The Derivation of Second and Fourth Order Differentiation Matrices - PowerPoint PPT Presentation

1/26 The Derivation of Second and Fourth Order Differentiation Matrices Chris Moody and Boden Hegdal ◭◭ ◮◮ ◭ ◮ Back Close

Introduction Differentiation matrices are a class of band matrices that use meth- ods of linear algebra to approximate derivatives. These matrices can 2/26 be derived a number of different ways. We will derive a second order differentiation matrix from the Taylor polynomial and a fourth order differentiation matrix from the Lagrange interpolating polynomial. ◭◭ ◮◮ ◭ ◮ Back Close

The Problem In many real life situations we don’t have a simple function to differ- entiate. We may have a set of discrete points and we need to plot their 3/26 derivative. Differentiation matrices are the perfect tool for this job. They take that set of discrete points and approximate the derivative as another set of discrete points. u i − 1 u i +1 u 1 u i u n x 1 x i − 1 x i x i +1 x n h ◭◭ ◮◮ ◭ ◮ Back Close

Taylor’s Theorem The standard Taylor series has the following form. 4/26 u ( x ) = u ( a ) + u ′ ( a )( x − a ) + u ′′ ( a ) 2! ( x − a ) 2 + · · · For convenience we will substitute a + h for x and then substitute x for a . u ( x + h ) = u ( x ) + u ′ ( x )( h ) + u ′′ ( x ) h 2 + · · · 2! ◭◭ ◮◮ ◭ ◮ Back Close

The Second Order Difference Equation from Taylor’s Theorem 5/26 The standard second-order finite difference approximation can be derived by considering the Taylor expansions of u ( x i +1 ) and u ( x i − 1 ). h h x i − 1 x i x i +1 Note: x i +1 = x i + h and x i − 1 = x i − h Using the Taylor series: u ( x i + h ) = u ( x i ) + u ′ ( x i ) h + u ′′ ( x i ) h 2 + · · · ◭◭ 2 i ◮◮ ◭ u ( x i − h ) = u ( x i ) − u ′ ( x i ) h + u ′′ ( x i ) h 2 + · · · ◮ 2 i Back Close

Difference Equation Subtracting one equation from the other, and eliminating higher or- der terms, we get: 6/26 ′ ( x i ) h u ( x i + h ) − u ( x i − h ) = 2 u u ′ ( x i ) = u ( x i + h ) − u ( x i − h ) 2 h . However, x i +1 = x i + h and x i − 1 = x i − h . ◭◭ u ′ ( x i ) = u ( x i +1 ) − u ( x i − 1 ) ◮◮ 2 h ◭ ◮ Back Close

How the Difference Equation Works The difference equation approximates the derivative at a point ( x i , u ( x i )) using the slope of the line that passes through the point 7/26 before it ( x i − 1 , u ( x i − 1 )) and the point after it ( x i +1 , u ( x i +1 )). u ′ ( x i ) = u ( x i +1 ) − u ( x i − 1 ) 2 h u 2 h ◭◭ ◮◮ ◭ x i − 1 x i x i +1 ◮ Back Close

Assuming Periodicity Before we can go any further we must assume that the function we are evaluating is periodic. We know that in most cassis it is not but it 8/26 is going to be necessary so that we can set up our system of equations. u 0 = u n u 1 = u n +1 u i − 1 u i +1 u n u 1 u i u n u 1 ◭◭ x 0 x 1 x i − 1 x i x i +1 x n x n +1 ◮◮ h ◭ ◮ Back Close

System of Equations Our previous result: 9/26 u ′ ( x i ) = u ( x i +1 ) − u ( x i − 1 ) 2 h Replacing u ′ ( x i ) with w i , u ( x i +1 ) with u i +1 , and u ( x i − 1 ) with u i − 1 gives us the standard second order difference equation. w i = u i +1 − u i − 1 2 h We will now use this difference equation to set up a system of equa- tions that represent the derivative of the function. ◭◭ ◮◮ To set up the system of equations we will set i = 1, 2, 3, . . . , n ◭ ◮ Back Close

System of Equations 10/26 w 1 = u 2 − u n 2 h w 2 = u 3 − u 1 2 h w 3 = u 4 − u 2 2 h . . . w N − 1 = u N − u N − 2 2 h w N = u 1 − u N − 1 2 h ◭◭ ◮◮ ◭ ◮ Back Close

Matrix Form The system takes the form 11/26 w = D u , where 0 1 / 2 0 0 . . . 0 0 − 1 / 2   − 1 / 2 0 1 / 2 0 . . . 0 0 0   0 − 1 / 2 0 1 / 2 . . . 0 0 0   . . . . . . .  ...  . . . . . . . . . . . . . . D = .     ... 0 0 0 0 0 1 / 2 0   ◭◭   0 0 0 0 . . . − 1 / 2 0 1 / 2   ◮◮ 1 / 2 0 0 0 . . . 0 − 1 / 2 0 ◭ ◮ Back Close

Example Lets look at an example using the following set of 8 discrete points (2,5),(4,15),(6,35),(8,65),(10,100),(12,145),(14,195), and (16,255). 12/26     w 1 5 0 1 / 2 0 . . . 0 − 1 / 2 w 2   15     − 1 / 2 0 1 / 2 . . . 0 0 w 3 35           = 1 0 − 1 / 2 0 . . . 0 0 w 4 65       . . . . .    ...    . . . . . w 5 100 . . . . . 2             w 6 145 0 0 0 . . . 0 1 / 2           w 7 195 1 / 2 0 0 . . . − 1 / 2 0     w 8 255 ◭◭ ◮◮ ◭ ◮ Back Close

