Solve $u_t+cu_x=0$ with $c>0$ when $x,t >0$ and contions $u(x,0)=h(x)$ with $x>0$ and $u(0,t)=g(t)$ with $t>0$.

Question

$u_t+cu_x=0$ show that $u(x,t)=f(ct-x)$.
With conditions, it shows that $$u(x,0)=f(-x)=h(x),$$ $$u(0,t)=f(ct)=g(t)$$ I dont know how to find what $f(ct-x)$ is and what the purpose of condition: $c>0$ when $x,t >0$ .

Answer 1

We compose the system of characteristics $\frac{dt}1=\frac{dx}c$. So we have $c dt - dx = 0$. And $c t - x$ is the first integral of system of characteristics. $f(c t - x)$ is the general solution of equation, where $f$ is function from the class $C^1$.

$u(x, t) = f(c t - x)$ is general solution. From boundary conditions we get:

1. $u(x, 0) = f(-x) = h(x)$, $x > 0$.
2. $u(0, t) = f(c t) = g(t)$, $t > 0$.

From (1) $f(z) = h(-z)$, $z< 0$. From $f(z) =g(\frac{z}c)$, $z > 0$.

So, we have the solution $u(x, t) = f(c t - x) = \left\{\begin{matrix} g(\frac{c t - x}c), & c t - x > 0,\\ h(x - c t), & c t - x < 0. \end{matrix}\right.$

If $c < 0$ your the first mixed problem will not have a unique solution, or it will have no solution.