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While doing some estimates for PDEs I came across the following equation:

$$ y'(t) = \alpha(t) + \left( \int_0^t y(\tau) \, d\tau\right)^\gamma, \qquad t \in [0,1] $$

where $\alpha \colon [0,1] \to \mathbb R$ is some given function and $\gamma \ge 1$ is a fixed real number. I have a non-negative, bounded function $y$ satisfying this relation and I would like to derive some pieces of information about it.

Of course I do not expect any risolutive formulas to hold, but hope is the last to die... do you have some hints? Any ideas on which kind of estimates I could derive?

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    $\begingroup$ there may be a solution for $\alpha\equiv 0$, is that of interest? $\endgroup$ – Carlo Beenakker Jan 13 '18 at 16:58
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    $\begingroup$ Thanks a lot for the comment. Sure it may be of interest, of course, though not fully resolutive (but maybe with some tricks I can work out some estimates also for the general $\alpha$). Could you please point me out some references? Thanks! $\endgroup$ – Romeo Jan 13 '18 at 17:00
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If I define $f(t)=\int_0^t y(\tau)d\tau$, I need to solve $$f''(t)=\alpha(t)+f(t)^\gamma.$$ For $\alpha\equiv 0$ this has the implicit solution

$${({\gamma}+1) f(t)^2 \left(c_1 {\gamma}+c_1+2 f(t)^{{\gamma}+1}\right)^2 \, _2F_1\left(\frac{1}{2},\frac{1}{{\gamma}+1};1+\frac{1}{{\gamma}+1};-\frac{2 f(t)^{{\gamma}+1}}{{\gamma} c_1+c_1}\right)^2}{}=\left(c_2+t\right)^2\,\left(c_1 {\gamma}+c_1\right) \left(c_1 ({\gamma}+1)+2 f(t)^{{\gamma}+1}\right)^2$$

Closed-form expressions in terms of special functions (Weierstrass $\wp$ and Jacobi elliptic function sn, courtesy of Mathematica) are possible for small integer $\gamma$:

$$f(t)=c_1 e^t+c_2 e^{-t},\;\;\gamma=1$$ $$f(t)=\sqrt[3]{6}\; \wp \left(\frac{t+c_1}{\sqrt[3]{6}};0,c_2\right),\;\;\gamma=2$$ $$f(t)=\pm\sqrt[4]{2} \sqrt{ic_1}\; \text{sn}\left(\left.\frac{(-1)^{3/4} \sqrt{\sqrt{2} \sqrt{c_1} t^2+2 c_2 \sqrt{2} \sqrt{c_1} t+c_2^2 \sqrt{2} \sqrt{c_1}}}{\sqrt{2}}\right|-1\right),\;\;\gamma=3$$

More generally, for constant $\alpha$, the implicit solution is $$\int_1^{f(t)} \left({2\alpha x+\frac{2}{\gamma+1}x^{\gamma+1}+c_1}\right)^{-1/2} \, dx=c_2+t$$ with explicit solutions

$$f(t)=-\alpha+c_1 e^t+c_2 e^{-t},\;\;\gamma=1$$ $$f(t)=\sqrt[3]{6} \;\wp \left(\frac{t+c_1}{\sqrt[3]{6}};-2 \sqrt[3]{6} \alpha,c_2\right),\;\;\gamma=2$$

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  • $\begingroup$ Thanks for your precious effort and for writing the implicit expression of the solution for vanishing $\alpha$. I will try to work on it to see if I find something... Do you believe there is possibility of deriving some Gronwall-like estimates from the equation? Maybe writing it as an ODE is a good point... In any case, thanks again for your interest! $\endgroup$ – Romeo Jan 13 '18 at 17:19
  • $\begingroup$ Does your last expression adapt also to piecewise constant $\alpha$? Maybe that could be very useful to me... (I am not really in this framework but maybe I could get somehow close to it...) Thanks again! $\endgroup$ – Romeo Jan 13 '18 at 17:29
  • $\begingroup$ @user-unknown: For piecewise constant $\alpha$ you just need to stitch together the constant solution by matching the end point function values and its first derivatives to obtain $c_1$ and $c_2$, if you want the solution to be $C^1$. $\endgroup$ – Hans Jan 13 '18 at 19:34
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I personally do not know much about these types of equations but for $\gamma$ = 1 this equation is an example of a Volterra integro-differential equation which has been heavily studied.

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    $\begingroup$ I thought a Volterra equation was a convolution equation, $y(t)=\alpha(t)+\int_0^t K(t-\tau)y(\tau)d\tau$ --- this doesn't seem to be of that form for any $\gamma$. $\endgroup$ – Carlo Beenakker Jan 14 '18 at 11:26

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