what is the fundamental solution of the shifted operator $ \Delta + \lambda^2 $, i.e, what the function $f$ satisfying the following equation $$ (\Delta + \lambda^2 )f(x) = \delta(x),$$ where $ \Delta = \frac{d^2}{dx^2}$, $\lambda \in \mathbb C$ and $\delta(x)$ is the dirac measure at $x=0$.
In our considered case, this amounts to prove that for any $g$ in $C^\infty(\mathbb R)$ the following identity holds \begin{equation} g(0)=\int_0^\infty(\Delta + \lambda^2 )f(x) \, g(x) d(x). \end{equation}
Someone can help me!!!
Thanks in advance
You can try to Fourier transform your equation, using $$\hat{f}(y)=\int_{-\infty}^\infty f(x)e^{ixy}dx$$ your equation becomes: $$(-y^2+\lambda^2)\hat f(y)=1$$ So: $$\hat f(y)=\frac{-1}{y²- \lambda²}$$ And you need to Fourier transform it back: $$f(x)=-\frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{ixy}}{y²- \lambda²}dy$$ To evaluate this, you can use the residue theorem with a semi-circular contour on the upper-half of the complex plane for $x>0$ and the lower-half for $x<0$. There are two poles at $y\pm \lambda$ with residue $\pm e^{\mp i\lambda x}/2\lambda$. Now you can set a contour which goes around the poles in different manners ; using a contour going below both poles, I get: $$f(x)=\frac{\sin(\lambda x)}{\lambda}\theta(x)$$ where $\theta$ is the Heaviside step function. It works if I am not mistaken. This is more a physicist's answer than a mathematician's one though