I have a question about a variational problem
I want to solve a Calculus of Variations Problem of the following form:
$\min_{a,b} \int_0^\infty \int_{-\infty}^\infty a(x,y)dxdy$
Subject to
$a(x,y)=D(x,y, b(x,y))+\beta \int_{-\infty}^\infty a(x', b(x,y)) f(x'|x) dx'$ for all $x \in \mathbb{R},y \in \mathbb{R}_+$
Let me explain the notation: $a$ and $b$ are the functions on variables $x$ and $y$ that I want to find, $D$ is a functional that depends on $x,y$ and function $b$, and $f(x'|x)$ is a conditional and known probability density function.
I want to solve this problem, get the Euler Lagrande equations but havent been able to do it. Let me show you what I got:
Small variation on $a(x,y)$ as $a(x,y)+\epsilon \eta(x,y)$ for any function $\eta(x,y)$. I define the lagrangian as a function of $\epsilon$:
$J^a(\epsilon)=\int_0^\infty \int_{-\infty}^\infty (a(x,y)+\epsilon \eta(x,y))+\lambda(x,y)\Big[(a(x,y)+\epsilon \eta(x,y))-D(x,y, b(x,y))-\beta \int_{-\infty}^\infty [a(x',b(x,y))+\epsilon \eta(x',b(x,y))] f(x'|x) dx'\Big]dxdy$
Now, I take the derivative with respect to $\epsilon$
$\int_0^\infty \int_{-\infty}^\infty \big[\eta(x,y)+\lambda(x,y)\eta(x,y)-\beta \lambda(x,y)\int_{-\infty}^\infty\eta(x', b(x,y))f(x'|x)dx'\big]dxdy=0$
So here sadly I cant factorize $\eta(x,y)$ inside the integral, cause I also have $\eta(x',b(x,y))$ in the inner integral. Im a bit stucked here to get the first order condition.
Now, Ive done the same for $b$, I consider a variation $b(x,y)+\epsilon \eta(x,y)$ for any function $\eta(x,y)$.
$J^b(\epsilon)=\int_0^\infty \int_{-\infty}^\infty a(x,y)+\lambda(x,y) \Big[a(x,y)-D(x,y,b(x,y)+\epsilon \eta(x,y))-\beta \int_{-\infty}^\infty a(x,b(x,y)+\epsilon \eta(x,y))f(x'|x)dx' \Big]dxdy$
Now taking the derivative with respect to $\epsilon$ and evaluate the lagrangian when $\epsilon=0$ yields me
$\frac{d J}{d \epsilon} |_{\epsilon=0}=\int_0^\infty \int_{-\infty}^\infty \lambda(x,y)[-\frac{\partial D(x,y,b(x,y))}{\partial b} \eta(x,y)-\beta \int_{-\infty}^\infty \frac{\partial a(x,b(x,y))}{\partial b} \eta(x,y) f(x'|x)dx']dxdy=0$
In this case I can have an Euler Lagrange Equation
$\int_0^\infty \int_{-\infty}^\infty \eta(x,y) \Big[ \lambda(x,y) [-\frac{\partial D(x,y,b(x,y))}{\partial b} -\beta \int_{-\infty}^\infty \frac{\partial a(x,b(x,y))}{\partial b} f(x'|x)dx'\Big] dxdy=0$
Since this is for every $\eta(x,y)$
$\lambda(x,y) [-\frac{\partial D(x,y,b(x,y))}{\partial b} -\beta \int_{-\infty}^\infty \frac{\partial a(x,b(x,y))}{\partial b} f(x'|x)dx']=0$ I get the following differential equation.
So, to conclude, I might be doing something wrong in the first part. Any help is appreciated
Thank you