I'm trying to solve the torsion problem for a rectangular cross-section of shape $2a$ by $2b$ in the x-y plane. The coordinate system is set up so that the origin is the centroid of the cross-section and so that the boundary of the rectangle is given by $-a≤x≤a$ and $-b≤y≤b$. The problem setup is illustrated here (ignore the asked question though, I understand that answer):
https://en.wikiversity.org/wiki/Introduction_to_Elasticity/Torsion_of_rectangular_sections
The governing partial differential equation is given by:
$$\Delta^2\phi(x,y)=-2\mu\theta$$ Or: $$\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = -2\mu\theta$$
And is subject to the following: $$ϕ(±a,y) = 0$$
$$ϕ(x,±b) = 0$$
$$-a≤x≤a$$
$$-b≤y≤b$$
$$a≥0$$
$$b≥0$$
i.e. $ϕ = 0$ on the boundary of the rectangular cross-section.
Unfortunately, I have not taken a class on partial differential equations and clearly need that to solve this problem. If someone could walk me through the process of solving this it would be greatly appreciated. Honestly, if anyone could even tell me anything about it it would be greatly appreciated.
What I do know is that this is a Poisson's equation in two dimensions with a constant source function and Dirichlet boundary conditions. I think I can use separation of variables, but I do not know how to choose my functions because it isn't homogeneous.
The solutions I have been able to find for this problem jump directly into Fourier series and I don't understand that jump. My general understanding of the Fourier series is that if your answer contains a trigonometric function with a constant multiplied by a variable inside it, then you can express the answer as an infinite sum of that function changing only multiples of the angle that satisfy the boundary conditions.
So, for a function $T(x,t)$:
$$T(x,t)=Csin(\sqrt \lambda x)$$
With the boundary condition:
$$T(a,t) = 0$$
We get, for the non-trivial solution $C\neq 0$:
$$sin(\sqrt \lambda a) = 0$$
$$\sqrt \lambda a= 0, \pi, 2\pi, ..., n\pi$$
$$\lambda = \frac{n^2\pi^2}{a^2}$$
Meaning:
$$T(x,t) = Csin(\frac{n\pi}{a}x)$$
Or, as a Fourier Series Representation:
$$T(x,t) = \sum_{n=1}^\infty C_n sin(\frac{n\pi}{a}x)$$
And we can find $C$ and $C_n$ using other boundary conditions. If this is wrong at all please let me know as well, but my issue is determining the actual function itself, $T(x,y)$ in the above example, $\phi(x,y)$ in my question. Thank you for all of your help.
One can change variables such that without loss of generality the domain is $[0,\pi] \times [0,\pi]$ and the forcing is $1$. This is a fairly straightforward exercise.
The appropriate Fourier series for $f(x,y)=1$ with these boundary conditions is $\frac{4}{\pi^2} \sum_{m,n \text{ both odd }} \frac{4}{mn} \sin(mx) \sin(ny)$. The $\frac{4}{mn}$ and the restriction to odd $m,n$ is obtained by evaluating the integrals $\int_0^\pi \int_0^\pi \sin(mx) \sin(ny) \cdot 1 dx dy$; the $\frac{4}{\pi^2}$ is obtained by evaluating the integrals $\int_0^\pi \int_0^\pi \sin(mx)^2 \cdot \sin(ny)^2 dx dy$.
The corresponding eigenvalues of the Laplacian are $-(m^2+n^2)$, so you obtain the series solution $\frac{4}{\pi^2} \sum_{m,n \text{ both odd}} -\frac{4}{mn(m^2+n^2)} \sin(mx) \sin(ny)$.
Note that there is a serious difficulty implementing this formula in practice, because of the fact that the forcing does not actually satisfy the boundary conditions.