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Completing squares reveals the geometry of the sheared quartic double-well gradient flow:The critical points satisfy $2(x+y)=0$ and $4x^3-2x+2y=0$, so they are $(0,0)$, $(1,-1)$ and $(-1,1)$. The Hessian matrix isAt the origin its determinant is $-8$, so $(0,0)$ is a saddle point, with $V=0$. At either other critical point, the determinant is $16$ and the upper-left entry is $10$, so it is a positive-definite matrix. They are strict local minima; the completed squares show that both are global minima, with $V=-1$.

A contour at level $c$ obeysThere are no contours for $c<-1$, two isolated minimum points for $c=-1$, and two separate closed ovals for $-1<c<0$. At $c=0$ the two lobes meet at the saddle point; near the origin their tangents are $y=(-1\pm \sqrt{2})x$. For $c>0$ one closed contour surrounds both minima.

Along the gradient flow, the gradient-flow dissipation identity isAt the stated initial point, $V(1,-1/2)=-3/4$. Its trajectory stays in the invariant sublevel set $V\le -3/4$. This set is compact because $V$ is a coercive function. It also excludes every point with $x=0$, where $V=y^2\ge 0$; continuity therefore keeps the trajectory in its positive-$x$ component.

To justify its limit rather than merely infer it from the sketch, integrate the dissipation identity: ${\int }_0^{\infty }\Vert \nabla V{\Vert }^{2}dt$ is finite. On this compact set the gradient and its time derivative are bounded, so $\Vert \nabla V{\Vert }^2$ is uniformly continuous. A nonnegative uniformly continuous function with finite integral tends to zero: otherwise separated intervals of a fixed positive height and width would force an infinite integral. Every accumulation point is consequently a critical point. In the positive-$x$ component below this energy level the only one is $(1,-1)$. Compactness then implies convergence, andThis is a compact gradient-flow trapping criterion; the drawn trajectory illustrates, rather than replaces, the convergence proof.

Expand the function into separated terms: $f=g(x)+h(y)$, where $g=x^2-x^4$ and $h=y^8-y^4$. ThenPut $a=2^{-1/2}$ and $b=2^{-1/4}$. The nine critical points are all pairs from $x\in \{0,\pm a\}$ and $y\in \{0,\pm b\}$. The diagonal Hessian matrix has entries $2-12x^2$ and $56y^6-12y^2$.

At $(0,\pm b)$ both entries are positive, giving strict local minima of value $-1/4$. At $(\pm a,\pm b)$ they have opposite signs, giving four saddle points of value zero. At the three points on $y=0$ the Hessian is degenerate, so its sign alone does not classify them. Use the higher-order test for separated extrema: $h(y)=-y^4+O(y^8)$ is negative for small nonzero $y$. At $(0,0)$ the positive $x^2$ term and negative $y^4$ term give a saddle. At $(\pm a,0)$, $g-1/4=-(x^2-1/2{)}^2$, so every sufficiently small nonzero displacement decreases $f$ and these are strict local maxima.

For the level curves, the exact zero set consists of the two parabolas $x=\pm y^2$ and the oval $x^2+y^4=1$. They cross at the four nondegenerate saddle points. Small positive levels enclose each maximum; small negative levels enclose each minimum. The degenerate central saddle has quartic narrowing, locally $x^2-y^4=c$. Far along the $x$ axis the function is negative, while far along the $y$ axis it is positive. These signs and zero curves determine the connectivity shown in the sketch.

Use the Fourier transform pair $\hat{q}(k)=\int q(x)e^{ikx}dx$, $q(x)=(2\pi {)}^{-1}\int \hat{q}(k)e^{-ikx}dk$. The point force transforms to $F$. With the dispersion relation $D$ defined above, the sheet equation becomesThe point-force radiation from a fluid-loaded sheet is therefore represented exactly byThe causal contour and outgoing acoustic square-root branch are fixed first with $\mathrm{Im}\omega <0$, then continued to the desired real frequency. This prescription fixes how poles and the branch points are passed.

