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Introduction to Numerical Solutions of Ordinary Differential Equations

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1 Introduction to Numerical Solutions of Ordinary Differential Equations
Larry Caretto Mechanical Engineering 501AB Seminar in Engineering Analysis October 29, 2003

2 Outline Review last class and homework
Midterm Exam November 4 covers material up to and including last week’s lecture Overivew of numerical solutions Initial value problems in first-order equations Systems of first order equations and initial value problems in higher order equations Boundary value problems Stiff systems and eigenvalues

3 Review Last Class Systems of equations Laplace transforms for ODEs
Combine to higher order equation Matrix approaches Reduction of order Laplace transforms for ODEs Transform ODE from y(t) to Y(s) Solve for Y(s) Rearrange solution Transform back to y(t)

4 Simultaneous Solution
Differentiate first (y1) equation to obtain derivative(s) of y2 that are present in second equation Solve first equation for y2 Substitute equations for y2 and its deriv-atives into the result of the first step Result is a higher order equation that can be solved (if possible) by usual methods

5 Matrix Differential Equations II
Matrix components in

6 Solving Assume that A(n x n) has n linearly independent eigenvectors
Eigenvectors are columns of matrix, X Define a new vector s = X-1y (y = Xs) Substitute y = Xs into the matrix differential equation Obtain equation for s using L = X-1AX that gives independent equations

7 Terms in Solution of With these definitions, si = Cielit + pi/li becomes s = EC + L-1p (E(0) = I)

8 Apply Initial Conditions on y(0)
Get y = Xs = XEC + XL-1p At t = 0, E = I, and y = y0 = initial y components, giving y0 = XIC + XL-1p Premultiply by X-1 to obtain X-1y0 = X-1XC + X-1XL-1p = C + L-1p Constant vector, C = X-1y0 - L-1p Result: y = XE [X-1y0 - L-1p] + XL-1p Homogenous(p = 0): y = XEX-1y0

9 Reduction of Order An nth order equation can be written as a system of n first order equations We can write a general nonlinear nth order equation as shown below Redefine y as z0 and define the following derivatives

10 Reduction of Order II Continue this definition up to zn-1
z0 to zn-1 are n variables that satisfy simultaneous, first-order ODEs

11 Reduction of Order III The main application of reduction of order is to numerical methods Numerical solutions of ODEs develop methods to solve a system of first order equations Higher order equations are solved by converting them to a system of first order equations

12 Simple Laplace Transforms
f(t) F(s) ctn cn!/sn+1 ceatsin wt ctx cG(x+1)/sx+1 ceat c/(s – a) ceatcos wt csin wt cw/(s2 + w2) ccos wt cs/(s2 + w2) Additional transforms in Table 5.9, pp of Kreyszig csinh wt cw/(s2 - w2) ccosh wt cs/(s2 - w2)

13 Transforms of Derivatives
The main formulas for differential equations are the Laplace transforms of derivatives These have transform of desired function, L [f(t)] = F(s) and the initial conditions, f(0), f’(0), etc.

14 Solving Differential Equations
Transform all terms in the differential equation to get an algebraic equation For a differential equation in y(t) we get the transforms Y(s) = L [y(t)] Similar notation for other transformed functions in the equation R(s) = L [r(t)] Solve the algebraic equation for Y(s) Obtain the inverse transform for Y(s) from tables to get y(t)

15 Partial Fractions Method to convert fraction with several factors in denominator into sum of individual factors (in denominator) Necessary to modify complicated solution for Y(s) into simpler functions whose transforms are known Special rules for repeated factors and for complex conjugates

16 Shifting Theorems Sometimes required for transform table
First theorem is defined for function f(t) with transform F(s) F(s - a) has inverse of e-atf(t) Example Y(s) = [2(s+1)+4]/[(s + 1)2 + 1] For F(s) = s/(s2 + w2); f(t) = cos wt and f(t) = cos wt for F(s) = w/(s2 + w2) Here Y(s+1) = 2(s+1)+4]/[(s + 1)2 + 1] + 4/[(s + 1)2 + 1] so y(t) = e-t times Y(s) inverse y(t) = e-t[2 cos t + 4 sin t]

