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


2 Answers 2


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.


Frankly the question makes no sense. This equation is second order in both $x$ and $t$ (it is a "wave equation") so any canonical form must be second order also, not a first order equation. What we can do is take new variables $p= x- ct$ and $q= x+ ct$. Then we have $\frac{\partial u}{\partial x}= \frac{\partial u}{\partial p}\frac{\partial p}{\partial x}+ \frac{\partial u}{\partial q}\frac{\partial q}{\partial x}= \frac{\partial u}{\partial p}+ \frac{\partial u}{\partial q}$ since $\frac{\partial p}{\partial x}= \frac{\partial q}{\partial x}= 1$. Similarly, $\frac{\partial u}{\partial t}$$= \frac{\partial u}{\partial p}\frac{\partial p}{\partial t}+ \frac{\partial u}{\partial q}\frac{\partial q}{\partial t}= -c\frac{\partial u}{\partial p}+ c\frac{\partial u}{\partial q}$. Taking the derivatives again, $\frac{\partial^2u}{\partial x^2}= \frac{\partial^2u}{\partial p^2}+ 2\frac{\partial u}{\partial pq}+ \frac{\partial^2 u}{\partial q^2}$ and $\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial p^2}- 2c^2\frac{\partial^2 u}{\partial pq}+ c^2\frac{\partial^2 u}{\partial q^2}$.

Putting those into $\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$, we have $c^2\frac{\partial^2 u}{\partial p^2}- 2c^2\frac{\partial^2 u}{\partial pq}+ c^2\frac{\partial^2 u}{\partial q^2}= c^2\left(\frac{\partial^2u}{\partial p^2}+ 2\frac{\partial u}{\partial pq}+ \frac{\partial^2 u}{\partial q^2}\right)$ which reduces to $-\frac{\partial^2 u}{\partial pq}= \frac{\partial^2 u}{\partial pq}$ or $4\frac{\partial u}{\partial pq}= 0$. That last is, of course, the same as $\frac{\partial u}{\partial pq}= 0$ which can then write as $\frac{\partial}{\partial p}\left(\frac{\partial u}{\partial q}\right)= 0$. The derivative of a function with respect to $p$ is $0$ means the function is constant with respect to $p$- it must be a function of $q$ only- but it could be any function of $q$, say $\frac{\partial u}{\partial q}= f(q)$. Integrating with respect to $q$, and writing the integral of $f(q)$ with respect to $q$, $F(q)$, we have $u(p, q)= F(q)+ G(p)$ were $G(p)$ is any arbitrary function of $p$ only (it is the "constant of integration" when we are integrating with respect to $q$, just as $f(q)$ was the "constant of integration when we are integrating with respect to $ p$). That is, the general solution to $\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$ is $u(x, t)= F(x- ct)+ G(x+ ct)$ where $F$ and $G$ can be any twice differentiable functions of a single variable.

Given that $u(x, 0)= f(x)$ we must have $u(x, 0)= F(x)+ G(x)= f(x).$ Given that $u_t(x, 0)= -cF'(x)+ cG'(x)= g(x),$ differentiating that first equation with respect to $x$, $u_x(x, 0)= F'(x)+ G'(x)= f(x)$ so $cF'(x)+ cG'(x)= cf(x)$ and adding the previous equation, $2cG'(x)= cf(x)+ g(x)$. $G(x)$ is the anti-derivative of $(cf(x)+ g(x))/2$ and then $F(x)= f(x)- G(x)$ there will still be one undetermined numerical constant.

  • $\begingroup$ It should be followed instruction to be solved. It means should reduce to first order in canonical form first. $\endgroup$
    – Vui Tinh
    Nov 15, 2017 at 8:31
  • $\begingroup$ @user247327 Of course, it is not necessary to use a first-order formulation to obtain d'Alembert's formula, but it is one possible method (see answer for details). $\endgroup$
    – EditPiAf
    Nov 17, 2017 at 8:19

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