omyojj
May25-11, 01:47 AM
I have a Sturm-Liouville system
\frac{d}{dx}p(x)\frac{du}{dx} - q(x)u(x)+\lambda \rho(x) u(x) = 0
with
p(x) = (1-x^2)^{2p}
q(x) = k^2
\rho(x) = (1-x^2)^{p-1}
(p,k^2 are positive real)
u(x) is defined on the interval (-A,A) where 0<A<=1.
Boundary condition that u(x) satisfies is u'(A)=u'(-A)=0
and I want it to be symmetric with respect to zero, i.e., u(x) = u(-x).
I don't think that this equation is solvable in a closed form for general k, p values.
However, I want to find or estimate the lowest eigenvalue λ together with a trial eigenfunction that gives good description of true solution.
So I considered a functional
K[u(x)] = \dfrac{\int_{-A}^{A} pu^{\prime 2} + qu^2 dx}{\int_{-A}^{A} \rho u^2 dx}
or
K[u(x)] = \dfrac{\int_{-A}^{A} (1-x^2)^{2p}u^{\prime 2} - k^2u^2 dx}{\int_{-A}^{A} (1-x^2)^{p-1} u^2 dx}
As a trial function I took
u(x;\alpha) = 1 - \frac{\alpha}{2}\left( x^2 - \frac{x^4}{2A^2} \right)
as an approximation to fourth-order in x.
Note that u(x;a) satisfies the boundary conditions.
According to variational principle, the absolute minimum of K is the lowest eigenvalue λ.
The problem is that I want to find a good trial function,
\alpha = \alpha(p, k^2)
that gives absolute minimum(stationary) of K[u]
My question is
1. How do you think I can find the value of α as a function of p, k that gives minimum of K[u(x)].
I think that procedures like
\frac{d}{d\alpha} K[u] = 0
is needed.
2. Is it better to do it by differentiating K[u] before performing integration?
When integrated first, Denominator and Numerator are expressed in Gaussian hypergeometric functions. (Confirmed it with the help of Wolfram alpha)
\frac{d}{dx}p(x)\frac{du}{dx} - q(x)u(x)+\lambda \rho(x) u(x) = 0
with
p(x) = (1-x^2)^{2p}
q(x) = k^2
\rho(x) = (1-x^2)^{p-1}
(p,k^2 are positive real)
u(x) is defined on the interval (-A,A) where 0<A<=1.
Boundary condition that u(x) satisfies is u'(A)=u'(-A)=0
and I want it to be symmetric with respect to zero, i.e., u(x) = u(-x).
I don't think that this equation is solvable in a closed form for general k, p values.
However, I want to find or estimate the lowest eigenvalue λ together with a trial eigenfunction that gives good description of true solution.
So I considered a functional
K[u(x)] = \dfrac{\int_{-A}^{A} pu^{\prime 2} + qu^2 dx}{\int_{-A}^{A} \rho u^2 dx}
or
K[u(x)] = \dfrac{\int_{-A}^{A} (1-x^2)^{2p}u^{\prime 2} - k^2u^2 dx}{\int_{-A}^{A} (1-x^2)^{p-1} u^2 dx}
As a trial function I took
u(x;\alpha) = 1 - \frac{\alpha}{2}\left( x^2 - \frac{x^4}{2A^2} \right)
as an approximation to fourth-order in x.
Note that u(x;a) satisfies the boundary conditions.
According to variational principle, the absolute minimum of K is the lowest eigenvalue λ.
The problem is that I want to find a good trial function,
\alpha = \alpha(p, k^2)
that gives absolute minimum(stationary) of K[u]
My question is
1. How do you think I can find the value of α as a function of p, k that gives minimum of K[u(x)].
I think that procedures like
\frac{d}{d\alpha} K[u] = 0
is needed.
2. Is it better to do it by differentiating K[u] before performing integration?
When integrated first, Denominator and Numerator are expressed in Gaussian hypergeometric functions. (Confirmed it with the help of Wolfram alpha)