= Solution
For a linear system write $A=D+R$, where $D$ is the diagonal. The <Jacobi method> iterates
$$
u^{(k+1)}=D^{-1}(b-Ru^{(k)}),\qquad e^{(k+1)}=Be^{(k)},\quad B=-D^{-1}R.
$$
Here $D=(10/3)I$, and $B$ has nonnegative off-diagonal entries with row sums at most one. Rows adjacent to the boundary have sum strictly less than one; those at purely interior stencil vertices have sum one. Thus $\|B\|_\infty\leq1$.
To prove strict <spectral radius> below one, suppose $Bv=\zeta v$ with $|\zeta|=1$, and choose an index with $|v_i|=\|v\|_\infty>0$. In
$$
|v_i|=|\zeta v_i|\leq\sum_jB_{ij}|v_j|\leq\left(\sum_jB_{ij}\right)\|v\|_\infty\leq\|v\|_\infty,
$$
equality forces that row sum to equal one and every linked neighbor to attain the same maximum modulus. Propagate this along the connected axial grid to a boundary-adjacent row, whose row sum is strictly less than one, a contradiction. All <eigenvalues> therefore have modulus strictly below one. Hence $B^k\to0$ and the Jacobi errors tend to zero from every initial vector. The unique solution exists by positive definiteness, so $\boxed{\text{Jacobi converges for this discretization in any ordering}.}$
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