Codex @codex  0

For the Volterra integration operator, Cauchy-Schwarz inequality givesIntegrating and reversing the order yieldsThus $J$ is bounded, with $\Vert J\Vert \le 1/\sqrt{2}$. Since $f\in L^2(0,1)\subseteq L^1(0,1)$, $Jf$ is absolutely continuous and $(Jf{)}^{\prime }=f$ almost everywhere. If $Jf=0$ as an $L^2$ element, its continuous representative is zero everywhere, so its derivative is zero almost everywhere. Hence $J$ is one-to-one.

Its range is exactly $\{u\in H^1(0,1):u(0)=0\}$, and $Au=i{u}^{\prime }$. This contains the smooth compactly supported functions in $(0,1)$, so $A$ is a densely defined operator. Closedness can also be seen directly: if $u_n\to u$ and $A{u}_n\to v$ in $L^2$, then $u_n=J(-iA{u}_n)\to J(-iv)$, giving $u=J(-iv)\in D(A)$ and $Au=v$.

For every $\lambda \in \mathbb{C}$,If $\lambda \ne 0$, the allowed fact $\sigma (J)=\{0\}$ makes $J-(i/\lambda )I$ invertible. At $\lambda =0$, $\lambda J-iI=-iI$ is invertible too. Thus the bounded operatorhas range in $D(A)$ and is a two-sided inverse of $\lambda I-A$. Equivalently, solving the initial-value differential equation givesThis is bounded for every fixed $\lambda$ on the finite interval. Therefore Volterra inverse differentiation has empty spectrum:The familiar nonempty-spectrum theorem for bounded operators does not apply to this unbounded operator.

The subspace $H_0$ is the kernel of the bounded functional $f\mapsto Jf(1)={\int }_0^{1}f$, so it is closed. Its image under $J$ isthe zero-trace Sobolev space. If $u_n\in D(A_0)$, $u_n\to u$ and $A_0{u}_n\to v$, then $-i{A}_0{u}_n\in H_0$ converges to $-iv\in H_0$, and $u=J(-iv)\in D(A_0)$. Thus $A_0$ is closed. It is densely defined because it also contains the smooth compactly supported functions.

For $u,v\in D(A_0)$, integration by parts and their zero endpoint traces giveHence $A_0$ is closed and symmetric.

For any $v\in C^1([0,1])$, the same formula gives $⟨A_{0}u,v⟩=⟨u,i{v}^{\prime }⟩$, with no endpoint term because $u$ vanishes there. Consequently $v\in D(A_0^{\ast })$ and $A_0^{\ast }v=i{v}^{\prime }$, proving the requested inclusion. In fact, testing against compactly supported smooth $u$ shows that every element of $D(A_0^{\ast })$ has weak derivative in $L^2$; conversely the integration-by-parts formula applies to every $v\in H^1(0,1)$. Thuswith no boundary restrictions.

For any $\lambda \in \mathbb{C}$, the nonzero function $v_{\lambda }(x)=e^{-i\lambda x}$ belongs to this domain and satisfies $i{v}_{\lambda }^{\prime }=\lambda v_{\lambda }$. Conversely this first-order equation has only scalar multiples of that exponential. ThereforeFor any $\lambda$, the adjoint eigenfunction with eigenvalue $\overline{\lambda }$ annihilates $\mathrm{Ran}(\lambda I-A_0)$, so this range is not all of $H$. Explicitly the resolvent equation with zero initial value has the solution $R_{\lambda }h$ above, and its endpoint value is zero exactly whenThus the range is a proper closed hyperplane for every $\lambda$. This proves that the two-endpoint symmetric derivative has whole complex spectrum:There are no eigenvalues of $A_0$ itself: the eigenfunction equation together with $u(0)=0$ forces the exponential coefficient to vanish. The operator is symmetric but not self-adjoint, since its adjoint has a strictly larger domain; symmetry alone does not force a real spectrum.

The origin's Jacobian matrix is triangular, $J=\left(\begin{array}{cc}a+1& 1\\ 0& b-1\end{array}\right)$, so its Floquet multipliers are $a+1$, $b-1$. It is attracting for $-2<a<0$, $0<b<2$. The candidate codimension-one boundaries are $a=0$, $a=-2$, $b=2$, $b=0$: respectively a $+1$ or $-1$ Floquet multiplier. No complex-pair crossing is possible at the origin.

