How to solve $u_t + u_x =0$ with $u_0(x) = \sin(\pi x)$ with Characteristics?

Question

Consider the initial-boundary value problem (IBVP) for the convection equation \begin{array} { l } { u _ { t } + u _ { x } = 0 \quad x \in [ a ( t ) , b ( t ) ] , t \in [ 0 , T ] } \\ { u ( x , 0 ) = u _ { 0 } ( x ) \quad x \in [ a ( 0 ) , b ( 0 ) ] } \end{array} with initial conditionsu $u_{ 0 } ( x ) = \sin ( \pi x )$ The boundaries $a(t)$ and $b(t)$ are determined by the characteristics that go through $(x = 0; t = 0)$ and $(x = 2; t = 0)$, respectively. Let $x _ { j } = j h$ with $j = 0,...,20$ and $h = 0.1$ .

Answer 1

I don't really understand the final sentence of your elaboration. The method of characteristics is a hand-solved method, not an iteration.

The characteristics are :

$$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}x}{1} = \frac{\mathrm{d}u}{0}$$ Taking any of the two first with the last one, yields : $$\frac{\mathrm{d}x}{1} = \frac{\mathrm{d}u}{0} \implies u_1 = u$$ Taking the first pair, yields : $$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}x}{1} \implies u_2 = x-t $$

Since $u_2$ is straight-forwardly expressed by $u$, simply enough, the solution will be given as $$u_1=F(u_2) \Rightarrow u(x,t) = F(x-t)$$ where $F$ is an adequatelly smooth function.

But $u(x,0) = u_0(x) = \sin(\pi x)$, which means :

$$u(x,0) = F(x) = \sin(\pi x)$$

Now, letting $x : = x-t$, we yield : $$u(x,t) = F(x-t) = \sin[\pi(x-t)]$$