Solve Eigenvalue Problem for ODE: Find Eigenvalue & Eigenfunctions

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Discussion Overview

The discussion revolves around an eigenvalue problem for a second-order ordinary differential equation (ODE) with specified boundary conditions. Participants explore the existence of eigenvalues and corresponding eigenfunctions, while addressing the implications of the characteristic equation and initial conditions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the eigenvalue problem and the characteristic equation, seeking to determine the existence of a single eigenvalue and its corresponding eigenfunctions.
  • Another participant points out a missing initial condition necessary for a complete Cauchy problem, questioning the clarity of the problem statement.
  • There is a discussion about the definition of an eigenvalue, with emphasis on the conditions that lead to non-trivial solutions.
  • Participants derive a system of equations for the constants in the general solution based on the boundary conditions.
  • One participant suggests that the eigenvalue could be λ = -1, prompting others to confirm or challenge this hypothesis.
  • Another participant notes that λ = -1 leads to identical equations, complicating the determination of unique constants.
  • There is a correction regarding the form of the eigenfunction when λ = -1, with a participant asserting that the solution should involve a double root, leading to a different expression for the eigenfunction.
  • Concerns are raised about the uniqueness of the solution when applying the boundary conditions, particularly regarding the implications of setting constants to zero.
  • Discussion includes the nature of eigenvalues and their relation to the uniqueness of solutions in the context of the eigenvalue problem.

Areas of Agreement / Disagreement

Participants express differing views on the correct form of the eigenfunction and the implications of the boundary conditions. There is no consensus on the uniqueness of the solution or the correctness of the proposed eigenvalue λ = -1, as some participants challenge earlier claims and provide corrections.

Contextual Notes

Participants highlight limitations in the problem statement, including the need for clear initial conditions and the implications of double roots in the characteristic equation. The discussion reflects uncertainty regarding the uniqueness of solutions based on the derived equations.

sigmund
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I have this eigenvalue problem:
[itex]\frac{\mbox{d}^2y}{\mbox{d}x^2}+\left(1-\lambda\right)\frac{\mbox{d}y}{\mbox{d}x}-\lambda y = 0 \ , \ x\in[0,1], \ \lambda\in\mathbb{R} [/itex]
[itex]y(0)=0[/itex]
[itex]\frac{\mbox{d}y}{\mbox{d}x}(1)=0[/itex]
Then, I have to show that there exists only one eigenvalue [itex]\lambda[/itex], and find this eigenvalue and write the corresponding eigenfunctions.

Thus far, I have solved the ODE's characteristic equation
[itex]r^2+(1-\lambda)r-\lambda=0[/itex].
This gives me two solutions
[itex]r=-1[/itex] and [itex]r=\lambda[/itex].
Thus the solution to the ODE is
[itex] y=c_1\exp(-x)+c_2\exp(\lambda x) \ , \ x\in\mathbb{R} \ , \ c_1, \ c_2\in\mathbb{R} [/itex].

Can I now conclude that because we only have real solutions to the characteristic equation, only one eigenvalue exists?

Secondly, I am not completely sure how to find the sought eigenvalue. I know how to find the eigenvalue when I have a solution that involves [itex]\sin[/itex] and [itex]\cos[/itex], but here I am not sure how to do it. Could anyone give me a hint?

Thirdly, when I have the eigenvalue there should not be any problems in writing the corresponding eigenfunctions, or is it?

I would appreciate any help. I am not looking for a solution to my homework problem, but any hints to the problem mentioned above are welcome.
 
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You miss text from the problem.In order for that to be a genuine Cauchy problem,you need to specify 2 initial conditions.Besides,you didn't use the one which u have posted...:wink:

Daniel.
 
An "eigenvalue" is a value of &lamba; such that the problem has a non-trivial solution. Have you determined what functions satisfy your other conditions?
(You have y(0)= 0 but then is see only [itex]\frac{dy}{dx}[/itex]. Did you mean [itex]\frac{dy}{dx}(0)= 0[/itex] or [itex]\frac{dy}{dx}(1)= 0[/itex]?)

The problem is a lot easier if it is [itex]\frac{dy}{dx}(0)= 0[/itex]!

Put x= 0 into the formulas for y and dy/dx and you get two equations to solve for c1 and c2. For what values of &lamda; can you NOT solve for specific value of c1 and c2?
 
He corrected his typo and now he can follow your advice for finding "lambda"...:approve:

Daniel.
 
Has anyone tried to calculate the eigenvalue? My suggestion is [itex]\lambda=-1[/itex]. Can anyone either confirm or refute this?
 
What is the system of equations for C_{1} & C_{2}...?

Daniel.
 
When I use the initial conditions, I get this system of equations:

[tex]c_1+c_2=0[/tex]
[tex]-c_1\exp(-1)+c_2\lambda\exp(\lambda)=0[/tex]

I then guess that [itex]\lambda=-1[/itex], and find out that with this eigenvalue you cannot solve the system for any particular value of [itex]c_1[/itex] and [itex]c_2[/itex].
 
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Yes,indeed,lambda=-1 makes the 2 equations identical.

Now find the eigenfunctions corresponding to [itex]\lambda=-1[/itex]

Daniel.
 
Deleted- DexterCioby beat me to it!
 
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  • #10
Halls,it has lambda =-1...:wink:

I "guess" -1 can be called elementary,huh...?

Daniel.
 
  • #11
Well, the eigenfunctions corresponding to [itex]\lambda=-1[/itex] must then be

[tex]y=c_1\exp(-x)+c_2\exp(-x)=(c_1+c_2)\exp(-x)=c\exp(-x),~~c_1,\,c_2,\,c\in\mathbb{R}[/tex]
 
  • #12
Yes.[itex]y(x)=C\exp(-x),C\in\mathbb{R}[/itex] is the awaited sollution.

Daniel.
 
  • #13
After I have talked to the teacher, I have realized that the eigenfunction written in post #11 is wrong. The reason is that when [itex]\lambda=-1[/itex], the characteristic equation has a double root [itex]r=-1[/itex]. Thus, the solution is [itex]y=c_1\exp(-x)+c_2x\exp(-x)[/itex], and NOT [itex]y=c\exp(-x)[/itex] as written earlier. Do you agree dextercioby?
 
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  • #14
sigmund said:
After I have talked to the teacher, I have realized that the eigenfunction written in post #11 is wrong. The reason is that when [itex]\lambda=-1[/itex], the characteristic equation has a double root [itex]r=-1[/itex]. Thus, the solution is [itex]y=c_1\exp(-x)+c_2x\exp(-x)[/itex], and NOT [itex]y=c\exp(-x)[/itex] as written earlier. Do you agree dextercioby?

I'm confused: To meet the initial condition y(0)=0, [itex]c_1[/itex] has to be zero. But if that's the case, then any value of [itex]c_2[/itex] meets the derivative at the boundary condition specified above and thus we loose uniqueness.
 
  • #15
Yes, that's the whole point of "eigenvalue". If λ is not an eigenvalue, then the equation Ax= &lamda;x has a unique solution: x= 0. If λ is an eigenvalue, then there exist an infinite number of solutions: the set of eigenvectors corresponding to a given eigenvalue is is a subspace of the original vectdor space.
 

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