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The wave equation is given by $\mu_{tt}=c^2 \mu_{xx}$ and (in one spacial dimension) can be reduced to $\mu_{\alpha\beta}=0$ by doing the following changes $\alpha=x-ct$ and $\beta=x+ct$.

From $\mu_{\alpha\beta}=0$ how do I show that the general solution $\mu(x,t)$ can be written as $\mu(x,t)=f(x-ct)+g(x+ct)$ where $f$ and $g$ are arbitrary functions?

Winther
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gulraiz safdar
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  • The title and presentation may be more to the point if the one-dimensional nature of your question (and the coincidences in the fact and its proof) are clearer, hence, my edits. – paul garrett Apr 10 '18 at 00:52
  • https://en.wikipedia.org/wiki/Wave_equation#General_solution ; https://math.stackexchange.com/questions/1439392/algebraic-approach-to-dalembert-wave-solution ; https://math.stackexchange.com/questions/2244000/how-are-these-results-obtained-in-dalemberts-solution-to-the-11-wave-equati – Winther Apr 10 '18 at 00:54
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    yes but I don't see how they proof that u(x,t) = f(x-ct)+g(x+ct) in those links? – gulraiz safdar Apr 10 '18 at 01:03

1 Answers1

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$$\mu_{\alpha\beta}=0$$ From the chain rule of differentiation: $$\mu_\alpha=\mu_x x_\alpha+\mu_t t_\alpha$$ $x=\frac12(\alpha+\beta)\quad;\quad t=\frac{1}{2c}(\beta-\alpha)$

$x_\alpha=\frac12\quad;\quad x_\beta=\frac12\quad;\quad t_\alpha=-\frac{1}{2c}\quad;\quad t_\beta=\frac{1}{2c}$

$$\mu_\alpha=\frac12\mu_x -\frac{1}{2c}\mu_t $$ $$\mu_{\alpha\beta}=\frac12(\mu_{xx}x_\beta+\mu_{xt}t_\beta) -\frac{1}{2c}(\mu_{tx}x_\beta+\mu_{tt}t_\beta)$$ $$\mu_{\alpha\beta}=\frac12\left(\frac12\mu_{xx}+\mu_{xt}\frac{1}{2c}\right) -\frac{1}{2c}\left(\frac12\mu_{tx}+\mu_{tt}\frac{1}{2c}\right)$$ After simplification : $$\mu_{\alpha\beta}=\frac14\left(\mu_{xx}-\frac{1}{c^2}\mu_{tt}\right)$$ Since $\mu_{\alpha\beta}=0$ : $$\frac14\left(\mu_{xx}-\frac{1}{c^2}\mu_{tt}\right)=0$$ $$c^2\mu_{xx}=\mu_{tt}$$

JJacquelin
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