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The Euclidean logarithmic norm is the largest eigenvalue of a matrix's Hermitian part. Differentiating $\Vert y{\Vert }_2^2$ along $y^{\prime }=Ay$ gives $\Vert e^{tA}{\Vert }_2\le e^{t{\mu }_2(A)}$. It is the least exponent for such a bound with prefactor one, as a first-order expansion at zero shows. It equals the spectral abscissa for a normal matrix, but need not do so otherwise. A scalar rational stability function does not generally inherit a bound by its value at this real number.

Let $d=\Delta x$, impose $u_0=u_{M+1}=0$, and write the method of lines system as $U^{\prime }=LU$, whereThus $D_2$ is a negative definite symmetric matrix and $D_1$ is a skew-symmetric matrix. Use the mesh-weighted Euclidean norm $\Vert U{\Vert }_d^2=d{\sum }_{m=1}^M|u_m{|}^2$. Discrete summation by parts yields the centered Dirichlet drift-diffusion energy identityConsequentlyThe same estimate controls perturbations and is uniform in the number of grid points and in the fixed drift coefficient. Finite-dimensional linear ODE theory guarantees existence, so this proves stability of a numerical method for the semidiscretization.

The factor $d^{1/2}$ simply rescales the vector norm and does not change the induced matrix norm. Equivalently the symmetric part is $(L+L^{\ast })/2=D_2$, whose largest eigenvalue is $-4d^{-2}{\sin }^2[\pi /(2(M+1))]<0$. This is the Euclidean logarithmic norm, rather than generally the spectral abscissa of a nonnormal matrix. No periodic Fourier mode assumption has been made: the zero endpoint terms are part of the proof. In particular positivity of both off-diagonal coefficients is not needed for this $L^2$ stability result.

By orthogonal diagonalization of a real symmetric matrix, write $A=Q\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)Q^T$, with orthogonal $Q$. Its matrix exponential has the same eigenvectors and positive eigenvalues $e^{t{\lambda }_j}$. Orthogonal invariance of the induced Euclidean norm gives the exact identityThis proves the requested inequality with equality. If a real number $\beta$ gave the bound for every $t\ge 0$, evaluating on a unit eigenvector for ${\lambda }_{\max }(A)$ at any $t>0$ would give $e^{t{\lambda }_{\max }}\le e^{t\beta }$, so $\beta \ge {\lambda }_{\max }$. Thus the stated exponent is the smallest possible. For a symmetric matrix the spectral abscissa and Euclidean logarithmic norm coincide, unlike the general nonsymmetric case in Question 1.

The spectral abscissa is the largest real part of a matrix's eigenvalues. It describes asymptotic exponential rates of its matrix exponential, but a non-normal matrix can also have transient amplification. It is generally distinct from the Euclidean logarithmic norm, which gives an immediate Euclidean energy-growth bound.