I used to solve differential algebraic equations using Lagrange polynomials.
Essentially you convert the differential equations into an algebraic system by discretizing the solution. The method is called Orthogonal Collocation on Finite Elements, and it was developed by chemical engineers.
The Lagrange polynomials were calculated at special knots that corresponded to Radau interior points, which work great for stiff systems.
It’s great for solving differential algebraic equations through pure matrix operations, no explicit integration like Runge Kutta.
(Well, it’s inplicit Runge Kutta).
The Lagrange polynomials form the normal basis of most Finite Elements Method (FEM) software. There are other polynomials which are used as well, but these are the workhorse of most solvers.
(proof-reading through HN is a mildly embarrassing process, sorry about that! I do go over these posts and proof-read them several times myself before publishing)
the last matrix before the appendix is not the identity matrix, right now the matrix is: \begin{bmatrix}
1 & 0 & 0 & \dots & 0\\
1 & 0 & 0 & \dots & 0\\
1 & 0 & 0 & \dots & 0\\
\vdots & \vdots & \vdots & \ddots &\vdots \\
1 & 0 & 0 & \dots & 1
\end{bmatrix}
I used to solve differential algebraic equations using Lagrange polynomials.
Essentially you convert the differential equations into an algebraic system by discretizing the solution. The method is called Orthogonal Collocation on Finite Elements, and it was developed by chemical engineers.
The Lagrange polynomials were calculated at special knots that corresponded to Radau interior points, which work great for stiff systems.
It’s great for solving differential algebraic equations through pure matrix operations, no explicit integration like Runge Kutta. (Well, it’s inplicit Runge Kutta).
The Lagrange polynomials form the normal basis of most Finite Elements Method (FEM) software. There are other polynomials which are used as well, but these are the workhorse of most solvers.
In the Polynomial Interpolation Theorem, you have r(x) = p(x) - r(x), but I think it should be q(x) = p(x) - r(x).
Fixed, thank you! (it's actually r(x)=p(x)-q(x))
(proof-reading through HN is a mildly embarrassing process, sorry about that! I do go over these posts and proof-read them several times myself before publishing)
the last matrix before the appendix is not the identity matrix, right now the matrix is: \begin{bmatrix} 1 & 0 & 0 & \dots & 0\\ 1 & 0 & 0 & \dots & 0\\ 1 & 0 & 0 & \dots & 0\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ 1 & 0 & 0 & \dots & 1 \end{bmatrix}
Thanks for noticing, I'll fix it shortly