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A Lyapunov function near an equilibrium is a continuously differentiable positive-definite function $V$, vanishing there, whose derivative along trajectories is nonpositive. The strict Second Lyapunov theorem says a negative-definite derivative implies local asymptotic stability.

For sharper bounds than the circular test $x^2+y^2$, put $z=(x,y{)}^T$, $r^2=z^{T}z$ andA direct multiplication gives $A^{T}P+PA=-2I$. Its eigenvalues are ${\lambda }_{\pm }=(13\pm \sqrt{13})/12>0$. Since the nonlinear vector field is $Az+r^{2}z$,For $0<V<1$, the derivative is strictly negative. The Lyapunov theorem proves asymptotic stability. In fact $V(0)<1$ gives $V(t)\le V(0)$ andso the solution stays bounded, exists for all positive time and converges exponentially to zero.

The ellipse $V=1$ is invariant. In polar coordinates, $\dot{\theta }=-(2{\cos }^2\theta +{\sin }^2\theta )<0$, so it is a nonconstant periodic orbit, not part of the basin. For $V(0)>1$, use $r^2\ge V/{\lambda }_{+}$ to obtain $\dot{V}\ge 2V(V-1)/{\lambda }_{+}$. This implies finite-time blow-up by comparison with the scalar equation. Thus the exact basin is $\{V<1\}$ and every point outside its closure escapes to unbounded norm within its maximal existence interval.

The largest open disc centred at zero lying in this exact basin has radiusThe smallest radius such that every point strictly outside its circle escapes is the largest radius on the invariant ellipse:These are optimal: a larger attraction disc contains a periodic-orbit point, and a smaller escape threshold leaves a periodic-orbit point in its exterior. This is an exact quadratic basin boundary.

The simpler radial derivative is $\frac{1}{2}(d/dt)r^2=r^2(r^2-1)-xy$. It only certifies attraction for $r<1/\sqrt{2}$ and strictly increasing norm for $r>\sqrt{3/2}$. Those are optimal thresholds for a sign test on that radial derivative, not for the actual attraction and escape regions. If “increases without bound” is read as requiring monotonic increase from the initial instant, the smallest such uniform radius is $\sqrt{3/2}$: at direction $x=y$, any smaller threshold permits an initially decreasing norm. The sharper $r_2$ above concerns eventual unbounded escape, which need not be monotone at every instant.

At an interior local maximum, the second derivatives in both coordinate directions are nonpositive: restrict the function to each coordinate line and apply the one-variable second-derivative test. Thus ${\nabla }^2\varphi \le 0$ there, contradicting ${\nabla }^2\varphi \ge ϵ>0$. Since a maximum exists on the closure, a maximum lies on the boundary.

The first-order Cauchy-Kovalevskaya theorem says that an analytic PDE system solved for the derivative normal to an analytic noncharacteristic initial hypersurface, with analytic initial data, has a unique local analytic solution near each point of that hypersurface. The general theorem likewise applies to higher-order equations solved for the highest normal derivative with analytic data for the lower normal derivatives.

Here the characteristic vector is $(x,y,a)$ and the normal to $z=0$ is $(0,0,1)$. Its normal component is $a$, so the theorem applies precisely when $a\ne 0$. The method of characteristics gives $\dot{x}=x$, $\dot{y}=y$, $\dot{z}=a$, $\dot{u}=u$. Starting from $(\xi ,\eta ,0)$ at parameter $s=0$ gives $x=\xi e^s$, $y=\eta e^s$, $z=as$, $u=e^{s}f(\xi ,\eta )$. Eliminating $s=z/a$ yieldsThis is locally defined wherever the arguments remain in the analytic domain of $f$, and verifies both PDE and initial data.

As $a\to 0$, the normal component vanishes and the characteristic curves become tangent to the initial plane. The crossing parameter $s=z/a$ and slopes $dx/dz=x/a$, $dy/dz=y/a$ become singular. At $a=0$, arbitrary analytic data are not admissible: already on the plane they must obey $x{f}_x+y{f}_y=f$. Even compatible data do not determine derivatives normal to the plane. This is failure of the noncharacteristic condition, not a failure of the analytic theorem.

The Rayleigh quotient is $R[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$, interpreted through the energy quadratic form for trial states in its form domain. Expanding in normalized energy eigenstates gives $R[\psi ]=\sum E_n|c_n{|}^2/\sum |c_n{|}^2\ge E_0$. If the ground state is unique, equality requires all coefficients outside its one-dimensional eigenspace to vanish. Thus equality means the same ground-state ray, allowing normalization and an overall phase.

Use the normalized trial function $\sqrt{2/a}\cos (\pi x/a)$ on $|x|<a/2$, zero outside. Its kinetic energy is ${\pi }^2{\hslash }^2/(2m{a}^2)$ and its mean square position is $c_0{a}^2$, where $c_0=1/12-1/(2{\pi }^2)$. The energy isMinimization gives $a^4={\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_0)$ andFor the first excited well state use $\sqrt{2/a}\sin (2\pi x/a)$, zero outside. Now the kinetic energy is $2{\pi }^2{\hslash }^2/(m{a}^2)$ and the position factor is $c_1=1/12-1/(8{\pi }^2)$. Minimizing gives $a^4=4{\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_1)$ andThis particular excited-state estimate is an upper bound. The trial function is odd and the oscillator ground state is even, so it is exactly orthogonal to the ground state. Its spectral expansion therefore contains only energies at least $E_1$, giving the same variational bound on that orthogonal subspace. An arbitrary trial state without this orthogonality would bound only $E_0$, not $E_1$. The compactly supported trial functions belong to the quadratic-form domain despite derivative jumps at the well edges; their kinetic energies are correctly computed as ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$.

