 Solution

Use von Neumann stability analysis on the infinite grid. For the Fourier mode $u_{m,k}=\hat{u}(t)e^{i(m\xi +k\eta )}$, the semidiscrete eigenvalue isIts real part is nonpositive for every frequency. The mode evolves by $\hat{u}(t)=e^{t{\lambda }_h}\hat{u}(0)$, whose modulus does not grow. The discrete Fourier transform and Parseval identity therefore yieldwith a bound independent of $h$, where $\Vert u{\Vert }_{{\ell }_h^2}^2=h^2{\sum }_{m,k}|u_{m,k}{|}^2$. Hence the semidiscrete method is stable. Frequencies with $\eta =0$ are nondissipative rather than unstable.

The same sign can be seen directly by summation by parts: the centred $x$ operator is skew-adjoint, and for the unitary $y$ shift $S_y$,This is an energy version of the dissipative second-order forward advection stencil. A subsequent time integrator would need its own stability condition; semidiscrete stability does not assert that every fully discrete method is stable.