{\Delta t} \, ds $$

For $\Delta t \to 0$ this shrinks $\sigma \to b(t)$ and gives $$ \varphi_t(x,t) = u(x, t;b(t)) \, b'(t) + \int\limits_0^{b} u_t(x,t;s) \, ds $$

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  • $\begingroup$ @Glacier: This is just differentiation under integral sign (see en.wikipedia.org/wiki/Differentiation_under_the_integral_sign) coupled with his condition on the line (40s). $\endgroup$ – Bombyx mori Jun 4 '14 at 21:05
  • $\begingroup$ Yeah I understood the part with condition on line $40s$ of the textbook. I will look more into the "Differentiation under the integral sign", as I forgot a lot about it. :| By the way, I never mentioned "40s" in my question; did you physically open up your Evans textbook too? :) $\endgroup$ – Cookie Jun 4 '14 at 21:10
  • $\begingroup$ @glacier was that answer ok? $\endgroup$ – mvw Jun 6 '14 at 9:35
  • $\begingroup$ Yes, I think your solution is fine. It's just a little long though. I actually figured it out when I wrote my own proof down on the paper, and it seemed to follow your proof. $\endgroup$ – Cookie Jun 6 '14 at 16:18

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