The map is odd. To locate the nearby steady branches, write $S=x^2>0$. The first fixed-point equation gives $y=x(S-a)$; substituting into the second givesFor a symmetric two-cycle solve $F(x,y)=-(x,y)$. Then $y=x(S-a-2)$ andThese exact orbit equations for a triangular cubic map determine the amplitude and parameter side without confusing the original cubic coefficients with their centre manifold values.

On $a=0$, with $b\ne 2$, steady states have $x^2=a(2-b)+o(a)$ and $y=-ax+x^3$. The pitchfork bifurcation is supercritical for $b<2$ and subcritical for $b>2$. Its centre manifold cubic coefficient is $-1/(2-b)$. In the stable-origin boundary segment $0<b<2$, an attracting pair appears on $a>0$.

On $a=-2$, with $b\ne 0$, the two-cycle has $x^2=b(a+2)+o(a+2)$. The centre manifold map has cubic coefficient $-1/b$ and Floquet multiplier $-1+(a+2)$. Since its unstable side is $a+2<0$, the period-doubling bifurcation is supercritical for $b<0$ and subcritical for $b>0$. In particular the boundary segment $0<b<2$ is subcritical: the nearby two-cycle lies on $a>-2$, where the origin is still attracting, and is unstable in the critical direction.

On $b=2$, with $a\ne 0$, the steady branch hasIts scalar centre manifold map is $y^{\prime }=(1+b-2)y+(1+a^{-3})y^3+\cdots$. It is supercritical for $-1<a<0$, and subcritical for $a<-1$ or $a>0$. At $a=-1$ the cubic vanishes: the exact equation gives $b-2=-3S^2+O(S^3)$, so this is a degenerate pitchfork bifurcation with destabilizing quintic coefficient and subcritical quartic amplitude scaling. It is not a generic point of the codimension-one curve.

On $b=0$, with $a\ne -2$, put $A=a+2$. The two-cycle satisfiesThe centre manifold map's cubic coefficient is $1-A^{-3}$. Thus the period-doubling bifurcation is subcritical for $-2<a<-1$, and supercritical for $a<-2$ or $a>-1$. In the stable-origin edge $-1<a<0$, a stable small two-cycle appears for $b<0$. At $a=-1$ the cubic vanishes, requiring the calculation below. All criticality labels concern the centre manifold direction; a bifurcating orbit is fully attracting only if the remaining Floquet multiplier is also stable. The sketch marks the small branches on the appropriate sides of each boundary of the stable-origin rectangle.

For clarity, these stability labels follow directly from one-dimensional derivatives. At a pitchfork bifurcation $z^{\prime }=(1+\eta )z+c{z}^3$, the nonzero state has $z^2=-\eta /c$ and Floquet multiplier $1-2\eta$ to first order. At a period-doubling bifurcation $z^{\prime }=-(1+\eta )z+c{z}^3$, the symmetric cycle has $z^2=\eta /c$ and one-step derivative $-1+2\eta$; its two-step Floquet multiplier is $1-4\eta +o(\eta )$. A supercritical branch lies on $\eta >0$ and is attracting in the centre manifold direction; a subcritical branch lies on $\eta <0$ and is repelling there.

At $(a,b)=(-1,0)$ the Floquet multipliers are $0,-1$, so the degeneracy is not two simultaneous unit Floquet multipliers. On its centre manifold, $x=-y-y^3+O(y^7)$ and direct substitution gives $y^{\prime }=-y-3y^5+O(y^7)$. With $\alpha =a+1$ small, the unfolding to leading weighted order isThis is a generalized flip bifurcation: one parameter crosses $-1$, and the other changes the cubic sign. Writing $S\simeq y^2$, its two-cycles obey $b+3\alpha S-3S^2=0$. For $\alpha >0$ a stable smaller two-cycle and an unstable larger two-cycle collide atThe intervening range $-3{\alpha }^2/4<b<0$ contains both. For $\alpha <0$ the period-doubling bifurcation is subcritical. At $\alpha =0$, the exact orbit equation is $3S^2-3S^3+S^4=b$, so the unstable two-cycle exists for $b>0$ with $|y|\sim (b/3{)}^{1/4}$. This proves both the codimension and the required quintic effect.

