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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 $F(z-vt)+G(z+vt)$.

Peter Mortensen
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Goldenalcheese
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4 Answers4

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

hft
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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.

Qmechanic
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  • Notes for later: Let $v=1$. $\quad \bar{G}(x,t)=\frac{1}{4}\theta(t^2!-!x^2)$. $\quad \Box\bar{G}(x,t)=-\delta(x)\delta(t)$. $\quad \Box=\partial_x^2-\partial_t^2$. $\quad G_{\rm ret}(x,t)=\frac{1}{2}\theta(t)\theta(t^2!-!x^2)=\frac{1}{2}\theta(t!-!|x|)$. – Qmechanic Dec 25 '22 at 10:55
  • 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 $t>0$. – Qmechanic Dec 26 '22 at 22:24
  • 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. – Qmechanic Dec 27 '22 at 09:40
  • $\quad \partial_{\mu}\theta_{\varepsilon}(t^2!-!x^2) = -\frac{1}{\pi}{\rm Im}\frac{2x_{\mu}}{x^2-t^2+i\varepsilon}$. $\quad \partial_{\mu}\partial_{\nu}\theta_{\varepsilon}(t^2!-!x^2) =\frac{1}{\pi}{\rm Im} \left(-\frac{2\eta_{\mu\nu}}{x^2-t^2+i\varepsilon}+\frac{4x_{\mu}x_{\nu}}{(x^2-t^2+i\varepsilon)^2}\right)$. $\quad \Box\theta_{\varepsilon}(t^2!-!x^2) =-\frac{4}{\pi}{\rm Im}\frac{i\varepsilon}{(x^2-t^2+i\varepsilon)^2} \longrightarrow -4\delta(x)\delta(t)$ for $\varepsilon \searrow 0^+$. – Qmechanic Dec 27 '22 at 12:05
  • Proof: $\quad \iint_{\mathbb{R}^2} \frac{dxdt}{\pi} f(x,t) {\rm Im}\frac{i\varepsilon}{(x^2-t^2+i\varepsilon)^2}$ $= \iint_{\mathbb{R}^2} \frac{dyds}{\pi} f(\sqrt{\varepsilon}y,\sqrt{\varepsilon}s) {\rm Im}\frac{i}{(y^2-s^2+i)^2}$ $\longrightarrow f(0,0) \iint_{\mathbb{R}^2} \frac{dyds}{\pi} {\rm Im}\frac{-i}{(s^2-a^2)^2}$ $=f(0,0) \iint_{\mathbb{R}^2} \frac{dyds}{\pi} {\rm Im}\frac{-i}{4a^2}\left(\frac{1}{s-a}-\frac{1}{s+a}\right)^2$ $=f(0,0) \int_{\mathbb{R}} \frac{dy}{\pi} {\rm Im}2\pi i \frac{-i}{4a^2}\left(0+\left.\frac{-2}{(t+a)}\right|_{t=a}+0\right)$ – Qmechanic Dec 27 '22 at 12:15
  • $=-f(0,0) \int_{\mathbb{R}} dy~{\rm Im} \frac{1}{2a^3}$ $=-f(0,0) \int_{\mathbb{R}} dy~{\rm Im} \frac{1}{2(y^2+i)^{3/2}}$ $=-f(0,0) {\rm Im} \left[ \frac{-iy}{2\sqrt{y^2+i}}\right]^{\infty}_{-\infty}$ $=f(0,0)$. $\Box$ – Qmechanic Dec 27 '22 at 12:29
  • $\quad\Box 2G_{\rm ret}(x,t) =\theta(t)\Box\theta(t^2!-!x^2) +\delta(t)\partial_t\theta(t^2!-!x^2)+\delta^{\prime}(t)\theta(t^2!-!x^2)$. All terms localize at the origin of spacetime. Terms odd in $t$ vanishes, which are almost all. $\quad 2\partial_tG_{\rm ret}(x,0) =\delta_{\eta}(0)\theta_{\varepsilon}(-x^2) =\frac{1}{\pi^2\eta}\arg(x^2+i\varepsilon) =\frac{1}{\pi^2\eta}{\rm arccot}\frac{x^2}{\varepsilon}\stackrel{?}{=}2\delta(x)$. $\quad \delta_{\varepsilon}(x) =\frac{1}{\pi\sqrt{2\varepsilon}}{\rm arccot}\frac{x^2}{\varepsilon}$. $\Rightarrow \pi\eta=\sqrt{\varepsilon/2}$. – Qmechanic Dec 27 '22 at 14:42
  • $\quad 2\partial_tG_{\rm ret}(x,0) =\theta_{\eta}(0)\theta_{\varepsilon}(-x^2)=\frac{1}{2}\theta_{\varepsilon}(-x^2)\sim 0$. $\quad 2\partial_t^2G_{\rm ret}(x,0) =\delta^{\prime}{\eta}(0)\theta{\varepsilon}(-x^2)+2\theta_{\eta}(0)\delta_{\varepsilon}(-x^2)=\delta_{\varepsilon}(-x^2)$. – Qmechanic Dec 28 '22 at 09:17
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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 $G$:

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

Claudio Saspinski
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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 $v$.

Mister Mak
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