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For an axisymmetric thin elastic plate, define ${\nabla }_r^2=r^{-1}{\partial }_r(r{\partial }_r)$. Uniform excess pressure and bending stiffness $B$ give $B{\nabla }_r^{4}h=p$. The roof is attached to undeformed rock at the front, giving clamped boundary conditions $h(R)=h_r(R)=0$, and regularity excludes singular displacement or curvature at $r=0$. A regular particular solution is $p{r}^4/(64B)$; adding a constant and a multiple of $r^2$ and enforcing the two edge conditions gives the axisymmetric clamped-plate deflectionIntegrating the magma intrusion volume,therefore yieldsHere the axisymmetric phrase requires the radial plate biharmonic operator; a Cartesian one-dimensional beam equation would not give the same pressure coefficient. Hydrostatic contributions are excluded as instructed.

For the early front, let $U_p=\dot{R}$ be the leading propagation speed. The fluid and fracture radii differ only within a local region of length $l_p\ll R$. In lubrication theory, the elastic pressure gradient near the front is of order $B{h}_f/l_p^5$, and the volume flux through the gap is of order $B{h}_f^4/(\mu l_p^5)$. In a frame following the fluid front that flux is of order $U_p{h}_f$. The viscous peeling of an elastic plate law is thereforeMatch the peeling region to the interior curvature: $h_f/l_p^2\sim {\kappa }_i$. The stated vapour-gap relation $h_f\simeq p_T{l}_p^4/(24B)$, with $p_T>0$ interpreted as the relevant pressure magnitude, then givesSubstitution gives the vapour-tip peeling of a magma intrusion speedWith constant supplied volume flux $Q$ and negligible initial intrusion volume, $V=Qt$. Integrating $R^{14}\dot{R}\sim B^{3/2}{Q}^{7/2}{t}^{7/2}/(\mu p_T^{1/2})$ gives $R^{15}\sim B^{3/2}{Q}^{7/2}{t}^{9/2}/(\mu p_T^{1/2})$, orOrder-one coefficients require the full local peeling profile. This early regime is a front-controlled asymptotic regime, not an assertion that a continuum peeling zone remains thin at arbitrarily small $t$.

When fracture toughness instead imposes fixed ${\kappa }_f>0$, equating it to the outer edge curvature gives the toughness-controlled magma intrusion lawIts pressure can be written in either useful formThe pressure decreases as the intrusion grows on this advancing branch.

If $V_{r0}$ is the reservoir volume before any magma was transferred to the initially negligible intrusion, volume conservation and negligible conduit storage give $V_r=V_{r0}-V$. With the supplied pressure-volume law and conduit conductance, the reservoir-fed magma intrusion obeysIn this law $E$ is an effective reservoir pressure-volume stiffness, with dimensions of pressure per volume. If the stated initial reservoir volume refers instead to the start of the late regime, when the intrusion already has volume $V_i$, replace $V_{r0}$ by the total $V_T=V_{r0}+V_i$ throughout. An independent value of $V_i$ would then be needed.

For an advancing solution, inflow ceases when the reservoir and intrusion pressures equalize. Its final volume is the larger positive root in $0<V<V_{r0}$ ofEquivalently, $y=\sqrt{V_{\infty }}$ solves $y^3-V_{r0}y+\mathcal{A}/E=0$. The larger root is essential: $\sqrt{V}(V_{r0}-V)$ rises from zero to a maximum $2V_{r0}^{3/2}/(3\sqrt{3})$ at $V=V_{r0}/3$ and then decreases. A nonempty interval of positive inflow exists only ifThere are then two positive equilibria $V_{-}<V_{r0}/3<V_{+}$. At equilibrium the derivative of the volume ordinary differential equation is $\beta E(V_{r0}-3V)/(2V)$, so $V_{-}$ is unstable and $V_{+}$ has asymptotic stability. An initial advancing intrusion with $V_{-}<V_i<V_{+}$ approaches $V_{\infty }=V_{+}$. The equality case has a double equilibrium and no positive-inflow interval; above the threshold there is no advancing equilibrium branch of this model.

The late formula $p\propto V^{-1/2}$ is singular at $V=0$ and cannot nucleate the intrusion: the earlier propagation regime must supply its finite seed. Nor does this fixed-edge-curvature propagation law prescribe how an irreversible fracture closes under negative inflow. The final-size answer is conditional on reaching the advancing late branch. In particular the reservoir is generally not emptied, and $\beta$ controls the approach rate rather than the equilibrium size.

Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

Eliminate $R=N-S-I$ to obtain the two-dimensional autonomous differential equationIts Jacobian matrix at the endemic equilibrium isSince $I^{\ast }>0$,The trace-determinant stability criterion shows that both eigenvalues have negative real part. The Endemic equilibrium of the SIR model with waning immunity is therefore locally asymptotically stable.

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

At a positive equilibrium, division by $N$ givesSince $b>d$, this has the positive solution

Write the vector field as $F(N)=N(b{e}^{-aN}-d)$. Its derivative isAt $N=N^{\ast }$, the first two terms cancel and $b{e}^{-a{N}^{\ast }}=d$, soThe one-dimensional linear stability analysis therefore shows that $N^{\ast }$ is locally asymptotically stable.

