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} - V\frac{\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

$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} - (\beta_h+V\beta_c)\theta_w+(\beta_h\theta_h+V\beta_c\theta_c) = 0$$

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 ?


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.

  • $\begingroup$ But you still have the unknown functions $\theta_h$ and $\theta_c$ in your PDE after this substitution?! $\endgroup$ – Christoph Jan 8 at 16:38
  • $\begingroup$ Also, there should be a factor $V$ in front of $\frac{\partial \theta_c}{\partial y}$. $\endgroup$ – Christoph Jan 8 at 16:43
  • $\begingroup$ @Christoph Made the edit, you were right about the $V$. Can't $(\beta_h\theta_h+V\beta_c\theta_c)$ be together considered a variable say $k$ which would be a function of $x$ and $y$ . And a second order linear PDE requires the coefficients to be the functions of the independent variables $x$ and $y$. $\endgroup$ – Indrasis Mitra Jan 8 at 16:48
  • $\begingroup$ @Christoph. Any suggestion on how tot tackle the problem if what i was thinking is wrong $\endgroup$ – Indrasis Mitra Jan 8 at 16:50
  • $\begingroup$ It is correct to have 5 boundary conditions. As the scale of $F$ and $G$ is free for a homogeneous equation, fixing the scales adds two additional equations, making 7 conditions for a state of 6 function values and derivatives plus one parameter $\mu$. // In the second approach either one of the cited BC is redundant or there is some other error, as 8 equations for 7 variables is usually not solvable. $\endgroup$ – LutzL Jan 24 at 12:36

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 $f \equiv F'$, $g \equiv G'$.

  • $\begingroup$ Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $\theta_w$ in the PDE of the 2nd step i am encountering a term $\int F(x) \mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ? $\endgroup$ – Indrasis Mitra Jan 9 at 4:38
  • $\begingroup$ To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $\int F(x) \mathrm{d}x$ ? $\endgroup$ – Indrasis Mitra Jan 9 at 4:45
  • $\begingroup$ Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear. $\endgroup$ – Christoph Jan 9 at 5:05
  • $\begingroup$ This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $\theta_w$. And then $\theta_h$ and $\theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $\mu$. Will that come out from the boundary conditions on the two ODEs? $\endgroup$ – Indrasis Mitra Jan 9 at 12:33
  • $\begingroup$ Yes, that's correct. From the boundary conditions on $\theta_h, \theta_c, \theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $\mu_n$, $n \in \mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $\theta_w(x,y) = \sum_n c_n e^{-\beta_h x} F_n(x) e^{-\beta_c y} G_n(y)$, with some coefficients $c_n \in \mathbb{R}$. $\endgroup$ – Christoph Jan 9 at 13:50

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.