Boundary Value Problems (Sect. 10.1). Two-point BVP. from physics. Comparison: IVP vs BVP. Existence, uniqueness of solutions to BVP. Particular case of BVP: Eigenvalue-eigenfunction problem. Two-point Boundary Value Problem. Definition A two-point BVP is the following: Given functions p, q, g, and constants x 1 < x 2, y 1, y 2, b 1, b 2, b1, b 2, find a function y solution of the differential equation y + p(x) y + q(x) y = g(x), together with the extra, boundary conditions, Remarks: b 1 y(x 1 ) + b 2 y (x 1 ) = y 1, b 1 y(x 2 ) + b 2 y (x 2 ) = y 2. Both y and y might appear in the boundary condition, evaluated at the same point. In this notes we only study the case of constant coefficients, y + a 1 y + a 0 y = g(x).
Two-point Boundary Value Problem. s of BVP. Assume x 1 x 2. (1) Find y solution of y + a 1 y + a 0 y = g(x), y(x 1 ) = y 1, y(x 2 ) = y 2. (2) Find y solution of y + a 1 y + a 0 y = g(x), y (x 1 ) = y 1, y (x 2 ) = y 2. (3) Find y solution of y + a 1 y + a 0 y = g(x), y(x 1 ) = y 1, y (x 2 ) = y 2. Boundary Value Problems (Sect. 10.1). Two-point BVP. from physics. Comparison: IVP vs BVP. Existence, uniqueness of solutions to BVP. Particular case of BVP: Eigenvalue-eigenfunction problem.
from physics. Problem: The equilibrium (time independent) temperature of a bar of length L with insulated horizontal sides and the bar vertical extremes kept at fixed temperatures T 0, T L is the solution of the BVP: T (x) = 0, x (0, L), T (0) = T 0, T (L) = T L, y insulation T 0 T L 0 x L x z insulation Boundary Value Problems (Sect. 10.1). Two-point BVP. from physics. Comparison: IVP vs BVP. Existence, uniqueness of solutions to BVP. Particular case of BVP: Eigenvalue-eigenfunction problem.
Comparison: IVP vs BVP. Review: IVP: Find the function values y(t) solutions of the differential equation y + a 1 y + a 0 y = g(t), together with the initial conditions y(t 0 ) = y 1, y (t 0 ) = y 2. Remark: In physics: y(t): Position at time t. Initial conditions: Position and velocity at the initial time t 0. Comparison: IVP vs BVP. Review: BVP: Find the function values y(x) solutions of the differential equation y + a 1 y + a 0 y = g(x), together with the initial conditions y(x 1 ) = y 1, y(x 2 ) = y 2. Remark: In physics: y(x): A physical quantity (temperature) at a position x. Boundary conditions: Conditions at the boundary of the object under study, where x 1 x 2.
Boundary Value Problems (Sect. 10.1). Two-point BVP. from physics. Comparison: IVP vs BVP. Existence, uniqueness of solutions to BVP. Particular case of BVP: Eigenvalue-eigenfunction problem. Existence, uniqueness of solutions to BVP. Review: The initial value problem. Theorem (IVP) Consider the homogeneous initial value problem: y + a 1 y + a 0 y = 0, y(t 0 ) = y 0, y (t 0 ) = y 1, and let r ± be the roots of the characteristic polynomial p(r) = r 2 + a 1 r + a 0. If r + r, real or complex, then for every choice of y 0, y 1, there exists a unique solution y to the initial value problem above. Summary: The IVP above always has a unique solution, no matter what y 0 and y 1 we choose.
