Interpolation (Shape Functions)

Size: px

Start display at page:

Download "Interpolation (Shape Functions)"

Erik Bryan
5 years ago
Views:

1 Mètodes Numèrics: A First Course on Finite Elements Interpolation (Shape Functions) Following: Curs d Elements Finits amb Aplicacions (J. Masdemont) Dept. Matemàtiques ETSEIB - UPC BarcelonaTech

2 Interpolation Problem The general solution of problem using Finite Elements has the following steps: Finite Elements Stp1: Discretize in elements Stp2: Write the variational equations Stp3: Build the Linear System Impose the BC Stp4: Get nodes solution Stp5: Extent solution to the Domain

3 Interpolation Problem 1D Let s assume that we have measured an unknown magnitude u = u(x) on a set of data points x = [x 1, x 2,., x N ] obtaining a set of values u = [u 1, u 2,., u N ]

4 Interpolation Problem 1D Let s assume that we have measured an unknown magnitude u = u(x) on a set of data points x = [x 1, x 2,., x N ] obtaining a set of values u = [u 1, u 2,., u N ] How can we guess the value of a new point?

5 Interpolation Problem 1D Let s assume that we have measured an unknown magnitude u = u(x) on a set of data points x = [x 1, x 2,., x N ] obtaining a set of values u = [u 1, u 2,., u N ] Actual function f x = x 3 3x 2 3sin(5x)

6 Interpolation Problem 1D To Interpolate a set of data measured points (x i, u i ) means to build a function P n (x) (usually a polynomial) passing through these points. That is u i = P n x i, i = 1 N Theorem: The interpolation problem has a unique solution when n = N 1. Idea of the proof: P n x = a 0 + a 1 x + a 2 x a n x n n + 1 N unknowns u i = P n x i, i = 1 N N linear equations Unique solution if the system is compatible determined. That means the interpolation polynomial has degree one less than the number of points

7 Shape Functions 1D

8 Shape Functions 1D Lagrange Polynomials (Shape functions) One (N-1) degree Polynomial for each measure point x i, i = 1,, N ψ i x = x x 1 x x 2.. x xi. x x N 1 x x N x i x 1 x i x 2.. xi x i. x i x N 1 x i x N Properties: 1.- Lagrange Polynomials are a base of the set P n [x] 2.- ψ i x j = 0 if i j 3.- ψ i x i = 1 Interpolation Polynomial (a linear combination of the shape functions) P n x = u 1 ψ 1 x + u 2 ψ 2 x + + u N ψ N x

9 Shape Functions 1D N = 2 only two points: x = x 1, x 2 and their measures u = u 1, u 2 ψ 1 x = x x 2 x 1 x 2 x = 1, 2 It is a straight line

10 Shape Functions 1D N = 2 only two points: x = x 1, x 2 and their measures u = u 1, u 2 ψ 1 x = x x 2 x 1 x 2 ψ 2 x = x x 1 x 2 x 1 x = 1, 2

11 Shape Functions 1D N = 2 only two points: x = x 1, x 2 and their measures u = u 1, u 2 ψ 1 x = x x 2 x 1 x 2 ψ 2 x = x x 1 x 2 x 1 u 2 P 1 x Interpolation Polynomial P 1 x = u 1 ψ 1 x + u 2 ψ 2 x u It is also a straight line (linear interpolation)

12 Shape Functions 1D N = 2 only two points: x = x 1, x 2 and their measures u = u 1, u 2 ψ 1 x = x x 2 x 1 x 2 ψ 2 x = x x 1 x 2 x 1 Interpolation Polynomial P 1 x = u 1 ψ 1 x + u 2 ψ 2 x Example: In the case x = 1, 2 and u = 0.7, 1.8 ψ 1 x = x 2 = (x 2) 1 2 ψ 2 x = x 1 = (x 1) 2 1 How to approximate the value for x = 1.5? P = 0.7ψ ψ = = 1. 25

13 Shape Functions 1D N = 3 three points: x = x 1, x 2, x 3 and measures u = u 1, u 2, u 3 ψ 1 x = x x 2 x x 3 x 1 x 2 x 1 x 3 ψ 2 x = x x 1 x x 3 x 2 x 1 x 2 x 3 ψ 3 x = x x 1 x x 2 x 3 x 1 x 3 x 2

