# Toronto Math Forum

## APM346-2015F => APM346--Home Assignments => HA4 => Topic started by: Victor Ivrii on October 10, 2015, 07:24:51 AM

Title: HA4-P5
Post by: Victor Ivrii on October 10, 2015, 07:24:51 AM
Problem 9: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.9 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.9)

Reflects new numeration of the problems
Title: Re: HA4-P5
Post by: Yeming Wen 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 $u=g(\eta)+f(\zeta)$.
Title: Re: HA4-P5
Post by: Victor Ivrii 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 ...
Title: Re: HA4-P5
Post by: Yeming Wen on October 15, 2015, 12:11:06 PM
Can we say $t=0\implies \zeta=\eta$?
Title: Re: HA4-P5
Post by: Victor Ivrii on October 15, 2015, 01:06:41 PM
Can we say $t=0\implies \zeta=\eta$?
Indeed (and v.v.) We did it after class Wed.