$3D$ Heat transfer equation with internal loss generation (zero boundary temperatures ) I am sincerely asking the analytical solution to the $3D$ Heat transfer equation with constant internal loss generation.



I don't know how to find the particular solution for the Poisson equation. The general solution of Laplace equation is simply.
$$\sum_{n,m}\sin\left(\frac{m\pi }{W}x\right)\cdot\sin\left(\frac{n\pi }{H}y\right)\cdot\sinh(k_{mn}z)$$
 A: So first of all I think you have a small error in the boundary conditions, I expect it should read $T(x,y,{\color{red} 0})=T(x,y,L)=0$.
Furthermore your "general solution" is a little bit odd: It satisfies the differential equation and the boundary conditions in $x$ and $y$ direction, but not in $z$.
To obtain a solution to the differential equation i would propose to start with a basis of eigenfunctions of the laplacian (together with the boundary conditions), that is:
$$
\varphi_{l,m,n}(x,y,z)=\sin\left(\frac{l\pi }{W}x\right)\cdot\sin\left(\frac{m\pi }{H}y\right)\cdot\sin\left(\frac{n\pi }{L}z\right)\\
\Delta\varphi_{l,m,n}=\underbrace{-\left((\tfrac{l\pi }{W})^2+(\tfrac{m\pi }{H})^2+(\tfrac{n\pi }{L})^2\right)}_{=\lambda_{l,m,n}}\varphi_{l,m,n}
$$
Next build up our solution from the eigenfunctions and fit the coefficients $a_{l,m,n}$ to satisfy the differential equation:
$$
u=\sum_{l,m,n}a_{l,m,n}\varphi_{l,m,n}\\
\Rightarrow \Delta u + C = 0\\
\Leftrightarrow \sum_{l,m,n}\lambda_{l,m,n}a_{l,m,n}\varphi_{l,m,n} + C = 0
$$
next we mulitply with $\varphi_{l',m',n'}$ and integrate over the domain
$$
\sum_{l,m,n}\lambda_{l,m,n}a_{l,m,n}\iiint\varphi_{l,m,n}\varphi_{l',m',n'}dxyz + C\iiint\varphi_{l',m',n'}dxyz = 0
$$
Since the eigenfunctions are orthogonal the first integral os only nonzero if $l=l',m=m',n=n'$, therefore the sum reduces to
$$
\lambda_{l',m',n'}a_{l',m',n'}\iiint\varphi_{l',m',n'}^2dxyz + C\iiint\varphi_{l',m',n'}dxyz = 0\\
\Rightarrow a_{l',m',n'}=-C\frac{\iiint\varphi_{l',m',n'}dxyz}{\lambda_{l',m',n'}\iiint\varphi_{l',m',n'}^2dxyz}\\
\Rightarrow a_{l',m',n'}=\begin{cases}-C\frac{64W^2H^2L^2}{\pi^5l'm'n'(H^2L^2l'^2+L^2W^2m'^2+H^2W^2n'^2)}&l',m',n' \mbox{ odd}\\
0&\mbox{otherwise}\end{cases}
$$
Which finally gives the solution as:
$$
u(x,y,z)=-C\sum_{l,m,n=0}\frac{64W^2H^2L^2\sin\left(\frac{(2l+1)\pi }{W}x\right)\cdot\sin\left(\frac{(2m+1)\pi }{H}y\right)\cdot\sin\left(\frac{(2n+1)\pi }{L}z\right)}{\pi^5(2l+1)(2m+1)(2n+1)(H^2L^2(2l+1)^2+L^2W^2(2m+1)^2+H^2W^2(2n+1)^2)}
$$
(Here I have rewritten the indices to automatically omit the even ones)
