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As title, given a general wave equation problem: $$\frac{\partial^2u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}+f(x,t)$$

with Dirichlet or Neumann boundary conditions and initial value $u(x,0)=g(x)$ how can I transform this problem in an hyperbolic system like $$ \frac{\partial \bf{c}}{\partial t}=\bf{A} \frac{\partial \bf{c} }{\partial x} +\bf{f}(x,t) $$

where $\bf{c}$ and $\bf{f}$ are vectors of two functions?

EditPiAf
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skdys
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1 Answers1

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First, we write the equality of mixed derivatives $$ (u_x)_t = (u_t)_x \, . $$ Second, we rewrite the wave equation as $$ (u_t)_t = a^2 (u_x)_x + f \, . $$ Therefore, setting ${\bf c} = (u_x, u_t)^\top$ and ${\bf f} = (0, f)^\top$, we have ${\bf c}_t = {\bf A} {\bf c}_x + {\bf f}$, where $$ {\bf A} = \left( \begin{array}{cc} 0 & 1 \\ a^2 & 0 \end{array} \right) . $$ For Dirichlet boundary conditions $u(x_0,t) = \alpha$, we set $u_t(x_0,t) = 0$. Neumann boundary conditions $u_x(x_0,t) = \beta$ are unchanged. The initial value $u(x,0)=g(x)$ is accounted for by setting $u_x(x,0)=g'(x)$.

Let us diagonalize ${\bf A}$ as follows: $$ {\bf A} = \underbrace{ \left( \begin{array}{cc} -1/a & 1/a \\ 1 & 1 \end{array} \right) }_{\bf P}\; \underbrace{ \left( \begin{array}{cc} -a & 0 \\ 0 & a \end{array} \right) }_{\bf \Lambda}\; \underbrace{ \frac{1}{2} \left( \begin{array}{cc} -a & 1 \\ a & 1 \end{array} \right) }_{{\bf P}^{-1}} . $$ Now, setting ${\bf d} = {\bf P}^{-1}{\bf c}$ and ${\bf h} = {\bf P}^{-1}{\bf f}$, one has ${\bf d}_t = {\bf \Lambda} {\bf d}_x + {\bf h}$.

EditPiAf
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  • Ok, this is the standard way. Can you please add how to obtain the system with ${\bf A} = \left( \begin{array}{cc} 0 & a \ a & 0 \end{array} \right) .$ in order to get two advenction equations with opposite directions? – skdys Jan 11 '18 at 23:21
  • @skdys It's done. – EditPiAf Jan 12 '18 at 09:36
  • @Harry49 What if we have Dirichlet boundary condition, how we can deal with the boundary conditions? – Gustave Oct 06 '18 at 10:48
  • @Gustave One should impose the boundary conditions on $u$ before differentiation w.r.t. $x$, $t$. If $u(0,t)=g(t)$, then $u_t(0,t)=\dot g(t)$. – EditPiAf Oct 06 '18 at 11:40