Consider the PDE $$u_{tt}=c^2u_{xx}$$ where $u(x,0)=\phi(x)$ and $u_t(x,0)=\psi(x)$. I'm asked to derive the D'Alembert solution, which I have done and found to be $$u(x,t)=\frac{1}{2}(\phi(x-ct)+\phi(x+ct))+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s) \, ds$$

I'm then asked to show that if $\phi(x)=0$ and $\psi(x)=\delta(x)$ then $$u(x,t)=\frac{1}{2c}\left(H(x+ct)-H(x-ct) \right)$$ where $H$ is the Heaviside function.

From my previous work I get that $$u(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct}\delta(s) \, ds$$

So in order to get the solution it seems as though it's the case that $\int_0^{x+ct}\delta(s) \, ds=H(x+ct)$.

Is this true? If so, why?

  • $\begingroup$ mathworld.wolfram.com/DeltaFunction.html (The Heaviside step function is the antiderivative of the delta "function") $\endgroup$ Commented Mar 22, 2017 at 16:05
  • $\begingroup$ Why is harder than is it true - so I will explain what can be considered true, the derivative of the Heaviside function is the Dirac delta function. $\endgroup$
    – Chinny84
    Commented Mar 22, 2017 at 16:05
  • 1
    $\begingroup$ By definition, delta is an infinitely tall rectangle with area $1$. As such, it's integral should jump by $1$ as it hits $x = 0$. Ergo, the step function. $\endgroup$
    – Kaynex
    Commented Mar 22, 2017 at 16:25

1 Answer 1


\begin{align} \int_{x-ct}^{x+ct}\delta(s)ds & = \int_{-\infty}^{x+ct}\delta(s)ds-\int_{-\infty}^{x-ct}\delta(s)ds \\ & = H(x+ct)-H(x-ct) \end{align}


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