At an endemic equilibrium, $I^{\ast }>0$. The equation $\dot{I}=0$ then givesSince also $S^{\ast }>0$, the equation $\dot{S}=0$ givesThis infective population is positive precisely whenwhich is exactly the assumed instability condition for the disease-free equilibrium. Thus the endemic equilibrium of the susceptible-infective model with exponential birth exists.

At this equilibrium the relations above reduce the Jacobian matrix toHenceBy the trace-determinant stability criterion, both eigenvalues have negative real part. The linearization stability theorem therefore proves that the endemic equilibrium is locally asymptotically stable.

For $\dot{x}=f(x)$, a fixed point $x^{\ast }$ satisfies $f(x^{\ast })=0$. It is Lyapunov stable when, for every $\epsilon >0$, there is $\delta >0$ such that

A Lyapunov function on a neighbourhood of $x^{\ast }$ is a continuously differentiable function $V$ such thatThe First Lyapunov theorem states that such a function makes $x^{\ast }$ Lyapunov stable. If $\dot{V}<0$ away from $x^{\ast }$, then $x^{\ast }$ is locally asymptotically stable.

To prove stability, choose a closed ball contained in the domain of $V$. On its boundary sphere $|x-x^{\ast }|=\epsilon$, compactness and positive definiteness giveContinuity at $x^{\ast }$ supplies $\delta <\epsilon$ such that $|x-x^{\ast }|<\delta$ implies $V(x)<m$. Along a trajectory, $V$ cannot increase. Such a trajectory therefore cannot first reach the boundary sphere, where its value would be at least $m$. This proves Lyapunov stability.

If $\dot{V}<0$ away from $x^{\ast }$, take a sufficiently small compact sublevel set of $V$. The LaSalle invariance principle says that every trajectory in it approaches the largest invariant subset of $\{\dot{V}=0\}$, which is just $\{x^{\ast }\}$. Hence the equilibrium is also asymptotically stable.

At an equilibrium, the second equation gives $x=y=:s$. The first equation then givesHence the fixed-point branches in the $(a,x)$ plane areandThe three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt{a}$, so the second bifurcation value is

The Jacobian matrix at a general point isOn the branch $P_0=(a^2,a^2)$ it is triangular, with eigenvaluesThus $P_0$ is unstable for $a<0$, asymptotically stable for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm \sqrt{a}$. Thenwhose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt{a}$ is therefore a saddle equilibrium for every $a>0$. The positive branch is a saddle for $0<a<1$ and asymptotically stable for $a>1$; close to $a=1$ it is a stable node.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitutionand append $\dot{a}=0$. The equations becomeThe centre subspace is $Y=X$. Seek the extended centre manifold for a parameter asThe centre-manifold invariance equation isTo second order, $\dot{X}=-aX+O(3)$, so comparison of coefficients givesand hence $A=C=0$, $B=1$. ThereforeSubstitution into the $X$ equation gives the reduced flowIts central branch $X=0$ is unstable for $a<0$ and stable for $a>0$, while the two nonzero branches $X\sim \pm \sqrt{a}$ for $a>0$ are unstable. Thus the bifurcation at zero is a subcritical pitchfork bifurcation with reversed normal-form parameter $\mu =-a$, exactly as recorded by the extended centre manifold of the 2023 Cambridge quadratic-product system.

The complete bifurcation diagram is therefore as follows. For $a<0$, only $P_0$ exists and is unstable. For $0<a<1$, $P_0$ is stable while both $P_{+}$ and $P_{-}$ are saddles. For $a>1$, $P_{+}$ is stable while $P_0$ and $P_{-}$ are saddles. At $a=1$, $P_0$ and $P_{+}$ cross and exchange stability, so the bifurcation is transcritical. This gives the bifurcation diagram of the 2023 Cambridge quadratic-product system.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll 1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt{a},\sqrt{a})$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt{a},\sqrt{a})$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional stable and unstable manifolds; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.

A resistive velocity term proportional to the speed squared removes energy at rate proportional to $|\omega {|}^3$. Its derivative at zero velocity vanishes, so it does not appear in the linearization of a dynamical system at rest. Nonlinear energy arguments can still prove asymptotic stability.

For real $\mu ,\beta$ and $\lambda >0$, writing $A=x+iy$ gives a gradient flow with potential $V=-(\mu +\beta )x^2/2-(\mu -\beta )y^2/2+\lambda (x^2+y^2{)}^2/4$. The origin has exponential asymptotic stability for $\mu <-|\beta |$, is algebraically attracting at equality, and is unstable above it. Nonzero stable equilibria are the real pair $\pm \sqrt{(\mu +\beta )/\lambda }$ for $\beta >0$, or the imaginary pair $\pm i\sqrt{(\mu -\beta )/\lambda }$ for $\beta <0$, when their squared amplitudes are positive. Differentiating the two real equations gives eigenvalues $-2(\mu +\beta ),-2\beta$ on the real branch and $-2(\mu -\beta ),2\beta$ on the imaginary branch. For $\beta =0$, $\mu >0$ gives a radially attracting circle with neutral phase, so individual equilibrium points have Lyapunov stability but are not individually asymptotically attracting.

For a real $2\times 2$ Jacobian matrix $J$, both eigenvalue have negative real part exactly whenThe linearization stability theorem then makes a hyperbolic equilibrium locally asymptotically stable.