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Let $u$ be a solution of the wave equation $u_{tt} = c^2u_{xx}$ on the domain $\Omega=(0,1)$, subject to homogeneous Neumann boundary conditions.

(a) Define $$E(t)=\int_{0}^{1}\big(u_t(t,x)^2+c^2u_x(t,x)^2\big)dx$$ Show that $E(t)$ is constant, i.e., it does not depend on the time $t$. That is, show that the derivative satisfies $dE(t)/dt = 0$.

(b) Use (a) to show that the only differentiable function satisfying the wave equation $u_{tt }= c^2u_{xx}$ such that $u(0,x) = 0$ and $u_t(0,x) = 0$ for all $x\in (0,1)$, as well as $u_x(t,0) = u_x(t,1) = 0$ for all $t > 0$ is the trivial function $u(t,x) \equiv 0$. Hint: Find $E(0)$ for a function with these initial conditions

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  • $\begingroup$ Shouldn't your energy integrad have a term in $u_t$? $\endgroup$ – Robert Lewis Sep 27 '17 at 3:48
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    $\begingroup$ Yes you are correct, fixing that now, sorry. $\endgroup$ – Peetrius Sep 27 '17 at 3:50
  • $\begingroup$ I edited your question to remove a seemingly spurious "\". Also answered it! $\endgroup$ – Robert Lewis Sep 28 '17 at 2:09
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In what follows, I assume without further comment that $u(t, x)$ is "sufficiently differentiable"; enough so to make the logic fly.

For part (a):

With

$u_{tt} = c^2 u_{xx} \tag 1$

and

$E(t) = \displaystyle \int_0^1 (u_t^2(t, x) + c^2 u_x^2(t, x))dx, \tag 2$

we may perform the "lite" computation of $\dot E(t) = E_t(t) = dE(t)/dt$, where I use the word "lite" since I am not going to worry about or otherwise vindicate differentiating under the integral sign with respect to $t$, an operation which would probably merit some justifying analysis and/or remarks in a more thorough treatment, as follows:

$\dfrac{dE(t)}{dt} = \dfrac{d}{dt}\displaystyle \int_0^1 (u_t^2(t, x) + c^2 u_x^2(t, x))dx= \int_0^1 \dfrac{d}{dt}(u_t^2(t, x) + c^2 u_x^2(t, x))dx$ $= \displaystyle \int_0^1 (2u_t(t, x)u_{tt}(t, x) + 2c^2u_x(t, x)u_{xt}(t, x))dx$ $ = \displaystyle 2\int_0^1 (u_t(t, x)u_{tt}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx; \tag 3$

using (1), the rightmost integral in (3) becomes

$\displaystyle \int_0^1 (u_t(t, x)u_{tt}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx$ $= \displaystyle \int_0^1 (c^2u_t(t, x)u_{xx}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx; \tag 4$

furthermore,

$\displaystyle \int_0^1 (c^2u_t(t, x)u_{xx}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx$ $= \displaystyle \int_0^1c^2 u_t(t, x)u_{xx}(t, x)dx + \int_0^1 c^2u_x(t, x)u_{xt}(t, x)dx; \tag 5$

as for the second integral on the right of (5), consider the identity

$(u_t(t, x) u_x(t, x))_x = u_{tx}(t, x) u_x(t, x) + u_t(t, s) u_{xx}(t, x), \tag 6$

which results from a simple application of the Leibniz rule for derivatives of products to $u_t(t, x) u_x(t, x)$; (6) may be re-arranged to yield

$u_{tx}(t, x) u_x(t, x) = (u_t(t, x) u_x(t, x))_x - u_t(t, s) u_{xx}(t, x), \tag 7$

which we integrate in $x$ over $[0, 1]$:

$\displaystyle \int_0^1 u_{tx}(t, x) u_x(t, x)dx = \int_0^1(u_t(t, x) u_x(t, x))_x dx - \int_0^1 u_t(t, s) u_{xx}(t, x)dx; \tag 8$

when $t \ge 0$, the leftmost integral on the right-hand side of (8) may be explicitly found:

$\displaystyle \int_0^1(u_t(t, x) u_x(t, x))_x dx= u_t(t, 1)u_x(t, 1) - u_t(t, 0)u_x(t, 0) = 0, \tag 9$

by virtue of the homogeneous Neumann boundary conditions $u_x(t, 1) = u_x(t, 0) = 0$, $t \ge 0$; thus, (8) becomes

$\displaystyle \int_0^1 u_{tx}(t, x) u_x(t, x)dx = - \int_0^1 u_t(t, s) u_{xx}(t, x)dx, \tag {10}$

in the light of which (5) yields

$\displaystyle \int_0^1 (c^2u_t(t, x)u_{xx}(t, x) + c^2u_x(t, x)u_{xt}(t, x))dx$ $= \displaystyle \int_0^1 c^2 u_t(t, x)u_{xx}(t, x)dx - \int_0^1 c^2 u_t(t, x) u_{xx}(t, x)dx = 0; \tag {11}$

now following the logic back from (5) to (3) we see that

$\dfrac{dE(t)}{dt} = \dfrac{d}{dt}\displaystyle \int_0^1 (u_t^2(t, x) + c^2 u_x^2(t, x))dx, \tag{12}$

for $t \ge 0$; the requisite result.

Now as for part (b):

Since $u(0, x) = 0$ we have

$u_x(0, x) = 0, \tag{13}$

and since we are given that

$u_t(0, x) = 0, \tag{14}$

we see that

$E(0) = 0; \tag{15}$

then by the result of part (a) we find

$E(t) = 0 \tag{16}$

for all $t \ge 0$. Since we (though perhaps tacitly) assume $u_t(t, x)$ and $u_x(t, x)$ are continuous, this implies

$u_t(t, x) = u_x(t, x) = 0 \tag{17}$

for $t \ge 0$; since the derivatives of $u(t, x)$ are both $0$, $u(t, x)$ is constant in both time and space for $t \ge 0$; since $u(x, 0) = 0$ it follows that

$u(x, t) = 0 \; \text{for} \; (x, t) \in [0, 1] \times [0, \infty), \tag{18}$

and the hypothesis proposed in part (b) is affirmed.

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    $\begingroup$ Wow, I managed to write up my answer and confirm what you wrote, but this is written up wonderfully that I now see for certain how (b) is true and how we check (a) thank you so much! $\endgroup$ – Peetrius Sep 29 '17 at 1:34
  • $\begingroup$ Thanks for the kind words. Any chance of adding a vote/acceptance to your written lauds? Cheers! $\endgroup$ – Robert Lewis Sep 29 '17 at 1:38
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    $\begingroup$ Of course! Don't worry I did, but because my rep is <15 it won't show it, but I gotcha :) still new and learning the ropes around stack exchange. $\endgroup$ – Peetrius Sep 29 '17 at 16:56
  • $\begingroup$ Many thanks good sir, my your reputation grow! $\endgroup$ – Robert Lewis Sep 29 '17 at 17:09

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