ODE/PDE- eighenvalues question

  • Thread starter Thread starter Roni1985
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on solving the second-order ordinary differential equation (ODE) X'' + λX = 0 with boundary conditions X(0) = X(2π) and X'(0) = X'(2π). The participant encounters a challenge when λ = 0, leading to the general solution X(x) = C1 + C2*x. After applying the boundary conditions, they derive the relationship C1 = C1 + 2πC2, prompting a question about the value of C2. The participant seeks clarification on whether C2 can be any value or must be zero, which determines the nature of the solution (trivial vs. nontrivial).

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with boundary value problems
  • Knowledge of eigenvalue problems in differential equations
  • Basic calculus and linear algebra concepts
NEXT STEPS
  • Explore the implications of boundary conditions on ODE solutions
  • Study the behavior of solutions as parameters approach limits, specifically λ → 0
  • Investigate nontrivial solutions in eigenvalue problems
  • Learn about Sturm-Liouville theory and its applications in solving differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, boundary value problems, and eigenvalue theory. This discussion is beneficial for anyone looking to deepen their understanding of ODEs and their solutions under specific conditions.

Roni1985
Messages
200
Reaction score
0

Homework Statement


X''+[tex]\lambda[/tex]*X=0 0<x<2pi

X(0)=X(2pi) and X'(0)=X'(2pi)


Homework Equations





The Attempt at a Solution


My only problem is the case when [tex]\lambda[/tex]=0
in such case the general solution is X(x)= C1+C2*x

Now, after applying the BCs, this is what I get:

C1=C2*2pi
and

C2=C2

now, what should C2 be ? zero or any value ?
if C2 is zero, we have the trivial solution only. However, if C2 is any number, we have a nontrivial solution.

how can I solve this question?


Thanks.
 
Physics news on Phys.org
After applying the first BC, I get C1=C1+C2*2*pi, not C1=C2*2*pi.
 
Roni1985 said:
C1=C2*2pi

This should be

[tex]C_1 = C_1 + 2\pi C_2.[/tex]

How does the solution with [tex]\lambda\neq 0[/tex] behave in the limit as [tex]\lambda \rightarrow 0[/tex]? Is this consistent with your solution for [tex]\lambda = 0[/tex]?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
4
Views
2K