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Write $z=(x,v)$ and let $\Phi (t,s,z)$ denote the Hamiltonian flow from time $s$ to time $t$. The characteristic curves satisfy Hamilton's equations:The Hamiltonian Liouville equation then reduces along each curve toThe signs and derivative variables here are those in the PDF.

The global characteristic flow for a Hamiltonian with bounded Hessian follows, for example, from $H\in C^2(\mathbb{R}\times {\mathbb{R}}^{2d})$ and, for every finite $T$,Thus the Hamiltonian vector field $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ is globally Lipschitz continuous in $z$ on each finite time interval and satisfies $|b(t,z)|\le A_T+L_T|z|$. The Picard-Lindelof theorem gives local existence and uniqueness, while the Gronwall inequality gives, for example,This excludes finite-time escape. There is a unique Hamiltonian flow for all finite forward and backward times, and $\Phi (s,t)$ is the inverse of $\Phi (t,s)$. These sufficient conditions are deliberately stronger than necessary.

The growth bound is $|F(t,x)|\le K(1+|x|)$. A characteristic curve through $(s,a)$ solvesThe $C^1$ field is locally Lipschitz in space, giving local existence and uniqueness. On either direction of a bounded time interval the Gronwall inequality gives $1+|X(t;s,a)|\le (1+|a|)e^{K|t-s|}$. Thus no trajectory can escape to infinity at finite time. On the resulting compact space-time region, local ordinary differential equation existence continues it, proving global definition for all $t\in \mathbb{R}$.

Uniqueness gives the flow composition law and inverse $X(s;t,\cdot )$. Differentiation with respect to $a$ solves the variational equation, making each map a $C^1$ diffeomorphism. Its positive Jacobian determinant isThese facts constitute the global characteristic flow under linear growth. A uniform global bound on $D_{x}F$ is not required; its bounds on each compact characteristic tube suffice.

The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.

If a single additive composite expansion is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-ϵ\log (2ϵ)$ and interpreting $x\log x=0$ at zero givesIt is zero at $x=0$, gives $2+O({ϵ}^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O({ϵ}^2)$ error uniformly through the nested regions.

Finally, near zero $H(X)=2X-\frac{8}{3}{X}^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first derivative but generally a second derivative diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y^{\prime \prime }$ multiplies that singular derivative to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y^{\prime }(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the initial value problem with $y(0)=y^{\prime }(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y^{\prime })$ with an integrable coefficient matrix; its integral equation and the Gronwall inequality force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.

Take the difference of two weak solutions, so $f={\psi }_0={\psi }_1=0$. At the stated $H^1(U_T)$ regularity, it is not justified simply to test by $u_t$, which need not belong to the spatial test space $H_0^1$. Instead use the antiderivative test underlying uniqueness of Sobolev weak solutions of a wave equation.

Fix $s\in (0,T]$ and defineThese are Bochner integrals in $H_0^1$. The function $v$ is an admissible space-time $H^1$ test, $v_t=-u$ before $s$, and $v(t)=z(s)-z(t)$. The time term in the weak formulation isbecause $u(0)=0$. Symmetry of $a^{ij}$ and integration by parts in time giveFor the lower-order part, integration by parts in space givesBounded coefficients, the Poincare inequality for $v$, and the Young inequality now implySince $Dv(t)=Dz(s)-Dz(t)$,For $s$ in a sufficiently short fixed interval, absorb the last term into the left-hand side. The Gronwall inequality applied to $\Vert u(s){\Vert }_2^2+(\theta /2)\Vert Dz(s){\Vert }_2^2$ proves $u=0$ on that interval. The equation gives continuity of $u_t$ in $H^{-1}$, so both initial traces at its endpoint are again zero. Repeating with the same uniform coefficient bounds covers $[0,T]$. Therefore

Let $X,Y$ be two continuous solutions on the same filtered probability space, driven by the same Brownian motion and with $X_0=Y_0$ almost surely. Write $D=X-Y$ and choose a common Lipschitz constant $L$ for $b,\sigma$. Introduce the stopping time ${\tau }_n=\inf \{t\ge 0:|X_t|+|Y_t|\ge n\}$. On $\{{\tau }_n=0\}$ the stopped difference is zero; before ${\tau }_n$, both coefficients and the difference are bounded.

