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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

Use the joint energy conventionOn the generator domain it is $-⟨Lf,g⟩$. The symmetric Feller semigroup has symmetric two-point measure ${\nu }_s(dx,dy)=\mu (dx)P_s(x,dy)$, with both marginals equal to $\mu$. Expanding the product of increments gives the finite-time identityfor real functions. This supplies a proof that works for a general symmetric Markov kernel, without assuming a diffusion chain rule.

For $a\ge b\ge 0$, Cauchy-Schwarz inequality givesInterchanging $a,b$ gives the same inequality in the other order. Apply it with $a=f(x),b=f(y)$, integrate against ${\nu }_s$, and take $s\downarrow 0$. This proves the power inequality for symmetric Markov energiesInitially take bounded nonnegative functions in the form domain. On their bounded range, the power maps are Lipschitz; their finite-time squared-increment energies are therefore bounded by a constant times that of $f$. This puts the powers in the form domain and makes both joint energy limits well defined. Truncation gives the corresponding extended-energy interpretation when needed. In particular the proof applies to the bounded positive functions used below.

Normalize the logarithmic Sobolev inequality asStart with a bounded $f$ bounded away from zero, and put $u(t)=P_{t}f$, $q=q(t)$ and $F(t)=\int u^{q}d\mu$. Positivity and preservation of constants keep $u$ within the same positive bounds. Symmetry realizes $P_t$ as a self-adjoint contraction semigroup on $L^2(\mu )$, so for $t>0$ its orbit lies in the generator domain and ${\partial }_{t}u=Lu$; this justifies the following differentiation away from zero.

Differentiate both the evolving function and exponent:Since $\log \Vert u{\Vert }_q=q^{-1}\log F$, it follows thatApply the logarithmic Sobolev inequality to $g=u^{q/2}$, and then the joint energy inequality:ThereforeThe specified exponent satisfies $q(0)=2$ and $q^{\prime }=4(q-1)/c_{\mathrm{LS}}$. Thus the right side is zero and the evolving norm is nonincreasing. Integrating from a positive time $\delta$ and letting $\delta \downarrow 0$ givesHere the initial limit follows from strong $L^2$ continuity, boundedness of $f$ and $P_{\delta }f$, and $q(\delta )\to 2$.

For a bounded nonnegative $f$, apply the estimate to $f+ϵ$ and let $ϵ\downarrow 0$. For arbitrary $f\in L^2(\mu )$, truncate $|f|$ to bounded nonnegative functions. Their images converge in $L^2$ to $P_t|f|$; an almost-everywhere convergent subsequence and Fatou lemma give the same $L^{q(t)}$ estimate for that limit. Positivity of the Markov kernel gives $|P_{t}f|\le P_t|f|$, also for complex $f$, soThis proves that the logarithmic Sobolev inequality implies hypercontractivity, with the numerical exponent matched to the stated joint-energy normalization.

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.

A linear operator $Z$ on a Banach space is called a dissipative operator whenOn a Hilbert space, using the inner product linear in its first argument,If $\mathrm{Re}⟨Zf,f⟩\le 0$, this proves the defining inequality. Conversely that inequality implies $2\lambda \mathrm{Re}⟨Zf,f⟩\le \Vert Zf{\Vert }^2$ for every $\lambda >0$; letting $\lambda \to \infty$ gives

For the next spectral assertion, dissipativity must be retained as a hypothesis. Let $Z$ be closed, densely defined and dissipative. For $\alpha >0$, the dissipative bound makes $\alpha I-Z$ injective. Its range is closed: if $(\alpha I-Z)f_n\to h$, the bound makes $f_n$ Cauchy, and then $Z{f}_n=\alpha f_n-(\alpha I-Z)f_n$ converges. Closedness produces a preimage of $h$. If the range is all of $H$, the same bound gives a bounded inverse, so $\alpha$ is in the resolvent set. If the closed range is proper, there is a nonzero vector in its orthogonal complement, and the definition of the adjoint of a densely defined operator givesThis proves that the positive spectrum of a dissipative operator is adjoint point spectrum:Without dissipativity the PDF's assertion is false. For example, multiplication by $x$ on $L^2(0,2)$ is bounded, closed and densely defined. The value $1$ is spectral: normalized functions supported closer and closer to $x=1$ make $\Vert (Z-I)f\Vert$ tend to zero. But $Z^{\ast }=Z$ has no eigenvector for eigenvalue $1$, since a square-integrable function supported on the singleton $\{1\}$ is zero.

For the given maximal-domain operator $Q$, finite-support sequences belong to $D(Q)$, so the domain is dense. Coordinatewise limits also show it is closed: if $f^{(j)}\to f$ and $Q{f}^{(j)}\to h$ in ${\ell }^2$, each defining coordinate equation passes to the limit, giving $Qf=h\in {\ell }^2$ and hence $f\in D(Q)$.

