Resolvent Equation PDE

Question

Let $\lambda>0$. Show that the fundament solution of the resolvent equation $(-\Delta+\lambda^2)u=f$ on $\mathbb{R}^3$ is given by $u(x)=\int_{\mathbb{R}^3} R(x-y)f(y)dy$ where $R(x)=\frac{1}{4\pi|x|}e^{-\lambda|x|}$ and $f\in C_c^2(\mathbb{R}^3)$. I know this looks very similar to the proof in THM 1 in section 2.2 Poission Equation solved in Evans PDE but how does the $\lambda$'s change this problem and completing the proof.

Answer 1

After Fourier transforming in $x$, the equation $(-\Delta+\lambda^2)u=f$ becomes $$ (|\xi|^2+\lambda^2)\hat{u}(\xi)=\hat{f}(\xi) \\ \hat{u}(\xi)=\frac{1}{\lambda^2+|\xi|^2}\hat{f}(\xi). $$ Therefore, $$ u(x) = \int_{\mathbb{R}^3} K(\lambda,x-y)f(y)dy, $$ where $$ K(\lambda,x)=\frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^3}\frac{1}{\lambda^2+|\xi|^2}e^{-i\xi\cdot x}d\xi \\= \frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^3}\left(\int_0^{\infty}e^{-t(\lambda^2+|\xi|^2)}dt\right)e^{-i\xi\cdot x}d\xi \\ =\frac{1}{(2\pi)^{3/2}}\int_0^{\infty}e^{-t\lambda^2}\left(\int_{\mathbb{R}^3}e^{-t|\xi|^2}e^{-i\xi\cdot x}d\xi \right)dt $$ Now the problem is reduced to a product of Fourier transforms of $1d$ Gaussians, and then an integral in $t$.