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Web bonus problem--Week 3

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Victor Ivrii:
From formula
$$u(x,t)=\frac{1}{2}\bigl( g(x+ct) +g(x-ct)\bigr)
\tag{1}$$
for the problem
$$
\left\{\begin{aligned}
&u_{tt}-c^2u_{xx}=0&&-\infty <x <\infty,\\
&u|_{t=0}=g(x),\\
&u_t|_{t=0}=0
\end{aligned}\right.
\tag{2}$$
derive
$$v(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct}g(x')\,dx'
\tag{3}$$
for the problem
$$
\left\{\begin{aligned}
&v_{tt}-c^2v_{xx}=0&&-\infty <x <\infty,\\
&v|_{t=0}=0,\\
&v_t|_{t=0}=g(x).
\end{aligned}\right.
\tag{4}$$
Also. from (4) for (3) derive (2) for (1).

Hint: prove that $v=\int _0^t u(x,t')\,dt'$.

Jingxuan Zhang:
Evaluated at $t=0$
$$\int _0^t u(x,t')\,dt'=0$$
whereas by assumption
$$(\int _0^t u(x,t')\,dt')_{t}=\int _0^t u_{t}(x,t')\,dt'+u(x,t)=g(x)\tag{5}$$
Also for each real $x$
$$ L(\int _0^t u(x,t')\,dt') = \int _0^t Lu(x,t')\,dt' + 2u_{t}(x,t) = 0\tag{6}$$
by formula $(1)$ and the equation in $(2)$. We thus recovered $(4)$ and uniqueness forces $v=\int _0^t u(x,t')\,dt'$. But then
$$\int _0^t u(x,t')\,dt' =\int _0^t \frac{1}{2}\bigl( g(x+ct') +g(x-ct')\bigr) dt' =
\frac{1}{2c} \bigl(\int _0^{x+ct}g(x')dx' -(-\int _{x-ct}^0 g(x')dx')\bigr)=\frac{1}{2c}\int_{x-ct}^{x+ct}g(x')\,dx'.\tag{7}$$

Someone else please do the Also.

Victor Ivrii:
In (5) I may agree that left-hand expression is equal to the right-hand expression, but the middle is a mystery

In (6) what is $L$ is also a mystery. 

Jingxuan Zhang:
Let me admit the middle of $(5)$ was not only mystery but also wrong. The reason is I incorrectly thought the integrand is dependent on $t$. I hope this should correct it:
$$(\int _0^t u(x,t')\,dt')_{t}=\frac{d}{dt}\int _0^t u(x,t')\,dt' =u(x,t) \tag{8}$$
simply by FTC.

Thus $(6)$ is also wrong, since really from above we have
$$(\int _0^t u(x,t')\,dt')_{tt}=u_{t}(x,t)=\int _0^t u(x,t')_{tt}\,dt'.\tag{9}$$

It is only after that do we have, if $Lu:=u_{tt}-c^{2}u_{xx}$,

$$ L(\int _0^t u(x,t')\,dt') = \int _0^t Lu(x,t')\,dt'= 0\tag{10}$$

by equation in $(2)$.

Victor Ivrii:
OK. Part 1 done

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