Why does $f(x,t)=f$(x-vt,0) imply the wave equation?

Conditions

• The function $f$ is twice differentiable.
• The wave propagates at a constant speed $v$ in the positive x-direction.

Reasoning, step by step

1. Treat $u = x-vt$ as an intermediate variable.
2. Compute the first partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \cdot 1$.
3. Compute the second partial derivative with respect to $x$: $\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial u^2}$.
4. Compute the first partial derivative with respect to $t$: $\frac{\partial f}{\partial t} = \frac{\partial f}{\partial u} \cdot (-v)$.
5. Compute the second partial derivative with respect to $t$: $\frac{\partial^2 f}{\partial t^2} = \frac{\partial^2 f}{\partial u^2} \cdot (-v) \cdot (-v) = v^2 \frac{\partial^2 f}{\partial u^2}$.
6. Divide the expression for $\frac{\partial^2 f}{\partial t^2}$ by $v^2$ and equate it to $\frac{\partial^2 f}{\partial x^2}$.

Example

The red box shows the process of computing the first- and second-order partial derivatives of f(x−vt,0) with respect to x and t, concluding with $\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 f(x,t)}{\partial t^2}$.

Common misconceptions

• Assuming that the chain rule only applies to single-variable functions.
• Forgetting that differentiating with respect to $t$ twice introduces $(-v)^2 = v^2$, not $-v^2$.

Watch the explanation

Explore next

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