Incorporating boundary conditions in the Finite Element Method (FEM)

[Vanilla Gorilla]

TL;DR Summary

As I understand it, the final step to solving a differential equation in FEM is to solve a matrix equation of the form $[K] [T] = [F]$, where $[K]$ is the stiffness matrix, while $[T]$ is the matrix of nodal temperature values. However, I am unsure as to what $F$ represents.

I have been watching Mike Foster's video series of the Finite Element Method for Differential Equations (FEM). In this episode, he solves a DE relating to temperature. As the final step, he gives the following equation: $$[K] [T] = [F]$$ In this equation, I understand that $[K]$ is the stiffness matrix, while $[T]$ is the matrix of nodal temperature values. However, I do not fully understand what $[F]$ is, and how to incorporate it into the method; I believe that it is something relating to boundary conditions, but I am unsure as to how to formulate it from said boundary conditions.

Any help is much appreciated!
P.S., I'm not always great at articulating my thoughts, so my apologies if this question isn't clear. Also, I know this isn't high school material, but I am currently in high school, which is why I made my level "Basic/high school level."

Note to moderators: I think that is the most appropriate forum for this post. However, if not, apologies, and please feel free to move at your own discretion. :)

 

[pasmith]

Expand the matrix equation in terms of its compoenents: $$K_{n,1}T_1 + \dots + K_{n,N}T_N = F_n, \quad n = 1, \dots, N.$$ If $T_1 = T(x_1)$ and $T_N = T(x_N)$ are known, they do not need to be solved for; substitute the known values and shift the corresponding terms to the right hand side as part of the source term $F$: $$K_{n,2}T_2 + \dots + K_{n,N-1}T_{N-1} = F_n - K_{n,1}T_1 - K_{n,N}T_N.$$ Now discard rows 1 and N of the matrix; we do not need them, as we only have $N - 2$ unknowns.

 

[Vanilla Gorilla]

I believe I understand that, but how do we get the source term $F$? That is where my confusion lies.

 

[pasmith]

THat comes from the differential equation; starting with $$\frac{d^2T}{dx^2} = -f$$ we multiply by a test function $v$ and integrate, using integration by parts on the left hand side: $$\left[ v(x)T'(x)\right]_a^b - \int_a^b T'(x)v'(x)\,dx = - \int_a^b f(x)v(x)\,dx$$ Now expand $T(x) = T_1\phi_1(x) + \dots + T_N\phi_N(x)$ where at each node $x_m$ $$\phi_n(x_m) = \begin{cases} 1 & n = m \\ 0 & n \neq m\end{cases}$$ so that $T(x_m) = T_m$, and choose $N$ convenient test functions $v_n$ such that $v_n(a) = v_n(b) = 0.$ We then get the system $$\left(\int_a^b \phi'_1(x) v'_m(x)\,dx\right) T_1 + \dots + \left(\int_a^b \phi'_N(x) v'_m(x)\,dx\right) T_N = \int_a^b f(x)v_m(x)\,dx$$ and so $$\begin{split} K_{m,n} &= \int_a^b \phi'_n(x) v'_m(x)\,dx \\ F_m &= \int_a^b f(x)v_m(x)\,dx\end{split}$$

 

[Vanilla Gorilla]

That makes more sense, but I still have a feq questions

Is $\phi$ representative of the shape functions?
How would the final system, shown below, end up looking in matrix form? $$ \begin{split} K_{m,n} &= \int_a^b \phi'_n(x) v'_m(x)\,dx \\ F_m &= \int_a^b f(x)v_m(x)\,dx\end{split}$$
In the orgininal equation, shown below, is it implied that we are solving for $T$ or $f$? $$\frac{d^2T}{dx^2} = -f$$