Toronto Math Forum
APM3462018S => APM346Tests => Term Test 1 => Topic started 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}

General solution for $ x >t $: $ u = \psi(x+2t) + \phi(x2t) $
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 $$ \phi(t) = \cos(\frac{t}{3}) + constant, t <0 $$
$$ \phi(t) = \cos(\frac{t}{3}) + constant, t<0 $$
$$ \phi(x  2t) = \cos(\frac{(x2t)}{3}) + constant $$
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{(x2t)}{3}) + 1 , t <x<2t $

should $\phi(t) = 3cos(\frac{1}{3}t) + constant $ ? And the answer for t<x<2t is 3$cos(\frac{x2t}{3}) 3$

Yeah you're right, looks like I forgot to write it with the chain rule

OK