# Toronto Math Forum

## APM346-2018S => APM346--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 07:05:01 PM

Title: P3
Post by: Victor Ivrii on February 15, 2018, 07:05:01 PM
Find  continuous solution to
\begin{align}
&u_{tt}-4u_{xx}=0, &&   t>0, x>-t,\tag{1}\\
&u|_{t=0}=0, && x>0,\tag{2}\\
&u_t|_{t=0}=0, && x>0,\tag{3}\\
&u_x|_{x=-t}= \sin(t), &&t>0.\tag{4}
\end{align}
Title: Re: P3
Post by: Ioana Nedelcu on February 15, 2018, 09:51:06 PM
General solution for $x >-t$: $u = \psi(x+2t) + \phi(x-2t)$

Given initial conditions, we have $\psi(x) = \phi(x) = 0$ so $$u = 0, x > 2t$$

Using the boundary conditions for $-t < x < 2t$, ie where $\phi(x), x<0$

$$\psi'(t) + \phi'(-3t) = \sin(t)$$
From previous solution $$\psi(x) = 0 \implies \psi'(x) = 0$$
So $$\phi(t) = -\cos(\frac{-t}{3}) + constant, t <0$$
$$\phi(t) = -\cos(\frac{t}{3}) + constant, t<0$$
$$\phi(x - 2t) = -\cos(\frac{(x-2t)}{3}) + constant$$

For $\phi$ to be constant at x = 2t, ie the same value from both sides of the characteristic equation:

$$0 = -\cos(0) + c \implies c = 1$$

So $u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t$
Title: Re: P3
Post by: Jilong Bi on February 15, 2018, 10:16:26 PM
should $\phi(t) = 3cos(-\frac{1}{3}t) + constant$ ? And the answer for -t<x<2t is 3$cos(\frac{x-2t}{3}) -3$
Title: Re: P3
Post by: Ioana Nedelcu on February 15, 2018, 11:09:40 PM
Yeah you're right, looks like I forgot to write it with the chain rule
Title: Re: P3
Post by: Victor Ivrii on February 22, 2018, 06:47:10 AM
OK