For the acoustic far field $x=r\cos \theta$, $y=r\sin \theta$, take $0<\theta <\pi$ bounded away from grazing and $k_{0}r\gg 1$. The method of steepest descent saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=i{k}_0\sin \theta$. The supplied saddle point rule, including its $\sin \theta$ factor, gives, provided the contour deformation crosses no poles,A convenient simplification, free of division by $T$, isEquivalently, when $T\ne 0$,The expression printed in the PDF is missing sound-speed factors for general dimensional $c_0$. It agrees with this result if $c_0=1$ in fully normalized units; when $c_0$ is retained as an arbitrary sound speed, the numerator needs $c_0^{-3/2}$ and the structural term needs $T/c_0^2$ in the last form. These factors arise respectively from cylindrical spreading, the pressure-density relation, and $k_s=\omega \cos \theta /c_0$.

A direct countercheck is the transparent-sheet limit $m=T=0$. The sheet jump condition then gives ${\hat{p}}_{+}(k,0)=F/2$, so the saddle point rule requiresThe printed expression, interpreted continuously after multiplying out its structural factor, instead gives $F(2c_0{)}^{-1}\sqrt{\omega /(2\pi r)}\sin \theta$ times the same phase. It differs by a factor $c_0^{3/2}$; for example it is eight times too large when $c_0=4$. This limit also verifies the normalization of the corrected density field independently of the sheet's elastic-sheet tension.

To decide about poles, track the roots of $\Delta (k,\omega )$ on the chosen square-root sheet and deform the original causal contour to the steepest descent contour. A root contributes a residue exactly when it lies in the region swept out by that deformation; its sign is fixed by the contour orientation. Branch cuts must be retained throughout this comparison. Which roots are crossed can depend on observation angle, producing a change of the modal contribution when a pole meets the deformation boundary. A saddle point approaching a pole or a grazing endpoint requires an approximation uniform in that limit, rather than the isolated saddle point formula above.

The crossed poles are the free fluid-sheet modes of the preceding solution. Real subsonic roots represent evanescent acoustic surface waves carrying energy along the sheet, with normal decay; complex continuations represent leaky or radiating modes. Their residues must be added to the saddle point sound when the causal contour selects them. The specification “no poles contribute” is therefore a substantive condition on the contour, not permission to ignore zeros of the dispersion relation.

Assume $m\ne 0$, since the trivial $m=0$ theory cannot fix $\beta$ or the vacuum expectation value. A zero-energy vacuum must satisfy both $V=0$ and stationarity in both real scalar directions. With the notation from the preceding solution, these become $H=H_x=H_y=0$, because the prefactor $|m{|}^4{e}^{x^2+y^2}$ is positive. The derivatives areThese conditions also show that the zero-energy stationary point must be real. If $y\ne 0$, the second equation gives $P=(3-{\beta }^2)/2$; the first then gives $x=-({\beta }^2+3)/(2\beta )$. Substituting into the definition of $P$ gives $x^2+y^2=2$. Butwhich is impossible. Thus $y=0$, without assuming a real vacuum in advance.

Set $t=x+\beta$ and $F=1+xt$. Since $t=0$ would give $H=1$, it cannot occur. The zero-energy equation gives $F=s\sqrt{3}t$, $s\in \{1,-1\}$. Stationarity givesCombining these equations gives $1=(s\sqrt{3}-x)t=t^2$. Writing $t=\epsilon \in \{1,-1\}$ producesThe condition $\beta >0$ leaves exactly $(s,\epsilon )=(1,1)$ and $(-1,1)$. The second branch is a saddle point, as its real-direction curvature is negative; the stability calculation in the next solution verifies this explicitly. The stable zero-energy Polonyi vacuum therefore selectsZero energy alone, without stationarity and stability, would not imply this parameter value. Even zero energy plus stationarity also admits $\beta =2+\sqrt{3}$ on the unstable branch.

