# Lecture Notes to Accompany. Scientific Computing An Introductory Survey. by Michael T. Heath. Chapter 10

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1 Lecture Notes to Accompany Scientific Computing An Introductory Survey Second Edition by Michael T. Heath Chapter 10 Boundary Value Problems for Ordinary Differential Equations Copyright c Reproduction permitted only for noncommercial, educational use in conjunction with the book. 1

2 Boundary Value Problems Side conditions prescribing solution or derivative values at specified points required to make solution of ODE unique In initial value problem, all side conditions specified at single point, say t 0 In boundary value problem (BVP), side conditions specified at more than one point kth order ODE, or equivalent first-order system, requires k side conditions For ODE, side conditions typically specified at two points, endpoints of interval [a, b], so we have two-point boundary value problem 2

3 Boundary Value Problems, continued General first-order two-point BVP has form y = f(t, y), a < t < b, with boundary conditions g(y(a), y(b)) = o, where f: R n+1 R n and g: R 2n R n Boundary conditions are separated if any given component of g involves solution values only at a or at b, but not both Boundary conditions are linear if of form B a y(a) + B b y(b) = c, where B a, B b R n n and c R n BVP is linear if both ODE and boundary conditions are linear 3

4 Example: Separated Linear BC Two-point BVP for second-order scalar ODE u = f(t, u, u ), a < t < b, with boundary conditions u(a) = α, u(b) = β, is equivalent to first-order system of ODEs y 1 y 2 = [ y 2 f(t, y 1, y 2 ) ], a < t < b, with separated linear boundary conditions [ 1 0 y1 (a) 0 0 y1 (b) α + = 0 0 y 2 (a) 1 0 y 2 (b) β ] 4

5 Existence and Uniqueness Unlike IVP, with BVP we cannot begin at initial point and continue solution step by step to nearby points Instead, solution determined everywhere simultaneously, so existence and/or uniqueness may not hold For example, u = u, 0 < t < b, with boundary conditions u(0) = 0, u(b) = β, with b integer multiple of π, has infinitely many solutions if β = 0, but no solution if β 0 5

6 Existence and Uniqueness, continued In general, solvability of BVP y = f(t, y), a < t < b, with boundary conditions g(y(a), y(b)) = o, depends on solvability of algebraic equation g(x, y(b; x)) = o, where y(t; x) denotes solution to ODE with initial condition y(a) = x for x R n Solvability of latter system is difficult to establish if g is nonlinear 6

7 Existence and Uniqueness, continued For linear BVP, existence and uniqueness are more tractable Consider linear BVP y = A(t) y + b(t), a < t < b, where A(t) and b(t) are continuous, with boundary conditions B a y(a) + B b y(b) = c Let Y (t) denote matrix whose ith column, y i (t), called ith mode, is solution to y = A(t)y with initial condition y(a) = e i Then BVP has unique solution if, and only if, matrix is nonsingular Q B a Y (a) + B b Y (b) 7

8 Existence and Uniqueness, continued Assuming Q is nonsingular, define and Green s function Φ(t) = Y (t) Q 1 G(t, s) = { Φ(t)Ba Φ(a)Φ 1 (s), a s t Φ(t)B b Φ(b)Φ 1 (s), t < s b Then solution to BVP given by y(t) = Φ(t) c + b a G(t, s) b(s) ds, This result also gives absolute condition number for BVP, κ = max{ Φ, G } 8

9 Conditioning and Stability Conditioning or stability of BVP depends on interplay between growth of solution modes and boundary conditions For IVP, instability is associated with modes that grow exponentially as time increases For BVP, solution is determined everywhere simultaneously, so there is no notion of direction of integration in interval [a, b] Growth of modes increasing with time is limited by boundary conditions at b, and growth of decaying modes is limited by boundary conditions at a For BVP to be well-conditioned, growing and decaying modes must be controlled appropriately by boundary conditions imposed 9

10 Numerical Methods for BVPs For IVP, initial data supply all information necessary to begin numerical solution method at initial point and step forward from there For BVP, we have insufficient information to begin step-by-step numerical method, so numerical methods for solving BVPs are more complicated than those for solving IVPs We consider four types of numerical methods for two-point BVPs: Shooting Finite difference Collocation Galerkin 10

