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I have the following different equation

$$f'(x) = f(x/2)$$

with $f(0)=10$.

What type of DE is this, and how would you solve it?

It seems $f(x)$ is likely to be some relative of $e^x$, since $f'(x) = f(x)$, which is close, but I don't even know what that type of DE is called with that $f(x/2)$ feature, so I'm not having any luck searching for a tutorial. The best candidate I've found was a "delay differential equation", but that seems more suited to $f(x-3)$ than $f(x/2)$.

YuiTo Cheng
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Glen
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  • You may try solve it with slope fields. – McCarter Jun 12 '20 at 23:08
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    Maybe taking $x = 2^t$ makes it into a DDE – Will Jagy Jun 12 '20 at 23:12
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    It's a functional differential equation, although that isn't much help. If the naming of difference and differential equations were parallel, it would be a geometric (functional) differential equation or a $q$-differential equation (except that $q$-differential has been applied to a modification of the derivative operator, not to modification of the argument). – Eric Towers Jun 12 '20 at 23:13
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    If we use Laplace Transform the problem is to solve the following functional equation: $$sF(s)-2F(2s)=10.$$ – DiegoMath Jun 12 '20 at 23:20
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    Try the power series method. And then prove that the solution is unique. – fedja Jun 12 '20 at 23:54
  • As people mentioned, a power series could work here. Let $f(x) = \sum a_n x^n$, then plugging in we obtain ( after some manipulation), $a_{n+1} = \frac{a_n}{2^n(n+1)}$ with $a_0 =10$. $a_n$ is then $\frac{a_0}{2^{0.5n(n+1) }(n+1)!}$. I'll try and see if this gives a 'nice function'. – Rohan Nuckchady Jun 13 '20 at 00:23

2 Answers2

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Notice

$${f'(x)=f\left(\frac{x}{2}\right)\Leftrightarrow f''(x) = \frac{1}{2}f'\left(\frac{x}{2}\right), f'''(x)=\frac{1}{2^2}f''\left(\frac{x}{2}\right)}$$

Plugging in some numbers we get ${f(0) = 10, f'(0) = 10, f''(0) = \frac{10}{2}, f'''(0) = \frac{1}{2^2}\left(\frac{10}{2}\right)=\frac{10}{2^3},f''''(0) = \frac{10}{2^6}}$... In general we see

$${f^{(n)}(0)=\frac{10}{2^{\left(\frac{n(n-1)}{2}\right)}}}$$

And now we can create a power series solution using a taylor series centered around $0$:

$${f(x)=10 + \sum_{n\geq 1}\frac{10x^n}{2^{\left(\frac{n(n-1)}{2}\right)}n!}}$$

Riemann'sPointyNose
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If you let: $x=2^t$ then remember we have: $dx=\ln(2)2^tdt$ and our original equation is: $$\frac{df(x)}{dx}-f(x/2)=0$$ so plugging in we get: $$\frac{1}{2^t\ln(t)}\frac{df(2^t)}{dt}-f(2^{t-1})=0$$ or we could write it as: $$\frac{df(2^t)}{f(2^{t-1})}=2^t\ln(t)dt$$ $$\int\frac{df(2^t)}{f(2^{t-1})}=2^t+C$$ I would assume the best way to approach this would be to assume that: $$f(2^{t-1})=f(2^t)g(2^t)$$ or something like this

Henry Lee
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