I asked the three-dimensional version of this problem here which lead to a trivial solution. I have now tried it in 2-D

$$\frac{\partial \theta_h}{\partial x} + b_h (\theta_h - \theta_w) = 0, \tag 1\\\\$$ $$\frac{\partial \theta_c}{\partial y} + b_c (\theta_c - \theta_w) = 0,\tag 2\\\\$$ $$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} + b_h (\theta_h - \theta_w) + V b_c (\theta_c - \theta_w) = 0 \tag 3$$

It is known that $\theta_h(0,y)=\theta_{hi}$ and $\theta_c(x,0)=\theta_{ci}$. We know that $b_h,b_c,V,\theta_{hi}, \theta_{ci}$ are constants greater than 0. Using this information and some manipulation on $(1),(2)$ we can write $(3)$ as follows:

$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} +( -b_h - V b_c )\theta_w +b_h^2 e^{-b_h x} \int e^{b_h x} \theta_w(x,y) \mathrm{d}x + Vb_c^2 e^{-b_c y}\int e^{b_c y} \theta_w(x,y)\mathrm{d}y = 0$$

which can be manipulated to:

$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} +( -b_h - V b_c )\theta_w + b_he^{-b_h x}\bigg(\theta_{hi}+b_h\int_0^x e^{b_h s}\theta_w (s,y)\mathrm{d}s\bigg) + Vb_ce^{-b_c y}\bigg(\theta_{ci}+b_c\int_0^y e^{b_c s}\theta_w (x,s)\mathrm{d}s\bigg)=0 \tag 4$$

$(4)$ is subjected to the following boundary conditions:

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

In mathematica code $(4)$ and the boundary conditions are:

eq = λh D[θw[x, y], {x, 2}] + λc V D[θw[x, y], {y, 2}] + (-bh - V bc) θw[x, y] + bh E^(-bh x) (θhi + bh Integrate[E^(bh s) θw[s, y], {s, 0, x} ]) + V bc E^(-bc y) (θci + bc Integrate[E^(bc s) θw[x, s], {s, 0, y} ]) == 0

bcx = {D[θw[x, y], x] == 0 /. x -> 0, D[θw[x, y], x] == 0 /. x -> 1}

bcy = {D[θw[x, y], y] == 0 /. y -> 0, D[θw[x, y], y] == 0 /. y -> 1}

Can eq. $4$ subjected to the given boundary conditions, be solved using the method of finite Fourier or any other integral transform ?

Some parameter values are:

bh=0.433;bc=0.433;λh = 2.33 10^-6; λc = 2.33 10^-6; V = 1;θhi = 1; θci = 0;

NOTE: In case the above b.c. causes problems, a solution with $\frac{\partial \theta_w(1,y)}{\partial x}=\frac{\partial \theta_w(x,1)}{\partial y}=0, \theta_w(x,0)=\theta_{hi}, \theta_w(0,y)=\theta_{ci}$ is also acceptable.

NOTE 2: I have observed that $(4)$ can be variable separated if the ansatz $\theta_w=e^{-b_h x}f(x)e^{-b_c y}g(y)$ is used. But this path lead me to nowhere. On applying SOV to the eqn. above 4 one gets: \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 $\mu \in \mathbb{R}$ where $F(x) := \int f(x) \, \mathrm{d}x$ and $G(y) := \int g(y) \, \mathrm{d}y$ with bc transformed as with boundary conditions as:

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

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

The non-homogeneous condition in $F$ is: $\beta_h e^{-\beta_c y}G'(y)F(0)=1$ . I could not proceed anymore.

  • $\begingroup$ Still, finite Fourier transform won't help. Terms like $\frac{\partial \theta_h}{\partial x}$ is on the way. Please try transforming the system and observing the output, and think about why the resulting system is still hard to solve. $\endgroup$
    – xzczd
    Jul 27, 2020 at 5:53
  • $\begingroup$ The equation 4 I wrote has no $\theta_h$ or other derivatives except $\theta_w$. The problem you answered before was similar. For start I iust wanted to solve for the theta w term using 4.Nevertheless, Thanks for the attention. $\endgroup$
    – Avrana
    Jul 27, 2020 at 6:05
  • 1
    $\begingroup$ In short, eliminating $θ_h$ in this way won't help, because the underlying difficulty caused by $\frac{∂θ_h}{∂x}$ is still there, your previous problem can be solved with finite Fourier transform, because it has integral in only $x$ direction. Notice I've only transformed in $y$ direction there, finite Fourier transform just can't be used in $x$ direction. $\endgroup$
    – xzczd
    Jul 27, 2020 at 6:13


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