BVP of ODE

BVP of ODE 1Boundary-Value Problems for OrdinaryDifferential EquationsNTNUTsung-Min HwangDecember 20, 2003Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 21 – Mathematical Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 – Finite Difference Method For Linear Problems . . . . . . . . . . . . . . . . . 152.1 – The Finite Difference Formulation . . . . . . . . . . . . . . . . . . . . . . 152.2 – Convergence Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 – Shooting Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 3The two-point boundary-value problems (BVP) considered in this chapter involve asecond-order differential equation together with boundary condition in the following form:⎧⎨y ′′ = f(x, y, y ′ )⎩y(a) = α, y(b) = β(1)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 3The two-point boundary-value problems (BVP) considered in this chapter involve asecond-order differential equation together with boundary condition in the following form:⎧⎨y ′′ = f(x, y, y ′ )⎩y(a) = α, y(b) = β(1)The numerical procedures for finding approximate solutions to the initial-value problems cannot be adapted to the solution of this problem since the specification of conditions involvetwo different points, x = a and x = b.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 3The two-point boundary-value problems (BVP) considered in this chapter involve asecond-order differential equation together with boundary condition in the following form:⎧⎨y ′′ = f(x, y, y ′ )⎩y(a) = α, y(b) = β(1)The numerical procedures for finding approximate solutions to the initial-value problems cannot be adapted to the solution of this problem since the specification of conditions involvetwo different points, x = a and x = b. New techniques are introduced in this chapter forhandling problems (1) in which the conditions imposed are of a boundary-value rather thanan initial-value type.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 41 – Mathematical TheoriesBefore considering numerical methods, a few mathematical theories about the two-pointboundary-value problem (1), such as the existence and uniqueness of solution, shall bediscussed in this section.Theorem 1 Suppose that f in (1) is continuous on the setand that ∂f∂y1.D = {(x, y, y ′ )|a ≤ x ≤ b, −∞ < y < ∞, −∞ < y ′ < ∞} ,∂fand are also continuous on D. If∂y′∂f∂y (x, y, y′ ) > 0 for all (x, y, y ′ ) ∈ D, and∂f∣∂y ′ (x, y, y′ )∣ ≤ M , ∀ (x, y, y′ ) ∈ D,2. a constant M exists, withthen (1) has a unique solution.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 5When the function f(x, y, y ′ ) has the special formf(x, y, y ′ ) = p(x)y ′ + q(x)y + r(x),the differential equation become a so-called linear problem. The previous theorem can besimplified for this case.Corollary 1 If the linear two-point boundary-value problemsatisfies⎧⎨y ′′ = p(x)y ′ + q(x)y + r(x)⎩y(a) = α, y(b) = β1. p(x), q(x), and r(x) are continuous on [a, b], and2. q(x) > 0 on [a, b],then the problem has a unique solution.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 6Many theories and application models consider the boundary-value problem in a specialform as follows.⎧⎨y ′′ = f(x, y)⎩y(0) = 0, y(1) = 0We will show that this simple form can be derived from the original problem by simpletechniques such as changes of variables and linear transformation. To do this, we begin bychanging the variable. Suppose that the original problem is⎧⎨y ′′ = f(x, y)⎩y(a) = α,y(b) = βwhere y = y(x). Now let λ = b − a and define a new variablet = x − ab − a = 1 (x − a).