11 Shooting Method In statement of two-point BVP, we are given value of u(a) If we also knew value of u (a), then we would have IVP that we could solve by methods previously discussed Lacking that information, we try sequence of increasingly accurate guesses until we find value for u (a) such that when we solve resulting IVP, approximate solution value at t = b matches desired boundary value, u(b) = β α a β b 11

12 Shooting Method, continued For given γ, value at b of solution u(b) to IVP u = f(t, u, u ), with initial conditions u(a) = α, u (a) = γ, can be considered as function of γ, say g(γ) Then BVP becomes problem of solving equation g(γ) = β One-dimensional zero finder can be used to solve this scalar equation 12

13 Example: Shooting Method Consider two-point BVP for second-order ODE u = 6t, 0 < t < 1, with BC u(0) = 0, u(1) = 1 For each guess for u (0), we integrate ODE using classical fourth-order Runge-Kutta method to determine how close we come to hitting desired solution value at t = 1 We transform second-order ODE into system of two first-order ODEs y 1 y 2 = y2 6t 13

14 Example Continued We try initial slope of y 2 (0) = 1 Using step size h = 0.5, we first step from t 0 = 0 to t 1 = 0.5 Classical fourth-order Runge-Kutta method gives approximate solution value at t 1 y (1) = y (0) + h 6 (k 1 + 2k 2 + 2k 3 + k 4 ) +2 = ( 1 0 ) =

15 Example Continued Next we step from t 1 = 0.5 to t 2 = 1, getting y (2) = ( ) 4 6 = 2, 4 so we have hit y 1 (1) = 2 instead of desired value y 1 (1) = 1 We try again, this time with initial slope y 2 (0) = 1, obtaining y (1) 0 = ( ) = 0.375,

16 Example Continued y (2) = ( ) 2 6 = + 2 0, so we have hit y 1 (1) = 0 instead of desired value y 1 (1) = 1 We now have initial slope bracketed between 1 and 1 We omit further iterations necessary to identify correct initial slope, which turns out to be y 2 (0) = 0: y (1) = ( 0 0 ) = ,

17 Example Continued y (2) = ( ) 3 6 = + 2 1, so we have indeed hit target solution value, y 1 (1) = st attempt. target. 2nd attempt

18 Multiple Shooting Simple shooting method inherits stability (or instability) of associated IVP, which may be unstable even when BVP is stable Such ill-conditioning may make it difficult to achieve convergence of iterative method for solving nonlinear equation Potential remedy is multiple shooting, in which interval [a, b] is divided into subintervals, and shooting is carried out on each Requiring continuity at internal mesh points provides BC for individual subproblems Multiple shooting results in larger system of nonlinear equations to solve 18

19 Finite Difference Method Finite difference method converts BVP into system of algebraic equations by replacing all derivatives by finite difference approximations For example, to solve two-point BVP with BC u = f(t, u, u ), a < t < b, u(a) = α, u(b) = β, we introduce mesh points t i = a + ih, i = 0, 1,..., n + 1, where h = (b a)/(n + 1) We already have y 0 = u(a) = α and y n+1 = u(b) = β, and we seek approximate solution value y i u(t i ) at each mesh point t i, i = 1,..., n 19

20 Finite Difference Method, continued We replace derivatives by finite difference quotients, such as and u (t i ) y i+1 y i 1 2h u (t i ) y i+1 2y i + y i 1 h 2, yielding system of equations y i+1 2y i + y i 1 h 2 = f t i, y i, y i+1 y i 1 2h to be solved for unknowns y i, i = 1,..., n ( ), System of equations may be linear or nonlinear, depending on whether f is linear or nonlinear 20

21 Finite Difference Method, continued In this example, system to be solved is tridiagonal, which saves on both work and storage compared to general system of equations This is generally true of finite difference methods: they yield sparse systems because each equation involves few variables 21

22 Example: Finite Difference Method Consider two-point BVP u = 6t, 0 < t < 1, with BC u(0) = 0, u(1) = 1 To keep computation to minimum, we compute approximate solution at one mesh point in interval [0, 1], t = 0.5 Including boundary points, we have three mesh points, t 0 = 0, t 1 = 0.5, and t 2 = 1 From BC, we know that y 0 = u(t 0 ) = 0 and y 2 = u(t 2 ) = 1, and we seek approximate solution y 1 u(t 1 ) 22