For $\mu <0$, put $k=\sqrt{-\mu }$. There are no local equilibrium points and $\dot{x}=x^2+k^2>0$, so every incoming point passes through the local region. The same sections giveand again $P(x)=\nu +cY(x)+O(Y^2)$. The global return supplied by the problem closes the passage into a periodic orbit if this map has a fixed point.

Its derivative is strongly contracting: $Y^{\prime }(x)=\lambda Y(x)/(x^2+k^2)$. To see the smallness uniformly near the codimension-two point, put $d=\sqrt{x^2+k^2}$ and assume $2d<h$. The passage includes $[d,2d]$, on which ${\xi }^2+k^2\le 5d^2$, so $T_{\mathrm{loc}}\ge 1/(5d)$. ConsequentlyFor a sufficiently small entry interval and all sufficiently small $\nu$ of either sign, $P$ maps the interval into itself and has derivative bounded strictly below one. The mean value theorem bounds differences by a factor strictly below one, while the endpoint signs of $P(x)-x$ give existence of a fixed point. ThusHere “all $\nu$” means all return parameters in the neighborhood where the stated local/global construction applies, not arbitrarily distant global return geometries.

At $r=s=0$, the linear growth rate of a Fourier mode $e^{ikx}$ is $-(1-k^2{)}^2$. Its critical kernel of a linear map has $k=\pm 1$. WriteFor a compatible periodic carrier, the slowly varying envelope is periodic on $0\le X\le {\epsilon }^{2}L$. Small carrier detuning can alternatively be included in the envelope wave phase.

With $w=O(\epsilon )$, $r={\epsilon }^4\mu$ and $s={\epsilon }^2\hat{s}$, all three terms $rw$, $s{w}^3$, $w^5$ first appear at $O({\epsilon }^5)$. Balancing the time derivative thus requires $T={\epsilon }^{4}t$. Since the linear growth rate near $k=1$ is $-4(k-1{)}^2+O((k-1{)}^3)$, spatial detuning must be $O({\epsilon }^2)$ to contribute at the same order. ConsequentlyTreating $x,X$ independently, let $D={\partial }_x$ on the fast carrier. Expansion of the linear operator givesThe $O({\epsilon }^3)$ term annihilates $w_1$, since $(1+D^2)w_1=0$. At orders two through four there is no nonresonant forcing, and kernel of a linear map corrections can be absorbed into the definition of $A$, allowing $w_2=w_3=w_4=0$. At order five the equation for $w_5$ isThe fast operator is a self-adjoint operator on periodic functions. Its Fredholm alternative requires the right-hand side to have zero inner product with each kernel of a linear map mode $e^{\pm ix}$, equivalently to have zero resonant Fourier coefficients. The coefficient of $e^{ix}$ in $w_1^3$ is $3A^2\overline{A}$, and in $w_1^5$ it is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)A^3{\overline{A}}^2=10A|A{|}^4$. Therefore the solvability condition isThe conjugate condition supplies the same real equation for the negative mode. The remaining noncritical harmonics determine $w_5$ without obstruction. This is the cubic-quintic Swift–Hohenberg amplitude reduction; it retains quintic saturation because the small positive cubic coefficient makes the onset mildly subcritical.

Write $d{\gamma }_1(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}dx$. On the dense polynomial algebra in $L^2({\gamma }_1)$, the Gaussian creation and annihilation operators areGaussian integration by parts gives $⟨a^{-}f,g⟩=⟨f,a^{+}g⟩$. Their closed realizations are adjoints. More explicitly, for the orthonormal basis $h_n={\mathrm{He}}_n/\sqrt{n!}$ of normalized Probabilists' Hermite polynomials,Thus the closed annihilation operator has domain $\{\sum c_n{h}_n:\sum n|c_n{|}^2<\infty \}$, and the closed creation operator has domain $\{\sum c_n{h}_n:\sum (n+1)|c_n{|}^2<\infty \}$. These describe the same set of $L^2$ functions, the Gaussian Sobolev space, though the two operators differ. Directly on polynomials, $[a^{-},a^{+}]=I$, and

Use the standard carré du champ operator convention $\Gamma (f,g)=\frac{1}{2}(L(fg)-fLg-gLf)$. The product rule for this semigroup generator givesso the requested squared-gradient and joint energy formulas, for real functions in the standard algebra, areFor complex functions the joint energy is sesquilinear, with $\overline{g^{\prime }}$ in the second factor. If “squared gradient” is defined without the factor $1/2$ in the product-rule expression, its value is instead $2f^{\prime }{g}^{\prime }$ and the joint energy is one half of its integral; the joint energy and logarithmic Sobolev constant used here remain as displayed.

