1
$\begingroup$

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?

$\endgroup$
0
2
$\begingroup$

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 $$ G(x) = -\frac12 e^{-|x|}. $$

$\endgroup$
3
  • $\begingroup$ @LutzLehmann. True. Will edit my answer. $\endgroup$
    – md2perpe
    Jul 23 at 13:09
  • $\begingroup$ It should be $A_-=A_+ = -1/2$, and our answers will coincide ;) $\endgroup$
    – LL 3.14
    Jul 23 at 16:27
  • 1
    $\begingroup$ @LL3.14. Fixed. $\endgroup$
    – md2perpe
    Jul 23 at 16:30
2
$\begingroup$

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 $$ u(x) = \mathcal F^{-1}\!\,\left(\frac{-1}{1+|2π x|^2} \right) = -\frac{e^{-|x|}}{2}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.