14 Shape Functions 1D N = 3 three points: x = x 1, x 2, x 3 and measures u = u 1, u 2, u 3 ψ 1 x = x x 2 x x 3 x 1 x 2 x 1 x 3 ψ 2 x = x x 1 x x 3 x 2 x 1 x 2 x 3 ψ 3 x = x x 1 x x 2 x 3 x 1 x 3 x 2 Interpolation Polynomial P 2 x = u 1 ψ 1 x + u 2 ψ 2 x + u 3 ψ 3 x Now it is a parabola (quadratic interpolation)

15 Shape Functions 1D Example: Consider x = 1,0,1 and u = 1.2, 0.7, 1.8. Compute u(0.2)? x 0 x 1 ψ 1 x = = 1 x(x 1) 2 x + 1 x 1 ψ 2 x = = (x + 1) (x 1) x + 1 x 0 ψ 3 x = = 1 x(x + 1) 2 P = 1.2ψ ψ ψ =

16 Shape Functions 1D Matlab code: x = 0:0.25:1; % measure points f x.^3-3*x.^2-3*sin(5*x); %inline exemple function y = f(x); %measured values figure() plot(x,y,'o','marker','o','markerfacecolor','red'); title('measured values'); % plot only measured values figure() plot(x,y,'o','marker','o','markerfacecolor','red'); hold on; xx = 0:0.01:1; % take many points for a better plot yy = f(xx); plot(xx, yy,'-.') title('true function'); % plot the function hold off

17 Shape Functions 1D Matlab code (continuation): figure() % Polyfit degree 4 function (interpolator) plot(x,y,'o','marker','o','markerfacecolor','red'); hold on; plot(xx, yy, '-.') p = polyfit(x,y,4); yyy = polyval(p, xx); plot(xx, yyy, '-.b') title('degree 4 approximation'); hold off

18 Shape Functions 1D Interpolation Results:

19 Shape Functions 1D Approximation: In case n < (N 1) there is NO solution (incompatible system) because the polynomial can not pass through all the points. In that case: mean square approx P n x = min( P n x f x )

20 Shape Functions 1D Higher interpolation: When N is big (>8) there is a better solution than using a polynomial of high degree. Runge s Phenomenon: f x = x 2 The red curve is the Runge function f x. The blue curve is a 5th-order interpolating polynomial (using six equally spaced interpolating points). The green curve is a 9th-order interpolating polynomial (using ten equally spaced interpolating points).

21 Shape Functions 1D Two possible approaches: Polygonal curve: Take the points in pairs and build a line segment (1 st order interp) in each case. Continuous but not derivable. Matlab code: %% Polygonal figure() x=0:0.1:1; y=f(x); plot(x,y,'o','marker','o','markerfacecolor','red'); hold on; plot(xx,yy,'-.') xxx=0:0.01:1; yyy=interp1(x,y,xx); %only substitution is possible plot(xxx,yyy,'b') hold off

22 Shape Functions 1D Two possible approaches: Spline curve (cubic): Take the points in groups of 2 and build a third order Polynomial (cubic interp) in each case imposing C 2 continuity. Matlab code (continuation): %% Spline figure() x=0:0.1:1; y=f(x); plot(x,y,'o','marker','o','markerfacecolor','red'); hold on; plot(xx,yy,'-.') xxx = 0:0.01:1; yyy = spline(x,y,xx); %only substitution is possible plot(xxx,yyy,'b') hold off

23 Shape Functions 2D

24 Shape Functions 2D How can we extend the interpolation problem to 2D? That is, how can we approximate the value of f x, y using a 2D polynomial P n x, y? We restrict to fixed shapes: Triangular element Numbered counterclockwise 3 e 1 2 Quadrilateral element

25 Triangular Shape Functions We need the equivalent to the Lagrange polynomials (2-dim) ψ i e x, y = c 1 + c 2 x + c 3 y (three unknowns) such that, it s value is 1 for vertex i, and 0 for the other two: For the first vertex: 1 = ψ 1 e x 1, y 1 = c 1 + c 2 x 1 + c 3 y 1 0 = ψ 1 e x 2, y 2 = c 1 + c 2 x 2 + c 3 y 2 0 = ψ 1 e x 3, y 3 = c 1 + c 2 x 3 + c 3 y 3 (x 3, y 3 ) (x 1, y 1 ) (x 2, y 2 ) 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 A c b c 1 c 2 c 3 = c = A\b (matlab solution)

