Radial solution of this PDE

596 Views Asked by At

can you guys give me an idea on how to begin solving this problem?

suppose $u(x,t)$ is smooth and solves

$u_{tt}(x,t)-4\Delta_{x}u(x,t)=0$, $x\in \mathbb{R}^{3}, t>0$

compute the radial solutions $u(x,t)=v(r,t), r=\sqrt{x_{1}^{2}+x_{1}^{2}+x_{3}^{2}}$

Also, is $\Delta_{x}$ a typo? should it say just $\Delta$?

1

There are 1 best solutions below

4
On

In spherical coordinates, the Laplacian for a radially symmetric function $f=f(r)$ takes the form:

$$\Delta f = \frac{1}{r}\frac{\partial^2}{\partial r^2}(rf).$$

So, if you search for a solution $v=v(r,t)$ then $$ v_{tt} - 4\Delta v = v_{tt} - \frac{4}{r}\frac{\partial^2}{\partial r^2}(rv)=0.$$

Multiplying by $r$, and noting that $rv_{tt} = (rv)_{tt}$, we have

$$ \frac{\partial^2}{\partial t^2}(rv) - 4 \frac{\partial^2}{\partial r^2}(rv)=0.$$

Changing variables: $u = rv$ it follows $$ u_{tt} - 4u_{rr} =0, $$ so that $u$ solves the 1D wave equation. General solutions to the 1D wave equation are

$u(r,t) = F(r-2t)+ G(r+2t),$

so we must have $$ v(r,t) = \frac{1}{r}(F(r-2t)+G(r+2t)), $$ where $F$ and $G$ are arbitrary.