Eigenfunction equation

I've been going round in circles with this problem for days:

Find the eigenvalues and associated normalised eigenfunctions of the operator L:

[tex]L_y = x^2 y'' + 2 xy' + \frac{y}{4}[/tex]

Boundary conditions [tex]y(1)=y(e)=0[/tex]

So what I've done:
substitute [tex]x = \exp(t)[/tex]

Then [tex]L_y = \frac{d^2y}{dt^2} + \frac{dy}{dt} + \frac{y}{4}[/tex] where the differentials are now y wrt t

The eigenvalue equation is [tex]L_y = X_n y[/tex] where [tex]X_n[/tex] are my eigenvalues

Then I solve [tex]y'' + y' + (1/4 - X_n)y = 0[/tex]

and get [tex]y = \exp (-1/2 t) (A \exp(+\sqrt(X_n)t) + B \exp(-\sqrt(X_n)t)[/tex]

Applying the boundary problems I find B = - A

and that [tex]X_n == 0[/tex]. Which doesn't seem right!

Am I on the right track here or have I missed the point totally!
 
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Can anyone help? Please say if i need to give more info on the problem
 

HallsofIvy

Science Advisor
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I've been going round in circles with this problem for days:

Find the eigenvalues and associated normalised eigenfunctions of the operator L:

[tex]L_y = x^2 y'' + 2 xy' + \frac{y}{4}[/tex]

Boundary conditions [tex]y(1)=y(e)=0[/tex]

So what I've done:
substitute [tex]x = \exp(t)[/tex]

Then [tex]L_y = \frac{d^2y}{dt^2} + \frac{dy}{dt} + \frac{y}{4}[/tex] where the differentials are now y wrt t

The eigenvalue equation is [tex]L_y = X_n y[/tex] where [tex]X_n[/tex] are my eigenvalues

Then I solve [tex]y'' + y' + (1/4 - X_n)y = 0[/tex]
And your boundary conditions are now y(0)= y(1)= 0.

and get [tex]y = \exp (-1/2 t) (A \exp(+\sqrt(X_n)t) + B \exp(-\sqrt(X_n)t)[/tex]

Applying the boundary problems I find B = - A

and that [tex]X_n == 0[/tex]. Which doesn't seem right!

Am I on the right track here or have I missed the point totally!
You're doing fine except that you are ASSUMING that Xn is not negative (since you are taking its square root). What happens if Xn is negative?
 
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