23 Example Continued Approximating second derivative by standard finite difference quotient at t 1 gives equation y 2 2y 1 + y 0 h 2 = f ( t 1, y 1, y 2 y 0 2h ) Substituting boundary data, mesh size, and right hand side for this example, or so that 1 2y (0.5) 2 = 6t 1, 4 8y 1 = 6(0.5) = 3, y(0.5) y 1 = 1/8 = which agrees with approximate solution at t = 0.5 that we previously computed by shooting method 23

24 Example Continued In a practical problem, much smaller step size and many more mesh points would be required to achieve acceptable accuracy We would therefore obtain system of equations to solve for approximate solution values at mesh points, rather than single equation as in this example 24

25 Collocation Method Collocation method approximates solution to BVP by finite linear combination of basis functions For two-point BVP with BC u = f(t, u, u ), a < t < b, u(a) = α, u(b) = β, we seek approximate solution of form u(t) v(t, x) = n i=1 x i φ i (t), where φ i are basis functions defined on [a, b] and x is n-vector of parameters to be determined 25

26 Collocation Method Popular choices of basis functions include polynomials, B-splines, and trigonometric functions Basis functions with global support, such as polynomials or trigonometric functions, yield spectral or pseudospectral method Basis functions with highly localized support, such as B-splines, yield finite element method 26

27 Collocation Method, continued To determine vector of parameters x, define set of n collocation points, a = t 1 < < t n = b, at which approximate solution v(t, x) is forced to satisfy ODE and boundary conditions Common choices of collocation points include equally-spaced mesh or Chebyshev points Suitably smooth basis functions can be differentiated analytically, so that approximate solution and its derivatives can be substituted into ODE and BC to obtain system of algebraic equations for unknown parameters x 27

28 Example: Collocation Method Consider again two-point BVP u = 6t, 0 < t < 1, with BC u(0) = 0, u(1) = 1 To keep computation to minimum, we use one interior collocation point, t = 0.5 Including boundary points, we have three collocation points, t 0 = 0, t 1 = 0.5, and t 2 = 1, so we will be able to determine three parameters As basis functions we use first three monomials, so approximate solution has form v(t, x) = x 1 + x 2 t + x 3 t 2 28

29 Example Continued Derivatives of approximate solution function with respect to t are given by v (t, x) = x 2 + 2x 3 t, v (t, x) = 2x 3 Requiring ODE to be satisfied at interior collocation point t 2 = 0.5 gives equation or v (t 2, x) = f(t 2, v(t 2, x), v (t 2, x)), 2x 3 = 6t 2 = 6(0.5) = 3 Left boundary condition at t 1 = 0 gives equation x 1 + x 2 t 1 + x 3 t 2 1 = x 1 = 0, and right boundary condition at t 3 = 1 gives equation x 1 + x 2 t 3 + x 3 t 2 3 = x 1 + x 2 + x 3 = 1 29

30 Example Continued Solving this system of three equations in three unknowns gives x 1 = 0, x 2 = 0.5, x 3 = 1.5, so approximate solution function is quadratic polynomial u(t) v(t, x) = 0.5t + 1.5t 2 At interior collocation point, t 2 = 0.5, we have approximate solution value u(0.5) v(0.5, x) = 0.125, which agrees with solution value at t = 0.5 obtained previously by other two methods

31 Galerkin Method Rather than forcing residual to be zero at finite number of points, as in collocation, we could instead minimize residual over entire interval of integration For example, for scalar Poisson equation in one dimension, u = f(t), a < t < b, with homogeneous boundary conditions u(a) = 0, u(b) = 0, subsitute approximate solution u(t) v(t, x) = n i=1 into ODE and define residual r(t, x) = v (t, x) f(t) = n i=1 x i φ i (t) x i φ i (t) f(t) 31

32 Galerkin Method, continued Using least squares method, we can minimize b F (x) = 1 r(t, 2 a x)2 dt by setting each component of its gradient to zero, which yields symmetric system of linear algebraic equations Ax = b, where b b a ij = a φ j (t)φ i (t) dt and b i = a f(t)φ i (t) dt, whose solution gives vector of parameters x More generally, weighted residual method forces residual to be orthogonal to each of set of weight functions or test functions w i, i.e., b r(t, x)w i(t) dt = 0, i = 1,..., n, a which yields linear system Ax = b, where now b b a ij = a φ j (t)w i(t) dt and b i = f(t)w i(t) dt, a whose solution gives vector of parameters x 32