λ(2)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 7That is, x = a + λt. Notice that t = 0 corresponds to x = a, and t = 1 corresponds tox = b. Then we definewith λ = b − a. This givesz(t) = y(a + λt) = y(x)z ′ (t) = d dt z(t) = d dt y(a + λt) = [ ddx y(x) ] [ ddt (a + λt) ]and, analogously,= λy ′ (x)z ′′ (t) = d dt z′ (t) = λ 2 y ′′ (x) = λ 2 f(x, y(x)) = λ 2 f(a + λt, z(t)).Likewise the boundary conditions are changed toz(0) = y(a) = α and z(1) = y(b) = β.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 9and the original equation is reduced toλ 2 f(a + λt, z) = 9 [ sin(1 + 3t)z + z 2] ,⎧⎨z ′′ (t) = 9 sin((1 + 3t)z) + 9z 2⎩z(0) = 3, z(1) = 7To reduce a two-point boundary-value problemto a homogeneous system, let⎧⎨z ′′ (t) = g(t, z)⎩z(0) = α, z(1) = βu(t) = z(t) − [α + (β − α)t]then u ′′ (t) = z ′′ (t), andu(0) = z(0) − α = 0 and u(1) = z(1) − β = 0Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 10Moreover,g(t, z) = g(t, u + α + (β − α)t) ≡ h(t, u).The system is now transformed intoExample 2 Reduce the system⎧⎨u ′′ (t) = h(t, u)⎩u(0) = 0, u(1) = 0⎧⎨z ′′ = [5z − 10t + 35 + sin(3z − 6t + 21)]e t⎩z(0) = −7, z(1) = −5to a homogeneous problem by linear transformation technique.Solution: Letu(t) = z(t) − [−7 + (−5 + 7)t] = z(t) − 2t + 7.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 11Then z(t) = u(t) + 2t − 7, andu ′′ = z ′′ = [5z − 10t + 35 + sin(3z − 6t + 21)]e t= [5(u + 2t − 7) − 10t + 35 + sin(3(u + 2t − 7) − 6t + 21)]e t= [5u + sin(3u)]e tThe system is transformed to⎧⎨u ′′ (t) = [5u + sin(3u)]e t⎩u(0) = u(1) = 0Example 3 Reduce the following two-point boundary-value problem⎧⎨y ′′ = y 2 + 3 − x 2 + xy⎩y(3) = 7, y(5) = 9to a homogeneous system.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 12Solution: In the original system, a = 3, b = 5, α = 7, β = 9. Let λ = b − a = 2 anddefine a new variablet = 1 (x − 3) =⇒ x = 2t + 3.2Let the function z(t) = y(x) = y(2t + 3). Thenz ′′ (t) = λ 2 y ′′ (2t + 3) = λ 2 f(2t + 3, u)The original problem is first transformed intoNext let= 4[z 2 + 3 − (2t + 3) 2 + (2t + 3)z]= 4[z 2 + 3z + 2tz − 4t 2 − 12t − 6]⎧⎨z ′′ (t) = 4[z 2 + 3z + 2tz − 4t 2 − 12t − 6]⎩z(0) = 7, z(1) = 9u(t) = z(t) − [7 + 2t], or equivalently, z(t) = u(t) + 2t + 7.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 13Thenu ′′ (t) = 4[(z + 2t + 7) 2 + 3(u + 2t + 7) + 2t(u + 2t + 7) − 4t 2 − 12t − 6]= 4[u 2 + 6tu + 17u + 4t 2 + 36t + 64].The original problem is transformed into the following homogeneous system⎧⎨u ′′ (t) = 4[u 2 + 6tu + 17u + 4t 2 + 36t + 64]⎩u(0) = u(1) = 0Theorem 2 The boundary-value problemhas a unique solution if ∂f∂y0 ≤ x ≤ 1 and −∞ < y < ∞.⎧⎨y ′′ = f(x, y)⎩y(0) = 0, y(1) = 0is continuous, non-negative, and bounded in the stripDepartment of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 14Theorem 3 If f is a continuous function of (s, t) in the domain 0 ≤ s ≤ 1 and−∞ < t < ∞ such that|f(s, t 1 ) − f(s, t 2 )| ≤ K|t 1 − t 2 |, (K < 8).Then the boundary-value problemhas a unique solution in C[0, 1].⎧⎨y ′′ = f(x, y)⎩y(0) = 0, y(1) = 0Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 152 – Finite Difference Method For Linear ProblemsWe consider finite difference method for solving the linear two-point boundary-value problemof the form⎧⎨ y ′′ = p(x)y ′ + q(x)y + r(x)⎩y(a) = α, y(b) = β.(4)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 152 – Finite Difference Method For Linear ProblemsWe consider finite difference method for solving the linear two-point boundary-value problemof the form⎧⎨ y ′′ = p(x)y ′ + q(x)y + r(x)⎩y(a) = α, y(b) = β.Methods involving finite differences for solving boundary-value problems replace each of thederivatives in the differential equation by an appropriate difference-quotient approximation.