The Itô formula and the zero expectation of the stopped square-integrable Itô integral yieldThe Gronwall inequality implies $\mathbb{E}{D}_{t\wedge {\tau }_n}^2=0$. Continuous solution paths are bounded on each compact time interval, so ${\tau }_n\uparrow \infty$ almost surely. Letting $n\to \infty$ proves $X_t=Y_t$ almost surely at each fixed time, and a countable dense set of times together with continuity gives indistinguishability of stochastic processes. Therefore the stochastic differential equation has pathwise uniqueness.

An $s$-stage Runge-Kutta method for $y^{\prime }=f(t,y)$ is defined byThe Butcher tableau records $A=(a_{ij})$, $b$ and $c$; internal consistency usually sets $c=Ae$, and first-order consistency requires $b^{T}e=1$. A strictly lower triangular $A$ gives an explicit method, while other stage dependencies generally require an implicit Runge-Kutta method. The stage solvability of an implicit Runge-Kutta method must be checked separately: if $f$ is globally Lipschitz continuous in $y$ with constant $L$, the stage fixed-point map is a contraction in the maximum stage norm whenever $hL{\max }_i{\sum }_j|a_{ij}|<1$. This is a sufficient small-step condition, not a restriction intrinsic to the definitions of A-stability or B-stability.

Linear stability begins with the Dahlquist test equation $y^{\prime }=\lambda y$. Solving the stages givesThe linear stability domain consists of test values for which the stage system is well defined and $|R(z)|\le 1$. A-stability means this includes the closed left half-plane. For $\mathrm{Re}\lambda <0$, this gives stability with no scalar decay-mode step restriction. For a normal matrix $L$ in $y^{\prime }=Ly$, unitary diagonalization of a normal matrix reduces the norm estimate to these scalar factors. For a non-normal matrix, eigenvalues alone do not establish a uniform norm bound: eigenvector conditioning and transient amplification matter. Mesh-uniform estimates for discretized PDEs must control operator powers, not only their spectra.

Every consistent explicit Runge-Kutta method has a nonconstant polynomial stability function and therefore cannot be A-stable, since that polynomial is unbounded on the negative real axis. For example, Forward Euler method has $R(z)=1+z$, stability disk $|1+z|\le 1$, and restriction $0\le hq\le 2$ for $\lambda =-q<0$. For a pure imaginary mode $\lambda =i\omega \ne 0$, $|1+ih\omega |>1$, so it is unstable for every positive step. The classical fourth-order Runge-Kutta method instead hasIt is stable on the imaginary axis exactly for $|q|\le 2\sqrt{2}$, an illustrative conditional stability range for oscillatory evolution.

Backward Euler method has $R(z)=(1-z{)}^{-1}$ and is A-stable. L-stability adds $R(z)\to 0$ as $|z|\to \infty$ in the left half-plane; backward Euler satisfies this and strongly damps unresolved rapidly decaying modes in a stiff differential equation. The implicit midpoint rule and the trapezoidal rule both have $R(z)=(1+z/2)/(1-z/2)$, so they are A-stable but not L-stable. They preserve the modulus of a pure imaginary test mode, but their stiff-decay limit is $-1$, leaving oscillatory numerical remnants. The three-stage Lobatto IIIA method has the fourth-order rational function found in Question 2 and likewise lacks L-stability. Thus high order and A-stability do not themselves imply efficient stiff damping.

