1
$\begingroup$

I am not specifically asking for a solution, but any reference on any method i could read about would be a big help. This I clarify as I am aware of the fact that MathOverflow only deals with research level questions. A starting point is all I aim to get from posting this question here

I have the following 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} The boundary conditions are :

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

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

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

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

$\beta_h,\beta_c,\lambda_h,\lambda_c,V$ are all constants $>0$

Attempt

The third equation can be written as $$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} = \frac{\partial \theta_h}{\partial x} + V\frac{\partial \theta_c}{\partial y} $$

Then, from the first two equations the following can be written :-

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

This resulting equation looks a lot like Laplace equation with integral source terms (if the LHS is taken into canonical form).

What kind of procedure should I take up to solve this problem? Is there any standard problem type that corresponds to this situation.

Attempt taken forward

It can be seen that the resulting equation can be variable separated if we consider the following ansatz:

$\theta_w(x,y) = e^{-\beta_h x} f(x) e^{-\beta_c y} g(y)$

with $F(x) := \int f(x) \, \mathrm{d}x$ and $G(y) := \int g(y) \, \mathrm{d}y$

to get:

\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}

for some separation constant $\mu$

$\endgroup$
1
$\begingroup$

More a long comment than an answer, but perhaps it could be useful to you: one reference I know that deals extensively with non-local equation (read "integrodifferential" and "with integral type boundary conditions") is the monograph of Chia-Ven Pao [1]. For example, elliptic integrodifferential equations are analyzed in chapter 3, §3.7, while boundary conditions of integral type (though for parabolic equations, and of Volterra type) are analyzed in chapter 4, §4.3 (this may be relevant for your later question here). The whole monograph is however devoted to non-classical boundary value problems so, even if it don't seem to suit your precise needs, I advice you to have a look at it. Note that

  • Pao develops the theory in Ḧolder spaces, i.e. $C^{k+\alpha}\equiv C^{(k,\alpha)}$, $k\in\Bbb N$ and $\alpha\in (0,1)$ and uses $\min$-$\max$-solutions constructed by iteration (convergence of the processes is proved by using maximum principles for the problem at hand): this implies that no complicated functional analysis is used.
  • Though been mathematically sound and rigorous, the book deals extensively with the analysis of models derived from biology, engineering and physics, which seems the stuff you (and perhaps the author of this answer...) are interested to.

[1] Chia-Ven Pao (1992), Nonlinear Parabolic and Elliptic Equations, Plenum Press, xv+777, MR1212084 Zbl 0777.35001.

$\endgroup$
  • 1
    $\begingroup$ Thanks. I would now try to get a hand on this text. Will revert back, with what I find. This is much appreciated. $\endgroup$ – Indrasis Mitra Mar 10 at 12:59
  • $\begingroup$ For the sake of completion, I have updated my question here with further modifications of the BC that I carried out, since I posted it. $\endgroup$ – Indrasis Mitra Mar 11 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.