Existence, uniqueness of solutions to BVP. Theorem (BVP) Consider the homogeneous boundary value problem: y + a 1 y + a 0 y = 0, y(0) = y 0, y(l) = y 1, and let r ± be the roots of the characteristic polynomial p(r) = r 2 + a 1 r + a 0. (A) If r + r, real, then for every choice of L 0 and y 0, y 1, there exists a unique solution y to the BVP above. (B) If r ± = α ± iβ, with β 0, and α, β R, then the solutions to the BVP above belong to one of these possibilities: (1) There exists a unique solution. (2) There exists no solution. (3) There exist infinitely many solutions. Existence, uniqueness of solutions to BVP. Proof of IVP: We study the case r + r -. The general solution is y(t) = c 1 e r - t + e r + t, c 1, R. The initial conditions determine c 1 and as follows: y 0 = y(t 0 ) = c 1 e r - t 0 + e r + t 0 y 1 = y (t 0 ) = c 1 r - e r - t 0 + r + e r + t 0 Using matrix notation, e r - t 0 e r + t 0 c1 r - e r -t 0 r + e r + t 0 = The linear system above has a unique solution c 1 and for every constants y 0 and y 1 iff the det(z) 0, where e r - t 0 e Z = r + t 0 c1 y0 r - e r - t 0 r + e r + t 0 Z =. y0 y 1. y 1
Existence, uniqueness of solutions to BVP. Proof of IVP: e r - t 0 e Recall: Z = r + t 0 r - e r - t 0 r + e r + t 0 Z c1 = y0 y 1. A simple calculation shows det(z) = ( r + r - ) e (r + +r - ) t0 0 r + r -. Since r + r -, the matrix Z is invertible and so c1 = Z 1 y0. y 1 We conclude that for every choice of y 0 and y 1, there exist a unique value of c 1 and, so the IVP above has a unique solution. Existence, uniqueness of solutions to BVP. Proof of BVP: The general solution is y(x) = c 1 e r - x + e r + x, c 1, R. The boundary conditions determine c 1 and as follows: y 0 = y(0) = c 1 +. y 1 = y(l) = c 1 e r - L + e r + L Using matrix notation, 1 1 c1 e r - L e r + L = The linear system above has a unique solution c 1 and for every constants y 0 and y 1 iff the det(z) 0, where 1 1 c1 y0 Z = Z =. e r - L e r + L y0 y 1. y 1
Existence, uniqueness of solutions to BVP. 1 1 Proof of IVP: Recall: Z = A simple calculation shows e r - L e r + L Z c1 det(z) = e r + L e r - L 0 e r + L e r - L. (A) If r + r - and real-valued, then det(z) 0. = We conclude: For every choice of y 0 and y 1, there exist a unique value of c 1 and, so the BVP in (A) above has a unique solution. (B) If r ± = α ± iβ, with α, β R and β 0, then det(z) = e αl( e iβl e iβl) y0 y 1. det(z) = 2i e αl sin(βl). Since det(z) = 0 iff βl = nπ, with n integer, (1) If βl nπ, then BVP has a unique solution. (2) If βl = nπ then BVP either has no solutions or it has infinitely many solutions. Existence, uniqueness of solutions to BVP. Find y solution of the BVP y + y = 0, y(0) = 1, y(π) = 1. Solution: The characteristic polynomial is The general solution is The boundary conditions are p(r) = r 2 + 1 r ± = ±i. y(x) = c 1 cos(x) + sin(x). 1 = y(0) = c 1, 1 = y(π) = c 1 c 1 = 1, free. We conclude: y(x) = cos(x) + sin(x), with R. The BVP has infinitely many solutions.
Existence, uniqueness of solutions to BVP. Find y solution of the BVP y + y = 0, y(0) = 1, y(π) = 0. Solution: The characteristic polynomial is The general solution is The boundary conditions are p(r) = r 2 + 1 r ± = ±i. y(x) = c 1 cos(x) + sin(x). 1 = y(0) = c 1, 0 = y(π) = c 1 The BVP has no solution. Existence, uniqueness of solutions to BVP. Find y solution of the BVP y + y = 0, y(0) = 1, y(π/2) = 1. Solution: The characteristic polynomial is The general solution is The boundary conditions are p(r) = r 2 + 1 r ± = ±i. y(x) = c 1 cos(x) + sin(x). 1 = y(0) = c 1, 1 = y(π/2) = c 1 = = 1. We conclude: y(x) = cos(x) + sin(x). The BVP has a unique solution.