17 Shifting Theorems II Second shifting theorem
Applies to e-asF(s), where F(s) is known transform of a function f(t) Inverse transform is f(t – a) u(t – a) where u(t – a) is the unit step function If we have e-ass/(s2 + w2), we recognize s/(s2 + w2) as F(s) for f(t) = cos wt Thus e-ass/(s2 + w2) is the Laplace transform for cos[wt(t – a)] u(t – a)

18 Numerical Analysis Problems
Numerical solution of algebraic equations and eigenvalue problems Solution of one or more nonlinear algebraic equations f(x) = 0 Linear and nonlinear optimization Constructing interpolating polynomials Numerical quadrature Numerical differentiation Numerical differential equations

19 Interpolation Start with N data pairs xi, yi
Find a function (polynomial) that can be used for interpolation Basic rule: the interpolation polynomial must fit all points exactly Denote the polynomial as p(x) The basic rule is that p(xi) = yi Many different forms

20 Newton Polynomials p(x) = a0 + a1(x – x0) + a2(x – x0)(x – x1) + a3(x – x0)(x – x1)(x – x2) + … + an-1(x – x0)(x – x1)(x – x2) … (x – xn-2) Terms with factors of x – xi are zero when x = xi Use this and rule that p(xi) = yi to find ai a0 = y0, a1 = (y1 – y0) / (x1 – x0) y2 = a0 + a1(x2 – x0) + a2(x2 – x0)(x2 – x1) Solve for a2 using results for a0 and a1

21 Newton Polynomials II y2 = a0 + a1(x2 – x0) + a2(x2 – x0)(x2 – x1)
Data determine coefficients Develop scheme known as divided difference table to compute ak

22 Divided Difference Table
x0 y0 a0 a1 x1 y1 a2 x2 y2  a3 x3 y3

23 Divided Difference Example
a0 a1 10 a2 20 40  a3 30 100

24 Divided Difference Example II
Divided difference table gives a0 = 0, a1 = 1, a2 = .1, and a3 = 1/600 Polynomial p(x) = a0 + a1(x – x0) + a2(x – x0)(x – x1) + a3(x – x0)(x – x1)(x – x2) = 0 + 1(x – 0) + 0.1(x – 0)(x – 10) + (1/600)(x – 0)(x – 10)(x – 20) = x + 0.1x(x – 10) + (1/600)x(x – 10)(x – 20) Check p(30) = (30)(20) + (1/600) (30)(20)(10) = = 100 (correct)

25 Constant Step Size Divided differences work for equal or unequal step size in x If Dx = h is a constant we have simpler results Fk = Dyk/h = (yk+1 – yk)/h Sk = D2yk/h2 = (yk+2 – 2yk-1 + yk)/h2 Tk = D3yk/h3 = (yk+3 – 3yk yk+1 – yk)/h3 Dnyk is called the nth forward difference Can also define backwards and central differences

26 Interpolation Approaches
When we have N data points how do we interpolate among them? Order N-1 polynomial not good choice Use piecewise polynomials of lower order (linear or quadratic) Can match first and or higher derivatives where piecewise polynomials join Cubic splines are piecewise cubic polynomials that match first and second derivatives (as well as values)

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29 Polynomial Applications
Data interpolation Approximation functions in numerical quadrature and solution of ODEs Basis functions for finite element methods Can obtain equations for numerical differentiation Statistical curve fitting (not discussed here) usually used in practice

30 Derivative Expressions
Obtain from differentiating interpolation polynomials or from Taylor series Series expansion for f(x) about x = a Note: d0f/dx0 = f and 0! = 1 What is error from truncating series?

31 Truncation Error If we truncate series after m terms
Terms used Truncation error, em Can write truncation error as single term at unknown location (derivation based on the theorem of the mean)

32 Derivative Expressions
Look at finite-difference grid with equal spacing: h = Dx so xi = x0 + ih Taylor series about x = xi gives f(xi + kh) = f[x0 + (i+k)h] = fi+k in terms of f(xi) = fi Compact derivative notation

33 Derivative Expressions II
Combine all definitions for compact series notation Use this formula to get expansions for various grid locations about x = xi and use results to get derivative expressions

34 Derivative Expressions III
Apply general equation for k = 1 and k = –1 Forward Backward

35 Derivative Expressions IV
Subtract fi+1 and fi-1 expressions Result called central difference expression

36 Order of the Error Forward and backward derivative have error term that is proportional to h Central difference error is proportional to h2 Error proportional to hn called nth order Reducing step size by a factor of a reduces nth order error by an