For the stable branch found above, $t=x+\beta =1$ and $s=1$. Hence the vacuum expectation value isin the stated Planck units. To verify that it is a vacuum rather than merely a zero-energy stationary point, evaluate the Hessian matrix. At either zero-energy stationary branch,Because $H$ and its first derivatives vanish there, the Hessian matrix of $V$ is just $|m{|}^4{e}^{x^2}$ times this Hessian matrix. For $s=1$, both eigenvalues are positive. This proves a strict local minimum in both real scalar directions. For $s=-1$, $H_{xx}<0$, so the alternative $\beta =2+\sqrt{3}$, $⟨z⟩=-\sqrt{3}-1$ is a saddle point and is excluded from the stable zero-energy Polonyi vacuum.

There is also a useful global check. Set $u=x-(\sqrt{3}-1)$ on the stable branch. Directly completing squares givesBoth remaining coefficients are positive. Thus $H\ge 0$ everywhere, with equality only at $u=y=0$. The positive exponential prefactor proves that this is the unique global minimum, not just a metastable vacuum.

Finally, at the stable vacuum $z+\beta =1$ and $D_{z}W=\sqrt{3}{m}^2$. Its supergravity auxiliary field hasThus the Minkowski vacuum breaks supersymmetry, even though its cosmological constant vanishes. If $m=0$, the potential is flat and the displayed tuned parameter and scalar value are not selected.

Set $z=x^2/\epsilon$. The differential operator simplifies because its first-derivative terms cancel:Thus the homogeneous solutions are $e^{\pm z}$ and the transformed forcing is $1/(4\epsilon z)$. For the decaying-sector variation of parameters solution choose the contour directionsThe first integrand decays at positive-real infinity and the second at positive-imaginary infinity. Join each endpoint to $x$ within the first quadrant, avoiding zero. This also fixes the exponentially small homogeneous contributions, which cannot be determined from an algebraic asymptotic expansion alone.

After $s=t^2/\epsilon$, write $A(z)={\int }_{+\infty }^z{e}^{-s}ds/s$ and $B(z)={\int }_{-\infty +i0}^z{e}^{s}ds/s$. Their derivatives are $A^{\prime }=e^{-z}/z$ and $B^{\prime }=e^z/z$. Direct differentiation of $y=(e^{z}A-e^{-z}B)/(8\epsilon )$ gives $y_{zz}-y=1/(4\epsilon z)$, verifying the original equation. In terms of the exponential integral, with the logarithm continued through the upper half-plane,The $-i\pi$ is essential: the contour for $B$ starts at argument $\pi$, not at real positive infinity. On the positive real axis this choice gives the imaginary part $\pi e^{-z}/(8\epsilon )$.

Repeated endpoint integration by parts gives $e^z{E}_1(z)\sim z^{-1}(1-z^{-1}+2!z^{-2}-\cdots )$ and the matching expansion of $e^{-z}(\mathrm{Ei}(z)-i\pi )$ with positive factorial coefficients in the upper-half-plane sector. Their odd inverse-power corrections cancel. Therefore the leading large-$x$ term in the requested quadrant isUse the convention that anti-Stokes lines have equal exponential magnitudes, $\mathrm{Re}(x^2/\epsilon )=0$. They are $\arg x=\pi /4+m\pi /2$. The Stokes lines are $\arg x=m\pi /2$. Continuing this same solution below the positive-real ray switches on $i\pi e^{-z}/(4\epsilon )$; continuing above the positive-imaginary ray switches on $i\pi e^z/(4\epsilon )$. These exponentials remain decaying up to the neighbouring anti-Stokes lines. Hence the maximal adjacent decay sector on this continuation isOn its two boundary rays the switched exponential has constant modulus and oscillates, so $y$ does not tend to zero. Within the sector the leading algebraic term above persists. The internal anti-Stokes line at $\pi /4$ presents no growing exponential for the chosen solution.

