eigenfunctions

Poirot

Banned
consider $$\frac{d^2}{dx^2}(xy) - λxy=0$$. Show eigenfunctions are $$y_{n}=\frac{\sin(n\pi x)}{x}$$. Boundary conditions are y(1)=0 and y regular at x=0

I integrated twice to obtain $$6xy=λx^3y+6Ax+6B$$ where A,B constants. I can't apply the condition y is regular because I don't know what it means pratically. Besides, I can't see how I can get the required solution from this equation.

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Ackbach

Indicium Physicus
Staff member
Your "twice integrating" is invalid, because you don't know what $y$ is, aside from being a function of $x$. I suggest letting a new function $u=u(x)=xy(x)$. Then the DE becomes $u''=\lambda u$. What are the solutions of that DE?

Can't help you with the "regular at $x=0$", other than it sounds like something to do with differentiability at the origin.

Poirot

Banned
It says, as a boundary condition, y(x) must be regular at the singular point x=0. I see my mistake integrating, I was treating y as a constant w.r.t x.

Ackbach

Indicium Physicus
Staff member
I've got a line on the "regularity" thing. Solve the DE I posted in my previous post. What do you get?

Poirot

Banned
ok so characteristic equation is m^2=λ, so 3 cases to consider. λ=0, λ<0 and λ>0. I can solve when I know boundary conditions.

HallsofIvy

Well-known member
MHB Math Helper
You are not asked to solve the equation- you are given a possible solution and asked to show that it does, in fact, satisfy the differential equation. That requires that you differentiate the given function, not integrate anything.

Poirot

Banned
You're wrong. It asks 'show that the eigenfuctions are .....' i.e you have to show there are no others.

Ackbach

Indicium Physicus
Staff member
You're wrong. It asks 'show that the eigenfuctions are .....' i.e you have to show there are no others.
I could be wrong, but I don't think you have to show uniqueness. Eigenvectors, for example, are not unique. In fact, a scalar times an eigenvector is an eigenvector. Similarly, in this case, at the very least, a constant times an eigenfunction is an eigenfunction. Moreover, I think you show that the functions $e^{in\pi x}/x$ are eigenfunctions.

Opalg

MHB Oldtimer
Staff member
Following Ackbach's comment #2 above, let $u=xy$. Then $u''=\lambda u$. That is an SHM equation with solution $u=A\cos(\omega x) + B\sin(\omega x)$, where $\omega = \sqrt{-\lambda}$.

Thus $y = \dfrac{A\cos(\omega x) + B\sin(\omega x)}x$. The boundary condition that $y$ is regular at $x=0$ means that $y$ should not go to infinity at $x=0$. That tells you that $A=0$. The other boundary condition $y(1)=0$ then tells you that $\omega=n\pi$ for some integer $n$. Therefore the eigenfunctions are $\dfrac{\sin(n\pi x)}x$, with corresponding eigenvalues $-n^2$.