37 Order of the Error Notation
Write the error term for nth error term as O(hn) Big oh notation, O, denotes order Recognizes that factor multiplying hn may change slightly with h First order forward First order backward Second order central

38 Higher Order Derivatives
Add fi+1 and fi-1 expressions f’’ is second-order, central difference

39 Higher Order Directional
We can get higher order derivative expressions at the expense of more computations Get second order forward and backward derivative expressions from fi+2 and fi-2, respectively Combine with previous expressions for fi+1 and fi-1 to eliminate first order error term

40 Specific Taylor Series
General equation k = 2 k = -2 k = 3 k = -3

41 Second Order Forward Subtract 4fi+1 from fi+2 to eliminate h2 term
Second order error

42 Second Order Backwards
Add 4fi-1 to –fi-2 to eliminate h2 term Second order error

43 Other Derivative Expressions
Can continue in this fashion Write Taylor series for fi+1, fi-1, fi+2, fi-2, fi+3, fi-3, etc. Create linear combinations with factors that eliminate desired terms Eliminate fi term to obtain central difference Keep only terms in fk with k  i for forward difference expressions Keep only terms in fk with k  i for forward difference expressions See results on page 271 of Hoffman

44 Order of Error Examples
Table 2-1 in notes shows error in first derivative for ex around x = 1 Using first- and second-order forward and second-order central differences Step h = 0.4, 0.2, and 0.1 Error ratio for doubling step size 4.01 to 4.02 for central differences 2.07 to 2.15 for first-order forward differences 4.32 to 4.69 for second-order forward

45 Roundoff Error Possible in derivative expressions from subtracting close differences Example f(x) = ex: f’(x)  (ex+h – ex-h)/(2h) and error at x = 1 is (e1+h – e1-h)/(2h) - e Second order error

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47 Numerical ODE Solutions
The initial value problem Euler method as a prototype for the general algorithm Local and global errors More accurate methods Step-size control for error control Applications to systems of equations Higher-order equations

48 The Initial Value Problem
dy/dx = f(x,y) (known) with y(x0) = y0 Basic numerical approach Use a finite difference grid: xi+1 – xi = h Replace derivative by finite-difference approximation: dy/dx  (yi+1 – yi) / (xi+1 – xi) = (yi+1 – yi) / h Derive a formula to compute favg the average value of f(x,y) between xi and xi+1 Replace dy/dx = f(x,y) by (yi+1 – yi) / h = favg Repeatedly compute yi+1 = yi + h favg

49 Notation xi is the value of the independent variable at point i on the grid Determined from the user-selected value of step size (or the series of hi values) Can always specify exactly the independent variable’s value, xi yi is the value of the numerical solution at the point where x = xi fi is derivative value found from xi and the numerical value, yi. I.e., fi = f(xi, yi).

50 More Notation y(xi) is the exact value of y at x = xi
Usually not known but notation is used in error analysis of algorithms f(xi,y(xi)) is the exact value of the derivative at x = xi e1 = y(x1) – y1 is the local truncation error This is error for one step of algorithm starting from known initial condition

51 Local versus Global Error
At the initial condition we know the solution, y, exactly First step introduces some error Remaining steps have single step error plus previous accumulated error Ej = y(xj) – yj is global truncation error Difference between numerical and exact solution after several steps This is the error we want to control

52 Euler’s Method Simplest algorithm, example used for error analysis, not for practical use Define favg = f(xi, yi) Euler’s method algorithm is yi+1 = yi + hifi = yi + hi f(xi, yi) Example dy/dx = x + y, y = 0 at x = 0 Choose h = 0.1 We have x0 = 0, y0 = 0, f0 = x0 + y0 = 0 x1 = x0 + h = 0.1, y1 = y0 + hf0 = 0

53 Euler Example Continued
Next step is from x1 = 0.1 to x2 = 0.2 f1 = x1 + y1 = = 0.1 y2 = y1 + hf1 = 0 + (.1)(.1) = .01 For dy/dx = x + y, we know the exact solution is (x0 + y0 + 1)ex-x0 – x – 1 For x0 = y0 = 0, y = ex – x – 1 Look at application of Euler algorithm for a few steps and compute the error

