Solve first order PDE system

I need help with this:

Write $u_{tt} = c^2u_{xx}$, $\quad u(x,0)=f(x)$, $\quad u_t(x,0)=g(x)$ as an initial value problem for the vector $(u_1, u_2)=(u_t,u_x)$. Reduce the system to the canonical form, $v_t +\Lambda v_x=cv + d$, and solve the problem.

Solve the mixed problem:$\quad u_{tt}-c^2u_{xx}=0, \quad$ for $\quad x>0, t>0$

$\quad\quad\quad\quad\quad\quad\quad\quad\ u(x,0) =f(x), \quad u_t(x,0)=g(x),\quad for \quad x>0$

$\quad\quad\quad\quad\quad\quad\quad\quad\ u_t(0,t) + au_x(0,t)=h(t),\quad t>0$

with $a$ = constant.

I really need help to solve this. I had no idea to how to convert to canonical form. Thanks

Solution 1:

We have $(u_t)_t = c^2 (u_x)_x$. Moreover, for $u$ sufficiently smooth, we have $(u_x)_t = (u_t)_x$. Therefore, we can write the first-order system of conservation laws ${\bf u}_t + {\bf A}{\bf u}_x = {\bf 0}$ (see e.g. this post), where ${\bf u} = (u_x,u_t)^\top$ and \begin{aligned} {\bf A} &= \left(\begin{array}{cc}0 & -1\\ -c^2 & 0\end{array}\right) \\ &= \left(\begin{array}{cc}1/c & -1/c\\ 1& 1\end{array}\right)\left(\begin{array}{cc}-c & 0\\ 0 & c\end{array}\right) \left(\begin{array}{cc}c/2 & 1/2\\ -c/2 & 1/2\end{array}\right) = {\bf P}\, {\bf \Lambda}\, {\bf P}^{-1} . \end{aligned} Now, setting ${\bf v} = {\bf P}^{-1} {\bf u}$, one obtains ${\bf v}_t + {\bf \Lambda}{\bf v}_x = {\bf 0}$. The solution to the initial value problem ${\bf v}(x,0) = (v_1^0(x),v_2^0(x))^\top$ is ${\bf v}(x,t) = (v_1^0(x+ct),v_2^0(x-ct))^\top$, where $$ \left(\begin{array}{c} v_1^0(\xi) \\ v_2^0(\xi) \end{array}\right) = {\bf P}^{-1} \left(\begin{array}{c} u_x(0,\xi) \\ u_t(0,\xi) \end{array}\right) = {\bf P}^{-1} \left(\begin{array}{c} f'(\xi) \\ g(\xi) \end{array}\right) , $$ and $f'$ is the derivative of $f$. Going back to the initial unknowns, we have ${\bf u}(x,t) = {\bf P}\, {\bf v}(x,t)$, i.e. \begin{aligned} u_x(x,t) &= \frac{1}{2}\left(f'(x+ct) + f'(x-ct)\right) + \frac{1}{2c}\left(g(x+ct) - g(x-ct)\right) ,\\ u_t(x,t) &= \frac{1}{2}\left(g(x+ct) + g(x-ct)\right) + \frac{c}{2}\left(f'(x+ct) - f'(x-ct)\right) . \end{aligned} Finally, $u$ is obtained by integration of $u_t$ with respect to $t$, $$ u(x,t) = \frac{1}{2}\left(f(x+ct) + f(x-ct)\right) + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s)\,\text{d}s \, , $$ and one recognizes the well-known d'Alembert's formula.

The same method applies to the mixed type problem (see this post where the case $h=0$ is solved). First, it is solved in the characteristic fields where the linear system of conservation laws is diagonal. Then, we go back to the initial unknowns.