I am stuck with a slightly more involved variant of the following, quite elementary I suppose, variable coefficient ordinary differential equation:

\begin{equation} i\frac{d}{dx} p + a(x)p(x) = 0 \,, \qquad p(0) = 1 \end{equation} where $a \in C^\infty([0,\beta])$ is complex valued (ideally no more conditions are required, what I probably need is that it is non-vanishing).

Of course if $a$ were a non-zero constant then the solution is $p(x) = e^{\,iax}$, but here I am not sure how to proceed. Eventually I'd need to understand the solution in the case where this is a system (so $a \in GL(n\mathbb{C})$ and so on) but for now I should concentrate on understanding the scalar - case.

Any hints as to where to read up on this or basic ideas would be hugely helpful to get started. I appologize in case this is too elementary, my ode - knowledge is unfortunately very sketchy and I hope I can change this going forward. Many thanks!

  • $\begingroup$ There's a common method for solving the scalar case, at least. The key phrase to search for is "integrating factor". $\endgroup$ – Antonio Vargas Feb 2 '14 at 16:14

Assume that $p$ satisfies the initial value problem \begin{equation} \frac{d}{dx} p -i a(x)p(x) = 0 \,, \qquad p(0) = 1. \end{equation} Then $$ \left(\exp\left(-i\int_0^x a(t)\,dt\right)p(x)\right)'= \exp\left(-i\int_0^x a(t)\,dt\right)\big(\frac{d}{dx} p -i a(x)p(x)\big) = 0, $$ and hence $\exp\left(-i\int_0^x a(t)\,dt\right)p(x)$ is constant, which implies that $$ \exp\left(-i\int_0^x a(t)\,dt\right)p(x)=\exp\left(-i\int_0^0 a(t)\,dt\right)p(0)=1, $$ and therefore $$ p(x)=\exp\left(i\int_0^x a(t)\,dt\right). $$


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