26 Triangular Shape Functions Explicitly ψ i e x, y = c 1,i + c 2,i x + c 3,i y, i = 1,2,3 c 1,i = x jy k x k y j 2A e c 2,i = y j y k 2A e c 3,i = x k x j 2A e where i, j, k = 1,2,3 cyclic and A e is the Area of Ω e Remember: Area= 1 2 det 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 (x 3, y 3 ) (x 1, y 1 ) (x 2, y 2 )

27 Triangular Shape Functions Once we have the shape functions ψ i e x, y, i = 1,2,3 (geometrically they are planes passing through one of the edges) finally, the interpolation value of a point ( x, y) is f x, y = f x 1, y 1 ψ 1 e ( x, y) + f x 2, y 2 ψ 2 e ( x, y) + f x 3, y 3 ψ 3 e ( x, y) The terms α i e = ψ i e ( x, y) are known as Barycentric Coordinates of ( x, y) (We ll see more in the practical sessions)

28 Triangular Shape Functions Example: The temperature of a thin triangular plate has been measured at the vertices T v 1 = 40 o, T v 2 = 90 o, T v 3 = 10 o. If the coordinates of the vertices are: v 1 = 0,0, v 2 = 5,0, v 3 = 2,3, estimate the temperature at the point p = (3,1)? Solution: ψ 1 e x, y = 1 0.2x 0.2y ψ 2 e x, y = 0 0.2x y ψ 3 e x, y = 0 0x y T p = 40α α α 3 = o Exercise: Compute the shape functions associated to the triangle v 1 = 1,1, v 2 = 3,2, v 3 = 2,4. Solution: ψ e 1 x, y = 1 (8 2x y) 5 ψ e 2 x, y = 1 ( 2 + 3x y) 5 α 1 = ψ e 1 3,1 α 2 = ψ e 2 3,1 α 3 = ψ e 3 3,1 ψ e 3 x, y = 1 ( 1 x + 2y) 5

29 Quadrilateral Shape Functions A similar situation is found when a rectangular element is used. ψ e i x, y = c 1 + c 2 x + c 3 y + c 4 xy (four unknowns) such that, it s value is 1 for vertex i, and 0 for the other three: For the first vertex: 1 = ψ e 1 x 1, y 1 = c 1 + c 2 x 1 + c 3 y 1 +c 4 x 1 y 1 0 = ψ e 1 x 2, y 2 = c 1 + c 2 x 2 + c 3 y 2 +c 4 x 2 y 2 0 = ψ e 1 x 3, y 3 = c 1 + c 2 x 3 + c 3 y 3 +c 4 x 3 y 3 0 = ψ e 1 x 4, y 4 = c 1 + c 2 x 4 + c 3 y 4 +c 4 x 4 y 4 (x 4, y 4 ) (x 3, y 3 ) Analogously for the other vertices (This is not correct for general quadrilateral elements) (x 1, y 1 ) (x 2, y 2 )

30 Quadrilateral Shape Functions General expressions for Rectangles: If we denote by a = x 2 x 1 b = y 2 y 1 Then ψ 1 e x, y = 1 ab (x x 2)(y y 2 ) (x (x 2, y 2 ) 1, y 2 ) b a (x 1, y 1 ) (x 2, y 1 ) ψ 2 e x, y = 1 ab (x x 1)(y y 2 ) ψ 3 e x, y = 1 ab (x x 1)(y y 1 ) ψ 4 e x, y = 1 ab (x x 2)(y y 1 )

31 Quadrilateral Shape Functions The reference bilinear quadrilateral element Ω R : [-1,1]x[-1,1] We can use matlab to compute the coeff. ψ j e x i, y i = c 1j + c 2j x i + c 3j y i +c 4j x i y i = δ ij punts=[ -1,-1; 1,-1; 1,1; -1,1]; A=[ones(4,1), punts(:,1), punts(:,2), punts(:,1).*punts(:,2)] b=[1;0;0;0]; C1= A\b %coeff of the fisrt shape function % exercise: do it for the other shape functions