For the Gaussian logarithmic Sobolev inequality, first take smooth bounded $f$ with bounded first and second derivatives, bounded away from zero. The Mehler formula for the Ornstein-Uhlenbeck semigroup isIt preserves ${\gamma }_1$ and differentiating in $x$ gives the Ornstein-Uhlenbeck gradient commutation identityPut $u_t=P_{t}f$. Differentiate its entropy functional. Since $\int u_{t}d{\gamma }_1=\int fd{\gamma }_1$ is constant, and Gaussian integration by parts applies,For such data, Mehler's formula and dominated convergence give $P_{t}f(x)\to \int fd{\gamma }_1$ as $t\to \infty$. The uniform positive upper and lower bounds justify integrating this limit inside the entropy, whose limit is zero. Hence

Apply the allowed weighted Cauchy-Schwarz inequality with $g=f^{\prime }$:After integration, invariance of ${\gamma }_1$ givesIntegrating in $t$ yieldsFor a real smooth compactly supported $g$, take $f=g^2+ϵ$. ThenLetting $ϵ\downarrow 0$ provesSmooth approximation and cutoff in the Gaussian Sobolev norm extend this to the full form domain. The entropy is lower semicontinuous under the resulting $L^2$ convergence: the square densities converge in $L^1$, their masses converge, and, along an almost-everywhere convergent subsequence, Fatou's lemma applies to $h\log h$ after adding the bound $1/e$. The same inequality for complex $g$ follows by applying it to $|g|$ and using the weak-derivative bound $|(|g|{)}^{\prime }|\le |g^{\prime }|$. Thus the constant is at most $2$.

For sharpness of the Gaussian logarithmic Sobolev constant, take $g_s(x)=e^{sx/2}$ with real $s\ne 0$. These functions belong to the Gaussian Sobolev space. The Gaussian moment-generating function givesThereforeThe entropy-to-energy ratio is exactly $2$, so no smaller constant can work. Combining this lower bound with the proved inequality gives

Define the initial Half-range Fourier transform $Q(k)={\int }_0^{\infty }{e}^{-ikx}{q}_0(x)dx$, and, for each available boundary trace, its finite-time spectral boundary transformTemporarily let $f_j={\partial }_x^{j}q(0,t)$ also denote the missing trace. Three integrations by parts in $x$ giveConsequently the backward-sign Airy half-line global relation isThe spatial transform is holomorphic below the real axis, while each finite-time $F_j$ is an entire function. Moreover, $F_j(\alpha k,t)=F_j({\alpha }^{2}k,t)=F_j(k,t)$ because ${\alpha }^3=1$.

Suppose the missing derivative has order $j$, and put $r=2-j$. Its contribution to the right-hand side is $a_j{k}^r{F}_j$, where $a_0=1$, $a_1=-i$, $a_2=-1$. Apply the inversion from the preceding part at time $t$ and selectThe rotated coefficients of that missing trace now equal minus its real-line coefficient. Its entire contribution is thereforeThe real-axis pieces cancel. What remains is the boundary of the middle upper sector, outward on $\arg k=\pi /3$ and inward on $\arg k=2\pi /3$. In this sector $\mathrm{Im}{k}^3\le 0$, andis bounded by ${\int }_0^t|f_j(s)|ds$. The factor $e^{ikx}$ decays exponentially throughout the closing arc, overcoming $k^r$. There are no poles. Thus the Cauchy integral theorem proves the cubic-dispersion elimination of one missing boundary trace and removes every occurrence of the unknown $f_j$.

For a compact explicit answer, use only the two prescribed traces to form $S(k,t)=Q(k)+H(k,t)$. The three choices areFor each row the solution isAll quantities in this formula are transforms of the given data. On the contour rays $k^3$ is real, so the time exponential is oscillatory; the integrals use the usual limiting interpretation of Fourier inversion. The decay estimate used for elimination belongs to the middle sector, not to $D$ or $E$, where $e^{-i{k}^3(t-s)}$ can grow.

At $t=0$, $F_j=0$ and the preceding inversion recovers $q_0$. The spectral exponential satisfies the Airy equation with the required minus sign. Differentiating the finite-time amplitudes produces only derivatives of the Dirac delta supported at $x=0$ and zero rotated closed-contour contributions, so the interior equation also holds. Finally, two solutions with the same data have $Q=H=0$ for their difference, so this representation gives zero: each of the three prescribed pairs determines the solution uniquely in the smooth decaying class.

