Eigenvalue Problem. 1 The First (Dirichlet) Eigenvalue Problem

Transcription

1 Eigenvalue Problem A. Salih Department of Aerospace Engineering Indian Institute of Space Science and Technology, Thiruvananthapuram July 06 The method of separation variables for solving the heat equation often leads transformation of PDEs to ODEs. The ODE thus obtained will be in the form of an eigenvalue problem. Four types of eigenvalue problems are commonly encountered as discussed below. The First Dirichlet Eigenvalue Problem Consider the second order differential equation with Dirichlet type boundary conditions d X + λx = 0, 0<x< a dx X0 = 0, X = 0 b The problem here is to find the values of λ and the nontrivial solutions Xx of. All the conditions in this problem are linear and homogeneous, and so any nonzero constant times a nontrivial solution Xx is essentially the same as Xx. This problem is called an eigenvalue problem. The Dirichlet eigenvalue problem involves the determination of a solution Xx of in a domain [0,] for some λ that satisfies the boundary conditions X0=X=0. This is a special case of more general problem called Sturm-iouville problem. The possible solutions of fall into the following three cases: Case λ = 0 If λ = 0, the differential equation a reduces to and its general solution is X = 0 Xx = c + c x where c and c are constants. The boundary condition X0 = 0 requires that c = 0. So Xx=c x. Further, the boundary condition X=0 requires that c = 0. So the eigenvalue problem has only trivial solution Xx 0 if λ = 0 and hence λ = 0 is not an eigenvalue. Case λ < 0

2 If λ < 0, we write λ = ω ω > 0 and the differential equation a becomes The condition X0=0 requires that Hence But the condition X=0 requires that X ω X = 0 Xx = c e ωx + c e ωx c = c Xx = c e ωx e ωx = c sinhωx c sinhω = 0 This implies that c = 0. Again we see that the eigenvalue problem has only trivial solution Xx 0 if λ < 0 and hence negative values of λ are not eigenvalues. Case 3 λ > 0 If λ > 0, we write λ = ω ω > 0 and the differential equation becomes X + ω X = 0 Xx = c cosωx + c sinωx The condition X0=0 requires that c = 0. Hence Xx = c sinωx Further, the boundary condition X = 0 requires that c sinω = 0 We see that c = 0 results in trivial solution of. Hence for nontrivial solution ω must be positive roots of the equation sinω = 0 from which we have ω = nπ So, except for the constant factor c Xx = sin nπx

3 and the corresponding values of λ are λ = ω = nπ Thus the discrete values λ n = nπ of λ for which problem has nontrivial solutions are called eigenvalues of that problem, and the solutions are the corresponding eigenfunctions. X n x = sin nπx 3 The Second Neumann Eigenvalue Problem The second eigenvalue problem Sturm-iouville problem that comes up frequently is with Neumann type boundary conditions d X + λx = 0, 0<x< 4a dx X 0 = 0, X = 0 4b This problem is called a Neumann eigenvalue problem. By the Neumann eigenvalue problem we mean the determination of a solution Xx of 4 in a domain [0,] for some λ that satisfies the boundary conditions X 0=X =0. The possible solutions of 4 fall into the following three cases: Case λ = 0 If λ = 0, the differential equation 4a has the general solution Xx = c + c x where c and c are constants. Application of boundary conditions requires the first derivative of X: X = c The boundary condition X 0=0 requires that c = 0. So Xx = c Further, the boundary condition X =0 automatically satisfied. So the eigenvalue problem 4 has a nontrivial solution if λ = 0 and hence λ 0 = 0 is an eigenvalue. 3

