I have the 3D Laplace equation:

$$\nabla^{3} T_w = 0$$

where $\nabla^{(3)}=(\frac{\partial^{2}}{\partial x^2}+\frac{\partial^{2}}{\partial y^2}+\frac{\partial^{2}}{\partial z^2})$ defined on $x \in [0,L]$, $y \in [0,l]$ and $z \in [-w,0]$.

with the following boundary conditions

$$\frac{\partial T_w(0,y,z)}{\partial x}=\frac{\partial T_w(L,y,z)}{\partial x}=0 \rightarrow Neumann$$

$$\frac{\partial T_w(x,0,z)}{\partial y}=\frac{\partial T_w(x,l,z)}{\partial y}=0 \rightarrow Neumann$$

$$\frac{\partial T_w(x,y,-w)}{\partial z}=p_h\bigg( T_w(x,y,-w) - \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T_w\mathrm{d}x \bigg) \rightarrow Convection$$

$$\frac{\partial T_w(x,y,0)}{\partial z}=p_c\bigg(\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T_w\mathrm{d}y -T_w(x,y,0) \bigg)\rightarrow Convection$$

**Contextual information**

*How the third and fourth boundary conditions were arrived at*

There are two additional equations

$$ \frac{\partial T_h}{\partial x} + \frac{b_h T_h}{L} = \frac{b_h T_w}{L} $$

gives $T_h = \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T_w\mathrm{d}x$

and

$$ \frac{\partial T_c}{\partial y} + \frac{b_c T_c}{l} = \frac{b_c T_w}{l} $$

which gives $T_c = \frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T_w\mathrm{d}y$

Now to get the boundary conditions, the following relation were used viz. $T_h$ and $T_c$ were substituted:

$$\frac{\partial T_w(x,y,-w)}{\partial z} = p_h\bigg(T_w(x,y,-w)-T_h \bigg)\rightarrow Convection$$

and

$$\frac{\partial T_w(x,y,0)}{\partial z} = p_c\bigg(T_c - T_w(x,y,0) \bigg)\rightarrow Convection$$

An additional piece of information that is available is

$T_h(0,y,-w) = T_{h,i}$ (a constant)

$T_c(x,0,0) = T_{c,i}$ (a constant)

$l,L,b_h,b_c,p_h,p_c$ are all constants $>0$

Contextual Informationto my original question to explain your query. $\endgroup$ – Indrasis Mitra Feb 10 at 16:26