### Author Topic: HA4-P5  (Read 3354 times)

#### Victor Ivrii

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##### HA4-P5
« on: October 10, 2015, 07:24:51 AM »
« Last Edit: October 17, 2015, 05:29:35 AM by Victor Ivrii »

#### Yeming Wen

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##### Re: HA4-P5
« Reply #1 on: October 15, 2015, 11:37:02 AM »
By the hint, we do a change of variable. Let $$x=\frac{1}{2}(\zeta+\eta)$$ $$t=\sinh(\frac{1}{2}(\zeta-\eta))$$
Then we compute $u_{\zeta}$ and $u_{\zeta\eta}$.
By chain rule,
$$u_\zeta=\frac{1}{2}u_x+u_t\cdot t_\zeta=\frac{1}{2}(u_x+u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)))$$
Similarly, $$u_{\zeta\eta}=\frac{1}{2}([u_x]_\eta+[u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)]_\eta)$$
$$[u_x]_\eta=\frac{1}{2}u_{xx}-\frac{1}{2}u_{xt}\cosh(\frac{1}{2}(\zeta-\eta))$$
$$[u_t\cdot \cosh(\frac{1}{2}(\zeta-\eta)]_\eta=[u_t]_\eta\cosh(\frac{1}{2}(\zeta-\eta))+u_t[\cosh(\frac{1}{2}(\zeta-\eta))]_\eta=\frac{1}{2}u_{tx}\cosh(\frac{1}{2}(\zeta-\eta))+u_{tt}[\cosh(\frac{1}{2}(\zeta-\eta))]^2-\frac{1}{2}u_t\sinh(\frac{1}{2}(\zeta-\eta))$$
Notice that $$[\cosh(\frac{1}{2}(\zeta-\eta))]^2=1+[\sinh(\frac{1}{2}(\zeta-\eta))]^2=t^2+1$$
We have $$u_{\zeta\eta}=\frac{1}{2}[\frac{1}{2}u_{xx}-\frac{1}{2}(t^2+1)u_{tt}-\frac{1}{2}tu_t]=\frac{1}{4}(u_{xx}-(t^2+1)u_{tt}-tu_t)$$
Combine with the original PDE, we have $u_{\zeta\eta}=0$. So $u=g(\eta)+f(\zeta)$.
« Last Edit: October 15, 2015, 12:13:14 PM by Yeming Wen »

#### Victor Ivrii

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##### Re: HA4-P5
« Reply #2 on: October 15, 2015, 12:05:12 PM »
It is $u$, not $\mu$. Also you need to find out $f,g$ from initial conditions. $t=0$ translates in $\xi,\eta$ coordinates as ...

#### Yeming Wen

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##### Re: HA4-P5
« Reply #3 on: October 15, 2015, 12:11:06 PM »
Can we say $t=0\implies \zeta=\eta$?

#### Victor Ivrii

Can we say $t=0\implies \zeta=\eta$?