Solutions to $-t\ f'(t)+(1+r t)\ f(t)=f(t/2)$

Question

Any thoughts on how to tackle this ODE $-t\ f'(t)+(1+r t)\ f(t)=f(t/2)$, $t\in[0,t_1]$, with boundary conditions $f(0)=0$ and $f(t_1)=s$, with $r,s$ and $t_1$ fixed constant.

Answer 1

One technique that may be worth investigating is assuming that $f$ is regular enough, and expressing it as a power series $f(t) = \sum_{k \geq 0} a_k t^k$. $$ \begin{align*} -t f'(t) + (1+rt)f(t) - f(t/2) & = -t \sum_{k\geq 0}(k+1)a_{k+1}t^k + \sum_{k \geq 0} a_k t^k + r\sum_{k \geq 0} a_k t^{k+1} - \sum_{k\geq 0}a_k 2^{-k} t^k\\ & = - \sum_{k\geq 1} k a_{k}t^{k} + \sum_{k \geq 1} a_k t^k + a_0 + r\sum_{k \geq 1} a_{k-1} t^{k} - \sum_{k\geq 1}a_k 2^{-k} t^k - a_0\\ & = \sum_{k \geq 1} (-ka_k + a_k + ra_{k-1} - 2^{-k}a_k)t^k \\ & = \sum_{k \geq 1} \left((1-k - 2^{-k})a_k + ra_{k-1} \right)t^k \\ \end{align*} $$ we want this to be zero so all the coefficients must vanish i.e. $a_k$ satisfies the recurrence $$ a_k = a_{k-1} \frac{r}{2^{-k} + k - 1} $$ But then we see that $a_1 = a_0 \times 2r = 0$ since $a_0 = f(0) = 0$. An immediate induction then concludes that $f(t)$ is identically zero. From there it would seem that there is no (series) solution to the problem as stated.

If however we have $a_0 = f(0) \neq 0$ then the recurrence can be worked out, with $$ a_k = a_0 \frac{r^{k}}{(k-1)!} \prod_{\ell = 1}^{k} \left( 1 + \frac{1}{2^{\ell+1}\ell} \right)^{-1} $$ (hopefully I didn't make a mistake). The product on the right is certainly unwieldy and I don't know of a better expression for it, but it clearly is convergent as $k \to \infty$ so as a rough approximation I would replace it by some crude constant $\kappa$. Hence, $$ a_k \approx a_0 \kappa \frac{r^k}{(k-1)!} $$ which gives an approximate solution of $f(t) \approx a_0 \kappa r t e^{rt}$. Imposing $f(t_1) = s$ we get $a_0 = s/(\kappa r t_1 e^{rt_1})$. This is not entirely consistent and there is some handwaving, but maybe that's a step in some direction.