Solving the equation $f'(x)=-x-xf(x/2)$

Question

I am a bit stuck at solving the equation $f'(x)=-x-xf(x/2)$, where $x$ is a real variable.

The argument $x/2$ in $f$ seems a bit hard to overcome.

Could anyone please provide some help?

Answer 1

Let's proceed without thinking.

Let us suppose that $f(x) = \sum_{n\geq0}a_nx^n$ is a solution which is analytic near zero. Then \begin{align} 0 = f'(x)+xf(x/2)+x &= \sum_{n\geq1}na_nx^{n-1} + \sum_{n\geq0}a_n\frac{x^{n+1}}{2^n} + x\\ &= \sum_{n\geq0}(n+1)a_{n+1}x^{n} + \sum_{n\geq1}a_{n-1}\frac{x^{n}}{2^{n-1}} +x \\ &= a_1+(2a_2+a_0+1)x+\sum_{n\geq2}\left(a_{n+1}(n+1) + \frac{a_{n-1}}{2^{n-1}}\right)x^n \end{align} We must then have that $a_1=0$, $a_2=-(1+a_0)/2$ and $$a_{n+1} =- \frac{a_{n-1}}{2^{n-1}(n+1)}$$ for all $n\geq2$. For each choice of $a_0$ there is exactly one solution to these recurrence equations, and it is clear that the series $\sum_{n\geq0}a_nx^n$ does converge, as the coefficients go to zero very very fast.

The odd coefficients are all zero. OTOH, it is convenient to rewrite the recurrence in the form $$a_{n+2}=-\frac{a_n}{2^n(n+2)},$$ now valid for all $n\geq2$. We see that $$a_{2\cdot2}=a_4 = (-1)\frac{a_2}{2^2\cdot 4}, \quad a_{2\cdot3}=a_6=(-1)^2\frac{a_2}{2^2\cdot 4\cdot 2^3\cdot 6}, \quad a_{2\cdot4}=a_8=(-1)^3\frac{a_2}{2^2\cdot 4\cdot 2^3\cdot 6\cdot 2^4\cdot 8}, \dots$$ and suggests that more generally $$a_{2n}=(-1)^{n}\frac{1+a_0}{2^{n^2}n! }$$ This gives you the solutions that are analytic near zero, which are of the form $$a_0+\sum_{n\geq1}(-1)^n\frac{1+a_0}{2^{n^2}n!}x^{2n} $$

Answer 2

Let us take a constant function $f(x)=-1$.