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I want to solve the three-dimensional laplacian $$\nabla^{2} T = 0$$ where $\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$ defined on $x\in[0,L],y\in[0,l],z\in[-w,0]$ with the following boundary conditions

$$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 \rightarrow Neumann$$

$$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0\rightarrow Neumann$$

$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) - \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg) \rightarrow Convection$$

$$\frac{\partial T(x,y,0)}{\partial z}=p_c\bigg(\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y -T(x,y,0) \bigg)\rightarrow Convection$$ $b_h,b_c,p_h,p_c,L,l$ are constants $>0$

Addendum The physical situation that this problem describes has two additional pieces of information $\rightarrow$ At $(0,y,-w)$ $\rightarrow$ $\frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x = T_{h,i}$ a constant

and

At $(x,0,0)$ $\rightarrow$ $\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y = T_{c,i}$ a constant

Attempt

I know that for the 3D Laplacian,, when separation of variables is applied the following holds

$k_x^{2} + k_y^{2} = k_z^{2}$

So if $X(x)Y(y)Z(z)$ is substituted as $T$ we can say $X \sim \cos(n\pi x)$, $Y \sim \cos(m \pi y)$. I know that Z can be some hyperbolic function like $[A_{m,n}\cosh(\pi \sqrt{n^2 + m^2}) + B_{m,n}\cosh(\pi \sqrt{n^2 + m^2})]$

But when the solution form is substituted onto the $z$ bc, what i get is just two homogeneous linear equations in $A_{m,n}$ and $B_{m,n}$. From this point, I cannot proceed.

Any help or guidance on how this problem can be tackled will be appreciated.

Attempt 2

As suggested by @DisintegratingByParts I need to divide the problem into two parts.

Let us take Problem 1.

As already mentioned,I have the additional info that:

At $(0,y,-w)$ $\rightarrow$ $\frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x = T_{h,i}$.

So we can write

$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) - \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg) $$

as

$$\frac{\partial T(0,y,-w)}{\partial z}=p_h\bigg( T(0,y,-w) - T_{h,i} \bigg)$$

So now i have the boundary condition (more like an edge condition in 3D) without the integral on the RHS. Can this BC be used along with the others to form a solution of Problem 1 or am i wrong in some way ?


On applying the third BC from Problem 1 but at $(x,y,-w)$ I arrive at

$$ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\sinh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) = p_h\bigg(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{m\pi y}{l})\cos(\frac{n\pi x}{L})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) - \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{A_{n,m}b_h}{b_h^2 + n^2 \pi^2 }\cos(\frac{m\pi y}{l})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg)\bigg[b_h\cos(\frac{n\pi x}{L}) + n\pi\sin(\frac{n\pi x}{L}) \bigg]\bigg) $$

This needs to be now solved for Fourier coefficients $A_{n,m}$. I know that we have to somehow use the property of orthogonality of the Eigen functions, but the problem pertains from the fact that $A_{n,m}$ exist on both sides of the equation.

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  • $\begingroup$ What do you mean by $\nabla^3$? $\endgroup$ – Vasily Mitch Feb 12 at 14:53
  • $\begingroup$ $\Delta$ or $\nabla^2$ stands for the Laplacian in any dimension. en.wikipedia.org/wiki/Laplace_operator . $\endgroup$ – DisintegratingByParts Feb 12 at 18:40
  • $\begingroup$ You have asked essentially the same question three times now, in this post, here and here. It is ridiculous to keep posting the same question but adding new context or details to the problem. $\endgroup$ – Mattos Feb 13 at 0:35
  • $\begingroup$ @Mattos I can understand that. The first time I asked it , I included an attempt after recommendation from comments and the question itself became too long, so I reposted a new question with more bqckground information. I am sorry if i have broken the community rules by doing this. $\endgroup$ – Indrasis Mitra Feb 13 at 0:55
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    $\begingroup$ @Mattos Could you provide some guiding points on the problem here ? $\endgroup$ – Indrasis Mitra Feb 14 at 4:21
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You can solve two problems and sum the two solutions:

Problem 1
$$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 \rightarrow Neumann$$

$$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0\rightarrow Neumann$$

$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) - \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg) \rightarrow Convection$$

$$ \frac{\partial T(x,y,0)}{\partial z} = 0. $$

Problem 2
$$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 \rightarrow Neumann$$

$$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0\rightarrow Neumann$$

$$ \frac{\partial T(x,y,-w)}{\partial z} = 0. $$

$$\frac{\partial T(x,y,0)}{\partial z}=p_c\bigg(\frac{e^\frac{-b_c y}{l}b_c}{l}\int e^\frac{b_c y}{l}T\mathrm{d}y -T(x,y,0) \bigg)\rightarrow Convection$$

The Problem 1 Form of solution is $$ T(x,y,z)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\cosh\left(\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}z\right). $$ The constants $A_{n,m}$ are determined as Fourier coefficients through the third condition at $z=-w$.

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  • $\begingroup$ Thanks. Now i get that since its a linear operator , the total solution can be a superposition. I have added an Attempt2 to my original problem which makes the BC free from the integral. If possible have a look. I have some reservations about it. $\endgroup$ – Indrasis Mitra Feb 15 at 5:23
  • $\begingroup$ @IndrasisMitra : I added more to help you put the final pieces into place. $\endgroup$ – DisintegratingByParts Feb 15 at 6:44
  • $\begingroup$ This was great. I will try this now. Actually, my dilemma was whether I should use the modified condition at $x=0,y,z=-w$ (which forms an edge of the whole 3D domain, not a face)or the condition at $x,y,z=-w$ (which forms a face). Is there some restrictions like the bc(s) for a 3D domain should be only planes ? $\endgroup$ – Indrasis Mitra Feb 15 at 8:12
  • $\begingroup$ @IndrasisMitra : If you're going to use separation of variables to solve your problem, then the boundaries must be surfaces that are defined by setting one of the variables to a constant. In Cartesian coordinates, such surfaces are planes. If my answer is sufficient, please press the check mark next to it to accept the answer. You can also +1 it if you are so inclined. $\endgroup$ – DisintegratingByParts Feb 15 at 15:10
  • $\begingroup$ +1'd it already. Also, I applied the b.c. at $(x,y,-w)$ viz. a surface (have added it to the end of the question). The difficulty i now face is that the constant $A_{n,m}$ appear on both sides of the resulting equation. Any hint you can provide to find these coeff.(s) ? You have been a great help anyway $\endgroup$ – Indrasis Mitra Feb 16 at 5:43

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