(4)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 152 – Finite Difference Method For Linear ProblemsWe consider finite difference method for solving the linear two-point boundary-value problemof the form⎧⎨ y ′′ = p(x)y ′ + q(x)y + r(x)⎩y(a) = α, y(b) = β.(4)Methods involving finite differences for solving boundary-value problems replace each of thederivatives in the differential equation by an appropriate difference-quotient approximation.2.1 – The Finite Difference FormulationDepartment of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 152 – Finite Difference Method For Linear ProblemsWe consider finite difference method for solving the linear two-point boundary-value problemof the form⎧⎨ y ′′ = p(x)y ′ + q(x)y + r(x)⎩y(a) = α, y(b) = β.(4)Methods involving finite differences for solving boundary-value problems replace each of thederivatives in the differential equation by an appropriate difference-quotient approximation.2.1 – The Finite Difference FormulationFirst, partition the interval [a, b] into n equally-spaced subintervals by pointsa = x 0 < x 1 < . . . < x n < x n = b.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 152 – Finite Difference Method For Linear ProblemsWe consider finite difference method for solving the linear two-point boundary-value problemof the form⎧⎨ y ′′ = p(x)y ′ + q(x)y + r(x)⎩y(a) = α, y(b) = β.(4)Methods involving finite differences for solving boundary-value problems replace each of thederivatives in the differential equation by an appropriate difference-quotient approximation.2.1 – The Finite Difference FormulationFirst, partition the interval [a, b] into n equally-spaced subintervals by pointsa = x 0 < x 1 < . . . < x n < x n = b. Each mesh point x i can be computed byx i = a + i ∗ h,i = 0, 1, . . . , n, with h = b − anwhere h is called the mesh size.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 16At the interior mesh points, x i , for i = 1, 2, . . . , n − 1, the differential equation to beapproximated satisfiesy ′′ (x i ) = p(x i )y ′ (x i ) + q(x i )y(x i ) + r(x i ). (5)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 16At the interior mesh points, x i , for i = 1, 2, . . . , n − 1, the differential equation to beapproximated satisfiesThe central finite difference formulaey ′′ (x i ) = p(x i )y ′ (x i ) + q(x i )y(x i ) + r(x i ). (5)y ′ (x i ) = y(x i+1) − y(x i−1 )2h− h26 y(3) (η i ), (6)for some η i in the interval (x i−1 , x i+1 ),Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 16At the interior mesh points, x i , for i = 1, 2, . . . , n − 1, the differential equation to beapproximated satisfiesThe central finite difference formulaey ′′ (x i ) = p(x i )y ′ (x i ) + q(x i )y(x i ) + r(x i ). (5)y ′ (x i ) = y(x i+1) − y(x i−1 )2h− h26 y(3) (η i ), (6)for some η i in the interval (x i−1 , x i+1 ), andy ′′ (x i ) = y(x i+1) − 2y(x i ) + y(x i−1 )h 2− h212 y(4) (ξ i ), (7)for some ξ i in the interval (x i−1 , x i+1 ), can be derived from Taylor’s theorem byexpanding y about x i .Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 16At the interior mesh points, x i , for i = 1, 2, . . . , n − 1, the differential equation to beapproximated satisfiesThe central finite difference formulaey ′′ (x i ) = p(x i )y ′ (x i ) + q(x i )y(x i ) + r(x i ). (5)y ′ (x i ) = y(x i+1) − y(x i−1 )2h− h26 y(3) (η i ), (6)for some η i in the interval (x i−1 , x i+1 ), andy ′′ (x i ) = y(x i+1) − 2y(x i ) + y(x i−1 )h 2− h212 y(4) (ξ i ), (7)for some ξ i in the interval (x i−1 , x i+1 ), can be derived from Taylor’s theorem byexpanding y about x i .Let u i denote the approximate value of y i = y(x i ).Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 16At the interior mesh points, x i , for i = 1, 2, . . . , n − 1, the differential equation to beapproximated satisfiesThe central finite difference formulaey ′′ (x i ) = p(x i )y ′ (x i ) + q(x i )y(x i ) + r(x i ). (5)y ′ (x i ) = y(x i+1) − y(x i−1 )2h− h26 y(3) (η i ), (6)for some η i in the interval (x i−1 , x i+1 ), andy ′′ (x i ) = y(x i+1) − 2y(x i ) + y(x i−1 )h 2− h212 y(4) (ξ i ), (7)for some ξ i in the interval (x i−1 , x i+1 ), can be derived from Taylor’s theorem byexpanding y about x i .Let u i denote the approximate value of y i = y(x i ). If y ∈ C 4 [a, b], then a finite differencemethod with truncation error of order O(h 2 ) can be obtained by using the approximationsDepartment of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 17y ′ (x i ) ≈ u i+1 − u i−12hfor y ′ (x i ) and y ′′ (x i ), respectively.andy ′′ (x i ) ≈ u i+1 − 2u i + u i−1h 2Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 17y ′ (x i ) ≈ u i+1 − u i−12handfor y ′ (x i ) and y ′′ (x i ), respectively. Furthermore, lety ′′ (x i ) ≈ u i+1 − 2u i + u i−1h 2p i = p(x i ), q i = q(x i ), r i = r(x i ).Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 17y ′ (x i ) ≈ u i+1 − u i−12handfor y ′ (x i ) and y ′′ (x i ), respectively. Furthermore, letThe discrete version of equation (4) is thenu i+1 − 2u i + u i−1h 2y ′′ (x i ) ≈ u i+1 − 2u i + u i−1h 2p i = p(x i ), q i = q(x i ), r i = r(x i ).= p iu i+1 − u i−12htogether with boundary conditions u 0 = α and u n = β.+ q i u i + r i , i = 1, 2, . . . , n − 1, (8)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 17y ′ (x i ) ≈ u i+1 − u i−12handfor y ′ (x i ) and y ′′ (x i ), respectively. Furthermore, letThe discrete version of equation (4) is thenu i+1 − 2u i + u i−1h 2y ′′ (x i ) ≈ u i+1 − 2u i + u i−1h 2p i = p(x i ), q i = q(x i ), r i = r(x i ).= p iu i+1 − u i−12h+ q i u i + r i , i = 1, 2, . . . , n − 1, (8)together with boundary conditions u 0 = α and u n = β. Equation (8) can be written in theform (1 + h )2 p i u i−1 − ( 2 + h 2 )q i ui +for i = 1, 2, . . . , n − 1.(1 − h )2 p i u i+1 = h 2 r i , (9)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 17y ′ (x i ) ≈ u i+1 − u i−12handfor y ′ (x i ) and y ′′ (x i ), respectively. Furthermore, letThe discrete version of equation (4) is thenu i+1 − 2u i + u i−1h 2y ′′ (x i ) ≈ u i+1 − 2u i + u i−1h 2p i = p(x i ), q i = q(x i ), r i = r(x i ).= p iu i+1 − u i−12h+ q i u i + r i , i = 1, 2, . . . , n − 1, (8)together with boundary conditions u 0 = α and u n = β. Equation (8) can be written in theform (1 + h )2 p i u i−1 − ( 2 + h 2 )q i ui +(1 − h )2 p i u i+1 = h 2 r i , (9)for i = 1, 2, . . . , n − 1. In (8), u 1 , u 2 , . . . , u n−1 are the unknown, and there are n − 1linear equations to be solved.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 17y ′ (x i ) ≈ u i+1 − u i−12handfor y ′ (x i ) and y ′′ (x i ), respectively. Furthermore, letThe discrete version of equation (4) is thenu i+1 − 2u i + u i−1h 2y ′′ (x i ) ≈ u i+1 − 2u i + u i−1h 2p i = p(x i ), q i = q(x i ), r i = r(x i ).