Deriving an Explicit Formula using Fundamental Solutions

Question

I am currently trying to find a Fundamental Solution to the Equation: $$\Delta\Phi+c\Phi=\delta_0$$ To find an explicit formula for $$\Delta u+cu=f$$ Now I know I am supposed to look for radially symmetric solutions to the ODE: $$\Delta u+cu=0$$ Then we can write $u$ in terms of $r$, and thus: $$\Delta u(r)+cu(r)=0\rightarrow u_{rr}+\frac{u_r}{r}+cu=0$$ Now the solution to this ODE is given by: $$u=c\cos(\sqrt{1+c}r)+d\sin(\sqrt{1+c}r), \>\>c,d\in\mathbb R$$ How do I continue? I don't really know how to proceed from here..

Answer 1

You did not specify a dimension to solve this problem, but I will assume that this is in $\mathbb{R}$.

Rather than trying to show the solution is radially invariant, you can instead just observe that the fundamental solution must solve

$$\Delta \Phi + c \Phi = \delta(x)$$

for $\Phi$. One way to do this is via the Fourier transform(being slightly handwavy), as $\mathcal{F}(\delta(x))=1$. This approach is pretty much exactly what Evan's does in his PDE book. We arrive at

$$[\mathcal{F}(\Phi)](\xi) = \frac{1}{c - \xi^2}$$

Hence, applying an inverse Fourier transform, we find

$$\Phi(x) = \frac{1}{(2 \pi)^{1/2}}\int_{\mathbb{R}}\left(\frac{1}{c - \xi^2}\right) e^{-i \xi x} \, d\xi$$

Now, OP does not list any restrictions on $c$, so for simplicity I will assume $c>0$ so that we can find the nicer inverse Fourier transform, we arrive at

$$\Phi = \begin{cases} \frac{\sin{(x\sqrt{c}) }}{2 \sqrt{c}} & x\geq 0\\ -\frac{\sin{(x\sqrt{c}) }}{2 \sqrt{c}} & x\leq 0 \end{cases}$$

Hence, in general, $u = \Phi * f$ solves the original problem.