Solving Differential Functional Equation $f(2x)=2f(x)f'(x)$
Find all functions satisfying $f(2x)=2f'(x)f(x)$
Under given condition, can't we find explicit solutions?
If one restrict oneself to functions which admit a power series expansion at $x = 0$, following is the list of solutions I found:
$$f(x) = \begin{cases} 0, & f(0) = 0, f'(0) = 0,\\ k \sin(x/k), & f(0) = 0, f'(0) = 1, f'''(0) < 0\\ x, & f(0) = 0, f'(0) = 1, f'''(0) = 0\\ k \sinh(x/k), & f(0) = 0, f'(0) = 1, f'''(0) > 0\\ k \exp(x/2k), & f(0) \ne 0, f'(0) = \frac12\\ \end{cases}$$
where $k$ is a non-zero constant.
Let $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$. In terms of $a_n$, the function equation becomes:
$$\begin{align} a_0 &= 2 ( a_1 a_0 )\tag{*0}\\ 2\,a_1 &= 2 ( 2 a_2 a_0 + a_1 a_1 )\tag{*1}\\ 2^2 a_2 &= 2 ( 3 a_3 a_0 + 2 a_2 a_1 + a_1 a_2 )\\ 2^3 a_3 &= 2 ( 4 a_4 a_0 + 3 a_3 a_1 + 2 a_2 a_2 + a_1 a_3 )\\ 2^4 a_4 &= 2 ( 5 a_5 a_0 + 4 a_4 a_1 + 3 a_3 a_2 + 2 a_2 a_3 + a_1 a_4)\\ 2^5 a_5 &= 2 ( 6 a_6 a_0 + 5 a_5 a_1 + 4 a_4 a_2 + 3 a_3 a_3 + 2 a_2 a_4 + a_1 a_5 )\\ &\;\vdots \end{align}$$
The first thing we notice is $(*0)$ implies:
$$a_0 (1 - 2a_1) = 0 \implies a_0 = 0 \quad\text{ or }\quad a_1 = \frac12$$
If $a_0 = 0$, then $(*1)$ implies $a_1 = 0$ or $1$. It is easy to see $a_1 = 0$ will lead us to $f(x) \equiv 0$. So let us assume $a_1 = 1$. The set of equations become: $$\begin{align} ( 2^1 - 3 ) a_2 &= 0\\ ( 2^2 - 4 ) a_3 &= 2 a_2 a_2 \iff 0 = 0 \text{ ( i.e. } a_3\;\text{ is arbitrary )}\\ ( 2^3 - 5 ) a_4 &= 3 a_3 a_2 + 2 a_2 a_3\\ ( 2^4 - 6 ) a_5 &= 4 a_4 a_2 + 3 a_3 a_3 + 2 a_2 a_4\\ &\;\vdots \end{align}$$ It is then easy to see:
- $a_n = 0$ for all even $n$.
- $a_3$ is arbitrary.
- once $a_3$ is fixed, $a_n$ is recursively fixed for all odd $n > 3$.
Depends on the sign of $a_3$, it leads us to solutions of the form $k \sin(x/k), x $ or k sinh(x/k)$.
If $a_0 \ne 0$, then $a_1 = \frac12$ and the set of equations becomes:
$$\begin{align} 2 a_2 a_0 &= 2^0\,a_1 - ( a_1 a_1 )\tag{*1}\\ 3 a_3 a_0 &= 2^1 a_2 - ( 2 a_2 a_1 + a_1 a_2 )\\ 4 a_4 a_0 &= 2^2 a_3 - ( 3 a_3 a_1 + 2 a_2 a_2 + a_1 a_3 )\\ 5 a_5 a_0 &= 2^3 a_4 - ( 4 a_4 a_1 + 3 a_3 a_2 + 2 a_2 a_3 + a_1 a_4)\\ &\;\vdots \end{align}$$ It is easy to see these will recursively fix the values of $a_n$ for all $n > 1$. This lead us to the $k \exp(x/2k)$ solution.
Assume that a solution $f(x)$ is analytic in some neighborhood of $x=0$ so that it can be represented as a converging power series $$ f(x)=a_0+a_1x+a_2x^2+\cdots. $$
If we assume that $a_0=f(0)\neq0$, then, by equating the power series representing the two sides of the given equation, we see that the remaining coefficients are uniquely determined. Comparing the coefficient of $x^\ell$ on both sides gives a linear equation for $a_{\ell+1}$ with a non-zero coefficient.
So if an analytic solution with a fixed value $f(0)=k\neq0$ exists, it is unique. An ansatz of the form $$f(x)=ke^{\lambda x}$$ as in rajb245's answer works, iff $\lambda=1/2k$. Therefore we have a one-parameter family of solutions $$ f(x)=f_k(x)=ke^{x/2k}. $$ If $a_0=0$ things become messy, as the proffered solutions $f(x)=x$ and $f(x)=\sin x$ prove. Unless I made a mistake, we get that then $a_1=0$ or $a_1=1$. In the latter case I got that we necessarily have $a_2=0$, but $a_3$ can be arbitrary, which goes together with the known solutions. Probably this approach can be taken further (most likely by somebody else)...
Again assuming analyticity, and taking advantage of mezhang's observation that the r.h.s. is the derivative of $f(x)^2$ we get by differentiating the given equation $k$ times we get the equations $$ 2^kf^{(k)}(2x)=D^{k+1}\left[f(x)^2\right]=\sum_{i=0}^{k+1}{k+1\choose i} f^{(i)}(x)f^{(k+1-i)}(x). $$ Plugging in $x=0$ gives then the system of equations $$ 2^kf^{(k)}(0)=\sum_{i=0}^{k+1}{k+1\choose i} f^{(i)}(0)f^{(k+1-i)}(0) $$ that holds for all $k$. As $f^{(k)}(0)=k!a_k$ we get $$ 2^k k! a_k=(k+1)!\sum_{i=0}^{k+1} a_ia_{k+1-i}. $$
Assuming (see above) $a_0=0,a_1=1$ this gives us $$ 2^kk!a_k=(k+1)!\left(2a_k+\sum_{i=2}^{k-1}a_ia_{k+1-i}\right). $$ With $k=2$ we get $a_2=0$ (confirming my hand calculations). With $k=3$ we get a tautology $48a_3=48a_3$ that does, indeed, leave the value of $a_3$ undetermined. For $k>3$ we have $2^kk!>2(k+1)!$, so we get an equation recursively determining $a_k$. Observe that a straightforward induction proves (after this start) that $a_k=0$ whenever $k$ is even. Thus the resulting function $f(x)$ is odd, and uniquely determined by the value of $a_3$. I don't know, if the solution is always an elementary function.
In the case $a_0=a_1=0$ the term $(k+1)!(a_1a_k+a_ka_1)=2(k+1)!a_ka_1$ is not there on the r.h.s., so we get the equations $$ 2^kk!a_k=(k+1)!\left(\sum_{i=2}^{k-1}a_ia_{k+1-i}\right) $$ for all $k$. Again we recursively see that this system has only the trivial solution $a_k=0$ for all $k$.