Solving $\frac{\partial}{\partial t} f =hf+ h \int \mathrm {d} i\, h f$

Question

I'm looking for the solution of partial differential equation $$\frac{\partial}{\partial t} f(i,t) =\left(a f(i,t) + b\int_0^\infty \mathrm {d} i\, h(i) f(i,t)\right)h(i)$$

Where $$f(i, 0)=1, \int_0^\infty\mathrm {d} i\, h(i)=1$$

Interesting special case: $h(i)=(i+1)^{-2}$, $a=-2$, $b=1$, want to know $g(t)=\int_0^\infty \mathrm{d}i\ h(i)f(i,t)$.

Background: $g(t)$ gives effect on environment of bird population at time $t$ under continuous version of bird population dynamics model described here

Answer 1

The Laplace transform of the function in question can be calculated rather easily. One can Laplace transform in the time coordinate to obtain

$$s\hat F(i,s)=f(i,0)+ah(i)\hat F(i,s)+bh(i)\int_0^\infty h(x)\hat F(x,s)dx$$

Solving for the $\hat F$ yields $$\hat F(i,s)=\frac{f(i,0)}{s-ah(i)}+\frac{bh(i)}{s-ah(i)}\int_0^\infty h(x)\hat F(x,s)dx$$ Integrate the equation against $h(i)$ and solve the now self-consistent equation to obtain the value of the inner product $\langle h,\hat F\rangle:=\int_0^\infty dx h(x)\hat F(x,s)$

$$\langle h,\hat F\rangle(s)=\frac{\int_0^\infty dx\frac{f(x,0)h(x)}{s-ah(x)}}{1-b\int_0^\infty dx\frac{h^2(x)}{s-ah(x)}}$$

For the specific values provided in the question, the integrals can be computed analytically:

$$\int_0^\infty \frac{h(x)}{s+2h(x)}dx=\frac{1}{\sqrt{2s}}\tan^{-1}\sqrt{\frac{2}{s}}$$

$$\int_0^\infty \frac{h^2(x)}{s+2h(x)}dx=\frac{1}{2}\left(1- \sqrt{\frac{s}{2}}\tan^{-1}\sqrt{\frac{2}{s}}\right)$$

which finally yields for the two functions in question

$$\langle{h,\hat F}\rangle=\frac{\tan^{-1}\sqrt{\frac{2}{s}}}{\sqrt{\frac{s}{2}}\left(1+\sqrt{\frac{s}{2}}\tan^{-1}\sqrt{\frac{2}{s}}\right)}$$

$$\hat F(i,s)=\frac{(i+1)^2}{2+s(i+1)^2}+\frac{1}{s(i+1)^2+2}\frac{\tan^{-1}\sqrt{\frac{2}{s}}}{\sqrt{\frac{s}{2}}\left(1+\sqrt{\frac{s}{2}}\tan^{-1}\sqrt{\frac{2}{s}}\right)}$$

Inverting these LT's seems to be at least upon a first glance, nontrivial, however a numerical inversion is always possible.