# Boundary condition for a bulk-surface and bulk-bulk diffusion reaction system

Consider this simple example below and the corresponding geometries. I simplified these equations from the real system.

Geometry 1

The first geometry is a sphere. Inside this sphere a species $$b(t,x,y,z)$$ diffuses and converts on the surface of sphere to another species $$s(t,x,y,z)$$ which in turn diffuses only on the surface and can convert back to $$b(t,x,y,z)$$. The equations are, here $$\Delta$$ is Laplacian operator:

$$\frac{\partial s(t,x,y,z)}{\partial t} = k_1\cdot s(t,x,y,z)\cdot b(t,x,y,z)- k_2\cdot s(t,x,y,z) + \Delta s(t,x,y,z) \tag{1}$$

$$\frac{\partial b(t,x,y,z)}{\partial t} = \Delta b(t,x,y,z)\tag{2}$$

Boundary condition:

$$\left.\frac{\partial d(t,x,y,z)}{\partial r}\right\rvert_{r = R} = -k_1\cdot s(t,x,y,z)\cdot b(t,x,y,z) + k_2\cdot s(t,x) \tag{3}$$

Question 1:

Why isn't the term on the right hand side of the boundary condition, equation 3, added to equation 2? I understand that to solve this system of PDEs we need a boundary condition, but I do not get why we do not add the growth term for $$s(t,x,y,z)$$ which is $$k_1\cdot s(t,x,y,z)\cdot b(t,x,y,z)$$ and the decay term for $$s(t,x,y,z)$$ which is $$- k_2\cdot s(t,x,y,z)$$ to equation 2. After all if the system was only ODE I would have written equation 1 and 2 as:

$$\frac{\partial s(t)}{\partial t} = k_1\cdot s(t)\cdot b(t)- k_2\cdot s(t) \tag{4}$$

$$\frac{\partial b(t)}{\partial t} = -k_1\cdot s(t)\cdot b(t)+ k_2\cdot s(t)\tag{5}$$

So why not do the same thing for a PDE?

Question 2

If the geometry instead of a bulk and surface is made of two bulk with a boundary between them, how should I write down the equation for the system and the boundary condition? In this case $$b(t,x,y,z)$$ and $$s(t,x,y,z)$$ diffuse in their own bulk geometry and convert to each other at the boundary between the two bulks. Could I write down

$$\frac{\partial s(t,x,y,z)}{\partial t} = k_1\cdot s(t,x,y,z)\cdot b(t,x,y,z)- k_2\cdot s(t,x,y,z) + \Delta s(t,x,y,z) \tag{6}$$

$$\frac{\partial b(t,x,y,z)}{\partial t} = -k_1\cdot s(t,x,y,z)\cdot b(t,x,y,z)+ k_2\cdot s(t,x,y,z) +\Delta b(t,x,y,z)\tag{7}$$

with boundary condition:

$$\left.\frac{\partial d(t,x,y,z)}{\partial r}\right\rvert_{r = R} = -k_1\cdot s(t,x,y,z)\cdot b(t,x,y,z) + k_2\cdot s(t,x) \tag{8}$$

$$\left.\frac{\partial s(t,x,y,z)}{\partial r}\right\rvert_{r = R} = k_1\cdot s(t,x,y,z)\cdot b(t,x,y,z) - k_2\cdot s(t,x) \tag{9}$$

If yes, why then?

I realized that if we add equation 3 to equation 2 then the time derivative of species $$b(t,x,y,z)$$ at time t and position (x,y,z) will depend on its diffusion at position (x,y,z) at time t and the flux at the boundary at time t which is physically incorrect because then the physical process is non-local that is a change at far away has immediate effect at another place. Yet, I do not know the answer to question 2. I think in this second geometry the two species $$s(t,x,y,z)$$ and $$b(t,x,y,z)$$ in equation 6 and 7 should only involve the diffusion operator (Laplacian) and their conversion at the boundary should reflect only in the boundary condition.