Show that this is a solution of this wave equation Given the wave equation
$\frac{1}{c^2} \frac{\partial^2}{\partial t^2}u=\Delta u$ where $\Delta u$ is the Laplacian operator and a function $g$ that's two times continuously differentiable,
show that $u(x,t)=g(x \cdot d-ct), d\in \mathbb{R}^3 \text{constant}, ||d||=1$ is a solution of the wave equation.
What I have so far:
$\frac{\partial^2}{\partial t^2}u(x,t)=c^2 \frac{\partial^2}{\partial t^2}g(x \cdot d-ct)$
So if I put this in the left side of the equation, the $c^2$ cancel each other. However the Laplacian operator of $g$ involves a part where you also have to differentiate two times for x so for me it doesn't seem equal. What do I have to do?
 A: The notation you chose for $$\frac{\partial^2}{\partial t^2} u (x, t) = c^2 \frac{\partial^2}{\partial t^2} g(\boldsymbol{x} \cdot \boldsymbol{d} -ct)$$
is somewhat misleading since here the differentiation is still to be done. A better way would be introducing $y(\boldsymbol x, t) := \boldsymbol{x} \cdot \boldsymbol{d} -ct$ and thus examining $g\big(y(\boldsymbol x, t)\big)$. Then,
\begin{align}
\frac{\partial^2}{\partial t^2} u (\boldsymbol x, t) &= \frac{\partial^2}{\partial t^2} g\big(y(\boldsymbol x, t)\big) \\
&= \frac{\partial}{\partial t} \bigg( g'(y) \frac{\partial y}{\partial t} \bigg) \\
& = \frac{\partial}{\partial t} \bigg( -c g'(y) \bigg)\\
&=-c \frac{\partial}{\partial t} g'(y) \\
& = -c g''(y) \frac{\partial y}{\partial t}  = c² g''(y). 
\end{align}
Given that $\boldsymbol{x}$ is a vector, we have to work a little more careful. Note that $\Delta = \nabla \cdot \nabla$, so the divergence of the gradient. Let's take the gradient of $g$ first:
\begin{align}
\nabla g\big( y(\boldsymbol x, t) \big) = g'(y) \nabla y (\boldsymbol x, t) 
=g'(y) \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix} .
\end{align}
Now we have to take the divergence of this result:
\begin{align} 
 \nabla \cdot \left[ g'\big((y(\boldsymbol x, t) \big) \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix} \right] &= \begin{pmatrix} \frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3} \end{pmatrix} \cdot \begin{pmatrix} g'(y) d_1 \\ g'(y) d_2 \\ g'(y) d_3 \end{pmatrix}  \\
&=g''(y) d_1 d_1 + g''(y) d_2 d_2 + g''(y) d_3 d_3 \\
&= g''(y)  \Vert \boldsymbol{d} \Vert_2^2 = g''(y) 1^2 = g''(y).
\end{align}