Nonlinear stability measures differences between solutions of the same equation. A one-sided Lipschitz condition isDifferentiating the squared difference and applying the Gronwall inequality gives $\Vert y(t)-\tilde{y}(t)\Vert \le e^{\nu (t-t_0)}\Vert y(t_0)-\tilde{y}(t_0)\Vert$. A dissipative vector field has $\nu \le 0$ and is contractive. A B-stable method preserves this property for every positive step size for which the stages are well defined:This includes time-dependent vector fields when dissipativity holds at each common time argument. Applying it to scalar linear dissipative fields shows that B-stability implies A-stability when the linear stages are well defined; the converse fails.

The standard sufficient theorem is algebraic stability implies B-stability, subject to stage solvability. Define $B=\mathrm{diag}(b)$ and $M=BA+A^{T}B-b{b}^T$. Algebraic stability of a Runge-Kutta method means $b_i\ge 0$ and $M$ is a positive semidefinite matrix. To prove the theorem, set $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$ and $F_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Expanding the output difference and using $D_i=d+h{\sum }_j{a}_{ij}{F}_j$ yields the Runge-Kutta contractivity identityDissipativity makes the first sum nonpositive, while positive semidefiniteness makes the last quadratic form nonnegative. The latter follows by factoring $M=Q^{T}Q$ and writing it as a sum of squared norms of linear combinations of the $F_i$. This proves the claimed contraction. It is a sufficient theorem; absence of algebraic stability is not, by itself, a proof of failure of B-stability for every representation.

Several examples make the distinctions concrete. For backward Euler, $b=1$, $A=(1)$, so $M=(1)$: it is both L-stable and B-stable. For implicit midpoint, $b=1$, $A=(1/2)$, so $M=(0)$: it is B-stable and preserves squared norms for a skew-Hermitian linear equation, but lacks stiff damping. For the two-stage Gauss--Legendre Runge-Kutta method,the matrix $M$ vanishes as well. The collocation theorem gives order four, and its stability function is exactly the same as the three-stage Lobatto IIIA function. Yet the Gauss method is algebraically stable, whereas that Lobatto method is not. Identical scalar linear stability functions need not imply identical nonlinear stability properties.

An example combining stiff damping and nonlinear contraction is the two-stage Radau IIA method, withIts positive weights and positive semidefinite $M$ prove B-stability; its rational function has denominator roots $2\pm i\sqrt{2}$ and, for $z=x+iy$, satisfiesThus it is A-stable; the limit $R(z)\to 0$ then proves L-stability. The standard Radau IIA collocation theorem gives order $2s-1$, here three. These properties explain its usefulness for stiff nonlinear equations.

Finally, A-stability alone does not imply B-stability. The trapezoidal rule fails B-stability even for the scalar dissipative equation $y^{\prime }=-y^3$. With $h=1$, its step map $T$ is uniquely defined byAt $y=\sqrt{2}$, $T(y)=0$. Implicit differentiation givesThus nearby starting values are separated by approximately twice their original distance after one step, although the exact flow is contractive. This example and the contractivity theorem show why scalar linear analysis, stiff damping and genuinely nonlinear norm estimates answer different stability questions.

Because $\Vert b^{\prime }{\Vert }_{\infty }<\infty$, the function $b$ is Lipschitz continuous. the Picard-Lindelof theorem, proved by iteration onconverges uniformly on every compact interval. The usual factorial estimate proves convergence for arbitrary interval length, and the Gronwall inequality proves uniqueness. Thus there is a unique global continuous solution.

For $w=W$, induction shows that $X_t^{(n)}$ is ${\mathcal{F}}_t$-measurable: its value uses only $(W_s{)}_{s\le t}$ and earlier iterates up to time $t$. The pointwise limit $X_t$ is therefore ${\mathcal{F}}_t$-measurable. Hence $X$ is adapted to $({\mathcal{F}}_t)$.

For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelof theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

Let $X$ and $Y$ be two solutions with the same initial value and Brownian motion, and put $Z=X-Y$. Itô formula givesStop when either process or the stochastic integral becomes large. Taking expectations, using the assumed one-sided Lipschitz bound, and then removing the localization givesThe Gronwall inequality yields $\mathbb{E}{Z}_t^2=0$. Thus $X_t=Y_t$ almost surely at every rational time, and path continuity makes the two processes indistinguishable. This proves pathwise uniqueness.