For the formal small-$\epsilon$ series, substitution and comparison of successive powers gives $a_0=1$ and $a_r=2r(2r-1)a_{r-1}$. Thus all coefficients areThis is a factorially divergent asymptotic expansion. The ratio of consecutive term magnitudes is $(2r+2)(2r+1)/|z{|}^2$. For optimal truncation, retain $r=0,\dots ,n-1$, choosing $2n=|z|+O(1)$, so the cutoff is at a least term.

To resolve the remainder near the positive-real Stokes line, use its exact Borel integral with a simple pole:Here ${\mathcal{C}}_{-}$ runs from zero to positive-real infinity below the pole at one. Partial fractions reproduce the exponential-integral solution, including its positive imaginary part. The finite geometric identity for $(1-t^2{)}^{-1}$ gives the displayed exact remainder after integrating the first $n$ monomials.

Write $z=2n+i\mu \sqrt{n}+\nu$, with real bounded $\mu$ and bounded $\nu$. For a Gaussian saddle coalescing with a simple pole, put $t=1+u/\sqrt{n}$. The exponent becomes $-u^2-i\mu u+O(n^{-1/2})$ and $dt/(1-t^2)=-du/(2u)+O(n^{-1/2})du$. ThereforeThe integral is its Cauchy principal value plus $i\pi$. If $F(\mu )=\mathrm{PV}{\int }_{\mathbb{R}}{e}^{-u^2-i\mu u}du/u$, differentiation gives $F^{\prime }(\mu )=-i\sqrt{\pi }{e}^{-{\mu }^2/4}$ and $F(0)=0$. Consequently $F(\mu )=-i\pi \mathrm{erf}(\mu /2)$, and $J_n=-i\pi \mathrm{erfc}(\mu /2)/2+O(n^{-1/2})$.

For $x=\rho e^{i\theta }$ with fixed $\rho >0$ and $\theta =O(\sqrt{\epsilon })$, $\mu /2=\sqrt{2}\rho \theta /\sqrt{\epsilon }+O(\epsilon )$. The error-function smoothing of a Stokes multiplier therefore gives the optimally truncated remainderThe error estimate is uniform for bounded $\theta /\sqrt{\epsilon }$. On the Stokes line the multiplier is half its switched-on value; above it the exponentially small term switches off, and below it switches on. This term is invisible at every algebraic order. Near the line the remainder can exceed the least retained term by a square-root factor because a pole meets the saddle point.

The printed auxiliary Gaussian hint has inconsistent constants. Direct saddle point expansion of its nonsingular integral gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$ under its printed scaling, rather than $\frac{1}{2}\sqrt{\pi /(2n)}{e}^{-{\mu }^2/8}$. The derivation above uses the correct saddle point normalization and independently fixes both the smoothing width and the contour sign.

Take $F=(F^{\ast }{)}^{\ast }$ with the usual proper closed convex assumptions, so the Fenchel-Moreau theorem identifies its conjugate with the supplied $F^{\ast }$. Put $z^k=u^k-\tau D^T{v}^k$. The first update is $w^{k+1}={\mathrm{prox}}_{F^{\ast }/\tau }(z^k/\tau )$. By Moreau decomposition, the second update isAlso $\tau (D^T{v}^k+w^{k+1})=u^k-u^{k+1}$, so the argument in the final update contains $u^{k+1}-\tau (D^T{v}^k+w^{k+1})=2u^{k+1}-u^k$. Eliminating the auxiliary variable givesThis is the primal-dual hybrid gradient method with extrapolation parameter one and the primal update performed first. Its saddle point function is $F(u)+⟨Du,v⟩-G(v)$, corresponding to the primal objective $F(u)+G^{\ast }(Du)$.

For proper lower semicontinuous convex functions and a nonempty saddle-point set in these finite-dimensional spaces, the standard convergence of primal-dual hybrid gradient theorem gives the sufficient parameter conditionHere $\Vert D{\Vert }_2$ is the operator norm, or largest singular value. For $D\ne 0$, one possible choice is $\tau =\sigma =\eta /\Vert D{\Vert }_2$ with $0<\eta <1$; for $D=0$ any positive steps satisfy the condition. The existence assumption is necessary: step sizes alone cannot guarantee convergence to a saddle point that does not exist.