54 Euler Example xi yi fi y(xi) E(xi) .1 .0052 .2 .01 .21 .0214 .0114 .3
.1 .0052 .2 .01 .21 .0214 .0114 .3 .031 .331 .04986 .01886 .4 .0641 .4641

55 Euler Example Plot

56 Error Propagation Behavior of Euler algorithm is typical of all algorithms for numerical solutions Error grows at each step We usually do not know this global error, but we would like to control it Look at local error for Euler algorithm Then discuss general relationship between local and global error

57 Taylor Series to Get Error
Expand y(x) in Taylor series about x = a Look at one step from known initial condition, a = x0, to x0 + h so x – a = h In ODE notation, dy/dx|0 = f(x0, y(x0))

58 Local Euler Error Result of Taylor series on last chart
Euler Algorithm Truncation Error This is only the Euler algorithm for the first step when we know f(x0,y(x0)) This gives the local truncation error Local truncation error for Euler algorithm is second order

59 Global Error We will show that a local error of order n, has a global error of order n-1 To show this consider the global error at x = x0 + kh after k algorithm steps Is approximately k times the local error If local error is O(hn), approximate global error after k steps is k O(hn)  kAhn A new step size, h/r, takes kr steps to get to the same x value

60 Global Error Concluded
Compare error for same x = kh with step sizes h and h/r Ex=kh(h)  kAhn Ex=kh(h/r) = krA(h/r)n When we reduce the step size by a factor of 1/r we reduce the error by a factor of 1/rn-1; this is the behavior of an algorithm whose error is order n-1

61 Euler Local and Global Error
Previously showed Euler algorithm to have second order local error Should have first order global error Results for previous Euler example at x = 1 with different step sizes Step size First step Final error h = 0.1 5.17x10-3 1.25 x10-1 h = 0.01 5.02 x10-5 1.35 x10-2 h = 0.001 5.00 x10-7 1.36 x10-3

62 Better Algorithms Seek high accuracy with low computational work
Could improve Euler accuracy by cutting step size, but this is not efficient Use other algorithms that have higher order errors Runge-Kutta methods typically used This is a class of methods that use several function evaluation methods per step

63 Second-order Runge Kutta
Huen’s method Modified Euler method

64 Fourth-order Runge Kutta
Uses four derivative evaluations per step

65 Comparison of Methods Look at Euler, Heun, Modified Euler and fourth-order Runge-Kutta Solve dy/dx = e-y-x with y(0) = 1 Compare numerical values to exact solution y = ln( ey0 + e-x0 – e-x) Look at errors in the methods at x = 1 as a function of step size Compare error propagation (increase in error as x increases)

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68 Some Basic Concepts A numerical method is convergent with the solution of the ODE if the numerical solution approaches the actual solution as h  0 (with increase in numerical precision at smaller h) Mainly a theoretical concept; algorithms in use are convergent Stability refers to the ability of a numerical algorithm to damp any errors introduced during the solution

69 More on Stability Finite difference equations in numerical algorithms, when iterated, may be numerically increase without bound Stability usually is obtained by keeping step size h small, sometimes smaller than the h required for accuracy For most ODEs stability is not a problem, but it is for stiff systems of ODEs and for partial differential equations

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72 Error Control How do we choose h?
Want to obtain a result with some desired small global error Can just repeat calculations with smaller h until two results are sufficiently close Can use algorithms that estimate error and adjust step size during the calculation based on the error

73 Runge-Kutta-Fehlberg
Uses two equations to compute yn+1, one has O(h5), the other O(h6) error Requires six derivative evaluations per step (same evaluations used for both equations) The error estimate can be used for step size control based on an overall 5th order error Cask-Karp version different coefficients

74 Runge-Kutta-Fehlberg II
See page 377 in Hoffman for algorithm Typical formula components below yn+1= yn + (16k1/ k2/12825 … k3 = hf(xn + 3h/8, yn + 3k1/32 + 9k2/32) Error = k1/ k3/4275 … hnew = hold|EDesired/Error|1/5 EDesired is set by user RKF45 code by Watts and Shampine

