I'm trying to show that a delta function $\delta(x - ct)$ is a distributional solution of the PDE $$ D_{(0,2)}u(x, t) = c^2 D_{(2,0)}u(x,t). $$ Here $D_{(i,j)}$ means $i$-th partial differentiation on the first variable and $j$-th partial differentiation on the second variable. I'm not strictly sure the "travling delta function" is actually a distributional solution. However, I think it could be intuitively.

The delta function may be expressed as a distribution: $\left< \delta_c, \phi \right> = \int_{-\infty}^\infty \phi(ct, t) dt$. This is motivated by the manipulations of the delta function in physics literatures like $$ \int\int \delta(x - ct) f(x, t) dx dt = \int f(ct, t) dt. $$

Now for the left hand side of the PDE, $$ \left< D_{(0,2)}\delta_c, \phi \right> = \left< \delta_c, D_{(0,2)} \phi \right> = \int_{-\infty}^\infty D_{(0,2)}\phi (ct, t) dt. $$ where $\phi$ is a test function defined on $\mathbb{R}^2$. For the right hand side, $$ c^2 \left< D_{(2,0)}\delta_c, \phi \right> = c^2 \left< \delta_c, D_{(2,0)} \phi \right> = c^2 \int_{-\infty}^\infty D_{(2,0)}\phi (ct, t) dt. $$ I'm stuck here. How can I show the two integrals are the same? Or is the delta function not a solution?


1 Answer 1


I think I understood the phenomenon :

$$\int_{-\infty}^\infty \phi(ct-h,t) dt = \frac{1}{c} \int_{-\infty}^\infty \phi(x,\frac{x+h}{c}) dx$$

by the variable change $x = ct-h$, so that :

$$\int_{-\infty}^\infty \frac{\partial^2}{\partial x^2} \phi(ct,t) dt = \lim_{h \to 0} \int_{-\infty}^\infty \frac{-\phi(ct-h,t)+2\phi(ct,t)-\phi(ct+h,t)}{h^2} dt$$ $$ = \frac{1}{c}\lim_{h \to 0} \int_{-\infty}^\infty \frac{-\phi(x,\frac{x+h}{c})+2\phi(x,\frac{x}{c})-\phi(x,\frac{x-h}{c})}{h^2} dx = \frac{1}{c^3}\int_{-\infty}^\infty \frac{\partial^2}{\partial t^2} \phi(x,x/c) dx$$ $$= \frac{1}{c^2}\int_{-\infty}^\infty \frac{\partial^2}{\partial t^2} \phi(c\tau,\tau) d\tau$$

as far as $\phi$ is a nice $C^\infty$ function with compact support allowing all these manipulations.

note : this is nothing else than saying $\delta''_{x=t}$ as $\delta_{x=t}$ are symmetric 2D distributions : $\langle \ \delta_{\displaystyle x=t}, \varphi_{\displaystyle x,t}\ \rangle = \langle \ \delta_{\displaystyle x=t}, \varphi_{\displaystyle t,x} \ \rangle$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .