### Author Topic: P3  (Read 1495 times)

#### Victor Ivrii ##### P3
« 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}

#### Ioana Nedelcu

• Full Member
•   • Posts: 29
• Karma: 3 ##### Re: P3
« Reply #1 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 $u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t$

#### Jilong Bi

• Jr. Member
•  • Posts: 8
• Karma: 9 ##### Re: P3
« Reply #2 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$cos(\frac{x-2t}{3}) -3$

#### Ioana Nedelcu

• Full Member
•   • Posts: 29
• Karma: 3 ##### Re: P3
« Reply #3 on: February 15, 2018, 11:09:40 PM »
Yeah you're right, looks like I forgot to write it with the chain rule

#### Victor Ivrii ##### Re: P3
« Reply #4 on: February 22, 2018, 06:47:10 AM »
OK