= p iu i+1 − u i−12h+ q i u i + r i , i = 1, 2, . . . , n − 1, (8)together with boundary conditions u 0 = α and u n = β. Equation (8) can be written in theform (1 + h )2 p i u i−1 − ( 2 + h 2 )q i ui +(1 − h )2 p i u i+1 = h 2 r i , (9)for i = 1, 2, . . . , n − 1. In (8), u 1 , u 2 , . . . , u n−1 are the unknown, and there are n − 1linear equations to be solved. The resulting system of linear equations can be expressed inthe matrix formAu = f, (10)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 18whereA =⎡−2 − h 2 q 1 1 − h p 2 11 + h p 2 2 −2 − h 2 q 2 1 − h p 2 2⎢⎣. . .. . .. . .⎤,1 + h p 2 n−2 −2 − h 2 q n−2 1 − h p ⎥2 n−2 ⎦1 + h p 2 n−1 −2 − h 2 q n−1. . .. . .. . .u =⎡⎢⎣u 1u 2.u n−2u n−1⎤, ⎥⎦and f =⎡⎢⎣h 2 r 1 − ( 1 + h 2 p ⎤1)αh 2 r 2.h 2 r ⎥n−2h 2 r n−1 − ( ⎦1 − h 2 p )n−1 βDepartment of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 19Since the matrix A is tridiagonal, this system can be solved by a special Gaussianelimination in O(n 2 ) flops.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 19Since the matrix A is tridiagonal, this system can be solved by a special Gaussianelimination in O(n 2 ) flops.Theorem 4 Suppose that p(x), q(x), and r(x) in (4) are continuous on [a, b], andq(x) > 0 on [a, b]. Then (10) has a unique solution provided that h < 2/L, whereL = max a≤x≤b |p(x)|.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 19Since the matrix A is tridiagonal, this system can be solved by a special Gaussianelimination in O(n 2 ) flops.Theorem 4 Suppose that p(x), q(x), and r(x) in (4) are continuous on [a, b], andq(x) > 0 on [a, b]. Then (10) has a unique solution provided that h < 2/L, whereL = max a≤x≤b |p(x)|.2.2 – Convergence AnalysisWe shall analyze that when h converges to zero, the solution u i of the discrete problem (8)converges to the solution y i of the original continuous problem (5).Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 19Since the matrix A is tridiagonal, this system can be solved by a special Gaussianelimination in O(n 2 ) flops.Theorem 4 Suppose that p(x), q(x), and r(x) in (4) are continuous on [a, b], andq(x) > 0 on [a, b]. Then (10) has a unique solution provided that h < 2/L, whereL = max a≤x≤b |p(x)|.2.2 – Convergence AnalysisWe shall analyze that when h converges to zero, the solution u i of the discrete problem (8)converges to the solution y i of the original continuous problem (5).y i satisfies the following system of equationsy i+1 − 2y i + y i−1h 2for i = 1, 2, . . . , n − 1.(− h2yi+1 − y i−112 y(4) (ξ i ) = p i2h)− h26 y(3) (η i )+ q i y i + r i ,(11)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 19Since the matrix A is tridiagonal, this system can be solved by a special Gaussianelimination in O(n 2 ) flops.Theorem 4 Suppose that p(x), q(x), and r(x) in (4) are continuous on [a, b], andq(x) > 0 on [a, b]. Then (10) has a unique solution provided that h < 2/L, whereL = max a≤x≤b |p(x)|.2.2 – Convergence AnalysisWe shall analyze that when h converges to zero, the solution u i of the discrete problem (8)converges to the solution y i of the original continuous problem (5).y i satisfies the following system of equationsy i+1 − 2y i + y i−1h 2(− h2yi+1 − y i−112 y(4) (ξ i ) = p i2h)− h26 y(3) (η i )for i = 1, 2, . . . , n − 1. Subtract (8) from (11) and let e i = y i − u i , the result is+ q i y i + r i ,(11)e i+1 − 2e i + e i−1h 2= p ie i+1 − e i−12h+ q i e i + h 2 g i , i = 1, 2, . . . , n − 1,Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 20whereg i = 112 y(4) (ξ i ) − 1 6 p iy (3) (η i ).Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

whereBVP of ODE 20g i = 112 y(4) (ξ i ) − 1 6 p iy (3) (η i ).After collecting terms and multiplying by h 2 , we have(1 + h )2 p ie i−1 − ( 2 + h 2 q i)ei +(1 − h )2 p ie i+1 = h 4 g i , i = 1, 2, . . . , n − 1.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

