DISCLAMER : I first should apologize, it could be that my question does not make much sense or could be imprecise. Just take this question as a naive question from a physics guy who is trying to understand what he is doing...

Is there a generic way to solve distributional equations like : $$ p(x)\cdot T(x)=1 $$ where $p$ would be a second order polynome $p(x)=(x-z_1)(x-z_2)$ for instance ?

I do know that the general solution to the distributional equation $x\cdot T(x)=1$ reads : $$ T(x)=\text{vp}\frac{1}{x}+\alpha\,\delta(x)\;,\quad\alpha\in\mathbb{C} $$ For instance, is there a way to treat the equation $p\cdot T=1$ "locally around the roots" of $p$ as we would treat the equation $x\cdot T=1$ and give a solution which would look like : $$ T(x)=\text{vp}\frac{1}{x-z_1}+\alpha_1\,\delta(x-z_1)+\text{vp}\frac{1}{x-z_2}+\alpha_2\,\delta(x-z_2) $$ Any advice/ressources is appreciated. Thanks by advance.


1 Answer 1


If the roots of $p$ are real and distinct, then the intuition suggests that $$T" =" \frac{1}{(x-x_1)(x-x_2)}.$$ We can rewrite it as $$T(x) "=" \frac{1}{x_2-x_1}\left(\frac{1}{x-x_1} - \frac{1}{x-x_2}\right).$$ Now we apply a standard workaround to avoid non-local-integrability of terms $\frac{1}{x-x_i}$, we consider principal values (and add add delta-functions arising from the non-uniqueness of the solution): $$T = \frac{1}{x_2-x_1}\left(PV\left(\frac{1}{x-x_1}\right) - PV\left(\frac{1}{x-x_2}\right)\right) + c_1\delta_{x_1} + c_2\delta_{x_2}.$$

If a root, say, $x_1$ has a non-zero imaginary part, then the function $x\to x-x_1$ is $C^\infty$ on $\Bbb R$ and is never zero there, so we can safely divide both parts of the equation (and therefore reduce our problem to a well-known one): $$(x-x_2) T = \frac{1}{x-x_1}.$$

Finally, if $x_1=x_2\in\Bbb R$, then neither of the cases above apply. Let us take for simplicity $x_1=x_2=0$. We need to guess a solution of the equation $x^2T=1$. Obviously, we need to play around $PV(1/x)$. Let us take $G = -\left(PV(1/x)\right)'$, then $$\langle x^2 G,\phi\rangle = \langle G,x^2\phi\rangle = \langle PV(1/x),x^2\phi'+2x\phi\rangle = $$ $$=\lim_{\varepsilon\to 0}\left( \int_{\varepsilon}^{+\infty}\frac{x^2\phi'(x)+2x\phi(x)}{x}dx+\int_{-\infty}^{\varepsilon}\frac{x^2\phi'(x)+2x\phi(x)}{x}dx\right)$$

$$ = \int_{\Bbb R}(x\phi'(x)+2\phi(x))dx=\int_{\Bbb R}\left( \left( x\phi(x) \right)'+\phi(x) \right)dx=\langle 1,\varphi\rangle,$$ hence $G$ is a solution of $x^2T=1$. We can conclude $$T = -\left(PV(1/x)\right)' + c_0\delta_0 + c_1\delta_{0}' .$$

  • $\begingroup$ Thank you very much for your very clear derivation. This is what I was looking for! $\endgroup$
    – dolun
    Mar 31, 2017 at 7:53
  • $\begingroup$ Can you explain how did you get equality $\int_{\Bbb R}\left( \left( x\phi(x) \right)'+\phi(x) \right)dx=\langle 1,\varphi\rangle$? $\endgroup$
    – Simurgh
    Dec 3, 2019 at 17:09
  • $\begingroup$ @Simurgh if support of $\phi$ is a subset of interval $[-R,R]$, then $\int_\Bbb R (x\phi(x))'dx =\int_{-R}^R (x\phi(x))'dx = R\phi(R)+R\phi(-R) = 0$. And by definition $\int_\Bbb R \phi(x)dx = \langle 1,\phi\rangle$. $\endgroup$ Dec 4, 2019 at 9:14
  • $\begingroup$ Excuse me, could you give me your email to ask more questions about distributions? @TZakrevskiy $\endgroup$
    – Simurgh
    Dec 4, 2019 at 16:17
  • $\begingroup$ @Simurgh I think it would be more productive if you asked your questions here on Math.SE - like this not only me, but other users would be able to see and answer your questions. Also, there are chat rooms (chat.stackexchange.com/…), you should be able to join them. $\endgroup$ Dec 4, 2019 at 16:26

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