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This is Exercise 2.7.1 from "Fourier Series and Integrals" by Mckean.

Consider the ordinary differential equation $$u''-u=-f.$$ In the book, they derived that $$u=e^{-|x|}\circ f=\int_{\mathbb{R}}e^{-|x-y|}f(y)dy$$ is a solution to the differential equation. My task is to show, by formal differentiation that $u$, indeed, solves the problem.

I begin by computing the second derivative of $u$. Since $$\frac{d}{dx}(f_1\circ f_2)=\frac{d}{dx}(f_1)\circ f_2=f_1\circ \frac{d}{dx}(f_2),$$ I believe we could represent $u''$ in three "different" ways. Namely $$u''=\frac{d^2}{dx^2}(e^{-|x|})\circ f=\frac{d}{dx}(e^{-|x|})\circ \frac{d}{dx}(f)=e^{-|x|}\circ \frac{d^2}{dx^2}(f).$$ Consider the first case:

Then we would have (using the distributive property of convolution) $u''=(e^{-|x|}-2\delta(x))\circ f=e^{-|x|}\circ f-2\delta(x)\circ f$. But going back to the equation, this would give us $$u''-u=e^{-|x|}\circ f-2\delta(x)\circ f-e^{-|x|}\circ f=-2\delta(x)\circ f=-f.$$ The Dirac delta function is either equal to $0$ or $\infty$. The only way for me to make sense of the expression $-2\delta(x)\circ f=-f$ is to think of $\delta(x)$ equals $0$ then. But this would mean we are, in fact, considering $$0\circ f=-f\iff -f=0\iff f=0.$$ So what we have been considering is the equation $$u''-u=0,$$ and found that one solution is $u=0$? That doesn't sound too interesting to me, and I am sure I have done something crazy somewhere in my argument. But I don't know where, though.

Second case:

Because of the property of the derivative of the convolution product, to consider this case shouldn't make any difference. But I did it anyway just to see if I would end up at the same result as for the first case. However I ended up not knowing what to do, unfortunately.

In the second case, we have $$\frac{d^2}{dx^2}(e^{-|x|}\circ f)=-\frac{x}{|x|}e^{-|x|}\circ f'.$$ Then this would give us $$-\frac{x}{|x|}e^{-|x|}\circ f'-e^{-|x|}\circ f=\int_{\mathbb{R}}-\frac{x-y}{|x-y|}e^{-|x-y|}f'(y)dy-\int_{\mathbb{R}}e^{-|x-y|}f(y)dy.$$ I don't know how to simplify the above expression.

Any help would be really appreciated, because I have no clue how to solve this problem. I would really like to see that $u''-u$ turns out to be $-f$, "no matter" what $f$ is (not just the $f=0$ case ). :)

Thank you!

  • I suppose that the equation is really $u''-u=-f$. Because $$\frac{d}{dx}|x|=sign(x)=2\theta(x)-1$$ Where $\theta(x)$ is the Heaviside step function, thus $$\frac{d}{dx}sign(x)=2\delta(x)$$ – Kiryl Pesotski Apr 15 '18 at 18:26
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    There's a factor of $1/2$ missing in $u$ (which you need to cancel the factor of $2$ in $2 \delta$); cf. also this question: https://math.stackexchange.com/questions/2386372/about-peakon-kernel – Hans Lundmark Apr 15 '18 at 19:11
  • @HansLundmark Ah thank you very much for catching $\frac{1}{2}$ is missing. There is certainly something I don't understand. :) If we would include $\frac{1}{2}$ then we would end up with $-\delta(x)\circ f$. When I read "show it is a solution to the differential equation", I thought I wanted to simplify the left-hand side of the equation to get the right-hand side. In this case; simplify $-\delta(x)\circ f$ to see it equals $-f$. But I cannot see how to simplify $-\delta(x)\circ f$ to get $-f$. –  Apr 15 '18 at 20:34
  • @HansLundmark Do you think this answers my latest question https://math.stackexchange.com/questions/1015498/convolution-with-delta-function? :) –  Apr 15 '18 at 20:54
  • Nevermind, I am stupid. Of course, that answers my question... –  Apr 15 '18 at 21:06
  • Indeed it does. – Hans Lundmark Apr 16 '18 at 04:41

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