whereBVP of ODE 20g i = 112 y(4) (ξ i ) − 1 6 p iy (3) (η i ).After collecting terms and multiplying by h 2 , we have(1 + h )2 p ie i−1 − ( 2 + h 2 q i)ei +(1 − h )2 p ie i+1 = h 4 g i , i = 1, 2, . . . , n − 1.Let e = [e 1 , e 2 , . . . , e n−1 ] T and |e k | = ‖e‖ ∞ .Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

whereBVP of ODE 20g i = 112 y(4) (ξ i ) − 1 6 p iy (3) (η i ).After collecting terms and multiplying by h 2 , we have(1 + h )2 p ie i−1 − ( 2 + h 2 q i)ei +(1 − h )2 p ie i+1 = h 4 g i , i = 1, 2, . . . , n − 1.Let e = [e 1 , e 2 , . . . , e n−1 ] T and |e k | = ‖e‖ ∞ . Then((2 + h 2 )q k ek = 1 + h ) (2 p k e k−1 + 1 − h )2 p k e k+1 − h 4 g k ,Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

whereBVP of ODE 20g i = 112 y(4) (ξ i ) − 1 6 p iy (3) (η i ).After collecting terms and multiplying by h 2 , we have(1 + h )2 p ie i−1 − ( 2 + h 2 q i)ei +(1 − h )2 p ie i+1 = h 4 g i , i = 1, 2, . . . , n − 1.Let e = [e 1 , e 2 , . . . , e n−1 ] T and |e k | = ‖e‖ ∞ . Then((2 + h 2 )q k ek = 1 + h ) (2 p k e k−1 + 1 − h )2 p k e k+1 − h 4 g k ,and, hence∣ 2 + h 2 q k∣ ∣ |ek |≤≤∣ 1 + h 2 p k∣ |e k−1| +∣ 1 − h 2 p k∣ |e k+1| + h 4 |g k |∣ 1 + h 2 p k∣ ‖e‖ ∞ +∣ 1 − h 2 p k∣ ‖e‖ ∞ + h 4 ‖g‖ ∞Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 21When q(x) > 0, ∀x ∈ [a, b] and h is chosen small enough so that | h 2 p i| < 1, ∀i, thenthe the above inequality inducesh 2 q k ‖e‖ ∞ ≤ h 4 ‖g‖ ∞ .Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 21When q(x) > 0, ∀x ∈ [a, b] and h is chosen small enough so that | h 2 p i| < 1, ∀i, thenthe the above inequality inducesh 2 q k ‖e‖ ∞ ≤ h 4 ‖g‖ ∞ .Therefore, we derive an upper bound for ‖e‖ ∞‖e‖ ∞ ≤ h 2 ( ‖g‖∞inf q(x)).Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 21When q(x) > 0, ∀x ∈ [a, b] and h is chosen small enough so that | h 2 p i| < 1, ∀i, thenthe the above inequality inducesh 2 q k ‖e‖ ∞ ≤ h 4 ‖g‖ ∞ .Therefore, we derive an upper bound for ‖e‖ ∞By the definition of g i , we have‖e‖ ∞ ≤ h 2 ( ‖g‖∞inf q(x)).‖g‖ ∞ ≤ 112 ‖y(4) (x)‖ ∞ + 1 6 ‖p(x)‖ ∞‖y (3) (x)‖ ∞ .Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 21When q(x) > 0, ∀x ∈ [a, b] and h is chosen small enough so that | h 2 p i| < 1, ∀i, thenthe the above inequality inducesh 2 q k ‖e‖ ∞ ≤ h 4 ‖g‖ ∞ .Therefore, we derive an upper bound for ‖e‖ ∞By the definition of g i , we have‖e‖ ∞ ≤ h 2 ( ‖g‖∞inf q(x)).‖g‖ ∞ ≤ 112 ‖y(4) (x)‖ ∞ + 1 6 ‖p(x)‖ ∞‖y (3) (x)‖ ∞ .Hence‖g‖ ∞inf q(x)is a bound independent of h.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 21When q(x) > 0, ∀x ∈ [a, b] and h is chosen small enough so that | h 2 p i| < 1, ∀i, thenthe the above inequality inducesh 2 q k ‖e‖ ∞ ≤ h 4 ‖g‖ ∞ .Therefore, we derive an upper bound for ‖e‖ ∞By the definition of g i , we have‖e‖ ∞ ≤ h 2 ( ‖g‖∞inf q(x)).‖g‖ ∞ ≤ 112 ‖y(4) (x)‖ ∞ + 1 6 ‖p(x)‖ ∞‖y (3) (x)‖ ∞ .Hence‖g‖ ∞inf q(x) is a bound independent of h. Thus we can conclude that ‖e‖ ∞ is O(h 2 ) ash → 0.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 223 – Shooting MethodsWe consider solving the following 2-point boundary-value problem:⎧⎨⎩y ′′ = f(x, y, y ′ )y(a) = α,y(b) = β(12)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 223 – Shooting MethodsWe consider solving the following 2-point boundary-value problem:⎧⎨⎩y ′′ = f(x, y, y ′ )y(a) = α,y(b) = β(12)The idea of shooting method for (12) is to solve a related initial-value problem with a guessfor y ′ (a), say z.