The fixed observer is the ray $V=0$. The physically selected zero-group velocity saddle defines the absolute wavenumber $k_0$ and absolute frequency ${\omega }_0$:For a smooth frequency branch with $D_{\omega }\ne 0$, the saddle condition is equivalently $D_k(k_0,{\omega }_0)=0$. The absolute growth rate isHere $k_0$ and ${\omega }_0$ can be complex. A physical saddle point is selected by the localized impulse response and deformation of its Fourier transform contour; an arbitrary algebraic stationary point need not determine absolute hydrodynamic instability.

Let $b=e^{i\theta }$ and use the method of steepest descents with phase $\Phi (z)=-z^2$. For $z=x+iy$, $\mathrm{Re}\Phi =y^2-x^2$ and $\mathrm{Im}\Phi =-2xy$. The simple saddle point is $z=0$: the real axis is its descending path and the imaginary axis its ascending path. The quadratic complex endpoint descent path from $b$ satisfies $\mathrm{Im}(z^2)=\sin 2\theta$ and can be parametrized bywith the square-root branch chosen continuously from $b$. Increasing $s$ reduces the real part of the exponent by $\lambda s$.

For $0\le \theta <\pi /2$, that endpoint path goes to positive real infinity. Deform the original contour into the real path from $-d$ through the saddle point towards $+\infty$, followed backwards along the endpoint descent path to $b$. The integrand is an entire function, so finite contour deformations are legitimate; connecting arcs in the far decaying sector vanish. If $T_b$ denotes the oriented integral from $b$ to that infinity, this givesThe last expression follows by setting $z^2=b^2+s$: $dz/ds=1/(2z)$, so the endpoint integral is $e^{-\lambda b^2}{\int }_0^{\infty }{e}^{-\lambda s}/[2z(s)]ds$. This derives its phase and orientation directly.

For $\pi /2<\theta \le \pi$, the descent path from $b$ goes to negative real infinity. Deform from $-d$ towards that infinity and then backwards along this endpoint path to $b$. There is no saddle point segment, anduses the same formula with the negative square-root branch. At $\theta =\pi /2$ the contour can run from $-d$ to zero and then up the imaginary axis. It gives a half-Gaussian contribution and the dominant imaginary endpoint term $i{e}^{\lambda }/(2\lambda )$. The change in the subdominant saddle point contribution is a Stokes phenomenon; it does not change the leading endpoint term on either side of that angle.

The Gaussian saddle and endpoint dominance by angle follows by comparing exponential sizes and gives all fixed-angle leading behaviours:At $\theta =\pi /4$, the endpoint term has magnitude $O({\lambda }^{-1})$, so the saddle point term of order ${\lambda }^{-1/2}$ still dominates. In the middle interval the endpoint term grows exponentially for $\pi /4<\theta <3\pi /4$, is oscillatory of order ${\lambda }^{-1}$ at $3\pi /4$, and decays exponentially for $3\pi /4<\theta <5\pi /6$. On the latter interval it nevertheless decays more slowly than the fixed-endpoint term. Equality of their exponential rates means $\cos 2\theta =1/2$ on the left half-plane, hence $\theta =5\pi /6$. At that angle both terms are required; their coefficient magnitudes are unequal, so their sum has no complete destructive cancellation.

These fixed-angle asymptotic expansions are not uniform if $\theta$ is allowed to vary with $\lambda$ near a dominance transition. The endpoint/saddle balance near $\pi /4$ involves $\lambda (\theta -\pi /4)=O(\log \lambda )$; the two-endpoint competition near $5\pi /6$ has an angular scale $O({\lambda }^{-1})$.

The following sketches show the contour orientations, the saddle point and both endpoints. The hyperbolic endpoint paths are followed backwards when returning from infinity to $b$; the dotted far-end connectors indicate the vanishing connecting arc.