I'm reading a paper where the following parabolic PDE is considered:

$u_t(x,t)=u_{xx}(x,t)+b(x)u_x(x,t)+\lambda(x)u(x,t)$, with boundary conditions $u_x(0,t)=qu(0,t) \text{ and } u(1,t)=\int_0^1 k(\xi)u(\xi,t)d\xi$ and initial condition $u(x,0)=u_0(x)$.

Here $q \in \mathbb{R}$, $b$ and $\lambda$ are $C^1$ functions and $k$ is infinitely differentiable.

The authors claim that any function $u$ that satisfies the integral equation $u=Fu$, is a solution to the PDE. Here

$Fu=\int_0^1 G(x,\xi,t,\tau)u_0(\xi)d\xi-\int_0^t\int_0^1 G_\xi(x,\xi,t,\tau)b(\xi)u(\xi,\tau)d\xi d\tau \\ + \int_0^t\int_0^1 G(x,\xi,t,\tau)\left[\lambda(\xi)-b_\xi(\xi)\right]u(\xi,\tau)d\xi d\tau \\ +\int_0^t G(x,1,t,\tau)(1+b(1))\int_0^1 k(\xi)u(\xi,t)d\xi d\tau-\int_0^t G(x,0,t,\tau)(b(0)+q)u(0,\tau)d\tau,$

where $G(x,\xi,t,\tau)=2\sum_{n=1}^\infty \cos(\lambda_n x)\cos(\lambda_n \xi) e^{-\lambda_n^2(t-\tau)}$ is the Green's function of the heat operator $\mathcal{L}u=u_t-u_{xx}$ with boundary conditions $u_x(0,t)=u(1,t)=0$ and $\lambda_n=(2n-1)\frac{\pi}{2}$.

I'm really having a hard time understanding that a function $u$ that satisfies $u=Fu$ is a solution to the PDE. For example, I can't even see how such a $u$ even satisfies the boundary conditions.

Edit: Additionally, the fourth integral contains $G(x,1,t,\tau)$ which is clearly zero. This adds more to my confusion.


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