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I need to solve the three-dimensional Laplacian $$\nabla^2T(x,y,z)=0 \tag {A}$$

on a cuboidal domain $x\in[0,a], y\in [0,b], z\in [0,c]$.

(A) is subjected to the following boundary conditions

$$T(0,y,z)=T_{w} \tag{1}$$ $$\frac{\partial T}{\partial x} \bigg \vert_{a,y,z}=0 \tag {2}$$ $$T(x,0,z)=T(x,b,z)=T(x,y,0)=T(x,y,c)=0 \tag{3}$$

I need some guidance in attempting this problem. I know that all the homogeneous Dirichlet conditions described by $(3)$ enables us to write a preliminary solution of the form: $T(x,y,z)=T_{nm}(x)\sin\bigg(\frac{n\pi y}{b}\bigg)\sin\bigg(\frac{m\pi z}{c}\bigg)$. Some help in determining the function $T_{nm}(x)$ is needed.


Attempt After applying Mattos's suggestion, I can write the following

$$T(x,y,z)=\bigg[A_{nm}\bigg(-\tanh(\sqrt{\gamma}a)\cosh{(\sqrt{\gamma}x)}+\sinh{(\sqrt{\gamma}x)}\bigg)\bigg]\sin\bigg(\frac{n\pi y}{b}\bigg)\sin\bigg(\frac{m\pi z}{c}\bigg)$$

On applying boundary condition $(1)$ $$T_w=\bigg[A_{nm}\bigg(-\tanh(\sqrt{\gamma}a)\bigg)\bigg]\sin\bigg(\frac{n\pi y}{b}\bigg)\sin\bigg(\frac{m\pi z}{c}\bigg)$$ Subsequently applying orthogonality, the follwowing is the result:

$$A_{nm} = \frac{-4 T_w \tanh(\sqrt{\gamma}a)}{nm\pi^2}(1-\cos{(n\pi)})(1-\cos{(m\pi)})$$

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    $\begingroup$ You need to solve $$X''=(\lambda+\mu)X=\gamma X$$ where $$\lambda=\left(\frac{n\pi}{c}\right)^{2}, \quad \mu=\left( \frac{m\pi}{b}\right)^{2}, \quad n,m\ge 1$$ according to $T_{x}(a,y,z)=0$. You should get something like $$X = C [- \tanh(\sqrt{\gamma} a) \cosh(\sqrt{\gamma} x) + \sinh(\sqrt{\gamma} x) ]$$ Once you have done this, write out the solution to $T$ as a Fourier series, use the condition $T(0,x,y) = T_{w}$ and apply the orthogonality conditions to compute the coefficients $A_{mn}$. $\endgroup$ – mattos Jul 12 at 3:57
  • $\begingroup$ @Mattos Thanks. I applied your suggestion. I have posted my attempt as an edit to the original question. Would really appreciate if you take a look and comment if I have followed your advice correctly. $\endgroup$ – MuadDib Jul 12 at 4:49

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