Boundary Value Problems (Sect. 10.1). Two-point BVP. from physics. Comparison: IVP vs BVP. Existence, uniqueness of solutions to BVP. Particular case of BVP: Eigenvalue-eigenfunction problem. Particular case of BVP: Eigenvalue-eigenfunction problem. Problem: Find a number λ and a non-zero function y solutions to the boundary value problem Remark: This problem is similar to the eigenvalue-eigenvector problem in Linear Algebra: Given an n n matrix A, find λ and a non-zero n-vector v solutions of Differences: A Av λ v = 0. { } computing a second derivative and v {a function y}. applying the boundary conditions.
Particular case of BVP: Eigenvalue-eigenfunction problem. Find every λ R and non-zero functions y solutions of the BVP Remarks: We will show that: (1) If λ 0, then the BVP has no solution. (2) If λ > 0, then there exist infinitely many eigenvalues λ n and eigenfunctions y n, with n any positive integer, given by ( nπ ) 2, ( nπx ) λ n = yn (x) = sin, L L (3) Analogous results can be proven for the same equation but with different types of boundary conditions. For example, for y(0) = 0, y (L) = 0; or for y (0) = 0, y (L) = 0. Particular case of BVP: Eigenvalue-eigenfunction problem. Find every λ R and non-zero functions y solutions of the BVP Solution: Case λ = 0. The equation is y = 0 y(x) = c 1 + x. The boundary conditions imply 0 = y(0) = c 1, 0 = c 1 + L c 1 = = 0. Since y = 0, there are NO non-zero solutions for λ = 0.
Particular case of BVP: Eigenvalue-eigenfunction problem. Find every λ R and non-zero functions y solutions of the BVP Solution: Case λ < 0. Introduce the notation λ = µ 2. The characteristic equation is The general solution is The boundary condition are p(r) = r 2 µ 2 = 0 r ± = ±µ. y(x) = c 1 e µx + e µx. 0 = y(0) = c 1 +, 0 = y(l) = c 1 e µl + e µl. Particular case of BVP: Eigenvalue-eigenfunction problem. Find every λ R and non-zero functions y solutions of the BVP Solution: Recall: y(x) = c 1 e µx + e µx and c 1 + = 0, c 1 e µl + e µl = 0. We need to solve the linear system 1 1 c1 0 = Z 0 e µl e µl c1 = 0 1 1, Z = 0 e µl e µl Since det(z) = e µl e µl 0 for L 0, matrix Z is invertible, so the linear system above has a unique solution c 1 = 0 and = 0. Since y = 0, there are NO non-zero solutions for λ < 0.
Particular case of BVP: Eigenvalue-eigenfunction problem. Find every λ R and non-zero functions y solutions of the BVP Solution: Case λ > 0. Introduce the notation λ = µ 2. The characteristic equation is The general solution is The boundary condition are p(r) = r 2 + µ 2 = 0 r ± = ±µi. y(x) = c 1 cos(µx) + sin(µx). 0 = y(0) = c 1, y(x) = sin(µx). 0 = y(l) = sin(µl), 0 sin(µl) = 0. Particular case of BVP: Eigenvalue-eigenfunction problem. Find every λ R and non-zero functions y solutions of the BVP Solution: Recall: c 1 = 0, 0, and sin(µl) = 0. The non-zero solution condition is the reason for 0. Hence sin(µl) = 0 µ n L = nπ µ n = nπ L. Recalling that λ n = µ 2 n, and choosing = 1, we conclude ( nπ λ n = L ) 2, ( nπx ) yn (x) = sin. L