4 Case λ < 0 If λ < 0, we write λ = ω ω > 0 and the differential equation 4a becomes and its derivative is given by The condition X 0=0 requires that Hence X ω X = 0 Xx = c e ωx + c e ωx X = c ωe ωx c ωe ωx c = c Xx = c e ωx + e ωx = c coshωx and X = c ω sinhωx But the condition X =0 requires that c ω sinhω = 0 This implies that c = 0 and thus the eigenvalue problem 4 has only trivial solution Xx 0 if λ < 0 and hence the eigenvalues cannot be negative. Case 3 λ > 0 If λ > 0, we write λ = ω ω > 0 and the differential equation becomes and its derivative is given by X + ω X = 0 Xx = c cosωx + c sinωx X = c ω sinωx + c ω cosωx The condition X 0=0 requires that c = 0. Hence Xx = c cosωx and X = c ω sinωx The boundary condition X =0 requires that c ω sinω = 0 We see that c = 0 results in trivial solution of 4. Hence for nontrivial solution ω must be positive roots of the equation sinω = 0 4

5 from which we have ω = nπ So, except for the constant factor c Xx = cos nπx and the corresponding values of λ are nπ λ = ω = Thus the discrete eigenvalues are λ n = nπ 5 and the corresponding eigenfunctions are given by X n x = cos nπx 6 3 The Third Mixed Dirichlet-Neumann Eigenvalue Problem The first version The third eigenvalue problem Sturm-iouville problem that comes up frequently is with the boundary conditions of the type d X + λx = 0, 0<x< 7a dx X0 = 0, X = 0 7b where the differential equation and the domain is same as in problem and the possible solutions of 7 fall into the following three cases: Case λ = 0 If λ = 0, the differential equation 7a has the general solution Xx = c + c x where c and c are constants. The boundary condition X0=0 requires that c = 0. So Xx = c x and X = c The boundary condition X =0 requires that c = 0. So the eigenvalue problem 7 has only a trivial solution if λ = 0 and hence λ = 0 is not an eigenvalue. 5

6 Case λ < 0 If λ < 0, we write λ = ω ω > 0 and the differential equation 7a becomes X ω X = 0 Xx = c e ωx + c e ωx and its derivative is given by X = c ωe ωx c ωe ωx The condition X0=0 requires that c = c Hence Xx = c e ωx e ωx = c sinhωx and X = c ω coshωx But the condition X =0 requires that c ω coshω = 0 This implies that c = 0 and thus the eigenvalue problem 7 has only trivial solution Xx 0 if λ < 0 and hence the eigenvalues cannot be negative. Case 3 λ > 0 If λ > 0, we write λ = ω ω > 0 and the differential equation becomes X + ω X = 0 Xx = c cosωx + c sinωx The condition X0=0 requires that c = 0. Hence Xx = c sinωx and X = c ω cosωx The boundary condition X =0 requires that c ω cosω = 0 We see that c = 0 results in trivial solution of 7. Hence for nontrivial solution ω must be positive roots of the equation cosω = 0 from which we have π ω = 6

7 So, except for the constant factor c Xx = sin and the corresponding values of λ are πx λ = ω = [ ] π Thus the discrete eigenvalues are λ n = [ ] π 8a and the corresponding eigenfunctions are given by X n x = sin πx 8b The second version This a variation of the third eigenvalue problem Sturm-iouville problem of the first version. with the boundary conditions of the type The possible solutions of 9 fall into the following three cases: Case λ = 0 If λ = 0, the differential equation 9a has the general solution d X + λx = 0, 0<x< 9a dx X 0 = 0, X = 0 9b Xx = c + c x where c and c are constants. Application of boundary conditions requires the first derivative of X: X = c The boundary condition X 0=0 requires that c = 0. So Xx = c The boundary condition X=0 requires that c = 0. So the eigenvalue problem 9 has only a trivial solution if λ = 0 and hence λ = 0 is not an eigenvalue. 7

