$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} =\beta_h e^{-\beta_h x} \int e^{\beta_h x} \theta_w(x,y) \, \mathrm{d}x + \beta_c e^{-\beta_c y} \int e^{\beta_c y} \theta_w(x,y) \, \mathrm{d}y$$

on a domain where $x,y$ vary between $0$ to $1$. The bc(s) are $$\frac{\partial \theta_w(0,y)}{\partial x}=\frac{\partial \theta_w(1,y)}{\partial x}=0 $$

$$\frac{\partial \theta_w(x,0)}{\partial y}=\frac{\partial \theta_w(x,1)}{\partial y}=0 $$

The two terms on the RHS come from solving the following two $$ \frac{\partial \theta_h}{\partial x} + \beta_h (\theta_h - \theta_w) = 0, \frac{\partial \theta_c}{\partial y} + \beta_c (\theta_c - \theta_w) = 0$$

which have bc(s) as $\theta_h(0,y)=1$ and $\theta_c(x,0)=0$. Then subsequently $\theta_h$ and $\theta_c$ were eliminated from the following $$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} = \frac{\partial \theta_h}{\partial x} + V\frac{\partial \theta_c}{\partial y} $$

to reach the top equation

So to start i had three coupled PDEs

\begin{eqnarray} \frac{\partial \theta_h}{\partial x} + \beta_h (\theta_h - \theta_w) &=& 0,\\ \frac{\partial \theta_c}{\partial y} + \beta_c (\theta_c - \theta_w) &=& 0,\\ \lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} - \frac{\partial \theta_h}{\partial x} - V\frac{\partial \theta_c}{\partial y} &=& 0 \end{eqnarray} and then i worked onto the top equation

Any advice on how to move forward with this problem ?

  • $\begingroup$ The integrals should not be definite here. This was done the very first time you asked this problem. $\endgroup$ – Dylan Feb 11 at 5:05
  • $\begingroup$ @Dylan This was completely my mistake and incomplete understanding due to which this happened. You are exactly right, thy should not be definite. It would be really helpful if you could suggest some way forward either on this formulation or the 3D formulation. I know that the ansatz $e^{-\beta_h x}Xe^{-\beta_c y}Y$ can make the top equation variable separated but the resulting eigen value problems are third order which can not be solved analytically. $\endgroup$ – Indrasis Mitra Feb 11 at 5:55

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