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Can the functional differential equation $$D'(t) = 2 D(2t) - D(t)$$ with initial condition $D(0)=0$ be solved in closed form? A Laplace transform yields the functional equation $$ (s+1)d(s) = d(s/2) $$ but this equation seems badly behaved ($d(0)$ indeterminate).

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    $D(t) = 0$ is a closed-form solution. Maybe you want a different initial condition? – 2017-01-18

2 Answers 2

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There are series solutions

$$ \eqalign{d(t) &= c \left( e^{-t} - 2 e^{-2t} + \frac{4}{3} e^{-4t} - \frac{8}{21} e^{-8t} + \ldots\right)\cr &= c \sum_{k=0}^\infty \frac{(-2)^k}{ \prod_{j=1}^k (2^j-1)} e^{-2^kt}\cr}$$

which all have $d(0) = 0$. These should converge for $t \ge 0$.

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    @SangchulLee Please undelete your solution. It's interesting to write it in terms of $q$-Pochhammer symbols. – 2017-01-19
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Let us relax the initial condition and look at a solution of the form

$$ D(t) = \sum_{k=-\infty}^{\infty} a_k \exp(-2^k t). $$

Plugging this to the equation and assuming that term-wise differentiation applies, we find that $(a_k)$ satisfies the recurrence relation $ (1-2^k) a_k = 2a_{k-1}$, whose solution can be written in terms of the $q$-Pochhammer symbol:

$$ a_k = \begin{cases} \dfrac{2^k}{(2,2)_k} a_0, & k \geq 0 \\ 0, & k < 0 \end{cases} $$

Consequently we can write

$$ D(t) = a \sum_{k=0}^{\infty} \frac{2^k}{(2;2)_k}\exp(-2^k t) $$

for some constant $a$.

p.s. As pointed out by Robert Israel, these solutions indeed satisfy the initial condition $D(0) = 0$ (which I did not expect before looking at his answer). Here is a simple confirmation:

\begin{align*} D(0) &= a \sum_{k=0}^{\infty} \frac{2^k}{(2;2)_k} \\ &= a \left(1 + \sum_{k=1}^{\infty} \frac{2^k - 1}{(2;2)_k} + \sum_{k=1}^{\infty} \frac{1}{(2;2)_k} \right) \\ &= a \left(1 - \sum_{k=1}^{\infty} \frac{1}{(2;2)_{k-1}} + \sum_{k=1}^{\infty} \frac{1}{(2;2)_k} \right) \\ &= 0. \end{align*}