I am trying to find the eigenvalues for the discretization matrix in the Poisson equation using the Chebyshev polynomials, i.e. $$ -u''(x) = f(x), x \in [0,1],\;\; u(0)=u(1)=0 $$ Discretize the spatial domain uniformly: $x_i = ih$ with $h=1/m, i \in \{0,1,\ldots,m\}$. Let $$A = \begin{bmatrix} 2&-1&0&\cdots&0\\ -1&2&-1&\cdots&0\\ \cdots&\cdots&-1&2&-1\\ 0&0&0&2&-1\\ \end{bmatrix}$$ Then the solution is approximated in finite difference by solving $$Au = f$$
My question is how to find the eigenvealues of A using the Chebyshev polynomials. I can show the recurrence relation for $u_{k+1}$ is a special case of the Chebyshev polynomials, i.e $$A[u^p_1,u^p_2, \cdots, u^p_{m-1}]^T \Rightarrow (A u^p)_j = -\frac{1}{h^2} \left(u_{j-1}^p -2u_j^p + u_{j+1}^p \right) = \lambda_p u_j^p$$ $$\Rightarrow u_{j+1}^p = (2-h^2 \lambda_p)u_j - u_{j-1}$$ is a special case of $$U_{j+1}(x) = 2xU_j(x)-U_{j-1}(z)$$ by setting $x = 1 - \frac{h^2 \lambda_p}{2}$. So with representation of the Chebyshev polynomial, I can have $$ u_j= U_j(1 - \frac{h^2 \lambda_p}{2}) = \frac{\sin\left[(j+1)\cos^{-1}\left(1 - \frac{h^2\lambda_p}{2}\right)\right]}{\sqrt{1 - \left(1 - \frac{h^2\lambda_p}{2}\right)^2}}$$ But how can we find the the eigenvalue from the above equation? We know that for Chebyshev polynomials $U_0(x)=1$, but I don't think $u_0 = 0 \neq U_0(1 - \frac{h^2 \lambda_p}{2})$.
Any help/hints will be greatly appreciated. Thanks.
I am not sure that your recurrence relation can be deal with by Chebychev polynomials. Indeed the term in front of the $u_j$ should be dependent on $j$ for that.
Rather, from the solutions of the continuous eigenvalue problem you could look for an eigenvector/value by plugging $v_p=e^{iqp}$ in the expression $-v_{p+1}+2v_p-v_{p-1}$ and get $$-v_{p+1}+2v_p-v_{p-1}=(2-e^{iq}-e^{-iq})v_p=2(1-\cos q)v_p=4\sin^2(q/2)v_p$$ To fulfil boundary conditions, set $w_p=\sin(qp)=\frac{1}{2i}(v_p-v_{-p})$ which verifies $w_0=w_{m}=0$ if $q=\frac{k\pi}{m}$ with $k=1,\ldots,m-1$. Then $w_p$ verifies $w_0=w_{m}=0$ and $$-w_{p+1}+2w_p-w_{p-1}=4\sin^2(\frac{k\pi}{2m})w_p,\qquad k=1,\ldots,m-1$$ which is exactly $Aw=4\sin^2(\frac{k\pi}{2m})w$ (your matrix $A$ is of size $m-1$). Thus you got your eigenvectors and values.