# problem with the ode $f' = a \delta (x) f$

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

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

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

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?

• Is there a reason you expect an ODE with a distribution for the RHS to have a unique solution? Mar 19, 2020 at 23:32
• 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. Mar 19, 2020 at 23:36
• 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. Mar 20, 2020 at 1:00
• @whpowell96 I met it in a quantum mechanical problem. It is basically my schroedinger equation. Mar 20, 2020 at 1:34
• @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. Mar 20, 2020 at 1:38

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 $$H(x)\delta(x)$$.