I am studying the wave equation out of Olver's textbook on PDEs and I am looking through the derivation of d'Alembert's solution.

This seems to come in two parts and I am not sure which parts are 'necessary' for the derivation and which are not.

So he starts with the following problem:

Solve $$ u_{tt} - c^2 u_{xx} = 0$$ subject to $u(0,x) = f(x)$ and $u_t(0,x) = g(x)$.

First he defines the operator $\partial_{tt} - c^2\partial_{xx}$ which he factorises into $(\partial_t - c \partial_x)(\partial t + c\partial_x)$.

So now if $(\partial_t - c \partial_x)u = 0$, then we have a solution to the wave equation and if $(\partial_t + c \partial_x)u = 0$ we again have a solution.

This gives the result that two wave equation has solutions of both left and right travelling waves.

He then proves that the theorem that every solution to the wave equation can be represented as the superposition of a left and right travelling wave.

The derivation of d'Alembert's formula then follows easily by imposing the solution $u(t,x) = p(x+ct) + q(x-ct)$ to the boundary conditions.

However, I am confused about why he does the work with the operators and what that shows us. I can't see it as much more than a guide for what the solution could be. If someone asked to prove d'Alembert's formula, which parts of this are necessary to give.

In particular, if someone asked me to find the solution to say $u_{tt} + 3u_{tx} + 2u_{xx}$, is it sufficient to realise that $(\partial_{t} + 2\partial_{x})((\partial_{t} + \partial_{x})u = 0$ and then say the solution must be of the form $u(x,t) = p(x-2t)+q(x-t)$?

  • $\begingroup$ Yes, good question. I asked something very similar myself years ago. $\endgroup$ – Giuseppe Negro May 18 '18 at 9:00
  • $\begingroup$ @GiuseppeNegro Thanks, but maybe you can summarise it for me without the linear algebra language that I am not too familiar/confident with? $\endgroup$ – PhysicsMathsLove May 18 '18 at 9:02
  • $\begingroup$ I edited the message. $\endgroup$ – Giuseppe Negro May 18 '18 at 9:07
  • $\begingroup$ Anyway, I think that the answer is affirmative in your case. The counterexample by Bob Terrell (linked question) is in some sense degenerate, which is not your case. For you, $D=\partial^2_t +3\partial_t\partial_x +2\partial^2_x$ decomposes in the product of two linearly independent operators: $A=\partial_t+2\partial_x, B=\partial_t+\partial_x$. This was not the case of Bob Terrell who had $D=\partial_x^2$. $\endgroup$ – Giuseppe Negro May 18 '18 at 9:10

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