= Solution
Let $S$ be the periodic forward-shift matrix, $(Su)_m=u_{m+1}$ with indices modulo $M$. The semidiscrete operator is
$$
u'=h^{-1}Au,\qquad A=-\frac32I+2S-\frac12S^2,\qquad h=\Delta x.
$$
For $\omega_\ell=e^{2\pi i\ell/M}$, $\ell=0,\ldots,M-1$, the vector $v^{(\ell)}=(1,\omega_\ell,\ldots,\omega_\ell^{M-1})^T$ satisfies $Sv^{(\ell)}=\omega_\ell v^{(\ell)}$. These vectors divided by $\sqrt M$ form the unitary <discrete Fourier transform> basis. Hence the normal matrix $A$ has <eigenvalues>
$$
\lambda_\ell=-\frac32+2e^{i\theta_\ell}-\frac12e^{2i\theta_\ell},\qquad
\operatorname{Re}\lambda_\ell=-\frac32+2\cos\theta_\ell-\frac12\cos2\theta_\ell=-(1-\cos\theta_\ell)^2\le0.
$$
Every Fourier coefficient is multiplied by $e^{t\lambda_\ell/h}$ of modulus at most one. <Parseval's identity> therefore proves
$$
\boxed{\|u(t)\|_{2,h}\le\|u(0)\|_{2,h},\qquad \|u\|_{2,h}^2=h\sum_{m=1}^M|u_m|^2.}
$$
The same estimate holds for differences of solutions and is uniform in the mesh. This is the required semidiscrete <stability of a numerical method>. The mean mode has eigenvalue zero and is conserved, consistently with periodic advection.
One can also verify the energy estimate without diagonalization. Since $S^*=S^{-1}$,
$$
A+A^*=-\frac12(2I-S-S^*)^2,qquad
\frac d{dt}\|u\|_2^2=-\frac1{2h}\|(2I-S-S^*)u\|_2^2\le0.
$$
The <dissipative second-order forward advection semidiscretization> thus damps nonconstant modes rather than amplifying them. This does not assert stability of an arbitrary further time-stepping method; its <absolute stability> region must contain the scaled eigenvalues. No nonnegativity condition on the periodic boundary value is needed: the printed trailing $t\ge0$ specifies the time range.
Back to article page
