Differential Equations

Linear Homogeneous Boundary Value Problems Dr. P. Dhanumjaya

Linear Homogeneous Boundary Value Problems – p.1/19

Eigenvalues and Eigenfunctions Consider the boundary value problem (BVP) consisting of the differential equation (1)

y 00 + P (x)y 0 + Q(x)y = 0, 0 < x < 1,

and the boundary conditions (2) a1 y(0) + a2 y

0

(0) = 0, b1 y(1) + b2 y 0 (1) = 0.

Linear Homogeneous Boundary Value Problems – p.2/19

Note 1 All homogeneous problems have the solution y = 0. This solution is called trivial solution which is of no interest. The significant question is whether there are other nontrivial solutions.

Linear Homogeneous Boundary Value Problems – p.3/19

Example 1 Find y = Φ(x) satisfying y 00 + y = 0, y(0) = 0, y(1) = 0. Solution. The general solution for the given differential equation is

y(x) = C1 sin x + C2 cos x, where C1 and C2 are any arbitrary constants.

Linear Homogeneous Boundary Value Problems – p.4/19

The first boundary condition y(0) = 0 gives C2 = 0. The second boundary condition y(1) = 0 gives C1 = 0. Therefore, the only solution is Φ(x) = 0, the trivial solution.

Linear Homogeneous Boundary Value Problems – p.5/19

Example 2 Find y = Φ(x) satisfying y 00 + π 2 y = 0, y(0) = 0, y(1) = 0. Solution. The general solution for the given differential equation is

y(x) = C1 sin πx + C2 cos πx, where C1 and C2 are any arbitrary constants.

Linear Homogeneous Boundary Value Problems – p.6/19

The first boundary condition y(0) = 0 gives C2 = 0. The function y = C1 sin πx,

is also satisfies the second boundary condition for any constant C1 .

Linear Homogeneous Boundary Value Problems – p.7/19

Hence, Φ(x) = C1 sin πx,

where C1 is arbitrary.

The given problem has a single infinite family of solutions, each is a multiple of sin πx.

Linear Homogeneous Boundary Value Problems – p.8/19

Example 3 Find y = Φ(x) satisfying y 00 + π 2 y = 0, y(0) + y(1) = 0, y 0 (0) + y 0 (1) = 0. Solution. The general solution for the given differential equation is

y(x) = C1 sin πx + C2 cos πx, where C1 and C2 are any arbitrary constants.

Linear Homogeneous Boundary Value Problems – p.9/19

The first boundary condition gives y(0) + y(1) = C2 − C2 = 0.

Thus, the first boundary condition imposes no restriction on the arbitrary constants C1 and C2 . The second boundary condition gives y 0 (0) + y 0 (1) = C1 π − C1 π = 0.

Thus, the second boundary condition also satisfied for all values of C1 and C2 .

Linear Homogeneous Boundary Value Problems – p.10/19

Therefore, the function y = Φ(x) is the most general solution of the boundary value problem. In this case, there are two infinite family of solutions, one proportional to Φ1 (x) = sin πx, and the other proportional to Φ2 (x) = cos πx.

Linear Homogeneous Boundary Value Problems – p.11/19

More generally, we consider a boundary value problem in which the differential equation contains an arbitrary parameter λ, for example y 00 + λy = 0, y(0) = 0, y(1) = 0. As we have seen λ = 1 the BVP has only the trivial solution.

Linear Homogeneous Boundary Value Problems – p.12/19

If λ = π 2 the BVP has non trivial solutions. This raises the question of what happens when λ takes on some other value. The solution of the differential equation depends on λ, so it is necessary to consider several cases.

Linear Homogeneous Boundary Value Problems – p.13/19

Case(i).

If λ = 0.

The general solution is y(x) = C1 x + C2 , where C1 and C2 are any arbitrary constants. The boundary condition y(0) = 0 gives C2 = 0. The boundary condition y(1) = 0 gives C1 = 0. The BVP has nonontrivial solution.

Linear Homogeneous Boundary Value Problems – p.14/19

Case(ii).

If λ < 0.

Let λ = −µ, so that µ > 0. The given differential equation rewritten as y 00 − µy = 0. The general solution is

y(x) = C1 e−µx + C2 eµx , where C1 and C2 are any arbitrary constants. Using the two boundary conditions, we get trivial solution y = 0.

Linear Homogeneous Boundary Value Problems – p.15/19

Case(iii).

If λ > 0.

The general solution is √ √ y(x) = C1 sin λx + C2 cos λx,

λ > 0.

The boundary condition y(0) = 0 gives C2 = 0. This gives √ y(x) = C1 sin λx.

Linear Homogeneous Boundary Value Problems – p.16/19

Using the second boundary condition y(1) = 0, we get √ 0 = C1 sin λ. For a nontrivial solution, we must have C1 6= 0. This gives √ sin λ = 0 = sin nπ.

This implies λ = n2 π 2 , n = 1, 2, 3, · · · ,

Linear Homogeneous Boundary Value Problems – p.17/19

Therefore, the BVP has nontrivial solution exist only for certain real values of λ namely 2

2

2

2 2

λ = π , 4π , 9π , · · · , n π , · · ·

The values of λ are called eigenvalues of the BVP. The corresponding nontrivial solutions y(x) = Φ(x) = sin nπx are called eigenfunctions.

Linear Homogeneous Boundary Value Problems – p.18/19

Example 4 Find the eigenvalues and eigenfunctions of y 00 + λy = 0, y(0) = 0, y(π) = 0. Solution. The eigenvalues are

λ = 1, 4, 9, · · · , n2 , · · ·

and the corresponding eigenfunctions are y(x) = sin nx, n = 1, 2, 3, · · · ,

Linear Homogeneous Boundary Value Problems – p.19/19