# Fundamental solution of a second order differential equation

Consider the differential operator $$L=u''-u\qquad \mathrm{in}\ \ \mathbb{R}.$$ Find the fundamental solution of the above operator.

Now, I guessed the fundamental solution to be $$E=e^{x}H(x)$$, where $$H(x)$$ is the Heaviside function. But, after finding its second weak derivative (in distribution sense), I get it as $$E+\delta-\delta'$$ where $$\delta'$$ is the dipole distribution and $$\delta$$ is the Dirac distribution. But this doesn't satisfy the operator, $$L$$. My question is am I taking the guess solution correctly? If not what should it be and how to think about getting these guess solutions?

The fundamental solution $$G$$ should satisfy

1. $$LG=0$$ on $$(-\infty,0)$$ and on $$(0,\infty)$$,
2. boundary conditions, usually $$\lim_{R\to \infty} G(\pm R)=0$$,
3. be continuous, so $$G(0+)=G(0-)$$,
4. $$G'(0+)-G'(0-)=1$$ (to make $$LG=\delta$$).

For $$Lu=u''-u$$ conditions 1 and 2 give $$G(x)=\begin{cases} A_- e^x & (x<0),\\ A_+ e^{-x} & (x>0). \end{cases}$$

Then conditions 3 and 4 give $$A_- = A_+$$ and $$-A_+ - A_-=1,$$ i.e. $$A_-=A_+=-\frac12$$ so we end up with $$G(x) = -\frac12 e^{-|x|}.$$

• @LutzLehmann. True. Will edit my answer. Jul 23, 2021 at 13:09
• It should be $A_-=A_+ = -1/2$, and our answers will coincide ;) Jul 23, 2021 at 16:27
• @LL3.14. Fixed. Jul 23, 2021 at 16:30

A fast way to solve $$u''-u=\delta_0$$ is to take the Fourier transform to get $$(-|2π y|^2-1)\,\mathcal F(u)(y) = 1$$, and so $$u(x) = \mathcal F^{-1}\!\,\left(\frac{-1}{1+|2π x|^2} \right) = -\frac{e^{-|x|}}{2}.$$