# 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 $-n^2$.