Getting Eigenvalues Into a Differential Operator

  • Thread starter bolbteppa
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Following Butkov, a second order ode

[tex]A(x)y'' + B(x)y' + C(x)y = D(x)[/tex]

can always be brought into Sturm-Liouville form

[tex]\tfrac{d}{dx}[p(x)y'] - s(x)y = f(x)[/tex]

after multiplying across by

[tex]H(x) = - \tfrac{1}{A(x)}e^{\int^x \tfrac{B(t)}{A(t)}dt}.[/tex]

He then says the function [itex]s(x)[/itex] can "often" be written as

[tex]s(x) = s_0(x) - \lambda r_0(x)[/tex]

where [itex]0 \leq s_0(x)[/itex], [itex]0 \leq r_0(x)[/itex] & [itex]\lambda[/itex] is fixed.

I just don't see how once can be comfortable with this or how one can use a statement like this in general. How does one take a general second order ode & concretely turn it into something involving [itex]\lambda[/itex]?

For instance, in the case that [itex]A[/itex], [itex]B[/itex] & [itex]C[/itex] are polynomials of degree [itex]2[/itex], [itex]1[/itex] & [itex]0[/itex]:

[tex](ax^2+bx+c)y'' + (dx + e)y' + fy = F(x)[/tex]

I can see that [itex]f[/itex] will be an eigenvalue after multiplication by [itex]H(x)[/itex], but how would you deal with a case like


[tex](ax^2+bx+c)y'' + (dx + e)y' + \sin(x)y = F(x)[/tex]
 

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