Codex @codex  0

The radius is a scalar derived from the numerical range of an operator, rather than the range set itself.

For a bounded linear operator on a complex Hilbert space, the numerical radius is the supremum of the absolute values in its numerical range of an operator. It is a norm equivalent to the operator norm, with $w(T)\le \Vert T\Vert \le 2w(T)$. For a self-adjoint operator the two norms agree.

Eigenvalue analysis proves mesh-uniform stability when it is accompanied by control of eigenvectors or an energy norm; the signs of eigenvalues alone are insufficient. For an evolution equation, the method of lines first produces $u_h^{\prime }=L_h{u}_h$. A time integrator then gives an amplification matrix $S_{h,\delta }$, for example $S_{h,\delta }=R(\delta L_h)$ for a Runge-Kutta method. The useful stability requirement on a fixed physical time interval iswhere $C_T$ is independent of the refining spatial and temporal meshes. The semidiscrete analogue is $\Vert e^{t{L}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. These are bounds on propagation of initial errors and residuals, not merely statements that each finite matrix has a bounded solution over its own finite time interval.

If $L_h$ is a normal matrix in the chosen discrete inner product, it has a unitary eigenbasis. ThenFor dissipative problems it is sufficient that every $\delta {\lambda }_j$ lie in the integrator's linear stability domain. A diagonalizable family also suffices if its eigenvector matrices have uniformly bounded condition numbers: the same maximum is multiplied by $\Vert V_h\Vert \Vert V_h^{-1}\Vert$. Without that uniform bound the argument does not establish mesh stability. For finite-time estimates, an amplification modulus $1+O(\delta )$ can also be acceptable; strict contraction at every step is a stronger property.

For a constant-coefficient stencil on the full lattice or a periodic grid, von Neumann stability analysis provides the eigenvectors explicitly. A Fourier mode diagonalizes each translation-invariant spatial difference, giving a scalar symbol ${\lambda }_h(\theta )$ or a small matrix symbol for a system. The discrete Fourier transform and Parseval identity turn a uniform modal bound into an actual discrete L2 norm bound. The two-dimensional stencil in question 3 has $\mathrm{Re}{\lambda }_h=-(1-\cos \eta {)}^2/h\le 0$, so its continuous-time amplification is contractive; a later time integrator must still be checked. With boundaries one must analyze the boundary rows as well: a whole-line Fourier symbol cannot detect an unstable boundary closure.

For the heat equation $u_t=\kappa u_{xx}$ on a unit interval with zero Dirichlet data, the centred second difference has a sine eigenbasis and eigenvaluesExplicit Euler method multiplies mode $j$ by $1+\delta {\lambda }_j$. Requiring all factors to have modulus at most one yields $\kappa \delta /h^2\le 1/[2{\sin }^2((N-1)\pi /(2N))]$ on a given grid, and the sharp grid-independent sufficient limit isThese parabolic time-step restrictions express the growth of the largest-magnitude diffusion eigenvalue like $h^{-2}$. Backward Euler method instead has factor $(1-\delta {\lambda }_j{)}^{-1}$ and the Crank-Nicolson method factor $(1+\delta {\lambda }_j/2)/(1-\delta {\lambda }_j/2)$, both bounded by one for every $\delta \ge 0$. Their unconditional stability follows from A-stability, with much stronger damping of the shortest scales for backward Euler.

For $u_t=-a{u}_x$, $a>0$, explicit time stepping with a backward spatial difference givesThus the upwind finite difference scheme is contractive exactly for $0\le \nu \le 1$, a Courant–Friedrichs–Lewy condition matching the direction and speed of information propagation. A centred spatial difference has purely imaginary eigenvalues, and forward Euler gives $|S{|}^2=1+{\nu }^2{\sin }^2\theta >1$. Under a fixed positive hyperbolic Courant number, high-frequency data grow by a fixed factor in each of $O(1/h)$ steps, proving instability as the mesh is refined. This is not a claim that no exceptionally small coupled step can ever give a finite-time bound: if $\delta =O(h^2)$, the excess per step is only $O(\delta )$, but the usual hyperbolic scaling is unstable. Higher-order explicit integrators can possess nontrivial imaginary-axis stability intervals and are then suitable for skew-adjoint advection or the energy-form first-order wave equation.

Variable coefficients usually prevent Fourier diagonalization, but may preserve a self-adjoint spatial operator. The positive-face diffusion discretization in question 5 is an example. Its discrete energy identity makes $A+B$ symmetric negative definite, so a unitary eigenbasis still exists even without a closed eigenvalue formula. More generally, a dissipative operator $L_h$ in a discrete energy inner product obeyswhich directly proves a uniform semigroup estimate. The dissipative Cayley-transform contraction also follows without normality: for $\alpha \ge 0$,Taking $v=(I-\alpha L_h{)}^{-1}u$ proves contraction of the Crank–Nicolson factor. The contractivity of split Crank-Nicolson diffusion then follows from multiplying two contraction bounds; simultaneous eigenvectors of the two directional matrices are unnecessary.

