Why Does the Eigenvalue Equation Use Positive Lambda in the Solution?

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Homework Help Overview

The discussion revolves around the eigenvalue equation \(\frac{\partial^2 \phi}{\partial x^2} + \lambda \phi = 0\) and the implications of using positive eigenvalues (\(\lambda > 0\)) in the context of finding solutions. Participants are analyzing the nature of the solutions and the reasoning behind the use of positive \(\lambda\) in the square root.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the rationale for using positive \(\lambda\) in the square root and discussing the implications of using negative values. There are mentions of the general nature of the solutions and the need for specific inputs to derive particular solutions.

Discussion Status

The discussion is active, with participants exploring different interpretations of the eigenvalue equation. Some guidance has been offered regarding the nature of the solutions, particularly the transition from trigonometric to exponential forms based on the sign of \(\lambda\). However, there is no explicit consensus on the interpretation of \(\lambda\) at this stage.

Contextual Notes

Participants are navigating the complexities of eigenvalue problems, particularly in the context of differential equations. There is an acknowledgment of the general solution form and the conditions under which specific solutions can be derived.

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[tex]\frac{\partial^2 \phi}{\partial x^2} + \lambda \phi = 0[/tex]

We have to analyze eigenvalues.

My prof. gave me this answer for [tex]\lambda>0[/tex] it is
[tex]c*sin(\sqrt(\lambda))x+d*cos(\sqrt(\lambda))x[/tex]

Shouldn't it be -[tex]\lambda[/tex] inside the squareroot? If not can someone explain how he got this?
 
Last edited:
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The problem you'er presenting is homogenic.
sin and cos are natural solutions for this kind of equations.
pay attention that if you use diff on sin twice, you get -sin.
same goes for cos.
c,d = constants
you need 2 inputs to get the private solution. otherwisw it's a general solution.
 
I know its a general solution but I don't understand how he got the positive [tex]\lambda[/tex] inside the squareroot.
 
Looks right to me.. for lambda > 0, the eigenvalue is imaginary.

Yes, it is [tex]\sqrt(\lambda)[/tex] or [tex]sqrt(/lambda))i[/tex]

should be square root of neg lambda in the first.. give me a while to get latex right..
 
Last edited:
Your prof was right.

If you put a minus sign in the square root, you would have to change the sin and cosine to exponential functions. I.e. [tex]C \exp(\sqrt{-\lambda}x)+D\exp(\sqrt{-\lambda}x).[/tex]

Carl
 

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