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In Mathematics / High School | 2025-07-03

Consider the boundary value problem

[tex]y'' + \lambda y = 0, x \in [0, \pi], y(0) = 0, y(\pi) = 0[/tex]

Let [tex]S_\lambda = {y_\lambda(x) | y_\lambda(x)[/tex] is an eigenfunction corresponding to [tex] \lambda}[/tex]. Then which of the following is/are FALSE?

(a) [tex]S_\lambda[/tex] is linearly independent (L.I.).
(b) The number of linearly independent functions in [tex]S_\lambda[/tex] is exactly one.
(c) If [tex]y_1(x) \in S_{\lambda_1}[/tex] and [tex]y_2(x) \in S_{\lambda_2}[/tex], where [tex] \lambda_1 \neq \lambda_2[/tex], then [tex]y_1[/tex] and [tex]y_2[/tex] are linearly independent.
(d) There exists [tex]\lambda[/tex] such that the number of linearly independent functions in [tex]S_\lambda[/tex] is more than one.

Asked by prophetessadams692

Answer (1)

To solve this problem, we need to analyze the given boundary value problem:
y ′′ + λ y = 0 , x ∈ [ 0 , π ] , y ( 0 ) = 0 , y ( π ) = 0
This is a second-order linear differential equation with boundary conditions. We are tasked with understanding the properties of the set S λ ​ , which consists of eigenfunctions corresponding to the eigenvalue λ .
Firstly, let's find the general solution to the differential equation y ′′ + λ y = 0 :

If λ = 0 , then the equation becomes y ′′ = 0 . The general solution is a linear function: y ( x ) = A + B x . The boundary conditions y ( 0 ) = 0 and y ( π ) = 0 would force A = 0 and B = 0 , giving a trivial solution.
If 0"> λ > 0 , set λ = k 2 . The differential equation becomes y ′′ + k 2 y = 0 with a general solution y ( x ) = A cos ( k x ) + B sin ( k x ) . The boundary conditions yield A = 0 and B sin ( kπ ) = 0 . For B not to be zero, k must be an integer, meaning k = n , where n is a positive integer.
Therefore, the eigenvalues are λ n ​ = n 2 and the corresponding eigenfunctions are y n ​ ( x ) = B sin ( n x ) .

Now, let's evaluate each statement:
(a) S λ ​ is linearly independent (L.I.).

An eigenfunction corresponding to a specific λ = n 2 is of the form y n ​ ( x ) = sin ( n x ) . Since there is only one solution for each λ n ​ , the set as defined does not contain multiple functions that we need to check for linear independence. Thus, this statement regarding linearly independence is not applicable in the typical sense of multiple functions forming linearly independent sets.

(b) The number of linearly independent functions in S λ ​ is exactly one.

For each λ = n 2 , there exists exactly one non-trivial solution ( y n ​ ( x ) = sin ( n x ) ). Thus, the number of linearly independent functions is indeed one.

(c) If y 1 ​ ( x ) ∈ S λ 1 ​ ​ and y 2 ​ ( x ) ∈ S λ 2 ​ ​ , where λ 1 ​  = λ 2 ​ , then y 1 ​ and y 2 ​ are linearly independent.

This is true because eigenfunctions corresponding to different eigenvalues of a Sturm-Liouville problem are orthogonal over the interval with respect to the weight function 1.

(d) There exists λ such that the number of linearly independent functions in S λ ​ is more than one.

This statement is false because for every eigenvalue λ = n 2 , there is exactly one non-trivial eigenfunction, so S λ ​ can not have more than one linearly independent function.

Therefore, the false statements are: (a) and (d) .

Answered by OliviaLunaGracy | 2025-07-06