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Finding eigenvalues and eigenfunctions

  • Thread starter jaejoon89
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  • #1
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Homework Statement



Given
X''(x) + lambda*X(x) = 0
X(0) = X'(0), X(pi) = X'(pi)

Find all eigenvalues and eigenfunctions.

Homework Equations



Case lambda = 0
Case lambda > 0
Case lambda < 0

The Attempt at a Solution



First case, X(x) = Ax + B but the function doesn't satisfy the boundary condition at pi.
Second case, lambda = k^2
X(x) = Acos(kx) + Bsin(kx)
X'(x) = Bkcos(kx) - Aksin(kx)
X(pi) - X'(pi) = Bsin(kpi) + Bk^2 sin(kpi)
=> 1+ k^2 = 0 so lambda = -1

a) I was told the eigenfunction for that is X(x) = Ce^x. How?

b) Also, is that just the lambda_0 or does it satisfy all possibly eigenfunctions (i.e. eigenvalue lambda_n)?
 

Answers and Replies

  • #2
Cyosis
Homework Helper
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This is an eigenvalue problem of the form [itex]D^2 f=-\lambda f[/itex], [itex]-\lambda[/itex] the eigenvalue. With [itex]D^2[/itex] being the operator that takes the derivative of f(x) with respect to x twice.

Now if Ce^x is an eigenfunction it needs to satisfy the eigenvalue problem. Plugging it in [itex]D^2 f=-\lambda f[/itex] yields [itex]Ce^x=-\lambda C e^x[/itex]. For which [itex]\lambda[/itex] is this equation satisfied? This should also answer question b.

Your argument to why [itex]\lambda=0[/itex] isn't an eigenvalue isn't entirely water tight (although [itex]\lambda=0[/itex] is indeed not an eigenvalue). For example for A=B=0, [itex]f(x)=Ax+B=0[/itex] satisfies the requirements just fine. Why is f(x)=0 not an eigenfunction?

For the second case you've written down the correct eigenfunction, [itex]f(x)=A \cos (\sqrt{\lambda}x)+B \sin (\sqrt{\lambda}x)[/itex]. Emphasis on [itex]\lambda>0[/itex]. You solve this equation for a [itex]\lambda<0[/itex]. Now you get a cosine and a sine with an imaginary argument, which can be written as real exponents, which is the eigenfunction for [itex]\lambda<0[/itex]. Try to find eigenvalues for [itex]\lambda>0[/itex] with regards to the second case.
 
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