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?

  • $\begingroup$ How is $d - ct$ supposed to work if $d \in \mathbb{R}^{3}$ and $c, t \in \mathbb{R}_{+}$? And is $x \in \mathbb{R}^{3}$? $\endgroup$ Mar 30, 2022 at 1:55
  • 1
    $\begingroup$ Hint. Calculate first $\frac{\partial}{\partial t}g(x\cdot d-ct)$ using the chain rule. Ditto: $\frac{\partial}{\partial x_i}g(x\cdot d-ct)$. Then the second derivatives (chain rule again). Looking at the equation in one-dimension is probably simpler as a start. $\endgroup$
    – Kurt G.
    Mar 30, 2022 at 6:44

1 Answer 1


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}


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