75 Other Error Controls Do fourth-order Runge-Kutta two times, with step sizes different by a factor of two and compare results Runge-Kutta-Verner is similar to Runge-Kutta-Fehlberg, but uses higher order expressions

76 Systems of Equations Any problem with one or more ODEs of any order can be reduced to a system for first order ODEs Last week we reduced a system of two second order equations to a system of four first order equations In this process the sum of the orders is constant Example from last week

77 Reduction of Order Example
Define variables y3 = dy1/dt and y4 = dy2/dt Then dy3/dt = d2y1/dt2 and dy4/dt = d2y2/dt2 Have four simultaneous first-order ODEs

78 Solving Simultaneous ODEs
Apply same algorithms used for single ODEs Must apply each step and substep to all equations in system Key is consistent determination of fi(x,y) All yi values in y must be available at the same x point when computing the fi E.g., in Runge-Kutta we must evaluate k1 for all equations before finding k2

79 Runge-Kutta for ODE System
– y(i) is vector of dependent variables at x = xi k(1), k(2), k(3), and k(4), are vectors containing intermediate Runge-Kutta results f is a vector containing the derivatives k(1) = hf(xi, y(i)) k(2) = hf(xi + h/2, y(i) + k(1)/2) k(3) = hf(xi + h/2, y(i) + k(2)/2) k(4) = hf(xi + h, y(i) + k(3)) y(i+1) = (k(1) + 2k(2) + 2k(3) + k(4))/6

80 ODE System by RK4 dy/dx = -y + z and dz/dx = y – z with y(0) = 1 and z(0) = -1 with h = .1 k(1)y = h[-y + z] = 0.1[-1 + (-1)] = -.2 k(1)z = h[y - z] = 0.1[1 - (-1)] = .2 k(2)y = h[-(y+ k(1)y/2) + z + k(1)z/2] = 0.1[ -( /2) + ( /2)] = -.18 k(2)z = h[(y+ k(1)y/2) –(z + k(1)z/2)] = 0.1[( )/2 - ( /2)] = .18

81 ODE System by RK4 II k(3)y = h[-(y+ k(2)y/2) + z + k(2)z/2] = 0.1[ -( /2) + ( /2)] = -.182 k(3)z = h[(y+ k(2)y/2) –(z + k(2)z/2)] = 0.1[( )/2 - ( /2)] = .182 k(4)y = h[-(y+ k(3)y) + z + k(3)z] = 0.1[ -( ) + ( )] = k(4)z = h[(y+ k(3)y) –(z + k(3)z)] = 0.1[ ( ) - ( )] = .1636

82 ODE System by RK4 III yi+1 = yi + (k(1)y + 2k(2)y + 2k(3)y + k(4)y )/6 = [ (–.2) + 2(–.18) + 2(–.182) + (–.1636)]/6 = .8187 zi+1 = zi + (k(1)z + 2k(2)z + 2k(3)z + k(4)z )/6 = –1 + [(.2) + 2(.18) + 2(.182) + (.1636)]/6 = –.8187 Continue in this fashion until desired final x value is reached No x dependence for f in this example

83 Numerical Software for ODEs
Usually written to solve a system of N equations, but will work for N = 1 User has to code a subroutine or function to compute the f array Input variables are x and y; f is output Some codes have one dimensional parameter array to pass additional information from main program into the function that computes derivatives

84 Derivative Subroutine Example
Visual Basic code for system of ODEs at right is shown below Sub fsub( x as Double, y() as Double, _ f() as Double ) f(1) = -y(1) + Sqr(y(2)) + y(3)*Exp(2*x) f(2) = -2 * y(1)^2 f(3) = -3 * y(1) * y(2) End Sub

85 Derivative Subroutine Example
Fortran code for system of ODEs at right is shown below subroutine fsub( x, y, f ) real(KIND=8) x, y(*), f(*) f(1) = -y(1) + sqrt(y(2)) +y(3)*exp(2*x) f(2) = -2 * y(1)**2 f(3) = -3 * y(1) * y(2) end subroutine fsub

86 Derivative Subroutine Example
C++ code for system of ODEs at right is shown below void fsub(double x, double y[], double f[]) { f[1] = -y[1] + sqrt(y[2]) + y[3]*exp(2*x); f[2] = -2 * y[1] * y[1]; f[3] = -3 * y[1] * y[2]; }


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