When $g(x)$ and $h(x)$ are given functions, can $f(x)^2+(g*f)(x)+h(x)=0$ be solved for $f(x)$ in closed form (at least with some restrictions to $g,h$)? (The $*$ is not a typo, it really means convolution as in $(f*g)(x)=\int\limits_{\mathbb R}f(y)g(x-y)\,dy$)
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1even when $g=0$ the equation does not always admits solutions analytic near 0 (take $h=Id$). it seems pretty complicated... – Albert Nov 29 '11 at 11:40
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Sounds like an integral equation to me... – J. M. ain't a mathematician Nov 29 '11 at 11:45
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@J.M.: in the most general sense yes, though I think that label usually means linearity in $f$. So, is this some kind of "no"? – Tobias Kienzler Nov 29 '11 at 12:40
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Not really. There is such a thing as a nonlinear integral equation after all... – J. M. ain't a mathematician Nov 29 '11 at 12:41
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@Glougloubarbaki: I'm fine with multi-valued, non-entire solutions, but I guess you're right about it being complicated. Maybe only special choices of $g,h$ permit analytic expressions... – Tobias Kienzler Nov 29 '11 at 12:42
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@J.M.: True. So, are there any chances on an exact solution or do I have to stick to iteration/numerics? – Tobias Kienzler Nov 29 '11 at 12:45
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Unfortunately I've no experience with nonlinears. Sorry... :( – J. M. ain't a mathematician Nov 29 '11 at 12:49
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3If you mean convolution, you should use $(fg)(x)$ (or $fg(x)$) instead of $f(x)*g(x)$ which makes (strictly spoken) no other sense then multiplication. – Dirk Nov 29 '11 at 12:50
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@Dirk: thanks, I fixed it – Tobias Kienzler Nov 29 '11 at 12:52
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@J.M. no problem, thanks for reminding me that I should search for integral equations – Tobias Kienzler Nov 29 '11 at 12:55
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1You should probably say what you mean by convolution as well. For example, $\int_{-\infty}^\infty f(x-t)g(t),dt$ or $\sum_{k=0}^x f(x-k)g(k)$ or what? – GEdgar Nov 29 '11 at 22:57
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@GEdgar You're right, I'll specify that I mean the $\int_{-\infty}^\infty$ one – Tobias Kienzler Nov 30 '11 at 06:13
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1So the variable $x$ is real? Are the values real? When you say "analytically" do you mean it should be an analytic function? Or is that a vague way of saying "in closed form"? – GEdgar Nov 30 '11 at 14:54
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@GEdgar: $x$ is real, $f,g,h$ are complex valued (in fact, at least $f$ is analytic), and "yes", with analytically I mean in closed form (so I'll reword that, thanks for pointing it out) – Tobias Kienzler Nov 30 '11 at 15:00
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This is a very interesting question? Have you considered posting bounty on it? – Boby Jan 20 '15 at 04:29