Lorentz invariance of the wave equation I want to show that the 2-d wave equation is invariant under a boost, so, the starting point is the wave equation
$$\frac{\partial^2\phi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} $$
and the Lorentz transformation:
$$t'=\gamma(t-\frac{v}{c^2}x) \\ x'=\gamma(x-vt)$$
My question is should I write $\displaystyle\frac{\partial}{\partial t}$ as a derivative with respect to $x'$ and $t'$ and then substitute?
Work done so far
$$\frac{\partial}{\partial t}=\frac{\partial }{\partial x'}\frac{\partial x'}{\partial t}+\frac{\partial}{\partial t'}\frac{\partial t'}{\partial t}=-\gamma v\frac{\partial}{\partial x'}+\gamma\frac{\partial}{\partial t'} $$
$$\frac{\partial^2}{\partial t^2}=\frac{\partial }{\partial t}\left(\frac{\partial}{\partial t} \right)=\frac{\partial}{\partial t}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)= \\ 
=-\gamma v\frac{\partial}{\partial x'}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)+\gamma\frac{\partial}{\partial t'}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)=\\ =
\gamma^2v^2\frac{\partial^2}{\partial x'^2}-2\gamma^2v\frac{\partial ^2}{\partial x'\partial t'}+\gamma^2\frac{\partial^2 }{\partial t'^2}$$
-Edit-
The same applies to $x$
$$\frac{\partial}{\partial x}=\frac{\partial }{\partial x'}\frac{\partial x'}{\partial x}+\frac{\partial }{\partial t'}\frac{\partial t'}{\partial x}=\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} $$
$$\frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left( \frac{\partial }{\partial x}\right)=\frac{\partial}{\partial x}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)= \\ =
\gamma\frac{\partial}{\partial x'}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)=\\=
\gamma^2\frac{\partial^2 }{\partial x'^2}-2\frac{\gamma^2v}{c^2}\frac{\partial^2 }{\partial x'\partial t' }+\frac{\gamma^2v^2}{c^4}\frac{\partial^2}{\partial t'^2}$$
Edit 2 with the hints given by nervxxx
The wave equation becomes
$$\frac{\gamma^2v^2}{c^2}\frac{\partial^2 \phi}{\partial x'^2}-\frac{2\gamma^2v}{c^2}\frac{\partial ^2 \phi}{\partial x'\partial t'}+\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}=\gamma^2\frac{\partial^2 \phi}{\partial x'^2}-2\frac{\gamma^2v}{c^2}\frac{\partial^2 \phi}{\partial x'\partial t' }+\frac{\gamma^2v^2}{c^4}\frac{\partial^2\phi}{\partial t'^2}$$
$$ \frac{\gamma^2 v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}+\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}=\gamma^2\frac{\partial^2 \phi}{\partial x'^2}+\frac{\gamma^2v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}$$
But I still don't get... since all $\gamma^2$ cancel
Final edit. done!
$$ \frac{\gamma^2 v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}-\gamma^2\frac{\partial^2 \phi}{\partial x'^2}=\frac{\gamma^2v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}-\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}$$
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} $$
$$ \left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{\partial^2 \phi}{\partial x'^2}=\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t'^2}$$
$$ \left( \frac{v^2}{c^2-v^2}\right)\frac{\partial ^2 \phi}{\partial x'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{\partial^2 \phi}{\partial x'^2}=\left( \frac{v^2}{c^2-v^2}\right)\frac{\partial ^2\phi}{\partial t'^2}\frac{1}{c^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t'^2} $$
$$\frac{\partial \phi^2}{\partial x'^2}\left(\frac{v^2}{c^2-v^2}- \frac{1}{1-\frac{v^2}{c^2}}\right)=\frac{\partial^2 \phi}{\partial t'^2}\frac{1}{c^2}\left(\frac{v^2}{c^2-v^2}- \frac{1}{1-\frac{v^2}{c^2}}\right) $$
$$ \frac{\partial^2\phi}{\partial x'^2}=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t'^2} $$
 A: First, your wave equation is wrong. You can see this from dimensional analysis. It should be
\begin{align}
\frac{\partial^2 \phi}{\partial t^2} = c^2 \frac{\partial^2 \phi}{\partial x^2}
\end{align}
[Edit (5/30/2020): The poster has edited the question to correct this mistake, so the above point no longer holds].
Second, you made a mistake in the cross terms for the $\partial^2 /\partial x^2$ term. The cross term should have the coefficient $-2\gamma^2 v/c^2$.
Third, use the fact that $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$.
You will get the desired result.
A: It's worth highlighting an important point that is skimmed over in the question: the wave equation stated is not, in general, invariant under a boost. It is Lorentz invariant only when the wave propgates with a velocity $c$. Specifically, for a
wave travelling with a velocity $v$ satisfying $$\frac{\partial^2\phi}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2\phi}{\partial t^2}$$ in certain coordinate system, the given wave equation is invariant under a Lorentz transformation to a frame travelling with velocity $\beta c$ given by $$\begin{align}t'&=\gamma\left(t-\frac{\beta x}{c}\right)\\x'&=\gamma(x-\beta c t)\end{align}$$ only when $v=\pm c$. This variable convention (specifically, the meaning of $v$) is different in the question.
