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I want to solve a PDE coupled with an ODE:

$$ \nabla^2 T(x,y)=0 $$

defined over a rectangular region $x\in[0,L], y\in[0,l]$ subjected to:

$$\frac{\partial T(0,y)}{\partial x}=\frac{\partial T(L,y)}{\partial x}=0 \tag 1$$

$$\frac{\partial T(x,0)}{\partial y}=\gamma \tag 2$$

and on the top face $(y=l)$:

$$\frac{\partial T(x,l)}{\partial y}=\beta \Bigg[T(x,l)-t\Bigg] \tag 3$$

$t$ is not a constant and the equation governing it is:

$$\frac{\partial t}{\partial x}+\alpha(t-T(x,l))=0 \tag 4$$

where it is known that $t(x=0)=0$. $\gamma, \beta$ are constants.

Attempt Considering the b.c. (1), a solution of the following form can be assumed:

$$T(x,y)=\sum_{n=0}^{\infty}(c_3 e^{ay} + c_4 e^{-ay})\cos\bigg(\frac{n\pi x}{L}\bigg)$$

where $a=\frac{n\pi}{L}$ is the separation constant. Then we have two non-homogeneous b.c.(s) i.e. (2) and (3).

So probably these two conditions can be used one-by-one to form two simultaneous linear equations (after applying orthogonality) with two unknowns $c_3$ and $c_4$.

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    $\begingroup$ You can solve (4) with the initial condition as $t(x,l) = \alpha e^{-\alpha x} \int \limits_0^x e^{\alpha \xi} T(\xi,l) \, \mathrm{d}\xi$, $x \in [0,L]$, and eliminate $t$ in (3). The PDE should be solved by separation of variables. $\endgroup$
    – Christoph
    Aug 25, 2020 at 9:38
  • $\begingroup$ @Christoph Thanks! I actually tried this trnasformation but I am finding it difficult to deal with the integral conditions in the b.c. $$\frac{\partial T(x,l)}{\partial y}=\beta(T(x,l)-\alpha e^{-\alpha x}\int_0^x e^{\alpha \zeta} T(\zeta,l)\mathrm{d}\zeta)$$ Can you elaborate a bit more ? $\endgroup$
    – Avrana
    Aug 25, 2020 at 10:54
  • $\begingroup$ @Christoph I have included an attempt to my original problem. $\endgroup$
    – Avrana
    Aug 25, 2020 at 11:38
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    $\begingroup$ Agreed, the hard part is to construct a function $g(x,y)$ which satisfies the boundary conditions, so that the difference $T-g$ satisfies a (non-homogeneous) Poisson equation with homogeneous boundary conditions. $\endgroup$
    – Christoph
    Aug 26, 2020 at 3:50
  • $\begingroup$ @Christoph I did try an alternative with a separation of variables in the form $T(x,y)=X(x)Y(y)+Xp(x)+Yp(y)$. After using bc(s) (1) and (2) I divided $T(x,y)$ into $T_0 (x,y)$ and $T_n (x,y)$ and then applied bc(3) (with the integral form that you suggested) to each part. Unfortunately, $Tn(x,y)$ came out as zero. $\endgroup$
    – Avrana
    Aug 26, 2020 at 5:01

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