33 Galerkin Method, continued Matrix resulting from weighted residual method is generally not symmetric, and its entries involve second derivatives of basis functions Both drawbacks overcome by Galerkin method, in which weight functions are chosen to be same as basis functions, i.e., w i = φ i, i = 1,..., n Orthogonality condition then becomes b r(t, x)φ i(t) dt = 0, i = 1,..., n, a or b b a v (t, x)φ i (t) dt = f(t)φ i(t) dt, i = 1,..., n a 33

34 Galerkin Method, continued Degree of differentiability can be reduced using integration by parts, which gives b b a v (t, x)φ i (t) dt = v (t)φ i (t) b a a v (t)φ i (t) dt = v (b)φ i (b) v (a)φ i (a) b a v (t)φ i (t) dt Assuming basis functions φ i satisfy homogeneous boundary conditions, so φ i (0) = φ i (1) = 0, orthogonality condition then becomes b b a v (t)φ i (t) dt = f(t)φ i(t) dt, i = 1,..., n, a which yields system of linear equations Ax = b, with b a ij = a φ j (t)φ i (t) dt and b i = f(t)φ i(t) dt, a whose solution gives vector of parameters x b A is symmetric and involves only first derivatives of basis functions 34

35 Example: Galerkin Method Consider again two-point BVP with BC u = 6t, 0 < t < 1, u(0) = 0, u(1) = 1 We will approximate solution by piecewise linear polynomial, for which B-splines of degree 1 ( hat functions) form suitable set of basis functions To keep computation to minimum, we again use same three mesh points, but now they become knots in piecewise linear polynomial approximation 1. φ φ φ

36 Example Continued Thus, we seek approximate solution of form u(t) v(t, x) = x 1 φ 1 (t) + x 2 φ 2 (t) + x 3 φ 3 (t) From BC, we must have x 1 = 0 and x 3 = 1 To determine remaining parameter x 2, we impose Galerkin orthogonality condition on interior basis function φ 2 and obtain equation 3 j=1 ( 1 0 φ j (t)φ 2 (t) dt ) x j = 1 0 6tφ 2(t) dt, or, upon evaluating these simple integrals analytically, 2x 1 4x 2 + 2x 3 = 3/2 36

37 Example Continued Substituting known values for x 1 and x 3 then gives x 2 = 1/8 for remaining unknown parameter, so piecewise linear approximate solution is u(t) v(t, x) = 0.125φ 2 (t) + φ 3 (t) We note that v(0.5, x) = 0.125, which again is exact for this particular problem 37

38 Example Continued More realistic problem would have many more interior mesh points and basis functions, and correspondingly many parameters to be determined Resulting system of equations would be much larger but still sparse, and therefore relatively easy to solve, provided local basis functions, such as hat functions, are used Resulting approximate solution function is less smooth than true solution, but becomes more accurate as more mesh points are used 38

39 Eigenvalue Problems Standard eigenvalue problem for second-order ODE has form with BC u = λf(t, u, u ), a < t < b, u(a) = α, u(b) = β, where we seek not only solution u but also parameter λ as well Scalar λ (possibly complex) is eigenvalue and solution u corresponding eigenfunction for this two-point BVP Discretization of eigenvalue problem for ODE results in algebraic eigenvalue problem whose solution approximates that of original problem 39

40 Example: Eigenvalue Problem Consider linear two-point BVP u = λg(t)u, a < t < b, with BC u(a) = 0, u(b) = 0 Introduce discrete mesh points t i in interval [a, b], with mesh spacing h and use standard finite difference approximation for second derivative to obtain algebraic system y i+1 2y i + y i 1 h 2 = λg i y i, i = 1,..., n, where y i = u(t i ) and g i = g(t i ), and from BC y 0 = u(a) = 0 and y n+1 = u(b) = 0 40

41 Example Continued Assuming g i 0, divide equation i by g i for i = 1,..., n, to obtain linear system Ay = λy, where n n matrix A has tridiagonal form A = 1 h 2 2/g 1 1/g /g 2 2/g 2 1/g /g n 1 2/g n 1 1/g n /g n 2/g n This standard algebraic eigenvalue problem can be solved by methods discussed previously 41

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