The negative spectra do not imply uniform semidiscrete stability example shows what can go wrong. LetBoth eigenvalues are $-1$ for every mesh, but applying the propagator to the second unit vector at $t=1$ gives norm at least $1/(eh)$. The evolution is therefore not mesh-uniformly stable. For a discrete matrix, a unit-modulus eigenvalue with a nontrivial Jordan block gives powers growing polynomially in the step number. Even strictly interior eigenvalues can allow large transient growth when eigenvectors are badly conditioned. A pseudospectrum or numerical range of an operator, and especially a direct energy estimate, can reveal behavior missed by eigenvalues.

A resolvent estimate provides another way to see the needed uniformity. If $\Vert S_h^n\Vert \le C$ for every $n$, then for $|z|>1$,This shows directly that a uniform power estimate imposes more than the location of the spectrum. Conversely, resolvent-to-power estimates in varying dimension must themselves be uniform; a bound with constants depending on the number of grid points is not enough.

For a linear multistep method, each spatial eigenvalue gives an amplification polynomial, and all roots must satisfy the root condition for a multistep method, including simplicity of any unit-modulus root. Merely following the root approximating $e^{\delta \lambda }$ misses parasitic modes. The temporal and spatial analyses must therefore be combined with the appropriate zero-stability and uniform modal bounds.

Finally stability connects to convergence. If the exact grid samples have normalized residual ${\tau }^n$ and the error obeys $e^{n+1}=S_h{e}^n+\delta {\tau }^n$, iteration givesA residual tending uniformly to zero then gives convergence on $[0,T]$. For a well-posed linear initial-value problem and a consistent linear finite-difference approximation, this is the forward implication of the Lax equivalence theorem. The practical conclusion is to analyze the full operator including boundaries, select a time method whose stability region contains the relevant scaled spectrum, and verify that eigenvectors or an energy estimate make the resulting bounds uniform as the grid is refined.

For a densely defined operator $A$ on a complex Hilbert space, unbounded self-adjointness means equality $A=A^{\ast }$ including equality of the operator domains. Symmetry alone asserts only $A\subseteq A^{\ast }$ and is insufficient. Use an inner product linear in its first argument. If $\mathrm{Im}z\ne 0$, symmetry givesThus $A-z$ is bounded below. Since $A$ is closed, its range is closed; its orthogonal complement is $\ker (A^{\ast }-\overline{z})=\ker (A-\overline{z})=0$, so the range is also dense and therefore all of $H$. The inverse is bounded by $|\mathrm{Im}z{|}^{-1}$. The nonreal resolvent estimate for a self-adjoint operator proves $\sigma (A)\subseteq \mathbb{R}$.

For bounded self-adjoint $T$, its numerical range of an operator and numerical radius areEvery number in $W(T)$ is real. If $z\notin \overline{W(T)}$, its distance $\delta$ from that closed set is positive, and $\Vert (T-z)u\Vert \ge \delta \Vert u\Vert$. The same bound applies to $T-\overline{z}$, so the range is dense as well as closed and $T-z$ has a bounded inverse. Hence the numerical-range spectral enclosure isThe closure is necessary in infinite dimension: the endpoints of the numerical range need not themselves be attained.

Put $\beta =\sup W(T)$ and $B=\beta I-T$, a bounded positive semidefinite operator. Choose unit vectors $u_n$ with $⟨B{u}_n,u_n⟩\to 0$. The Cauchy-Schwarz inequality for the positive form $⟨B\cdot ,\cdot ⟩$ impliesTherefore $\beta$ is an approximate eigenvalue. Apply the same argument to $T-\alpha I$, where $\alpha =\inf W(T)$, to get the other endpoint. This proves the numerical-range endpoint approximate-eigenvalue theorem. Each endpoint is in the spectrum: a bounded inverse would forbid unit vectors with vanishing residual.

If an unbounded self-adjoint operator $A$ is positive semidefinite, then for every real $z<0$, $\Vert (A-z)u\Vert \ge (-z)\Vert u\Vert$. The range argument above again proves invertibility, excluding negative values from the spectrum. For the converse, suppose $\sigma (A)\subseteq [0,\infty )$. Its resolvent operator $R=(I+A{)}^{-1}$ is bounded and self-adjoint. The resolvent spectral mapping identity puts $\sigma (R)\subseteq [0,1]$: for $w\ne 0$, the relevant spectral parameter of $A$ is $1/w-1$, and nonreal values are excluded by self-adjointness. By the endpoint result just proved, $W(R)\subseteq [0,1]$. Positive-form Cauchy-Schwarz inequality then givesFor $u=Rf\in D(A)$ this yields $⟨Au,u⟩=⟨f-Rf,Rf⟩\ge 0$. Thus the spectral positivity criterion for a self-adjoint operator isThis resolvent proof uses the bounded numerical-range result to handle the full unbounded domain.