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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, 2022 at 13:48
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    $\begingroup$ It means one spatial dimension and one time dimension, @PeterMortensen $\endgroup$
    – hft
    Mar 26, 2022 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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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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  • $\begingroup$ Notes for later: Let $v=1$. $\quad \bar{G}(x,t)=\frac{1}{4}\theta(t^2\!-\!x^2)$. $\quad \Box\bar{G}(x,t)=-\delta(x)\delta(t)$. $\quad \Box=\partial_x^2-\partial_t^2$. $\quad G_{\rm ret}(x,t)=\frac{1}{2}\theta(t)\theta(t^2\!-\!x^2)=\frac{1}{2}\theta(t\!-\!|x|)$. $\endgroup$
    – Qmechanic
    Dec 25, 2022 at 10:55
  • $\begingroup$ D'Alembert's formula: $\quad 2u(x,t)= u(x\!+\!t,0)+u(x\!-\!t,0) +\int_{x-t}^{x+t} \! dx^{\prime}~u_t(x^{\prime},0) $ $=\int_{\mathbb{R}} \! dx^{\prime}~\delta(|t|\!-\!|x\!-\!x^{\prime}|)u(x^{\prime},0) +{\rm sgn}(t)\int_{\mathbb{R}} \! dx^{\prime}~\theta(|t|\!-\!|x\!-\!x^{\prime}|)u_t(x^{\prime},0)$ for $t\neq 0$. $\quad u(x,t) = \int_{\mathbb{R}} \!dx^{\prime}~\partial_tG_{\rm ret}(x\!-\!x^{\prime},t) u(x^{\prime},0) +\int_{\mathbb{R}} \!dx^{\prime}~G_{\rm ret}(x\!-\!x^{\prime},t) u_t(x^{\prime},0) $ for $t>0$. $\endgroup$
    – Qmechanic
    Dec 26, 2022 at 22:24
  • $\begingroup$ Notes for later: $\quad {\rm sgn}_{\eta}(t)=\frac{2}{\pi}\arctan\frac{t}{\eta}$. $\quad \theta_{\eta}(t)=\frac{1}{\pi}{\arg}(-t+i\eta)=\frac{1}{\pi}{\rm arccot}\frac{-t}{\eta}$. $\quad \delta_{\eta}(t)=\frac{1}{\pi}{\rm Im}\frac{1}{t-i\eta}$. $\quad \delta^{\prime}_{\eta}(t)=-\frac{1}{\pi}{\rm Im}\frac{1}{(t-i\eta)^2}$. Retarded Greens function $\quad G_{\rm ret}(x,t)=\frac{1}{2}\theta(t)\theta(t^2\!-\!x^2)$. Regularization: $\quad G_{{\rm ret},\eta,\varepsilon}(x,t)=\frac{1}{2}\theta_{\eta}(t)\theta_{\varepsilon}(t^2\!-\!x^2)$. Here $\eta \sim \sqrt{\varepsilon}$ but that not necessary. $\endgroup$
    – Qmechanic
    Dec 27, 2022 at 9:40
  • $\begingroup$ $\quad \partial_{\mu}\theta_{\varepsilon}(t^2\!-\!x^2) = -\frac{1}{\pi}{\rm Im}\frac{2x_{\mu}}{x^2-t^2+i\varepsilon}$. $\quad \partial_{\mu}\partial_{\nu}\theta_{\varepsilon}(t^2\!-\!x^2) =\frac{1}{\pi}{\rm Im} \left(-\frac{2\eta_{\mu\nu}}{x^2-t^2+i\varepsilon}+\frac{4x_{\mu}x_{\nu}}{(x^2-t^2+i\varepsilon)^2}\right)$. $\quad \Box\theta_{\varepsilon}(t^2\!-\!x^2) =-\frac{4}{\pi}{\rm Im}\frac{i\varepsilon}{(x^2-t^2+i\varepsilon)^2} \longrightarrow -4\delta(x)\delta(t)$ for $\varepsilon \searrow 0^+$. $\endgroup$
    – Qmechanic
    Dec 27, 2022 at 12:05
  • $\begingroup$ Proof: $\quad \iint_{\mathbb{R}^2} \frac{dxdt}{\pi} f(x,t) {\rm Im}\frac{i\varepsilon}{(x^2-t^2+i\varepsilon)^2}$ $= \iint_{\mathbb{R}^2} \frac{dyds}{\pi} f(\sqrt{\varepsilon}y,\sqrt{\varepsilon}s) {\rm Im}\frac{i}{(y^2-s^2+i)^2}$ $\longrightarrow f(0,0) \iint_{\mathbb{R}^2} \frac{dyds}{\pi} {\rm Im}\frac{-i}{(s^2-a^2)^2}$ $=f(0,0) \iint_{\mathbb{R}^2} \frac{dyds}{\pi} {\rm Im}\frac{-i}{4a^2}\left(\frac{1}{s-a}-\frac{1}{s+a}\right)^2$ $=f(0,0) \int_{\mathbb{R}} \frac{dy}{\pi} {\rm Im}2\pi i \frac{-i}{4a^2}\left(0+\left.\frac{-2}{(t+a)}\right|_{t=a}+0\right)$ $\endgroup$
    – Qmechanic
    Dec 27, 2022 at 12:15
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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, 2022 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, 2022 at 16:41
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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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