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I was given the following PDE to solve $$u_t(x,y,z,t)=k\cdot\Delta u(x,y,z),\\u_x(0,y,z,t)=0,\quad u_y(x,0,z,t)=0,\quad u_z(x,y,0,t)=0\\ u_x(L,y,z,t)=0,\quad u_y(x,H,z,t)=0,\quad u_z(x,y,W,t)=0\\ u(x,y,z,0)=\alpha(x,y,z)$$ and find its behavior as $t\to\infty$.

I managed to find the solution: $$u(x,y,z,t)=\sum_{l=1}^\infty\sum_{m=1}^\infty\sum_{n=1}^\infty A_{nml}\cos\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{H}\right)\cos\left(\frac{l\pi z}{W}\right)e^{-\lambda_{nml}kt}$$ which agrees with the text, but I found $A_{nml}$ as the following $$A_{nml}=\frac{8}{LHW}\int_0^W\int_0^H\int_o^L\alpha(x,y,z)\cos\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{H}\right)\cos\left(\frac{l\pi z}{W}\right)\,dxdydz$$ while the book uses orthogonalization and expresses it as $$A_{nml}=\frac{\int_0^W\int_0^H\int_o^L\alpha\cos\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{H}\right)\cos\left(\frac{l\pi z}{W}\right)\,dxdydz}{\int_0^W\int_0^H\int_o^L\alpha\cos^2\left(\frac{n\pi x}{L}\right)\cos^2\left(\frac{m\pi y}{H}\right)\cos^2\left(\frac{l\pi z}{W}\right)\,dxdydz},\\ \\ \alpha=\sum_{l=1}^\infty\sum_{m=1}^\infty\sum_{n=1}^\infty A_{nml}\cos\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{H}\right)\cos\left(\frac{l\pi z}{W}\right)$$ which I can't seem to conclude if it is equivalent to my solution or not.

Also as $t\to\infty$ wouldn't $u\to 0$ due to the $e^{-\lambda_{nml}kt}$ term?

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The sums in your main solution should iterate from $0$, since $0$ is an eigenvalue too. After that your answer is correct. The book has a typo: the denominator should be

$$\int_0^W\int_0^H\int_0^L\cos^2\left(\frac{n\pi x}{L}\right)\cos^2\left(\frac{m\pi y}{H}\right)\cos^2\left(\frac{l\pi z}{W}\right)\,dxdydz =\frac{LHW}{8}$$ if $n,m,l$ are nonzero, which is exactly what you have (note that this is equal to $\frac{LHW}{2^k}$ with $k$ of the numbers $n,m,l$ nonzero, so $A_{0,0,0}=\frac{1}{LHW}\int_0^W\int_0^H\int_0^L\alpha(x,y,z)\,dxdydz$ in particular). If we had $\alpha=\frac{1}{2}$, we know that $u=\frac{1}{2}$, but the book would give $A_{0,0,0}=1$, while it should be $A_{0,0,0}=\frac{1}{2}$.

Note that $e^{-\lambda_{nml}kt}=1$ for $\lambda_{0,0,0}=0$, so $u\to A_{0,0,0}$ for $t\to\infty$.

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