Sturm-Liouville Problem: conditions over the coefficients

1. Jan 8, 2010

libelec

For a Sturm-Liouville problem, as in:

$$\frac{d}{{dx}}\left[ {r(x)\frac{{dy}}{{dx}}} \right] + [\lambda p(x) + q(x)]y = 0$$

I've read in several books that one assumes that p, q and r are continuous, real-valued and bounded in the interval I, r' is continuous, and p>0.

But I've never seen or understood the reason why we make this assumption. What's the difference? Can't p be complex-valued? Why can't q be unbounded?

2. Jan 9, 2010

HallsofIvy

Staff Emeritus
If any of p, q, or r are not continuous, you are going to have serious difficulties with extending solutions over a points of discontinuity. If r is not differentiable, then you are going to have difficulties with d/dx(r(x)dy/dx). As for "p> 0", I have not seen that. If r(x)= 0 anywhere on the given interval (and Sturm-Liouville problems are always boundary value problems defined on some interval) you will have a singularity. We must have $r(x)\ne 0$ which means, since it is continuous, it is of one sign. Since we can always multiply the entire equation by -1, we can assume that r(x)> 0, not p(x)> 0.

Perhaps you are confusing this with the equivalent form
$$\frac{d}{{dx}}\left[ {p(x)\frac{{dy}}{{dx}}} \right] + [q(x)+ \lambda r(x)]y = 0$$

Last edited: Jan 10, 2010
3. Jan 9, 2010

libelec

OK, but why are we having a singularity if r(x) = 0 for some x in the boundary interval? Is it because of Lagrange's exponential factor's form, $$\mu (x) = e^{\int\limits_{x_0 }^x {\frac{{Q(s)}}{{P(s)}}ds} }$$ ?

And why do we ask this coefficients to be real-valued? Is it for practical reasons, like the DE being related to a real problem that doesn't need to be complex valued?

Thanks very much for the rest.

4. Jan 9, 2010

LCKurtz

I think it is all tied up in the eigenfunction theory. If the leading coefficient is zero, you will have singularities as Halls has pointed out. The coefficient of lambda is the weight function in the definition of the inner product, so you want it positive. I haven't gone through it lately, but I believe that if the coefficients are not real you will have problems defining the inner product, having the eigenvalues be real, and getting orthogonality of the eigenfunctions.

5. Jan 9, 2010

libelec

But doesn't using the complex inner product for functions solve that problem?

6. Jan 10, 2010

HallsofIvy

Staff Emeritus
It's not so much a matter of being "real valued" as of being "finite"!

Any differential equation can be reduced, eventually, to "dX/dt= F(X,t)" where X may be a vector valued function. For example, the equation $p(x,t)d^2x/dt^2+ q(x,t)dx/dt+ r(x,t)= 0$ can be reduced by letting y= dx/dt so that $p(x,t) dy/dt+ q(x,t)y+ r(x,t)= 0$ which can be reduced to $dX/dt= F(X,t)$ with X= (x(t), y(t)) and F= (y, (-r(x,t)- q(x,t)y)/p(x,t)). If p(x,t) is ever 0, that "blows up".