Solve $u_{t}+u_{x}=0$ with $x,t>0$ and $u(x,0)=f(x), u(0,t)=g(t)$ using method of characteristics

Question

I am asked to solve the following PDE using the method of characteristics:

Solve $u_{t}+u_{x}=0$ with $x,t>0$ and $u(x,0)=f(x), u(0,t)=g(t)$ using method of characteristics.

So the equation can be seen as $(1,1) \cdot (\delta_t , \delta_x)u$ which means that along the vector (1,1) the function is constant and the curves in the direction (1,1) are defined by:

$\frac{dt}{dx} = \frac{1}{1} \Rightarrow t= x + c \Rightarrow c = t-x$

So the the general solution is $u(x,t)=h(t-x)$ for some arbitrary function $h$ but I don't know how involve the initial conditions.

I had thought of using an indicator, but I do not know how well how to do it, thank you.

Answer 1

We know that $u(x,t) = h(x-t)$. Notice that $x,t > 0$. The initial conditions give us $h(x) = f(x)$ for $ x>0$, so $h(x-t) = f(x-t)$ for $x>t$. Also, $h(-t) = g(t)$, so $h(x-t) = g(t-x)$ for $t>x$.

Thus $\begin{equation*} u(x,t) = \begin{cases} f(x-t) & x>t\\ g(t-x) & x<t \end{cases} \end{equation*}$