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Part I

Let us say I need to determine Fourier coefficient at the end of the solution to a PDE. On applying the non-homogeneous BC, I arrive at the following

$$\sum_{n=1}^{\infty} A_{n}\sinh(n\pi c)\sin\bigg(\frac{n\pi x}{a}\bigg)=f(x)$$

We have two cases that needs to be evaluated now:

  1. $f(x) = 2e^{-\frac{x}{3}}$
  2. $f(x) = \frac{4}{3} (x-x^2)$

Part II

For $\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0$ defined on $x \in [0,a]$ and $y \in [0,b]$

with the BC as

$T(0,y) = T(a,y) = T_a$

$\frac{\partial T(x,0)}{\partial y} = k_1\bigg[e^{-s_1y}\bigg(T_1 + s_1\int_0^y e^t T(x,t)\mathrm{d}t\bigg) - T(x,0)\bigg]$

$\frac{\partial T(x,b)}{\partial x} = 0$


For Part I After applying @Mattos suggestions

For $f(x) = 2e^{-\frac{x}{3}}$

$$\sum_{n=1}^{\infty} A_{n}\sinh(n\pi c)\int_0^a\sin\bigg(\frac{n\pi x}{a}\bigg)\sin\bigg(\frac{k\pi x}{a}\bigg)\mathrm{d}x=2\int_0^ae^{-\frac{x}{3}}\sin\bigg(\frac{k\pi x}{a}\bigg)\mathrm{d}x$$

The integrand on the LHS has values only for $n=k$ (From orthogonality)

$$A_k\sinh(n\pi c)\bigg(\frac{a}{2}\bigg)=2\int_0^a e^{\frac{-x}{3}}\sin\bigg(\frac{k\pi x}{a}\bigg)\mathrm{d}x$$

Since, $k$ is a dummy integer it can be replaced by arbitrary integer $n$

$$A_n = \frac{\frac{4}{a}\bigg[\int_0^ae^{\frac{-x}{3}}\sin\bigg(\frac{n\pi x}{a}\bigg)\mathrm{d}x\bigg]}{\sinh(n\pi c)}$$

Similarly, for $f(x) = \frac{4}{3} (x-x^2)$

$$A_n = \frac{\frac{8}{3a}\bigg[\int_0^a x\sin\bigg(\frac{n\pi x}{a}\bigg)\mathrm{d}x - \int_0^a x^2\sin\bigg(\frac{n\pi x}{a}\bigg)\mathrm{d}x\bigg]}{\sinh(n\pi c)}$$

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    $\begingroup$ Multiply both sides of the equality by $\sin(m \pi x/a)$ and integrate over the domain i.e \begin{align} \sum_{n \ge 1} \tilde{A}_{n}\sin\left(\frac{n \pi x}{a}\right)&=f(y) \\ \sum_{n \ge 1}\tilde{A}_{n}\sin\left(\frac{n \pi x}{a}\right)\sin\left(\frac{m \pi x}{a}\right)&=f(y)\sin\left(\frac{m \pi x}{a}\right) \\ \implies \sum_{n \ge 1} \tilde{A}_{n}\int_{\Omega} \sin\left(\frac{n \pi x}{a}\right)\sin\left(\frac{m \pi x}{a}\right)dx &= \int_{\Omega} f(y) \sin \left( \frac{m \pi x}{a} \right) dx \end{align} then check cases $m = n \ne 0$, $m \ne n$. Note that it should be $f(x)$ not $f(y)$. $\endgroup$ – Mattos Feb 28 at 4:45
  • $\begingroup$ @Mattos Thanks . I have edited the problem to change the mistakes. Also, there was another problem I dealt with and would like to know what you think. Will the same procedure you suggested above work ? Should then I have to expand $e^{-\frac{x}{3}}$ as some sort of series ? $\endgroup$ – Anakin Feb 28 at 6:41
  • $\begingroup$ @Mattos I apologise, if i broke any community rule by adding a separate question to the same problem. If so please let me know. $\endgroup$ – Anakin Feb 28 at 8:20
  • $\begingroup$ A few things. First, did you manage to compute the coefficients for the two different functions? Secondly, where did the new equation come from (the one you edited in to your original post)? How was it derived? It seems a bit arbitrary. Third, what was the original PDE you started with? I'm guessing it is something like Laplaces equation with Dirichlet boundary conditions. $\endgroup$ – Mattos Feb 28 at 12:03
  • $\begingroup$ @Mattos Yes I managed to find those coefficients ,after reading your comment then relating it to trigonometric orthogonal functions. Yes, you are right I ought to post my worked out solutions here $\endgroup$ – Anakin Feb 28 at 12:40

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