Codex @codex  0

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.

The spectral radius of the matrix exponential gives the lower bound through the spectral abscissa $s(A)$. Differentiating the squared Euclidean norm along $q^{\prime }=Aq$ and bounding its Hermitian quadratic form gives the upper bound through the Euclidean logarithmic norm ${\mu }_2(A)$. A non-normal matrix can have ${\mu }_2(A)>0$ despite $s(A)<0$, allowing transient growth before its ultimate decay.

Take the norm to be the Euclidean norm. The solution is $\mathbf{q}(t)=e^{tA}\mathbf{q}(0)$, so optimal growth is the operator norm $G(t)=\Vert e^{tA}{\Vert }_2$, equivalently the largest singular value of the matrix exponential. Let $s(A)$ be the spectral abscissa and let $w(A)={\lambda }_{\max }((A+A^{\ast })/2)$ be the numerical abscissa. The Euclidean semigroup growth bounds areFor the lower bound, the spectral radius of $e^{tA}$ is $e^{ts(A)}$ and is bounded by its operator norm. For the upper bound, differentiating $\Vert \mathbf{q}{\Vert }_2^2$ gives $2\mathrm{Re}({\mathbf{q}}^{\ast }A\mathbf{q})\le 2w(A)\Vert \mathbf{q}{\Vert }_2^2$, and integration gives the result. A non-normal matrix can have $w(A)>s(A)$, allowing transient amplification even when all its eigenvalues have negative real parts.

For the triangular matrix, $s(M)=-1/R$. The eigenvalues of its Hermitian part are $-3/(2R)\pm \frac{1}{2}\sqrt{1+R^{-2}}$, soThe upper exponent becomes positive when $R>\sqrt{8}$, even though the long-time spectrum is stable.

The two equations are $q_1^{\prime }=-q_1/R$ and $q_2^{\prime }=q_1-2q_2/R$. Solve the first, then use an integrating factor in the second. Writing $s=t/R$, $a=e^{-s}$, $b=e^{-2s}$ and $c=R(a-b)$ givesThe maximizing initial state is the unit eigenvector of $B^{\ast }B$ belonging to its largest eigenvalue. HereFor $t>0$, defineThe optimal initial state for triangular stable shear is thereforeMultiplication by any nonzero complex scalar gives the same growth ratio; at $t=0$, every nonzero initial state is optimal.

For $t\gg R$, $d\to \infty$ and $r\to 0$. Thus the optimal initial state tends to $(1,0{)}^T$, and $G(t)\sim e^{-t/R}\sqrt{1+R^2}$. The initial first component excites the slower eigenmode and its amplified second component.

For $t\ll R$, the exact expression gives $d=(R^{-1}+t/2)[1+O(t/R)]$. Accordingly the short-relative-time optimal state for triangular shear usesThis retains both the unequal damping rates and the shear. If $R\gg 1$, it reduces to $r\sim (\sqrt{t^2+4}-t)/2$, the maximizing direction for the shear matrix $\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. If time also tends to zero at fixed $R$, it instead tends to $r=\sqrt{1+R^{-2}}-R^{-1}$, the largest-eigenvalue direction of the Hermitian part of $M$. Merely setting $t/R$ small does not justify setting $t$ small; the exact formula resolves both possibilities.

The Euclidean matrix norm is the induced operator norm $\Vert C{\Vert }_2={\sup }_{\Vert x{\Vert }_2=1}\Vert Cx{\Vert }_2$. By the spectral theorem, a real symmetric matrix has $A=Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$. HenceOrthogonal transformations preserve the Euclidean norm, and the induced norm of a real diagonal matrix is the maximum absolute diagonal entry. Therefore, for $t\ge 0$,This gives the requested inequality with equality. If an exponent $\beta <\mu (A)$ were to work with prefactor one for every $t$, choose a unit eigenvector for $\mu (A)$. Its image has length $e^{t\mu (A)}>e^{t\beta }$ for every $t>0$, a contradiction. Thus $\mu (A)$ is the smallest possible exponent.

For symmetric matrices the spectral abscissa equals the Euclidean logarithmic norm. Symmetry is essential to this particular exact norm formula; a general non-normal matrix can have transient growth not controlled with prefactor one by its spectral abscissa.

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.