Answer The vector w contains the y values of the derivative approximation. The x values are the same as the x values of the points we wanted to 13/26 take a derivative of.     w 1 − 60 w 2 7 . 5     w 3 12 . 5         w 4 16     =     w 5 20         w 6 23 . 75         w 7 27 . 5     ◭◭ w 8 − 47 . 5 ◮◮ ◭ ◮ Back Close

Graph 14/26 300 (x,y) 250 y=2x (x,w) 200 150 100 50 0 −50 −100 5 10 15 x ◭◭ ◮◮ ◭ ◮ Back Close

Fourth Order Differentiation Matrix 15/26 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 x ◭◭ ◮◮ u i − 2 u i − 1 u i +1 u i +2 u 1 u i u n ◭ x 1 x i − 2 x i − 1 x i x i +1 x i +2 x n ◮ h Back Close

Lagrange Interpolating Polynomial 16/26   5 5 x − x l � � P ( x ) =  y k   x k − x l  k =1 l =1 l � = k       5 5 5 x − x l x − x l x − x l � � � P ( x ) =  y 1 +  y 2 +  y 3       x 1 − x l x 2 − x l x 3 − x l    l =1 l =1 l =1 l � =1 l � =2 l � =3     5 5 x − x l x − x l � � +  y 4 +  y 5     x 4 − x l x 5 − x l   ◭◭ l =1 l =1 l � =4 l � =5 ◮◮ ◭ ◮ Back Close

Fitting the Curve 1 17/26 0.5 0 −0.5 −1 0 1 2 3 4 5 6 x ◭◭ ◮◮ u i − 2 u i − 1 u i +1 u i +2 u 1 u i u n ◭ x 1 x i − 2 x i − 1 x i x i +1 x i +2 x n ◮ h Back Close

Fourth Order Polynomial 18/26 ( x − x j − 1 )( x − x j )( x − x j +1 )( x − x j +2 ) P j ( x ) = ( x j − 2 − x j − 1 )( x j − 2 − x j )( x j − 2 − x j +1 )( x j − 2 − x j +2 ) u j − 2 ( x − x j − 2 )( x − x j )( x − x j +1 )( x − x j +2 ) + ( x j − 1 − x j − 2 )( x j − 1 − x j )( x j − 1 − x j +1 )( x j − 1 − x j +2 ) u j − 1 ( x − x j − 2 )( x − x j − 1 )( x − x j +1 )( x − x j +2 ) + ( x j − x j − 2 )( x j − x j − 1 )( x j − x j +1 )( x j − x j +2 ) u j ( x − x j − 2 )( x − x j − 1 )( x − x j )( x − x j +2 ) + ( x j +1 − x j − 2 )( x j +1 − x j − 1 )( x j +1 − x j )( x j +1 − x j +2 ) u j +1 ( x − x j − 2 )( x − x j − 1 )( x − x j )( x − x j +1 ) + ( x j +2 − x j − 2 )( x j +2 − x j − 1 )( x j +2 − x j )( x j +2 − x j +1 ) u j +2 ◭◭ ◮◮ ◭ ◮ Back Close

Fourth Order Differentiation Equation 19/26 � 1 j ( x j ) = 1 12 u j − 2 − 2 3 u j − 1 + 2 3 u j +1 − 1 � P ′ 12 u j +2 h ◭◭ ◮◮ ◭ ◮ Back Close

System of Equations � 1 20/26 � w 1 = 1 12 u N − 1 − 2 3 u N + 2 3 u 2 − 1 12 u 3 h � 1 w 2 = 1 12 u N − 2 3 u 1 + 2 3 u 3 − 1 � 12 u 4 h � 1 � w 3 = 1 12 u 1 − 2 3 u 2 + 2 3 u 4 − 1 12 u 5 h . . . � 1 � w N − 1 = 1 12 u N − 3 − 2 3 u N − 2 + 2 3 u N − 1 12 u 1 h � 1 ◭◭ w N = 1 12 u N − 2 − 2 3 u N − 1 + 2 3 u 1 − 1 � 12 u 2 ◮◮ h ◭ ◮ Back Close

Fourth Order Differentiation Matrix 21/26 ...   1 / 12 − 2 / 3 w 1 u 1     ... − 1 / 12 1 / 12 w 2 u 2       ... ...   w 3 u 3 2 / 3       = 1 . .     ... ... . .   . . 0       h  .   .  ... ... . .   . . − 2 / 3           ...   w N − 1 u N − 1 − 1 / 12 1 / 12       ... w N u N 2 / 3 − 1 / 12 ◭◭ ◮◮ ◭ ◮ Back Close

Differentiation Using the Fourth Order Matrix 22/26 (2 , 2 . 5), (4 , . 5), (6 , . 75), (8 , 2 . 7), (10 , . 6), (12 , . 6), (14 , 2 . 7), (16 , . 75) ...     w 1 2 . 5   1 / 12 − 2 / 3 w 2 ... − 1 / 12 . 5 1 / 12       w 3 . 75     ... ...   2 / 3       = 1 w 4 2 . 7     ... ...   0       w 5 . 6 2     ... ...   − 2 / 3     w 6   . 6     ...   − 1 / 12 1 / 12     w 7 2 . 7       ... 2 / 3 − 1 / 12 w 8 . 75 ◭◭ ◮◮ ◭ ◮ Back Close