Web bonus problem--Week 3

[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

[Adam Gao]

The second part of the question asks us to prove (2), starting with the premises (1), (3), and (4).

First prove the hint:
$$
v = \int_{0}^{t} u(x,t')\,dt'.
$$
Define:
$$
g(x') = \frac{d}{dx'}G(x');
$$
We already defined:
$$
u(x,t) = \frac{1}{2}(g(x+ct) + g(x-ct)
$$
from (1) and
$$
v(x,t) = \frac{1}{2c} \int_{x-ct}^{x+ct}g(x')\,dt'
$$
from (3). Then:
\begin{gather*}
\int_{0}^{t} u(x,t')dt' = \frac{1}{2} \Bigl(\int_{0}^{t} g(x + ct')dt' + \int_{0}^{t} g(x - ct')\,dt'\Bigr)\\
= \frac{1}{2} \Bigl(\frac{1}{c}G(x+ct) - \frac{1}{c}G(x) + \frac{1}{-c}G(x-ct) - \frac{1}{-c}G(x)\Bigr)\\
= \frac{1}{2c} \Bigl(G(x + ct) - G(x-ct)\Bigr) = \frac{1}{2c} \int_{x-ct}^{x+ct}g(x')dt' = v(x,t)
\end{gather*}
Now I have proved the hint and I will implement it to help prove $(2)$. From the hint: $v_t = u$.

Thus $v_{tt} = u_{t}$. We know from (4):

$v_{tt} = c^{2}v_{xx}$

Thus $u_{t} = c^{2}v_{xx}$

Thus $u_{tt} = c^{2}v_{xxt}$

Let us find $v_{xxt}$. (We hope to prove the first condition of $(2)$.)

Previously we have shown $v(x,t) = \frac{1}{2c}(G(x+ct) - G(x-ct))$

Thus $v_x = \frac{1}{2c}(G'(x+ct) - G'(x-ct)) = \frac{1}{2c}(g(x+ct) - g(x-ct))$

Thus $v_{xx} = \frac{1}{2c}(g'(x+ct) - g'(x-ct))$

Thus $v_{xxt} = \frac{1}{2}g''(x+ct) + g''(x-ct)$

Let us find $u_{xx}$ in hopes that it is equal to $v_{xxt}$. This will help complete our proof.

$u_{x} = \frac{1}{2}(g'(x+ct) + g'(x-ct)$

Thus $u_{xx} = \frac{1}{2}g''(x+ct) + g''(x-ct) = v_{xxt}$.

Thus $u_{tt} = c^{2}u_{xx}$.

So $u_{tt} - c^{2}u_{xx} = 0$, proving that the first condition of $(2)$ must follow from $(1)$, $(3)$, and $(4)$.

To prove that the rest of $(2)$ follows, we don't even need $(3)$ or $(4)$. Plug $t=0$ into $u(x,t)$ and its partial derivative $u_t$ to prove those conditions.

[Victor Ivrii]

Right. What is $G$? Note, how I reformatted the first half of your post

[Adam Gao]

$G(x') = \int g(x') dx'$
or
$g(x') = \frac{d}{dx'} G(x')$