The proof for this statement is found through a slight modification of the calculus performed in the question. We have
$$
\begin{align}
\frac{\partial^2\phi}{\partial x^2}&=\frac{\gamma^2\beta^2}{c^2}\frac{\partial^2\phi}{\partial t'^2}+\gamma^2\frac{\partial^2\phi}{\partial x'^2}-\frac{2\gamma^2\beta}{c}\frac{\partial^2\phi}{\partial x'\partial t'};\\
\frac{\partial^2\phi}{\partial t^2}&=\gamma^2\frac{\partial^2\phi}{\partial t'^2}+\gamma^2\beta^2c^2\frac{\partial^2\phi}{\partial x'^2}-2\gamma^2\beta c\frac{\partial^2\phi}{\partial x'\partial t'}.
\end{align}
$$
Substituting these into the original wave equation, we find the following equation in terms of boosted coordinates
$$
\left(\frac{\gamma^2}{v^2}-\frac{\gamma^2\beta^2}{c^2}\right)\frac{\partial^2\phi}{\partial t'^2}=\left(\gamma^2-\frac{\gamma^2\beta^2c^2}{v^2}\right)\frac{\partial^2\phi}{\partial x'^2}-\left(\frac{2\gamma^2\beta}{c}-\frac{2\gamma^2\beta c}{v^2}\right)\frac{\partial^2\phi}{\partial x'\partial t'}.
$$
Clearly, the wave equation is invariant only if the mixed-partials term disappears. We have
$$\frac{2\gamma^2\beta}{c}=\frac{2\gamma^2\beta c}{v^2}\Rightarrow v=\pm c.$$
Performing this substitution for $v$ in the wave equation in primed coordinates, the original form with wave velocity $c$ is restored:
$$\frac{\partial^2\phi}{\partial x'^2}=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t'^2}.$$ This tells us that the speed of the wave observed in the boosted frame is also $c$, which is consistent with the principle that the speed of light measured by the boosted observer is also $c$.
A: What I do bears no different from others but may be slightly more concise. First let $\tau=ct$ and
$$
L=\gamma(v)\begin{pmatrix}
1 & -v/c\\-v/c & 1
\end{pmatrix}
$$
Then
$$
\begin{pmatrix}
 \dfrac{\partial}{\partial x'} \\[3pt]
 \dfrac{\partial}{\partial \tau'}
\end{pmatrix}
=L
\begin{pmatrix}
 \dfrac{\partial}{\partial x} \\[3pt]
 \dfrac{\partial}{\partial \tau}
\end{pmatrix}
$$
and
$$
\begin{align*}
\begin{pmatrix}
 \dfrac{\partial^2\phi}{\partial x'^2} & \dfrac{\partial^2\phi}{\partial x'\partial\tau'}\\[3pt]
 \dfrac{\partial^2\phi}{\partial x'\partial\tau'} & \dfrac{\partial^2\phi}{\partial \tau'^2}
\end{pmatrix}=&\begin{pmatrix}
\dfrac{\partial}{\partial x'} \\[3pt]
\dfrac{\partial}{\partial \tau'}
\end{pmatrix}
\begin{pmatrix}
 \dfrac{\partial}{\partial x'}
 \dfrac{\partial}{\partial \tau'}
\end{pmatrix}\phi
=\left[L\begin{pmatrix}
 \dfrac{\partial^2}{\partial x^2} & \dfrac{\partial^2}{\partial x\partial\tau}\\[3pt]
 \dfrac{\partial^2}{\partial x\partial\tau} & \dfrac{\partial^2}{\partial \tau^2}
\end{pmatrix}L^\mathrm{T}\right]\phi\\
=&\gamma^2(v)
\begin{pmatrix}
 \dfrac{\partial^2\phi}{\partial x^2}-\dfrac{2v}{c}\dfrac{\partial^2\phi}{\partial x\partial\tau}+\dfrac{v^2}{c^2}\dfrac{\partial^2\phi}{\partial \tau^2} & * \\[3pt]
 * & \dfrac{v^2}{c^2}\dfrac{\partial^2\phi}{\partial x^2}-\dfrac{2v}{c}\dfrac{\partial^2\phi}{\partial x\partial\tau}+\dfrac{\partial^2\phi}{\partial \tau^2}
\end{pmatrix}
\end{align*}
$$
So
$$
\frac{\partial^2\phi}{\partial x^2}=\frac{\partial^2\phi}{\partial \tau^2}
\Longleftrightarrow\frac{\partial^2\phi}{\partial x^2}+\frac{v^2}{c^2}\frac{\partial^2\phi}{\partial \tau^2}=\frac{v^2}{c^2}\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial \tau^2}
\Longleftrightarrow\frac{\partial^2\phi}{\partial x'^2}=\frac{\partial^2\phi}{\partial \tau'^2}
$$