32 Quadrilateral Shape Functions The shape functions for the reference element Ω R, are: Or equivalently:

33 Quadrilateral Shape Functions Example: The temperature of at the vertices of the reference quadrilateral has been measured as T v 1 = 10 o, T v 2 = 20 o, T v 3 = 30 o, T v 4 = 40 o. Estimate the temperature at the point p = ( 0.3,0.5)? Solution: ψ 1 R p =0.1625; ψ 2 R p =0.0875; ψ 3 R p =0.2625; ψ 4 R p = ; T p = T 1 ψ 1 R p + T 2 ψ 2 R p + T 3 ψ 3 R p + T 4 ψ 4 R p = o

34 Quadrilateral Shape Functions For a General Quadrilateral element usually the shape function is NOT a 2 degree polynomial. Because of that it is not easy to compute these functions, but we still can compute the Barycentric Coordinates. How to compute α i k = ψ i k (P)?

35 Quadrilateral Shape Functions Change from the Reference quadrilateral to [0,1]x[0,1] λ = ξ + 1 2, μ = η (-1,1) (1,1) (0,1) (1,1) η ξ (-1,-1) (-1,1) μ λ (0,0) (1,0) Shape functions on [0,1]x[0,1]: ψ 1 λ, μ = (1 λ)(1 μ) ψ 2 λ, μ = λ(1 μ) ψ 3 λ, μ = λμ ψ 4 λ, μ = (1 λ)μ

36 Quadrilateral Shape Functions General Quadrilateral: Isoparametric transformation (x, y) = ψ 1 ξ, η v 1 + ψ 2 ξ, η v 2 + ψ 3 ξ, η v 3 + ψ 4 ξ, η v 4 Change to another quadrilateral using the shape functions, for simplicity we will use here the rectangle [0,1]x[0,1]: (0,1) (1,1) P(λ, μ) v 4 μ = 1 P(λ, μ) v 3 λ = 1 μ λ λ = 0 (0,0) (1,0) v 2 v 1 μ = 0 P(λ, μ) = (1 λ)(1 μ) v 1 + λ(1 μ) v 2 + λμ v λ μ v 4 α 1 α 2 α 3 α 4

37 Quadrilateral Shape Functions We can compute them using λ, μ ε 0,1, as a parametrization of the quadrilateral edges. P = (1 λ)(1 μ) v 1 + λ(1 μ) v 2 + λμ v λ μ v 4 Rearranging the terms, the previous equation can be written as: a + λb + μc + λμd = 0 where a = v 1 P, b = v 2 v 1, c = v 4 v 1, d = v 1 v 2 + v 3 v 4

38 Quadrilateral Shape Functions We impose that the vector w = Q λ, 1 Q λ, 0 and the vector u = P Q λ, 0 are parallel, i.e. w u = 0 λ = 0 From the isometric transformation we obtain: w = v 4 v 1 + λ v 3 v 4 + v 1 v 2 = c + λd u = P v 1 + λ v 1 v 2 = (a + λb) v 1 v 4 μ = 1 μ = 0 Q(λ, 1) w v 3 P(λ, μ) u Q(λ, 0) λ = 1 v 2

39 Quadrilateral Shape Functions If this two vectors are parallel a + λb c + λd = 0 or b d λ 2 + a d + b c λ + a c = 0 which is a quadratic equation that can be solved for λ. Hint: All the 2D vectors are extended with a zero third component in order to define the cross product. Analogously a similar equation is found for μ.

40 Quadrilateral Shape Functions Example: Given a quadrilateral defined by vertices v 1 = 0,0, v 2 = 5, 1, v 3 = 4,5, v 4 = (1,4) compute the barycentric coordinates of point p=(3,2). Sol. λ = , μ = 0.5 α = (examples using Matlab will be shown in practices)

Curve Fitting and Interpolation

Chapter 5 Curve Fitting and Interpolation 5.1 Basic Concepts Consider a set of (x, y) data pairs (points) collected during an experiment, Curve fitting: is a procedure to develop or evaluate mathematical