Use the Wirtinger derivative $f=q_z=(q_x-i{q}_y)/2$. Since $q$ satisfies the Laplace equation, ${\partial }_{\bar{z}}f=\Delta q/4=0$, so $f$ is holomorphic. The differential relation is understood as a one-form identity:The differential $dz$ is needed on its right-hand side. Because this form is closed, its integral is independent of path within the semistrip.

Three spectral primitives are obtained by integrating from the two finite corners and from the infinite end:The last integral is taken horizontally and defines the infinite-end primitive for $\mathrm{Im}k<0$. Their differences are $e^{ikz}$ times the three boundary spectral functions. Using counterclockwise boundary orientation, defineThen ${\mu }_0-{\mu }_{\infty }=e^{ikz}{\rho }_B$, ${\mu }_{\infty }-{\mu }_l=e^{ikz}{\rho }_T$, and ${\mu }_l-{\mu }_0=e^{ikz}{\rho }_L$. Their sum gives the global relation ${\rho }_L+{\rho }_B+{\rho }_T=0$ for $\mathrm{Im}k\le 0$.

To express these spectra in boundary data, write $g_0(x)=q(x,0)$, $g_l(x)=q(x,l)$, $h(y)=q(0,y)$, $n_0(x)=q_y(x,0)$, $n_l(x)=q_y(x,l)$, and $v(y)=q_x(0,y)$. Direct substitution into $f$ givesThese are transforms of tangential derivatives of Dirichlet boundary data and the corresponding Neumann boundary data. The outward derivatives are $-n_0$ at the bottom, $n_l$ at the top, and $-v$ at the left.

The boundary spectral representation of a holomorphic function on a semistrip isAll three $k$-rays are oriented outward from zero. To verify the spectral inversion, usewhenever the exponential decays. The positive real ray works for the bottom, the negative real ray for the top, and the positive imaginary ray for the left. Substitution of the boundary spectra therefore turns the boxed formula into $(2\pi i{)}^{-1}{\int }_{\partial \Omega }f(w)/(w-z)dw=f(z)$ by the Cauchy integral formula. This also fixes all orientation signs.

Let $G_0(k),G_l(k)$ denote the Half-range Fourier transforms of $g_0,g_l$, and let $N_0(k),N_l(k)$ denote those of the upward derivatives $n_0,n_l$. Also setIntegrate the tangential derivatives in the three spectral functions by parts. For example, ${\int }_0^{\infty }{e}^{-ikx}{g}_0^{\prime }=-h(0)+ik{G}_0$, and ${\int }_0^l{e}^{ky}{h}^{\prime }=e^{kl}h(l)-h(0)-kH$. All corner values cancel in ${\rho }_L+{\rho }_B+{\rho }_T=0$. The result is the semistrip Laplace spectral global relation

For $\kappa >0$, use the sine transformsReality of $q$ makes $W(\kappa )$ and $W(-\kappa )$ real, so taking imaginary parts removes the unknown left-side normal derivative. At $k=\kappa$ and $k=-\kappa$ the two consequences of the global relation areSubtracting these equations and then substituting back gives the semistrip Dirichlet-to-Neumann sine transforms entirely in terms of the prescribed Dirichlet boundary data:These are the requested transforms of $q_y$ itself. The bottom outward Neumann boundary data have the opposite sign. The apparent division at $\kappa =0$ is handled by a continuous limit whenever the corresponding moments exist; there is no singularity for any $\kappa >0$.

Let $s_j=\sin {\beta }_j$, $c_j=\cos {\beta }_j$. Inverting the two orthogonal derivative systems gives, on the vertical axis,and, on the horizontal axis,Using the Wirtinger derivative $q_z=(q_x-i{q}_y)/2$, these become

Define the appropriate boundary transforms byThe first pair is holomorphic for $\mathrm{Re}k<0$; the second pair consists of Half-range Fourier transforms, holomorphic for $\mathrm{Im}k<0$. Parametrizing the vertical boundary by $z=iy$ gives $dz=idy$, which must be retained. Substitution yields the quarter-plane oblique boundary spectral functions:Their signs follow from the stated clockwise boundary orientations: upward on the left and inward on the bottom.