# Method of Characteristics and Initial Value Problem

$u_t + 3u_x = 2t$, $u(x,0)=\sin(x/2)$.

I used the method of characteristics to get the answer, $u(x,t)=t^2 + 2\sin^{-1}(x-3t)$.

Does this satisfy the initial condition? I checked for the first equation and it does; however I do not think it satisfies the intial value when $t=0.$ Am I correct in saying so?

• Yes you are corect to saying that it does not satisfy the IC. It is clear when you put $t=0$ in the solution you have found. Jan 16 '14 at 3:48

The solution is $u=\frac{2}{3}\cos(\frac{x}{2})+\sin(\frac{x-3t}{2})-\frac{2}{3}\cos(\frac{x-3t}{2})$.
I think your u(x,t) is not correct: apart from some factors of 2 different, I have $\sin$ and not $\sin^{-1}$.
Also, worth recognizing that you can solve this with just change of variables, $x \rightarrow x + 3 t$.
answer for above question is $$u(x,t)=t^2+\sin(x−3t/2)$$