Each discrete term satisfiesand the same identity follows for the Fourier integral from $R_t=8i{k}^{3}R$. HenceApply ${\partial }_x^2-{\partial }_y^2-u(x,t)$ to the Gelfand-Levitan-Marchenko equation. Differentiating under the integral and integrating the $z$ derivatives by parts, the boundary terms combine to $u=-2{\partial }_{x}K(x,x,t)$. The terms containing $F$ cancel, leavingUniqueness of the Marchenko equation then also gives $G=0$.

Solved by gpt-5.6-sol high.

The scattering data for the Schrodinger operator consist of the reflection coefficient $R(k,t)$ on the continuous spectrum, the negative discrete eigenvalues $-{\kappa }_n^2$, and their norming constants $c_n$. Isospectrality givesComparing the asymptotic plane waves in ${\psi }_t=A\psi$ gives

Solved by gpt-5.6-sol high.

The Lax equation $L_t=[A,L]$ withreduces, after expanding the commutator, to the KdV equationThus the evolution is isospectral: the eigenvalues of $L$ are constant. The fourth derivative shown in the converted question is a transcription error.

Solved by gpt-5.6-sol high.

For $\mu \gg 1$, trajectories rapidly approach the cubic nullclineand then move slowly along its attracting outer branches $|x|>1$. For $0<F<1$, the equilibrium lies on the repelling middle branch. Slow motion to a fold, a rapid jump to the opposite outer branch, and repetition produce a relaxation oscillation. For $F>1$, the equilibrium lies on the attracting right branch; trajectories undergo any necessary fast jump and then drift to that equilibrium, so no relaxation cycle remains.

Solved by gpt-5.6-sol high.

Eliminating $y$ givesFor $\mu \ll 1$, take $x=F+A\cos (t+\delta )$ over one fast period. The oscillator energy satisfiesand averaging givesThus, if $F^2<1$, amplitudes approachproducing a stable weakly nonlinear limit cycle. If $F^2>1$, every nonzero small amplitude loses energy and trajectories approach the stable equilibrium.

Solved by gpt-5.6-sol high.

On each side of the stated parallelogram, substitution into the outward normal component ofshows that the leading $L^3$ terms point inward; the remaining terms are $O(L)$, so the region is positively invariant for sufficiently large $L$.

There is one equilibrium,Its Jacobian has determinant one and trace $\mu (1-F^2)$. If $F^2<1$ the equilibrium is unstable. A trajectory starting nearby remains in the compact trapping region, and its omega-limit set contains no stable equilibrium. The Poincare-Bendixson theorem therefore supplies a periodic orbit.

Solved by gpt-5.6-sol high.

Write $\sinh (x\Phi )=(e^{x\Phi }-e^{-x\Phi })/2$. The global maximum of $\Phi$ is the nonstationary endpoint $t=1$, whereEndpoint Laplace's method therefore givesThe second exponential has its maximum at $t=1/2$ and is only of order $x^{-1/2}{e}^{x/8}$, exponentially smaller. Hence

Solved by gpt-5.6-sol high.

Let $\Phi (t)=2t^4-t^2$. Its stationary points on $[0,1]$ are the endpoint $0$ and the interior point $1/2$, withThe endpoint stationary-phase contribution to ${\int }_0^1{e}^{ix\Phi (t)}dt$ isand the interior contribution isThe nonstationary endpoint contributes only $O(x^{-1})$. Taking imaginary parts gives

Solved by gpt-5.6-sol high.

With the other coordinates fixed, the smooth part of the objective has derivative at ${\beta }_1=0$The subdifferential of the ${\ell }^1$ penalty at zero is $\gamma [-1,1]$. ThereforeThe stated condition $|2{\kappa }_1-{\kappa }_2|\le \gamma$ is stronger than the needed condition $|2{\kappa }_1-{\kappa }_2|\le n\gamma$, and hence places zero in this subdifferential. The one-variable exponential loss is strictly convex because the mismatch set is nonempty, so the minimizer is unique:

Solved by gpt-5.6-sol high.

For a convex function $f$, its subdifferential isFor $f(x)=\gamma |x|$,The defining inequality shows directly that $0\in \partial f(x)$ exactly when $x$ is a global minimizer.

Strict convexity meansfor distinct $x,y$ and $0<\lambda <1$. Two distinct minimizers would make the midpoint have a strictly smaller value, so a minimizer is unique.

Solved by gpt-5.6-sol high.

At the optimal terminal wealth, normalized marginal utility defines an equivalent martingale measure $Q$ and hence a marginal utility priceThis is an arbitrage-free price, so $p_Q(Y)\in I$. Concavity of utility implies that the finite-quantity indifference bid does not exceed its marginal price:Consequently

Solved by gpt-5.6-sol high.

Let $u(x)$ be maximal expected utility without the claim. The utility indifference price $\pi (Y)$ is determined byFor claims $Y_1,Y_2$, mix optimal portfolios for their indifference problems with weights $\lambda$ and $1-\lambda$. Concavity of $U$ shows that buying $\lambda Y_1+(1-\lambda )Y_2$ at pricegives expected utility at least $u(x)$. Since maximal utility decreases with purchase price,Thus $\pi$ is concave.

Solved by gpt-5.6-sol high.

If $G\ge 0$ were an arbitrage gain, then $X_1^{\ast }+tG$ would be attainable from the same initial wealth for every $t>0$. Since $U$ is increasing,almost surely, with strict inequality on a set of positive probability under the usual strictly increasing meaning of utility. This contradicts optimality of $X_1^{\ast }$. Hence existence of an optimizer implies no-arbitrage.

Solved by gpt-5.6-sol high.

A risky-asset portfolio $\phi$ financed through the risk-free asset has zero-cost discounted gainIt is an arbitrage when $G_{\phi }\ge 0$ almost surely and $\mathbb{P}(G_{\phi }>0)>0$.

Solved by gpt-5.6-sol high.

The assertion is true by the maximum principle for harmonic functions. A direct reduction to part (i) is also possible: for $\epsilon >0$, setThen $\Delta {\varphi }_{\epsilon }=4\epsilon >0$, soBecause $\Omega$ is bounded, ${\varphi }_{\epsilon }\to \varphi$ uniformly as $\epsilon \downarrow 0$. Taking the limit yields

Solved by gpt-5.6-sol high.

The assertion is false. On the open unit disk take, for $0<\epsilon <1$,Its relevant fourth derivatives satisfyNevertheless, since $x^4+y^4\le (x^2+y^2{)}^2$, every point with $0<x^2+y^2\le 1$ hasThus the maximum lies strictly inside $\Omega$.

Solved by gpt-5.6-sol high.

The assertion is false. On the open unit disk, letThen $\Delta \varphi =-4<0$, but the unique maximum is $phi(0,0)=0$, whereas $phi=-1$ on the boundary.

Solved by gpt-5.6-sol high.

The assertion is true. Continuity and compactness give a maximum of $\varphi$ on $\overline{\Omega }$. At an interior maximum the Hessian matrix is negative semidefinite, so its trace satisfies $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. The maximum must therefore occur on $\partial \Omega$. This is the strict form of the maximum principle for subharmonic functions.

Solved by gpt-5.6-sol high.