I'm trying an analytical solution for a 3D Laplace Equation and just wanted to know if I can use the symbolic computational functions of *Mathematica* to calculate the Fourier coefficients

**context**

(Please wear with the length)

The last step of solving a three-dimensional Laplace equation ($ nabla ^ 2T = 0 $, from where $ nabla ^ 2 = frac { partial ^ 2} { partial x ^ 2} + frac { partial ^ 2} { partial y ^ 2} + frac { partial ^ 2} { partial z ^ 2} $) with BC (s) as

$ T (0, y, z) = T (L, y, z) = T_a $,

$ T (x, 0, z) = T (x, l, z) = T_a $,

$ frac { partial T (x, y, 0)} { partial z} = p_c bigg (e ^ {- b_cy / l} left[T_{ci} + frac{b_c}{l}int_0^y e^{b_cs/l}T(x,s,z)dsright] – T (x, y, 0) bigg) $

$ frac { partial T (x, y, w)} { partial z} = p_h bigg (e ^ {- b_hx / L} left[T_{hi} + frac{b_h}{l}int_0^x e^{b_hs/l}T(s,y,z)dsright] – T (x, y, w) bigg) $

When using a solution form

$ T (x, y, z) = sum_ {m, n = 1} ^ { infty} (A_ {nm} e ^ { gamma z} + B_ {nm} e ^ {- gamma z}) sin ( frac {n pi x} {L}) sin ( frac {m pi y} {l}) + T_a. $

I have to rate the Fourier coefficients $ A_ {nm} $ and $ B_ {nm} $ using the following equation obtained after applying the $ z = 0 $ BC

$$

frac {1} {p_c} sum_ {n, m = 1} ^ infty sin ( frac {n pi x} {L}) sin ( frac {m pi y} {l}) gamma (A_ {nm} -B_ {nm}) = delta e ^ { frac {-b_cy} {l}} + PQ + RS rightarrow ( mathrm {1})

$$

$ P = sum_ {n, m = 1} ^ infty frac {b_c ^ 2} {b_c ^ 2 + (m pi) ^ 2} (A_ {nm} + B_ {nm}) sin ( frac {n pi x} {L}) sin ( frac {m pi y} {l}) $

$ Q = sum_ {n, m = 1} ^ infty frac {b_c m pi} {b_c ^ 2 + (m pi) ^ 2} (A_ {nm} + B_ {nm}) sin ( frac {n pi x} {L}) cos ( frac {m pi y} {l}) $

$ R = sum_ {n, m = 1} ^ infty frac {b_c m pi} {b_c ^ 2 + (m pi) ^ 2} (A_ {nm} + B_ {nm}) e ^ { frac {-b_c y} {l}} $

$ S = sum_ {n, m = 1} ^ infty (A_ {nm} + B_ {nm}) sin ( frac {n pi x} {L}) sin ( frac {m pi .) y} {l}) $

Here ,$ gamma ^ 2 = ( frac {n pi} {L}) ^ 2 + ( frac {m pi} {l}) ^ 2 $ and $ c, delta = (T_ {ci} -T_a) $ are constants.

Of course, there would be another relationship that would be a linear equation in $ A $ and $ B $ Finally, the values of $ A_ {nm} $ and $ B_ {nm} $ (of the $ z = w $ BC) separately. My question relates to the way I should handle it $ e ^ { frac {-b_c y} {l}} $ Here.

**attempt**

I can multiply both sides of the equation $ mathrm {(1)} $ With $ int_0 ^ L sin ( frac {k pi x} {L}) $ and $ int_0 ^ l sin ( frac {j pi y} {l}) $ and then use the principle of orthogonality to finally arrive

For some $ n = k $ and $ m = j $, $ mathrm {(1)} $ becomes

$$

frac {1} {p_c} frac {L} {2} frac {l} {2} gamma (A_ {kj} -B_ {kj}) = delta e ^ { frac {-b_c y} {l}} int_0 ^ L sin ( frac {k pi x} {L}) mathrm {d} x int_0 ^ l sin ( frac {j pi y} {l}) mathrm {d} y + underbrace { frac {b_c ^ 2} {b_c ^ 2 + (j pi) ^ 2} (A_ {kj} + B_ {kj}) frac {L} {2} frac { l} {2}} _ {P} – overbrace {0} ^ {Q} + underbrace { sum_ {n, m = 1} ^ infty frac {b_c m pi} {b_c ^ 2 + ( m pi) ^ 2} (A_ {nm} + B_ {nm}) e ^ { frac {-b_cy} {l}} int_0 ^ L sin ( frac {k pi x} {L}) mathrm {d} x int_0 ^ l sin ( frac {j pi y} {l}) mathrm {d} y} _ {R} – overbrace {(A_ {kj} + B_ {kj} ) frac {L} {2} frac {l} {2}} ^ {S}

$$

**Fourier extension of $ e ^ { frac {-b_c y} {l}} $ in the break $ y in [0,l]$**

I expanded $ e ^ { frac {-b_c y} {l}} $

$$ e ^ { frac {-b_c y} {l}} = frac {(1-e ^ {- b_c})} {b_c} + sum_ {r = 1} ^ { infty} bigg[frac{2b_c(1-e^{-b_c})}{(b_c)^2 + (2rpi)^2}cosbigg(frac{2rpi y}{l}bigg) + frac{4rpi(1-e^{-b_c})}{(b_c)^2 + (2rpi)^2}sinbigg(frac{2rpi y}{l}bigg)bigg]$$

I think I have to replace that (1) $ to the $ e ^ { frac {-b_c y} {l}} $

Suggestions or comments would be very helpful and would be very happy.

When you see (1) $, you can see how, when the extension of $ e ^ { frac {-b_c y} {l}} $ this will be replaced $ R $ The term would result in a triple summation, which complicates the way. Can there be an alternative way to perform these calculations in? *Mathematica*, I know that the PDE solvers can use discretization schemes to get a result. I hoped that I could use it *Mathematica* to help me in finding these Fourier coefficients in the last step.