Upper and Lower bound eigenvalues Sturm Liouville problem

[redarrows]

I have 2 questions that need to be solve:

01. Find upper and lower bound for the k-th eigenvalue $$\lambda_{k}$$ of the problem $$((1+x^2)u')'-xu+\lambda(1+x^2)u$$ for $$0< x< 1$$ with boundary conditions $$u(0)=0$$ and $$u(1)=0$$

02. Find a lower bound for the lowest eigenvalue of the problem
$$((1+x^2)u')'+\lambda(1+x^2)u=0$$ for $$0 < x < 1$$ with boundary conditions $$u(0)=0$$ and $$u'(1)=0$$

many thanks and looking forward from anyone. please put the steps as well.

 

[lippyka]

Answer 1:The upper and lower bounds for the k-th eigenvalue \lambda_k can be obtained by examining the Sturm-Liouville problem on the given domain. Using the Rayleigh quotient, we can obtain the lower bound for the k-th eigenvalue as follows: \lambda_k \geq \frac{\int_0^1 (1+x^2)u^2_{k-1}dx}{\int_0^1 (1+x^2)u^2_{k-1}dx} = \frac{\int_0^1 (1+x^2)u^2_{k-1}dx}{\int_0^1 u^2_{k-1}dx} Since the boundary conditions are satisfied for all possible solutions and the integrals have positive integrands, the lower bound is always greater than or equal to zero. The upper bound can then be obtained by solving the following equation: \lambda_k \leq \frac{\int_0^1 (1+x^2)u^2_{k-1}dx - \int_0^1 xu_{k-1}u'_{k-1}dx}{\int_0^1 (1+x^2)u^2_{k-1}dx} Answer 2:The lower bound for the lowest eigenvalue of the given problem can be obtained using the Rayleigh quotient. We can obtain the lower bound for the lowest eigenvalue as follows: \lambda_0 \geq \frac{\int_0^1 (1+x^2)u^2_0 dx}{\int_0^1 (1+x^2)u^2_0 dx} = \frac{\int_0^1 (1+x^2)u^2_0 dx}{\int_0^1 u^2_0 dx} Since the boundary conditions are satisfied for all possible solutions and the integrals have positive integrands, the lower bound is always greater than or equal to zero.