Put $g_n=2^n{f}_n$ and $h=Qf$. For $n\ge 1$,Since ${\sum }_{n\ge 1}{2}^{-n}|h_n|<\infty$ by Cauchy-Schwarz inequality, $g_n$ has a finite limit ${\ell }_f$. Its increments are square summable as well. Direct summation givesThe original inner-product series converges because $Qf,f\in {\ell }^2$. Taking $N\to \infty$ therefore givesThus $Q$ is dissipative, including its full maximal domain; the limiting boundary term has not been discarded.

To decide generation, solve $(\alpha I-Q)f=h$ for arbitrary $h\in {\ell }^2$ and fixed $\alpha >0$. The first coordinate gives $g_0=f_0=h_0/(\alpha +1)$, and the rest giveBecause all denominators are at least one,Hence $g_n$ is bounded, so $f_n=2^{-n}{g}_n$ belongs to ${\ell }^2$. The coordinate equations give $Qf=\alpha f-h\in {\ell }^2$, proving $f\in D(Q)$ and surjectivity of $\alpha I-Q$. Together with density, closedness and dissipativity, the Lumer-Phillips theorem proves the contraction generator from an exponentially weighted difference operator conclusion:

The domain is crucial. The formal transposed matrix suggests a positive-eigenvalue solutionwhich is square summable because the products converge. It is not a true adjoint eigenvector: the admissible test vector $f_n=2^{-n}$ has $Qf=(-1,0,0,\dots )$, so $⟨Qf,u⟩=-\overline{u_0}$, while $\alpha ⟨f,u⟩=\alpha \overline{u_0}{\sum }_n{4}^{-n}{\prod }_{k<n}(1+\alpha 4^{-k})$. These cannot agree for $\alpha >0$ unless $u_0=0$. This exemplifies how boundary terms exclude formal adjoint eigenvectors, and explains why a calculation using only finite-support test vectors would give the wrong generation answer.

A strongly continuous semigroup on the complex Banach space $E$ is a family $T_t\in \mathcal{L}(E)$ with $T_0=I$, $T_{s+t}=T_s{T}_t$ and $\Vert T_{t}f-f\Vert \to 0$ as $t\downarrow 0$ for every $f\in E$. Its semigroup generator is the linear operatorFor $f\in D(Z)$, boundedness of $T_t$ and the semigroup identity giveThus semigroup commutation with its unbounded generator meansThere is no assertion that $Z$ is defined on every vector. The orbit is differentiable on this domain, with $\frac{d}{dt}{T}_{t}f=T_{t}Zf$.

A contraction semigroup additionally satisfies $\Vert T_t\Vert \le 1$ for every $t\ge 0$. For $\mathrm{Re}\lambda >0$, define the Bochner integralStrong continuity gives a continuous integrand, and its norm is bounded by $e^{-t\mathrm{Re}\lambda }\Vert f\Vert$, so the integral exists andChanging variables in $T_h{L}_{\lambda }f$ yieldsAfter subtracting $L_{\lambda }f$ and dividing by $h$, the limit is $\lambda L_{\lambda }f-f$. ConsequentlyFor $f\in D(Z)$, integrate the derivative of $e^{-\lambda t}{T}_{t}f$ over $[0,\infty )$. Its limit at infinity is zero by the contraction bound, and its value at zero is $f$. ThereforeThus this is also the Laplace-transform formula for a semigroup resolvent, with $L_{\lambda }=(\lambda I-Z{)}^{-1}$.

To prove that $Z$ is a closed operator, suppose $f_n\in D(Z)$, $f_n\to f$ and $Z{f}_n\to g$. Fix one $\lambda$ with positive real part. The identity just proved and boundedness of $L_{\lambda }$ giveIts range lies in $D(Z)$, and applying $\lambda I-Z$ shows $Zf=g$. Thus the graph is closed. Linearity follows directly from linearity of the defining difference quotient, so $Z$ is a closed linear operator.

The closedness conclusion also holds without the contraction assumption. For a general strongly continuous semigroup, the Uniform boundedness principle gives a uniform operator bound on each compact time interval. Integrating the generator-orbit identity gives $T_t{f}_n-f_n={\int }_0^t{T}_{s}Z{f}_{n}ds$. Under $f_n\to f$ and $Z{f}_n\to g$, pass to the limit to obtain $T_{t}f-f={\int }_0^t{T}_{s}gds$. Divide by $t$ and let $t\downarrow 0$; strong continuity makes the right side converge to $g$, so $f\in D(Z)$ and $Zf=g$.

The material derivative follows surfactant carried by the moving interface. The term $-C{\nabla }_s\cdot u_s$ is dilution or concentration by tangential area expansion or contraction. The term $-C(u\cdot n){\nabla }_s\cdot n$ is the additional area change produced by normal motion of a curved interface. The term $D_s{\nabla }_s^{2}C$ represents surface diffusion, while $-k(C-C_0)$ relaxes the concentration towards a reservoir value through exchange. All these terms refer to concentration per surface area, as in surfactant transport with exchange relaxation.

