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The first solution is

$$f(0^+) = \exp\left(a \int_{0^-}^{0^+} dx \delta (x)\right) f(0^-) = \exp(a) f(0^-) .$$

The second solution is

$$ f(0^+) - f(0^-) = a \int_{0^-}^{0^+} dx \delta (x) f(x) = \frac{a}{2}(f(0^+ ) +f(0^-)) ,$$

which leads to

$$f(0^+) = \frac{1 + a/2}{1 - a/2} f(0^-) . $$

The two approaches yield different results.

Which one is right? I am biased on the second one. But can anyone justify it and point out the flaw in the first one?

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  • $\begingroup$ Is there a reason you expect an ODE with a distribution for the RHS to have a unique solution? $\endgroup$
    – whpowell96
    Commented Mar 19, 2020 at 23:32
  • $\begingroup$ For the first one, you should get $f(0^+)=e^af(0^-)$. I also think that the second solution is correct. The first one is fishy because it involves the integral $\int_{0^-}^{0^+}\frac{f'}{f}dx$, which to me doesn't give $\ln f(0^+)-\ln f(0^-)$ in this particular case. $\endgroup$
    – Batominovski
    Commented Mar 19, 2020 at 23:36
  • $\begingroup$ The ODE has no non-trivial solution in distribution. This is tantamount to asking "What is the meaning of the product $H\delta$?" The answer is that that product has no meaning. $\endgroup$
    – Mark Viola
    Commented Mar 20, 2020 at 1:00
  • $\begingroup$ @whpowell96 I met it in a quantum mechanical problem. It is basically my schroedinger equation. $\endgroup$
    – Beamer
    Commented Mar 20, 2020 at 1:34
  • $\begingroup$ @MarkViola No non-trivial solution in distribution? Could you make it more precise? It arises in a quantum mechanical problem. The second solution yields physically reasonable results. The equation is relevant in physics, although how to interpret it is a problem. $\endgroup$
    – Beamer
    Commented Mar 20, 2020 at 1:38

1 Answer 1

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Heuristically and naively, one might be tempted to write $f'(x)=0$ for $x\ne0$. Hence we would deduce that for some constants $A$ and $B$, $f(x)=A$ for $x<0$ and $f(x)=B$ for $x>0$.

But then we would have $f(x)=A+(B-A)H(x)$ which would suggest that for $a\ne 0$

$$(B-A)\delta(x) =a(A+(B-A)H(x))\delta(x)$$

Inasmuch as there is no meaning assignable to the distribution of the product $H(x)\delta(x)$, the only possible solution is the trivial solution $f(x)\equiv 0$

Let us attempt a solution to the original ODE by using a regularization approach. Proceeding, we let $f_n(x)$ be defined by the ODE

$$f_n'(x)=a\delta_n(x)f_n(x)$$

where $\delta_n(x)$ is any regularization of $\delta(x)$. The solution is

$$f_n(x)=f_{-\infty}\,e^{a\int_{-\infty}^x \delta_n(t)\,dt}$$

whereupon letting $n\to \infty$ reveals that

$$f(x)=f_{-\infty}e^{aH(x)}\tag 1$$

NOTE:

We remark that the function $f$, as given by $(1)$, is discontinuous and not uniquely defined at $0$. Furthermore, the candidate for its distributional derivative,

$$f'(x) =a\underbrace{f_{-\infty}e^{aH(x)}}_{\text{a discontinuous function at}\,0}\,\times \underbrace{ \delta(x)}_{\text{The Dirac Delta}}$$

is meaningless.

Another way to see this, is to observe that $(1)$ implies that $f$ can be written as the discontinuous function

$$f(x)=\begin{cases}f_{-\infty}&,x<0\\\\f_{-\infty}e^a&,x>0\end{cases}\tag2$$

And if $(2)$ is so, we can write

$$f(x)=f_{-\infty}\left(1+(e^a-1)H(x)\right)\tag3$$

But from $(3)$ the distributional derivative of $f$ is

$$f'(x)=f_{-\infty }(e^a-1)\delta(x)$$

Then, the original ODE would imply that $(e^a-1)\delta(x)=\left(1+(e^a-1)H(x)\right)\delta(x)$

which is meaningless due to the appearance of $H(x)\delta(x)$.

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