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Notes on some numerical schemes

Nasser M. Abbasi

May 8, 2020   Compiled on May 8, 2020 at 7:11pm

Contents

1 Introduction
 1.1 Centered difference for 1D wave PDE

1 Introduction

These are notes to describe some numerical schemes.

1.1 Centered difference for 1D wave PDE

Here we want to solve \(u_{tt}=c^{2}u_{xx}\) in 1D finite domain \(0<x<L\) and \(t>0\), with boundary conditions \(u\left ( 0,t\right ) =f\left ( t\right ) ,u\left ( L,t\right ) =g\left ( t\right ) \) and initial position \(u\left ( x,0\right ) =\alpha \left ( x\right ) \) and initial velocity \(\frac{\partial u\left ( x,0\right ) }{\partial t}=\beta \left ( x\right ) \).

Centered difference is used.

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Figure 1:Centered difference scheme used in time and space

At each internal node \(j\) we have the following finite difference represenation of \(u_{tt}=c^{2}u_{xx}\)

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Figure 2:FDM at each node

To handle initial conditions, initial velocity is used to solve for \(u_{j}^{-1}\)

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Figure 3:Solving for \(u^{-1}_j\)

This gives all the information needed to find the matrices to use. Let \(k=\Delta t\). From Eq(1) \begin{align*} \frac{u_{j}^{1}-u_{j}^{-1}}{2k} & =\alpha \\ u_{j}^{-1} & =u_{j}^{1}-2k\alpha \end{align*}

Substituting this in Eq(2) gives\begin{align*} \frac{\left ( u_{j}^{1}-2k\alpha \right ) -2u_{j}^{0}+u_{j}^{1}}{k^{2}} & =c^{2}\frac{u_{j-1}^{0}-2u_{j}^{0}+u_{j+1}^{0}}{h^{2}}\\ 2u_{j}^{1} & =\frac{k^{2}c^{2}}{h^{2}}\left ( u_{j-1}^{0}-2u_{j}^{0}+u_{j+1}^{0}\right ) +2u_{j}^{0}+2k\alpha \\ u_{j}^{1} & =\frac{1}{2}\frac{k^{2}c^{2}}{h^{2}}\left ( u_{j-1}^{0}-2u_{j}^{0}+u_{j+1}^{0}\right ) +u_{j}^{0}+k\alpha \end{align*}

Therefore for \(n=1\) only and for \(j=1\cdots N\) where \(N\) is number of nodes\[\begin{pmatrix} u_{1}^{1}\\ u_{2}^{1}\\ u_{3}^{1}\\ u_{4}^{1}\\ u_{5}^{1}\end{pmatrix} =\frac{1}{2}\frac{k^{2}c^{2}}{h^{2}}\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 1 & -2 & 1 & 0 & 0\\ 0 & 1 & -2 & 1 & 0\\ 0 & 0 & 1 & -2 & 1\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} u_{1}^{0}\\ u_{2}^{0}\\ u_{3}^{0}\\ u_{4}^{0}\\ u_{5}^{0}\end{pmatrix} +\begin{pmatrix} u_{1}^{0}\\ u_{2}^{0}\\ u_{3}^{0}\\ u_{4}^{0}\\ u_{5}^{0}\end{pmatrix} +k\alpha \] Where \(\begin{pmatrix} u_{1}^{0}\\ u_{2}^{0}\\ u_{3}^{0}\\ u_{4}^{0}\\ u_{5}^{0}\end{pmatrix} \) is known and comes from boundary and initial conditions. \(u_{1}^{0}\) is left B.C. and \(u_{N}^{0}\) comes from right B.C. and \(u_{2}^{0}\cdots u_{N-1}^{0}\) comes from initial conditions \(u\left ( x,0\right ) \). Now, for \(n=2\) or higher times\begin{align*} \frac{u_{j}^{n-1}-2u_{j}^{n}+u_{j}^{n+1}}{k^{2}} & =c^{2}\frac{u_{j-1}^{n}-2u_{j}^{n}+u_{j+1}^{n}}{h^{2}}\\ u_{j}^{n-1}-2u_{j}^{n}+u_{j}^{n+1} & =\frac{k^{2}c^{2}}{h^{2}}\left ( u_{j-1}^{n}-2u_{j}^{n}+u_{j+1}^{n}\right ) \\ u_{j}^{n+1} & =\frac{k^{2}c^{2}}{h^{2}}\left ( u_{j-1}^{n}-2u_{j}^{n}+u_{j+1}^{n}\right ) +2u_{j}^{n}-u_{j}^{n-1} \end{align*}

In Matrix form\[\begin{pmatrix} u_{1}^{n+1}\\ u_{2}^{n+1}\\ u_{3}^{n+1}\\ u_{4}^{n+1}\\ u_{5}^{n+1}\end{pmatrix} =\frac{k^{2}c^{2}}{h^{2}}\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 1 & -2 & 1 & 0 & 0\\ 0 & 1 & -2 & 1 & 0\\ 0 & 0 & 1 & -2 & 1\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} u_{1}^{n}\\ u_{2}^{n}\\ u_{3}^{n}\\ u_{4}^{n}\\ u_{5}^{n}\end{pmatrix} +2\begin{pmatrix} u_{1}^{n}\\ u_{2}^{n}\\ u_{3}^{n}\\ u_{4}^{n}\\ u_{5}^{n}\end{pmatrix} -\begin{pmatrix} u_{1}^{n-1}\\ u_{2}^{n-1}\\ u_{3}^{n-1}\\ u_{4}^{n-1}\\ u_{5}^{n-1}\end{pmatrix} \] So to find \(u_{j}^{n+1}\) we need to know the last time step solution and also the solution for the step before that.

Small Mathematica was written to implement the above scheme. Here is screen show of the GUI for one example.

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Figure 4:1D FDM for wave pde

The Mathematica notebook of the above is here

The following is an animation of \(u_{tt}=4u_{xx}\) with fixed ends, and string length \(L=5\) with initial position given by \(8x\frac{\left ( L-x\right ) ^{2}}{L^{3}}\) and zero initial velocity. This uses \(\Delta t=0.02\) seconds and \(30\) grid points over the length \(5\).




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