Differential equation combined with functional equation which looks simple

Question

The equation is $$f'(x)=cf(x/2).$$ The problem emerges when I'm trying to deal with a partial differntial equation $$u_{t}(x,t)=u_{xx}(x,t/2).$$ Either using separation of variables: $$u(x,t)=X(x)T(t)\rightarrow X(x)T'(t)=X''(x)T(t/2)$$ so that $$X''(x)-cX(x)=0, T'(t)-cT(t/2)=0$$ or doing Fourier transform: $$\hat{u}_t(s,t)+s^2\hat{u}(s,t/2)=0$$ eventually reduces the problem to solving $$f'(x)=cf(x/2).$$

But the equation does not seem easy to solve. I can only prove the existence of solutions using the Euler method. The numerical simulation looks like

The other function plotted for comparison is the exponential function.

Does a solution of closed form exist to the equation? If not, how do we solve the original PDE?

Answer 1

To find the analytic solution, we can use a power series expansion around $x=0$ in the usual way: $$ 0 = f'(x) - c f(x/2) = \sum_{n=0}^{\infty} \left( f^{(n+1)}(0)-\frac{c}{2^n}f^{(n)}(0) \right) \frac{x^n}{n!}, $$ and then equating coefficients gives $$ f^{(n+1)}(0)=\frac{c}{2^n}f^{(n)}(0). $$ It is then simple to prove by induction that $$ f^{(n)}(0) = \frac{c^n}{2^{1+2+3+\dotsb+(n-1)}} f(0) = \frac{c^n}{2^{n(n-1)/2}} f(0), $$ so $$ f(x) = f(0) \sum_{n=0}^{\infty} \frac{c^n}{2^{n(n+1)/2}} \frac{x^n}{n!}. $$

Answer 2

You can show pretty easily that $f$ is an entire function. Indeed, by taking an arbitrary $ K>0$, and setting $M_n=\|f^{(n)}\|_{\infty, [-K,K]}$, by differentiating the equation n times, we get the relation $M_n \leq \frac{c}{2^{n-1}}M_{n-1}$ for all positive integers $n$. This inequality yields $M_n \leq \frac{c^n}{2^{\frac{n(n-1)}{2}}}M_0$. This shows that $M_n$ converges to $0$ and thus by using taylor estimations, $f$ is entire.

Now writing :

\begin{equation*} f(x) = \sum_{k=0}^{\infty} a_k x^k \end{equation*}

We get by substituting in the equation: \begin{equation*} \sum_{k=0}^{\infty}(k+1) a_{k+1}x^k = \sum_{k=0}^{\infty} \frac{c}{2^k} x^k \end{equation*}

Which yields $a_{k+1} = \frac{c}{k+1}\cdot\frac{a_k}{2^k}$. Therefore, $a_k= \frac{c^k}{k!} \cdot \frac{f(0)}{2^{\frac{k(k-1}{2}}}$. Which gives you the function you are looking for.