In the steady bubble frame, the spherical interface has $u\cdot n=0$. Write $C=C_0+C^{\prime }$ with uniform $C_0$ and regard both $u$ and $C^{\prime }$ as first order in the rising speed. Advection of $C^{\prime }$ and its product with surface dilation are second order. Hence the linear steady equation isSpherical symmetry and linearity leave only the tangential projection of the imposed translation vector: $u_s=A{I}_{s}U$. An azimuthal term $n\times U$ would change sign under reflection and is excluded for this non-chiral rising bubble. The degree-one forcing does not excite other spherical-harmonic degrees in the linear problem.

On $r=a$, differentiating $n=x/r$ givesEach component of $n$ is a degree-one spherical harmonic, soOne may also obtain the coefficient from $n\cdot n=1$: its surface Laplacian gives $n\cdot {\nabla }_s^{2}n=-|{\nabla }_{s}n{|}^2=-2/a^2$, while rotational symmetry makes the vector Laplacian parallel to $n$. Expanding $I_{s}U=U-n(U\cdot n)$ then yieldsThe candidate $C^{\prime }=BU\cdot n$ therefore solves the linear transport equation ifIts spherical mean is zero. This is the dipolar surfactant distribution. Assuming $k{a}^2+2D_s>0$, its consistency condition isDiffusion or exchange must restore uniform concentration faster than the interfacial flow redistributes it: this holds with a small surface Péclet number, fast exchange, or sufficiently immobilized surface velocity. These same conditions justify neglecting the perturbation's advection. Small capillary number and sufficiently small buoyancy-induced shape distortion are separate requirements for a spherical interface.

Take $+$ to mean the outer liquid, $-$ the bubble interior and $n$ outward from the bubble. The general interfacial stress balance with variable surface tension isFor $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$, its tangential part is $-{\nabla }_s\gamma ={\gamma }_{1}B{I}_{s}U/a$. Thus the requested form isThis positive parameter measures the strength of the restoring Marangoni stress relative to viscous stress, including the relaxation of concentration by diffusion and exchange.

The far velocity $-U$ is produced by the constant harmonic vector potential $\Phi =U$. The two independent decaying degree-one modes are the force-carrying vector potential $\alpha aU/r$ and the scalar dipole $\chi =\beta a^{3}U\cdot \nabla (1/r)$. They are harmonic outside the bubble and have precisely the symmetry of uniform translation. The Unscaled Papkovich–Neuber representation gives, at arbitrary radius,The velocity at $r=a$ agrees with the printed expression. However, the original PDF's traction formula has a factor-of-two error. Direct differentiation with $\sigma =-pI+\mu (\nabla u+\nabla u^T)$ givesnot the printed prefactor $12\mu /a$. This is traction of translating-sphere Papkovich–Neuber potentials. The $\alpha$ term follows immediately from the Stokeslet stress, and the scalar-potential part gives $6\mu \beta [U-3(U\cdot n)n]/a$; neither contribution contains the extra factor two.

With this consistent traction, no penetration requires $\beta -\alpha =1/2$, tangential velocity requires $A=-(1+\alpha +\beta )$, and tangential stress balance requires $\beta =A\lambda$. Solving,For comparison, taking the printed $12\mu /a$ literally would instead giveThose solve the supplied algebraic pair but not the Newtonian stress calculated from the supplied potentials. The factor-six result is the physically consistent translating surfactant-coated bubble solution.

As $\lambda \to 0$, $A\to -1/2$ and $\beta \to 0$: the clean bubble has a mobile interface and zero tangential traction. As $\lambda \to \infty$, $A\to 0$, $\alpha \to -3/4$ and $\beta \to -1/4$: the interface becomes immobilized and the exterior flow approaches that past a no-slip rigid sphere. The limiting tangential traction is $-3\mu I_{s}U/(2a)$, a useful check on the correct normalization.

The normal dynamic traction isapart from an arbitrary constant pressure. In the normal stress balance on a translating bubble, its degree-one part is balanced by the buoyant hydrostatic pressure difference and by the variation $-2{\gamma }_1{C}^{\prime }/a$ of normal capillary stress. A uniform internal gas pressure balances the constant Laplace-pressure term but cannot by itself balance this angular variation. If $U$ is upwards and $\Delta \rho$ is liquid minus bubble mass density, the degree-one balance can be writtenIt fixes $U=\Delta \rho g{a}^2(1+2\lambda )/[3\mu (1+3\lambda )]$, tending to the clean-bubble and rigid-sphere rise speeds. A degree-one radial displacement translates a sphere rather than changing its curvature, so a supposed dipolar shape distortion cannot replace this force balance.