[proofplan]
We prove convergence of both the Gauss-Seidel and Jacobi methods for the five-point Poisson scheme by verifying the hypotheses of the relevant convergence theorems. The system matrix $A$ from the five-point stencil satisfies $-A$ is SPD (as established by the [Symmetry and Negative Definiteness of $A$](/theorems/1367) theorem). We work with the equivalent system $\tilde{A}x = \tilde{b}$ where $\tilde{A} = -A$ is SPD. For Gauss-Seidel, the [Gauss-Seidel Convergence for SPD Matrices](/theorems/1391) theorem applies immediately. For Jacobi, we verify that $2\tilde{D} - \tilde{A}$ is SPD using the explicit structure of the five-point stencil, then apply the [Jacobi Convergence for SPD-like Matrices](/theorems/1392) theorem.
[/proofplan]
[step:Reformulate the system so that the coefficient matrix is SPD]
The five-point finite difference discretisation of the Poisson equation $-\Delta u = f$ produces the linear system $Au = b$, where the matrix $A$ arises from the five-point stencil. By the [Symmetry and Negative Definiteness of $A$](/theorems/1367), $A$ is symmetric and negative definite: $A = A^\top$ and $x^\top Ax < 0$ for all $x \neq 0$.
Multiplying through by $-1$, the equivalent system $(-A)u = -b$ has coefficient matrix $\tilde{A} := -A$, which is symmetric positive definite. The iterative methods for $Au = b$ and $\tilde{A}u = -b$ have the same convergence properties (the iteration matrices are identical up to the sign absorbed by the right-hand side, which does not affect $\rho$).
[guided]
Why can we simply negate the system? The Jacobi and Gauss-Seidel splittings depend only on the coefficient matrix, not the right-hand side. If $A = D + L_0 + U_0$, then $\tilde{A} = -A = (-D) + (-L_0) + (-U_0) = \tilde{D} + \tilde{L}_0 + \tilde{U}_0$. The Jacobi iteration matrix for $\tilde{A}$ is $I - \tilde{D}^{-1}\tilde{A} = I - (-D)^{-1}(-A) = I - D^{-1}A$, which is the same as the Jacobi iteration matrix for $A$. The same holds for Gauss-Seidel. So the spectral radii are identical, and convergence for $\tilde{A}$ implies convergence for $A$.
[/guided]
[/step]
[step:Apply the Gauss-Seidel convergence theorem for SPD matrices]
Since $\tilde{A} = -A$ is symmetric positive definite, the [Gauss-Seidel Convergence for SPD Matrices](/theorems/1391) theorem applies directly: the Gauss-Seidel method for the system $\tilde{A}u = -b$ (equivalently, $Au = b$) converges.
[/step]
[step:Verify $2\tilde{D} - \tilde{A}$ is SPD and apply the Jacobi convergence theorem]
For the Jacobi method, we apply the [Jacobi Convergence for SPD-like Matrices](/theorems/1392) theorem, which requires both $\tilde{A}$ and $2\tilde{D} - \tilde{A}$ to be SPD.
The diagonal of $\tilde{A} = -A$ consists of the entries $\tilde{a}_{ii} = -a_{ii}$. In the five-point stencil, each diagonal entry of $A$ is $-4/h^2$ (the centre coefficient), so $\tilde{a}_{ii} = 4/h^2 > 0$ and $\tilde{D} = \frac{4}{h^2}I$.
Compute:
\begin{align*}
2\tilde{D} - \tilde{A} = \frac{8}{h^2}I - (-A) = \frac{8}{h^2}I + A.
\end{align*}
We need to show that $\frac{8}{h^2}I + A$ is SPD. Since $A$ is symmetric, $\frac{8}{h^2}I + A$ is symmetric. For positive definiteness, we use the eigenvalue structure: by the [Explicit Eigenvalues of $A$](/theorems/1368), the eigenvalues of $A$ are
\begin{align*}
\lambda_{k,l} = \frac{-4}{h^2}\left(\sin^2\frac{k\pi h}{2} + \sin^2\frac{l\pi h}{2}\right), \qquad k, l = 1, \ldots, N,
\end{align*}
where $h = 1/(N+1)$. These satisfy $\lambda_{k,l} < 0$ for all $k, l$ (since the sine terms are positive). The most negative eigenvalue occurs at $k = l = N$:
\begin{align*}
|\lambda_{k,l}| = \frac{4}{h^2}\left(\sin^2\frac{k\pi h}{2} + \sin^2\frac{l\pi h}{2}\right) \leq \frac{4}{h^2}\left(1 + 1\right) = \frac{8}{h^2},
\end{align*}
with equality only if $\sin^2\frac{k\pi h}{2} = \sin^2\frac{l\pi h}{2} = 1$, i.e., $k\pi h/2 = \pi/2$, which requires $kh = 1$, i.e., $k = N + 1$. But $k \leq N$, so the inequality is strict:
\begin{align*}
|\lambda_{k,l}| < \frac{8}{h^2} \quad \text{for all } k, l = 1, \ldots, N.
\end{align*}
The eigenvalues of $\frac{8}{h^2}I + A$ are $\frac{8}{h^2} + \lambda_{k,l} = \frac{8}{h^2} - |\lambda_{k,l}| > 0$. Therefore $\frac{8}{h^2}I + A = 2\tilde{D} - \tilde{A}$ is SPD.
By the [Jacobi Convergence for SPD-like Matrices](/theorems/1392) theorem, since $\tilde{A}$ and $2\tilde{D} - \tilde{A}$ are both SPD, the Jacobi method converges.
[guided]
The Jacobi case requires more work than Gauss-Seidel because the [Jacobi Convergence for SPD-like Matrices](/theorems/1392) theorem has the additional hypothesis $2\tilde{D} - \tilde{A}$ SPD.
For the five-point stencil, this reduces to showing $\frac{8}{h^2}I + A$ is SPD, i.e., that $\frac{8}{h^2}$ exceeds the magnitude of every eigenvalue of $A$. This is a spectral condition: $\frac{8}{h^2} > \rho(-A)$, where $\rho(-A) = \max_{k,l}|\lambda_{k,l}|$.
Using the [Explicit Eigenvalues of $A$](/theorems/1368), the largest eigenvalue magnitude is:
\begin{align*}
\rho(-A) = \frac{4}{h^2}\max_{k,l}\left(\sin^2\frac{k\pi h}{2} + \sin^2\frac{l\pi h}{2}\right).
\end{align*}
The maximum of $\sin^2\frac{k\pi h}{2}$ over $k = 1, \ldots, N$ is $\sin^2\frac{N\pi h}{2} = \sin^2\frac{N\pi}{2(N+1)}$. Since $\frac{N}{N+1} < 1$, this is $\sin^2\frac{N\pi}{2(N+1)} < \sin^2\frac{\pi}{2} = 1$. Therefore $\rho(-A) < \frac{4}{h^2}(1 + 1) = \frac{8}{h^2}$, confirming the strict positivity of the eigenvalues of $\frac{8}{h^2}I + A$.
In fact, the margin is small: as $N \to \infty$, $\sin^2\frac{N\pi}{2(N+1)} \to 1$, so $\rho(-A) \to \frac{8}{h^2}$ from below. The condition $2\tilde{D} - \tilde{A}$ SPD is "barely" satisfied, reflecting the fact that Jacobi convergence for the Poisson problem is slow (the spectral radius of $H_J$ approaches $1$ as $h \to 0$).
[/guided]
[/step]