The symmetrized covariant derivative equation is ${\nabla }_a{\xi }_b+{\nabla }_b{\xi }_a=\psi g_{ab}$. Lower ${\xi }_b=g_{bc}{\xi }^c$ and expand both derivatives. Using the Christoffel symbols of the metric and their symmetry, the coefficient of ${\xi }^c$ simplifies byHenceThis is the coordinate formula for the Lie derivative ${\mathcal{L}}_{\xi }g=\psi g$, proving equivalence with the Conformal Killing vector field condition.

Let all spatial derivatives be evaluated at $x=mh$, $t=nk$. Expand the two adjacent flux contributions symmetrically. The bracket in the scheme isFor clarity, the $h^2$ terms are $a{u}_{xx}+a^{\prime }{u}_x$; odd orders cancel under $h\mapsto -h$. Under four spatial derivatives the next coefficient is $a{u}_{xxxx}/12+a^{\prime }{u}_{xxx}/6+a^{\prime \prime }{u}_{xx}/8+a^{(3)}{u}_x/24$. Time expansion gives $u(x,t+k)=u+k{u}_t+O(k^2)$. Since $u_t=(a{u}_x{)}_x$ and $k=\mu h^2$,This stronger smooth-data residual implies the requested $O(h^3)$ consistency bound. Even a less detailed Taylor expansion with an $O(h^3)$ spatial remainder would suffice for the printed order. Uniform bounds on the needed derivatives over the region under consideration make the estimate uniform in the interior mesh index; the zero boundary values enter the scheme exactly.

Start from $I_0(\lambda )=1/\lambda$. Differentiation under the integral sign givesThis interchange is justified locally around any $\lambda >0$ by the dominated convergence theorem: the differentiated integrands are bounded by an integrable polynomial times $e^{-\lambda x/2}$. Successive derivatives of ${\lambda }^{-1}$ giveThis is also the integer case of the Gamma function integral after scaling the integration variable.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

For the stated order of the two solutions, define their Wronskian byTaking a derivative cancels the two mixed products: $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$. Substituting the second-order linear differential equation for each second derivative cancels the $q$ terms and leavesThis is the Abel identity. On an interval with continuous coefficients, linear independence means $W(x_0)\ne 0$; hence the Wronskian never vanishes anywhere on that interval.

For real parameters, work on $x>0$ so arbitrary real powers are unambiguous. Taking derivatives givesThe residual after substitution isWhen $\gamma \ne 0$, $x^{\gamma }$ ranges over $(0,\infty )$. Evaluating at the zeros of the cosine first forces the sine coefficient to vanish; then the cosine coefficient vanishes identically. If $\gamma =0$, the nonzero constant $\cos 1$ cannot satisfy the equation. Thus the necessary and sufficient conditions areUnder these conditions the normalized coefficient is $p=(1-\gamma )/x$. The Abel identity gives $W=C{x}^{\gamma -1}$. Choose $C=\gamma$ and apply reduction of order on an interval where $\cos (x^{\gamma })\ne 0$:The substitution $u=x^{\gamma }$ evaluates the integral. After continuation across the cosine zeros,This is a power substitution for an oscillatory second-order equation: in the variable $u=x^{\gamma }$, the equation reduces to the harmonic oscillator equation.

For a constant matrix, the matrix exponential series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a operator norm. Termwise derivatives therefore giveHence $\mathbf{x}(t)=e^{tM}{\mathbf{x}}_0$ solves the initial value problem. For this particular matrix, direct multiplication gives $M^2=-I$. Splitting the matrix exponential into its even and odd powers provesThis is the matrix exponential when the square is minus the identity. If ${\mathbf{x}}_0=(x_0,y_0{)}^{\mathsf{T}}$, the general solution isThe constants in part (a) are $\alpha =x_0/5$ and $\beta =y_0+2x_0/5$, which makes the two forms identical.

The function $g(t)=|t{|}^{3/2}$ is differentiable everywhere: for $t\ne 0$, its derivative is $(3/2)\mathrm{sgn}(t)\sqrt{|t|}$, while at zero the difference quotient is $\mathrm{sgn}(t)\sqrt{|t|}\to 0$.

The function $k(t)=|t{|}^{1/2}$ is differentiable for $t\ne 0$, with derivative $\mathrm{sgn}(t)/(2\sqrt{|t|})$. At zero its difference quotient is $\mathrm{sgn}(t)/\sqrt{|t|}$, which has no finite limit. Applying the coordinatewise-sum argument givesAt those points,where the first coefficient is zero at $x=0$. At any point with $y=0$, the restriction to the vertical coordinate line already fails to be differentiable.