Dtyjlui

I have the following three 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} - Vfrac{partial theta_c}{partial y} &=& 0
end{eqnarray}

From the first and second equation i expressed $frac{partial theta_h}{partial x}$ and $frac{partial theta_c}{partial y}$ in terms of $theta_w$. Then i substituted these in the third equation to yield

$$lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - (beta_h+Vbeta_c)theta_w+(beta_htheta_h+Vbeta_ctheta_c) = 0$$

The third PDE turns out to be a second order linear Elliptic PDE as $lambda_h$,$lambda_c$ and $V$ are all positive constants. I have reached a canonical form for this second order PDE. This PDE is defined on a rectangle with Neumann conditions. I plan to do the following next;

1. Calculate $theta_w(x,y)$ from the second order PDE.


2. Plug them in
   the first two to obtain $theta_h$ and $theta_c$
   

Am i following a correct approach or is there any subtlety i am over-looking ?

Attempt

The boundary conditions for the problem are as follows:

The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$.

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

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

$$theta_h(0,y)=1 $$$$theta_c(x,0)=0$$

After the suggestions from @Christoph here i have the following two linear third order differential equations:

begin{eqnarray}
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}

Both these ODEs now need to be converted to individual Boundary value problems using the BC(s). On substituting

$$theta_w(x,y) = e^{-beta_h x} F(x) e^{-beta_c y} G(y)$$ into the give BC(s), i arrive at the following

$$e^{-beta_cy}F(0)G(y)=1$$
$$e^{-beta_hx}F(x)G(0)=0$$
$$e^{-beta_cy}G(y)[F'(0)-beta_hF(0)]=0$$
$$e^{-beta_cy}e^{-beta_h}G(y)[F'(1)-beta_hF(1)]=0$$
$$e^{-beta_hx}F(x)[G'(0)-beta_cG(0)]=0$$
$$e^{-beta_hx}e^{-beta_c}F(x)[G'(1)-beta_cG(1)]=0$$

Following this (keeping in mind that exponential cannot attain a 0 value), i arrive at the following simplifications:

$$G(0)=0$$
$$G'(0)=0$$
$$frac{G'(1)}{G(1)}=beta_c$$
$$frac{F'(0)}{F(0)}=beta_h$$
$$frac{F'(1)}{F(1)}=beta_h$$

The ODEs are of third order and although i have six BC(s), on decoupling the BC(s) i get just 5. Am i misunderstanding something or is there some other way ?

Attempt 2

As @Christoph advised I made the following changes:
$$bar{{theta_h}}(x,y):=theta_h(x,y)-1$$
and the ansatz
$$theta_w(x,y)=e^{-beta_hx}f(x)e^{-beta_cy}g(y)$$ such that $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$

The third order linear DEs we arrive at still remain the same.
For figuring out the b.c.(s), the ansatz became:
$$theta_w(x,y)=e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$

But the boundary conditions now take the following form

For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$

For $G$:
$$G(0)=0$$
$$frac{G''(0)}{G'(0)}=beta_c$$
$$frac{G''(1)}{G'(1)}=beta_c$$

Now i have three b.c. (s) for each boundary value problem viz. $F$ and $G$.
Each BVP (one each of $F$ and $G$) now involve one Dirichlet and two Robin type b.c.

Here are a few hints:

1. Solve the two first-order PDEs for $theta_h, theta_c$ as functions of $theta_w$:
   begin{eqnarray}
   theta_h(x,y) &=& beta_h e^{-beta_h x} int e^{beta_h x} theta_w(x,y) , mathrm{d}x,\
   theta_c(x,y) &=& beta_c e^{-beta_c y} int e^{beta_c y} theta_w(x,y) , mathrm{d}y.
   end{eqnarray}




2. Eliminate $theta_h, theta_c$ in the second-order PDE to obtain the following equation for $theta_w$:
   begin{eqnarray}
   0 &=& e^{-beta_h x} left( lambda_h e^{beta_h x} frac{partial^2 theta_w}{partial x^2} - beta_h e^{beta_h x} theta_w + beta_h^2 int e^{beta_h x} theta_w , mathrm{d}x right) +\
   && + V e^{-beta_c y} left( lambda_c e^{beta_c y} frac{partial^2 theta_w}{partial y^2} - beta_c e^{beta_c y} theta_w + beta_c^2 int e^{beta_c y} theta_w , mathrm{d}y right).
   end{eqnarray}




3. Use separation of variables with the ansatz $theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$. You should obtain two linear third-order ODEs with constant coefficients for $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$:
   begin{eqnarray}
   lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
   V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
   end{eqnarray}
   with some separation constant $mu in mathbb{R}$.




4. From the boundary conditions on $theta_h, theta_c, theta_w$ you should obtain conditions on $F$, $G$. Once the BVPs with the third-order ODEs are solved, compute $f equiv F'$, $g equiv G'$.