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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?

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2 Answers 2

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

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  • $\begingroup$ @LutzLehmann. True. Will edit my answer. $\endgroup$
    – md2perpe
    Commented Jul 23, 2021 at 13:09
  • $\begingroup$ It should be $A_-=A_+ = -1/2$, and our answers will coincide ;) $\endgroup$
    – LL 3.14
    Commented Jul 23, 2021 at 16:27
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    $\begingroup$ @LL3.14. Fixed. $\endgroup$
    – md2perpe
    Commented Jul 23, 2021 at 16:30
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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 $$ u(x) = \mathcal F^{-1}\!\,\left(\frac{-1}{1+|2π x|^2} \right) = -\frac{e^{-|x|}}{2}. $$

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