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With $A=D+L+U$, the Gauss-Seidel method solves $(D+L)x^{(k+1)}=b-U{x}^{(k)}$. It uses each newly computed coordinate immediately in the remaining triangular solve. It is successive over-relaxation with parameter one, and converges from every start when $A$ is symmetric positive definite.

Order the interior grid values into a vector. The coefficient matrix has diagonal $4+c{h}^2$ and off-diagonal entries $-1$ at horizontal and vertical neighbours, with boundary values already zero. Every row is strictly diagonally dominant because $c>0$. The standard convergence theorem for strict diagonal dominance gives convergence of both the Jacobi method and the Gauss-Seidel method. The matrix is also symmetric positive definite.

For $D=(4+c{h}^2)I$, the weighted Jacobi method is $u^{(r+1)}=u^{(r)}+\omega D^{-1}(b-A{u}^{(r)})$, with iteration matrix $H_{\omega }=I-\omega D^{-1}A$. The sine-product grid eigenvectors giveThese weighted Jacobi eigenvalues for a square Dirichlet grid determine convergence. All eigenvalues of $A$ are positive. Thus for fixed $h$ convergence requires $0<\omega <2(4+c{h}^2)/a_{\max }$, where $a_{\max }=8{\cos }^2(\pi h/2)+c{h}^2$. As $h\downarrow 0$, this upper bound tends to $1$ from above. Any fixed $\omega >1$ therefore fails on sufficiently fine grids; $\omega \le 0$ also fails. Consequently $0<\omega \le 1$ is necessary for all mesh sizes, and it is sufficient, since $a_{\max }<2(4+c{h}^2)$ for every finite grid.

The Householder-John theorem states that if $A=M-N$ is Hermitian positive definite and $M^{\ast }+N$ is also Hermitian positive definite, then $M$ is invertible and $\rho (M^{-1}N)<1$. The associated stationary iteration $x^{(k+1)}=M^{-1}N{x}^{(k)}+M^{-1}b$ therefore converges from every starting vector. For designing a method, choose a splitting with easily solved $M$ and verify the positive-definiteness condition on $M^{\ast }+N$; it is a sufficient certificate and does not require computing every iteration-matrix eigenvalue.

For completeness, put $C=M^{\ast }+N=M^{\ast }+M-A$. If $Mv=0$ for $v\ne 0$, then $v^{\ast }Cv=-v^{\ast }Av<0$, a contradiction. If $M^{-1}Nv=zv$, then $Av=(1-z)Mv$. The case $z=1$ contradicts invertibility of $A$. Taking scalar products and using Hermitian symmetry giveshence $|z|<1$. The usual spectral-radius convergence criterion now proves the stated conclusion.

For the given scheme take the matrix splittingSince $A$ is real symmetric, $U=L^T$, and since it is positive definite, every diagonal entry in $D$ is positive. Thusis positive definite exactly for $0<\omega <2$. The theorem applies, and the iteration matrix has spectral radius below one. The unique solution $x_{\ast }=A^{-1}b$ is its fixed point, so $x^{(k)}-x_{\ast }=(M^{-1}N{)}^k(x^{(0)}-x_{\ast })\to 0$. ThereforeThe scheme is successive over-relaxation; at $\omega =1$ it becomes the Gauss-Seidel method.

For a splitting $A=D+L+U$, successive over-relaxation updatesIt reduces to the Gauss-Seidel method at $\omega =1$. If $A$ is symmetric positive definite, the Householder-John theorem applies to $M=D/\omega +L$, $N=(1/\omega -1)D-U$, since $M^T+N=(2/\omega -1)D$ is positive definite for $0<\omega <2$.