Let $u$ and $v$ be two weak solutions with the same initial value, and put $w=u-v$. Bilinearity givesPair with $w$. The first transport term vanishes because $P_{N}u$ is divergence free. Part a and the Young inequality giveThereforeThe coefficient is integrable because $v\in L^2(0,T;V)$. Since $w(0)=0$, the Gronwall inequality gives $w=0$ on $[0,T]$. The global weak solution is unique.

For $h=f_1-f_2$, $h\prime =\sqrt{1+x}(f_1+f_2)h$ and $h(0)=0$. On every compact subinterval the coefficient is continuous, so Gronwall inequality gives $h=0$; hence equality holds throughout $[0,a)$.

Take the $L^2$ inner product of the temperature equation with ${\theta }_m$. The spectral projection disappears against ${\theta }_m\in {\tilde{H}}_m$, and part (a)(iii) cancels transport. ThusThe periodic Poincare inequality, part (i), and the Cauchy-Schwarz inequality implyThe Gronwall inequality therefore gives, for $0\le t\le T$,This defines a bound $K_0(T)$ independent of $m$, and part (i) then givesIntegrating the energy identity and using the same bound on its right-hand side gives

It remains to estimate the time derivative. For $\phi \in H_{\mathrm{per}}^1$, the Fourier projection ${\Pi }_m$ is a contraction in $H^1$, and the skew identity from part (a) givesThe sobolev embedding theorem and the periodic elliptic estimate for the Stokes operator bound $\Vert u_m{\Vert }_{\infty }$ by $C|A{u}_m|$. Moreover,The already obtained bounds therefore implywith $K_0^{\prime }(T)$ independent of $m$.

Let $(u_1,{\theta }_1)$ and $(u_2,{\theta }_2)$ be two weak solutions with the same initial data, and set $U=u_1-u_2$ and $\Theta ={\theta }_1-{\theta }_2$. The diagnostic Stokes equation givesSubtracting the temperature equations yieldsPair this equation with $\Theta$. The term transported by $u_1$ vanishes by the skew-symmetry of incompressible transport, while the other nonlinear term satisfiesThe forcing difference is at most $C\Vert \Theta {\Vert }_2^2$. ConsequentlyThe coefficient is integrable on $[0,T]$ because ${\theta }_2\in L^2(0,T;H^1)$. Since $\Theta (0)=0$, the Gronwall inequality gives $\Theta =0$, and the Stokes equation then gives $U=0$. The weak solution is unique.

Before the difference $Z$ hits zero, its diffusion coefficient has magnitude at most $KZ$ and its drift is at least $-LZ$. Itô's formula bounds the drift of $f_{ϵ}(Z)$ by $(L+K^2)f_{ϵ}(Z)$. Its stochastic integrand is bounded by $K/(4ϵ)$, making expectation legitimate on finite horizons. The Gronwall inequality and the value $f_{ϵ}(0)=1/ϵ$ then exclude finite-time contact from a strictly positive starting difference.

For the weak energy solution of a variable-coefficient wave equation, uniqueness can be proved without using $u_t$ as a spatial $H_0^1$ test function. For a difference with zero data, fix $s$ and test by $v(t)={\int }_t^{s}u(r)dr$ for $t<s$, extended by zero afterwards. Put $z(t)={\int }_0^{t}u(r)dr$, so $v(t)=z(s)-z(t)$. Integration by parts in time for the symmetric principal form, and in space for the first-order terms, giveFor short enough intervals the last term is absorbed. The Gronwall inequality forces $u=0$, and repetition proves uniqueness on the full interval. This antiderivative test is valid at the stated space-time $H^1$ regularity, unlike an unqualified direct energy test by $u_t$.