Solution of a distributional differential equation

Question

I have to find the solutions in $\mathcal{D}'(\mathbb{R})$ of the differential equation $T'+rT=\delta_0$, where $r\in C^\infty(\mathbb{R})$ and $\delta_0$ is the Dirac distribution. I have tried with an integration factor and several of the methods that I know for ODEs but I am not able to reach anything clear.

Could somebody give me a hint?

Thanks in advance.

Answer 1

The integrating factor way:

$$\frac{d}{dx} \left ( e^{R(x)} T(x) \right ) = e^{R(x)} \delta_0(x) \\ e^{R(b)} T(b) - e^{R(a)} T(a)=\int_a^b e^{R(x)} \delta_0(x) dx = \begin{cases} 0 & 0 \not \in [a,b] \\ e^{R(0)} & 0 \in [a,b] \end{cases}.$$

Here $R$ is any antiderivative of $r$ (but the same one all around).

Note that there will be a discontinuity at $0$ so any IVP should strictly speaking be posed at a point other than $0$.

Another way is to use the fact that the delta function is guaranteed to be confined to the highest derivative in the LHS, so you can assume $T'=\delta_0+f$ and so $T=H+F$ where $H$ is the Heaviside function. Then the original equation becomes

$$\delta_0+F'+rH+rF=\delta_0 \\ F'+rF=-rH$$

which now basically makes sense without distribution theory (provided you do not insist that the equation holds in the classical sense at $0$).

Answer 2

First note that multiplication of a differential equation with a nowhere zero smooth function keeps the solution set unchanged. (It's an equivalence relation.)

Let $R$ be a primitive function to $r$ and multiply the differential equation with $e^R > 0$: $$e^R T' + r e^R T = e^R \delta_0.$$

Now the left hand side can be rewritten as a derivative of a product: $$e^R T' + r e^R T = (e^R T)'$$ and the function in the right hand side can be eliminated: $$e^R \delta_0 = e^{R(0)} \delta_0$$ since $f\,\delta_0 = f(0)\,\delta(0).$

Thus our differential equation can be written as $$(e^R T)' = e^{R(0)} \delta_0.$$

Taking the distributional antiderivative of both sides gives us $$e^R T = e^{R(0)} H_0 + C,$$ where $H_0$ is the Heaviside step function (with the step at $x=0$) and $C$ is some constant (which really is to be interpreted as the constant distribution $\phi \mapsto C \langle 1, \phi\rangle$).

Now we can multiply with $e^{-R}$: $$T = e^{-R} e^{R(0)} H + C e^{-R} = e^{-(R-R(0))} H + C e^{-R}.$$