Second order linear differential equation with distributions: $T''+zT=0$

Question

I have to solve the following second order linear differential equation, where T is a distribution: $$T''+zT=0$$ Where z is a fixed complex number. The approach I have to follow is similar to the one of the answer to this question: differential equation with distributions .

Answer 1

Let $a = z^{1/2}.$ The differential equation can then be written as $(D+ia)(D-ia)T = 0,$ where $D$ is the differentiation operator. We multiply this with the nowhere vanishing integrating factor $e^{iax}$: $$e^{iax}(D+ia)(D-ia)T = 0.$$

We can now do the rewrite $e^{iax}(D+ia)\bullet = De^{iax}\bullet,$ which gives $De^{iax}(D-ia)T = 0.$

We can now take the antiderivative: $e^{iax}(D-ia)T = A,$ where $A$ is a constant, i.e. $(D-ia)T = A e^{-iax}.$

Then we multiply this equation with $e^{-iax}$ giving $e^{-iax} (D-ia) T = A e^{-i2ax}$ or $De^{-iax}T = Ae^{-i2ax},$ which after taking the antiderivative results in $e^{-iax}T = A \frac{1}{-i2a} e^{-i2ax} + B,$ where $B$ is a constant.

Thus, we get $T = A \frac{1}{-i2a} e^{-iax} + B e^{iax},$ where $A$ and $B$ are constants.

As you can see, when $a=0$ we get a problem. Solve the equation for that case separately.