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I am dealing with the one-dimensional spatial wave equation $$\frac{\partial^2 \phi}{\partial z^2}-\frac{1}{v^2}\frac{\partial^2\phi}{\partial t^2}=0,$$ where $\phi=\phi(z,t)$ is required to solve.

According to the algebraic approach on Wikipedia (essentially change of variables), we obtain the general solution should be in the form: $$F(z-vt)+G(z+vt)\tag{1}$$

Looking back to the wave equation, there is a trivial solution: $$\phi(z,t)=(a+bz)(c+dt)\tag{2}$$ where $\{a,b,c,d\}$ are arbitrary constants. But it seems that this solution (2) is not compatible with $F(z-vt)+G(z+vt)$.

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    $\begingroup$ What does "...in 1+1D" mean? Can you elaborate? $\endgroup$ Mar 26 at 13:48
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    $\begingroup$ It means one spatial dimension and one time dimension, @PeterMortensen $\endgroup$
    – hft
    Mar 26 at 16:56

4 Answers 4

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Looking back to the wave equation, there is a trivial solution: $$\phi(z,t)=(a+bz)(c+dt)\tag{2}$$ where $\{a,b,c,d\}$ are arbitrary constants. But it seems that this solution is not compatible with $F(z-vt)+G(z+vt)$?

This solution is compatible with the F + G form.

It is maybe easiest to see this by defining $p = z + vt$ and $q = z - vt$.

Then substituting and rearranging, we find: $$ \phi = \left[\frac{ac}{2} + p\left(\frac{bc}{2} + \frac{ad}{2v}\right) + p^2 \left(\frac{bd}{4v}\right)\right] + $$ $$\left[\frac{ac}{2} + q\left(\frac{bc}{2} - \frac{ad}{2v}\right) - q^2\left(\frac{bd}{4v}\right)\right] = G(p) + F(q) $$

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Let's test the quadratic functions: $$F(z-vt) = \alpha(z-vt)+ \beta(z-vt)^2$$ $$G(z+vt) = \gamma(z+vt)+ \delta(z+vt)^2$$ $$F + G = (\alpha+\gamma)z + v(\gamma-\alpha)t + (\beta+\delta)z^2 + v^2(\delta+\beta)t^2 + 2vzt(\delta-\beta)$$

Equating:

$(\alpha+\gamma) = bc$
$v(\gamma-\alpha) = ad$
$2v(\delta-\beta)=bd$
$\delta+\beta = 0$

After setting $\delta = -\beta$ for the last equation, and making $d = v \implies$
$b = 4\delta$
$a = \gamma-\alpha$
$c = \frac{\alpha+\gamma}{4\delta}$

But the independent term is missing, what can be solved by modifying $F$ and $G$:

$$F(z-vt) = -\frac{\alpha^2}{4\delta} + \alpha(z-vt)+ \beta(z-vt)^2$$ $$G(z+vt) = \frac{\gamma^2}{4\delta} + \gamma(z+vt)+ \delta(z+vt)^2$$

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  • $\begingroup$ What is the conclusion? $\endgroup$ Mar 26 at 13:47
  • $\begingroup$ @PeterMortensen substituting the constants a,b,c,d with their expressions in $\alpha, \beta, \gamma, \delta$ in the original OP trivial solution, we get $F + G$ $\endgroup$ Mar 26 at 16:41
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Hint: One can actually decompose OP's trivial solution (2) to be on the form (1). Try e.g. to introduce light-cone coordinates $z^{\pm}=z\pm vt$. The decomposition (1) is unique up to an additive constant.

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From the standpoint of physics, Form (2) is a solution of the wave equation, but does not depend on $v$, so this is a standing wave.

As @hft and @Qmechanic have pointed out, you can force Form (2) into Form (1) where $v$ appears. This means you can decompose the standing wave into two waves traveling in opposite directions at velocity $v$.

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