8 Case λ < 0 If λ < 0, we write λ = ω ω > 0 and the differential equation 9a becomes and its derivative is given by The condition X 0=0 requires that Hence But the condition X=0 requires that X ω X = 0 Xx = c e ωx + c e ωx X = c ωe ωx c ωe ωx c = c Xx = c e ωx + e ωx = c coshωx c coshω = 0 This implies that c = 0 and thus the eigenvalue problem 9 has only trivial solution Xx 0 if λ < 0 and hence the eigenvalues cannot be negative. Case 3 λ > 0 If λ > 0, we write λ = ω ω > 0 and the differential equation becomes and its derivative is given by X + ω X = 0 Xx = c cosωx + c sinωx X = c ω sinωx + c ω cosωx The condition X 0=0 requires that c = 0. Hence Xx = c cosωx The boundary condition X=0 requires that c cosω = 0 We see that c = 0 results in trivial solution of 9. Hence for nontrivial solution ω must be positive roots of the equation cosω = 0 8

9 from which we have ω = π So, except for the constant factor c Xx = cos and the corresponding values of λ are πx λ = ω = [ ] π Thus the discrete eigenvalues are λ n = [ ] π 0a and the corresponding eigenfunctions are given by X n x = cos πx 0b 4 The Fourth Periodic Eigenvalue Problem Consider the second order differential equation with the boundary conditions d X + λx = 0, 0<x< a dx X0 = X, X 0 = X b This problem is called a periodic eigenvalue problem. By the periodic eigenvalue problem we mean the determination of a solution Xx of in a periodic domain [0,] for some λ that satisfies the periodic boundary conditions b. This is a special case of more general problem called periodic Sturm-iouville problem. The possible solutions of fall into the following three cases: Case λ = 0 If λ = 0, the differential equation a has the general solution Xx = c + c x 9

10 where c and c are constants. The boundary condition X0= X requires that c = 0 and Xx=c. So the eigenvalue problem has a nontrivial solution if λ = 0 and hence λ 0 = 0 is an eigenvalue with a corresponding eigenfunction. Case λ < 0 If λ < 0, we write λ = ω ω > 0 and the differential equation a becomes The condition X0=X gives and condition X 0=X gives X ω X = 0 Xx = c e ωx + c e ωx c + c = c e ω + c e ω c c = c e ω c e ω On simplification the above two equations gives e ω = e ω which cannot be satisfied by any nonzero values of ω and hence λ < 0 are not eigenvalues of. Case 3 λ > 0 If λ > 0, we write λ = ω ω > 0 and the differential equation becomes The condition X0=X gives and condition X 0=X gives X + ω X = 0 Xx = c cosωx + c sinωx c = c cosω + c sinω c = c cosω c sinω For nonzero values of c and c, on simplification the above two equations yields sinω = 0 and cosω = 0

11 both of which are satisfied if Therefore the discrete eigenvalues are ω = nπ λ n = nπ and we obtain two linearly independent eigenfunctions cos nπx and sin nπx corresponding to the same eigenvalue λ n. In conclusion, the solution of the periodic eigenvalue problem is the following infinite sequence of eigenvalues λ 0 = 0, λ n = nπ a and the corresponding eigenfunctions X 0 x =, X n x = cos nπx, sin nπx b It may be noted that for the eigenvalue problems previously considered Dirichlet, Neumann, and mixed Dirichlet-Neumann, we saw that there exists only one linearly independent eigenfunction X n corresponding to the eigenvalue λ n, which is called an eigenvalue of multiplicity one or a simple eigenvalue. On the contrary, for the periodic eigenvalue problem, the eigenfunctions cosnπx/ and sinnπx/ correspond to the same eigenvalue nπ/. Thus, this eigenvalue is of multiplicity two. Summary We give a summary of the results of various eigenvalue problems:. First Dirichlet eigenvalue problem: X0 = X = 0 = X n x = sin nπx. Second Neumann eigenvalue problem: X 0 = X = 0 = X n x = cos nπx n = 0,,,

12 3. Third mixed Dirichlet-Neumann eigenvalue problem I: X0 = X = 0 = X n x = sin πx 4. Third mixed Dirichlet-Neumann eigenvalue problem II: X 0 = X = 0 = X n x = cos πx 5. Fourth periodic eigenvalue problem: X0 = X, X 0 = X = X 0 =, X n x = cos nπx nπx, sin