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 223 – Shooting MethodsWe consider solving the following 2-point boundary-value problem:⎧⎨⎩y ′′ = f(x, y, y ′ )y(a) = α,y(b) = β(12)The idea of shooting method for (12) is to solve a related initial-value problem with a guessfor y ′ (a), say z. The corresponding IVP⎧⎨ y ′′ = f(x, y, y ′ )⎩ y(a) = α, y ′ (a) = zcan then be solved by, for example, Runge-Kutta method.(13)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 223 – Shooting MethodsWe consider solving the following 2-point boundary-value problem:⎧⎨⎩y ′′ = f(x, y, y ′ )y(a) = α,y(b) = β(12)The idea of shooting method for (12) is to solve a related initial-value problem with a guessfor y ′ (a), say z. The corresponding IVP⎧⎨ y ′′ = f(x, y, y ′ )⎩ y(a) = α, y ′ (a) = zcan then be solved by, for example, Runge-Kutta method. We denote this approximatesolution y z and hope y z (b) = β.(13)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 223 – Shooting MethodsWe consider solving the following 2-point boundary-value problem:⎧⎨⎩y ′′ = f(x, y, y ′ )y(a) = α,y(b) = β(12)The idea of shooting method for (12) is to solve a related initial-value problem with a guessfor y ′ (a), say z. The corresponding IVP⎧⎨ y ′′ = f(x, y, y ′ )⎩ y(a) = α, y ′ (a) = zcan then be solved by, for example, Runge-Kutta method. We denote this approximatesolution y z and hope y z (b) = β. If not, we use another guess for y ′ (a), and try to solve analtered IVP (13) again.(13)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 223 – Shooting MethodsWe consider solving the following 2-point boundary-value problem:⎧⎨⎩y ′′ = f(x, y, y ′ )y(a) = α,y(b) = β(12)The idea of shooting method for (12) is to solve a related initial-value problem with a guessfor y ′ (a), say z. The corresponding IVP⎧⎨ y ′′ = f(x, y, y ′ )⎩ y(a) = α, y ′ (a) = zcan then be solved by, for example, Runge-Kutta method. We denote this approximatesolution y z and hope y z (b) = β. If not, we use another guess for y ′ (a), and try to solve analtered IVP (13) again. This process is repeated and can be done systematically.(13)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Letφ(z) = y z (b) − β.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Letφ(z) = y z (b) − β.Now our objective is simply to solve the equation φ(z) = 0.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Letφ(z) = y z (b) − β.Now our objective is simply to solve the equation φ(z) = 0. Hence secant method canbe used.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Letφ(z) = y z (b) − β.Now our objective is simply to solve the equation φ(z) = 0. Hence secant method canbe used.☞ How to compute z?Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Letφ(z) = y z (b) − β.Now our objective is simply to solve the equation φ(z) = 0. Hence secant method canbe used.☞ How to compute z?Suppose we have solutions y z1 , y z2 with guesses z 1 , z 2 and obtain φ(z 1 ) and φ(z 2 ).Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Letφ(z) = y z (b) − β.Now our objective is simply to solve the equation φ(z) = 0. Hence secant method canbe used.☞ How to compute z?Suppose we have solutions y z1 , y z2 with guesses z 1 , z 2 and obtain φ(z 1 ) and φ(z 2 ).If these guesses can not generate satisfactory solutions, we can obtain another guessz 3 by the secant methodz 2 − z 1z 3 = z 2 − φ(z 2 )φ(z 2 ) − φ(z 1 ) .Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 23☞ Objective: select z, so that y z (b) = β.Letφ(z) = y z (b) − β.Now our objective is simply to solve the equation φ(z) = 0. Hence secant method canbe used.☞ How to compute z?Suppose we have solutions y z1 , y z2 with guesses z 1 , z 2 and obtain φ(z 1 ) and φ(z 2 ).If these guesses can not generate satisfactory solutions, we can obtain another guessz 3 by the secant methodz 2 − z 1z 3 = z 2 − φ(z 2 )φ(z 2 ) − φ(z 1 ) .In generalz k − z k−1z k+1 = z k − φ(z k )φ(z k ) − φ(z k−1 ) .Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 24☞ Special BVP:⎧⎨⎩y ′′ = u(x) + v(x)y + w(x)y ′y(a) = α,y(b) = β(14)where u(x), v(x), w(x) are continuous in [a, b].Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 24☞ Special BVP:⎧⎨⎩y ′′ = u(x) + v(x)y + w(x)y ′y(a) = α,y(b) = β(14)where u(x), v(x), w(x) are continuous in [a, b].Suppose we have solved (14) twice with initial guesses z 1 and z 2 , and obtainapproximate solutions y 1 and y 2 ,Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 24☞ Special BVP:⎧⎨⎩y ′′ = u(x) + v(x)y + w(x)y ′y(a) = α,y(b) = β(14)where u(x), v(x), w(x) are continuous in [a, b].Suppose we have solved (14) twice with initial guesses z 1 and z 2 , and obtainapproximate solutions y 1 and y 2 , hence⎧⎨ y ′′ ′1 = u + vy 1 + wy 1⎩ y 1 (a) = α, y ′ and1 (a) = z 1⎧⎨⎩y 2 ′′ = u + vy 2 + wy 2′y 2 (a) = α, y 2 ′ (a) = z 2Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 24☞ Special BVP:⎧⎨⎩y ′′ = u(x) + v(x)y + w(x)y ′y(a) = α,y(b) = β(14)where u(x), v(x), w(x) are continuous in [a, b].Suppose we have solved (14) twice with initial guesses z 1 and z 2 , and obtainapproximate solutions y 1 and y 2 , hence⎧⎨ y ′′ ′1 = u + vy 1 + wy 1⎩ y 1 (a) = α, y ′ and1 (a) = z 1⎧⎨⎩y 2 ′′ = u + vy 2 + wy 2′y 2 (a) = α, y 2 ′ (a) = z 2Now lety(x) = λy 1 (x) + (1 − λ)y 2 (x)for some parameter λ,Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 24☞ Special BVP:⎧⎨⎩y ′′ = u(x) + v(x)y + w(x)y ′y(a) = α,y(b) = β(14)where u(x), v(x), w(x) are continuous in [a, b].Suppose we have solved (14) twice with initial guesses z 1 and z 2 , and obtainapproximate solutions y 1 and y 2 , hence⎧⎨ y ′′ ′1 = u + vy 1 + wy 1⎩ y 1 (a) = α, y ′ and1 (a) = z 1⎧⎨⎩y 2 ′′ = u + vy 2 + wy 2′y 2 (a) = α, y 2 ′ (a) = z 2Now lety(x) = λy 1 (x) + (1 − λ)y 2 (x)for some parameter λ, we can showy ′′ = u + vy + wy ′Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 25andy(a) = λy 1 (a) + (1 − λ)y 2 (a) = αDepartment of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 25andy(a) = λy 1 (a) + (1 − λ)y 2 (a) = αWe can select λ so that y(b) = β.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 25andy(a) = λy 1 (a) + (1 − λ)y 2 (a) = αWe can select λ so that y(b) = β.β = y(b) = λy 1 (b) + (1 − λ)y 2 (b)= λ(y 1 (b) − y 2 (b)) + y 2 (b)Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 25andy(a) = λy 1 (a) + (1 − λ)y 2 (a) = αWe can select λ so that y(b) = β.β = y(b) = λy 1 (b) + (1 − λ)y 2 (b)⇒ λ == λ(y 1 (b) − y 2 (b)) + y 2 (b)β − y 2 (b)(y 1 (b) − y 2 (b))Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 25andy(a) = λy 1 (a) + (1 − λ)y 2 (a) = αWe can select λ so that y(b) = β.β = y(b) = λy 1 (b) + (1 − λ)y 2 (b)⇒ λ == λ(y 1 (b) − y 2 (b)) + y 2 (b)β − y 2 (b)(y 1 (b) − y 2 (b))In practice, we can solve the following two IVPs (in parallel)⎧⎨⎩y ′′ = u(x) + v(x)y + w(x)y ′y(a) = α, y ′ (a) = 0and⎧⎨⎩y ′′ = u(x) + v(x)y + w(x)y ′y(a) = α, y ′ (a) = 1Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003

BVP of ODE 26i.e.,⎧⎪⎨⎪⎩y ′ 1 = y 3y 3 ′ = y 1 ′′ = u + vy 1 + wy 1 ′ = u + vy 1 + wy 3y 1 (a) = α, y 3 (a) = y 1 ′ (a) = 0and⎧⎪⎨⎪⎩y 2 ′ = y 4y 4 ′ = u + vy 2 + wy 4y 2 (a) = α, y 4 (a) = 1to obtain approximate solutions y 1 and y 2 , then compute λ and form the solution y.Department of Mathematics – NTNU Tsung-Min Hwang December 20, 2003