The interface is material. Applying the kinematic boundary condition to $y-\eta (x,t)=0$ gives, exactly,The disturbance velocities and $\eta$ are small. Products of two disturbance quantities are second order, and a Taylor expansion moves their evaluation to $y=0$ at linear order. HenceThe extra upper-fluid term is advection of the displaced interface by the base uniform flow.

Use the Unsteady Bernoulli equation separately on each side, with constant mass density $\rho$. There are no restoring terms specified at this internal interface, and its pressure is continuous. Choose the time-dependent potential gauges so that the undisturbed pressure is $p_0$ on both sides. Subtracting the base values from Bernoulli's principle gives, at linear order,The quadratic disturbance-velocity terms have been discarded, while the cross term with the base velocity is first order and must remain. Equating the perturbation pressures yields the third condition:

The total velocity fields on the two sides areEach base flow is divergence-free. Incompressibility therefore makes both disturbance velocity potentials satisfy the Laplace equation:At the fixed horizontal walls, impermeability requires zero vertical velocity:The discontinuity in tangential base velocity at the interface is a vortex sheet. The assumption of irrotational flow is understood separately in each open fluid region; it cannot hold distributionally across that sheet when $U\ne 0$.

The complex velocity vanishes where $e^{-i\alpha }{\zeta }^2-e^{i\alpha }=0$, provided its denominator is nonzero. Thus $\zeta =\pm e^{i\alpha }$. Both lie on the unit circle, and ${\zeta }^2-a^2\ne 0$ since $a<1$. Mapping these points into the physical plane givesEquivalently, the two stagnation points are the opposite points $(A\cos \alpha ,B\sin \alpha )$ and $(-A\cos \alpha ,-B\sin \alpha )$ on the ellipse. The conformal map preserves the zeros of the complex velocity here because its derivative is finite and nonzero on the boundary.

At large $|\zeta |$, $z=\zeta +O({\zeta }^{-1})$, so the map tends to the identity. Its exterior inverse has the expansion $\zeta =z-a^2/z+O(z^{-3})$. Therefore the physical complex potential of a two-dimensional potential flow is $w(z)=W(\zeta (z))$, withIts leading term is unchanged, so the far-field velocity still has magnitude $U$ and inclination $\alpha$.

The chain rule gives the complex velocity:The parameter $\zeta$ denotes the unique exterior preimage of $z$. If an expression solely in $z$ is desired, takewhere the exterior holomorphic square root is chosen so that $\sqrt{z^2-4a^2}\sim z$ at infinity. Its branch points lie inside the ellipse. This is uniform potential flow around an ellipse.

Parameterize the unit circle by $\zeta =e^{i\theta }$. Under the Joukowski map,Consequently its image is the ellipsewith horizontal and vertical semi-axesBoth are positive because $0<a<1$. The horizontal axis is the major axis. The derivative $dz/d\zeta =1-a^2/{\zeta }^2$ is nonzero for $|\zeta |\ge 1$. Also, equality of two images implies $({\zeta }_1-{\zeta }_2)(1-a^2/({\zeta }_1{\zeta }_2))=0$; two distinct exterior points cannot satisfy the second factor since $|{\zeta }_1{\zeta }_2|>1>a^2$. Thus the exterior map is one-to-one, mapping the exterior of the unit circle conformally onto the exterior of the ellipse.

A uniform flow with physical velocity $(U\cos \alpha ,U\sin \alpha )$ has complex velocity $U{e}^{-i\alpha }$ and incident potential $f(\zeta )=U{e}^{-i\alpha }\zeta$. Apply the Milne-Thomson circle theorem with $R=1$:Indeed, on $\zeta =e^{i\theta }$ it is $W=2U\cos (\theta -\alpha )$, which is real, so the boundary is a streamline. At infinity, $W^{\prime }(\zeta )\to U{e}^{-i\alpha }$, verifying the specified incoming uniform flow.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in planar potential flow. Suppose $f(\zeta )$ is an incident complex potential of a two-dimensional potential flow with no singularities on or inside $|\zeta |=R$. Thenis a complex potential of a two-dimensional potential flow outside the circle, with the same exterior singularities and their strengths as the incident flow; the added image singularities lie inside the excluded disk. On $|\zeta |=R$, $R^2/\bar{\zeta }=\zeta$, so $W=f+\bar{f}$ is real. Its imaginary part, the streamfunction, is constant on the circle, proving the no-penetration boundary condition. The theorem gives the usual zero-circulation construction; a separately prescribed circulation can be added by its logarithmic vortex potential.

On the translating circular cylinder $r=a$, the preceding velocity potential and streamfunction giveUsing the boundary integral for kinetic energy of planar potential flow,ThereforeAs a direct check, $|\mathbf{u}{|}^2=U^2{a}^4/r^4$, and integration over $r>a$ gives the same result.

The fluid kinetic energy has the same dependence on body speed as the kinetic energy of an extra mass $M$ carried by the body. Thus $M$ is its added mass, here equal to the mass of displaced fluid per unit cylinder length. For a cylinder of physical mass per length $m$, the combined kinetic energy is $\frac{1}{2}(m+M)U^2$. Accelerating the body requires accelerating the surrounding fluid, producing the inertial reaction $-M\dot{U}$ along the direction of motion. This is the added mass of a circular cylinder.

The product rule, the Laplace equation and $\mathbf{u}=\nabla \varphi$ giveFirst truncate the exterior fluid domain by a large circle $C_R$. Let ${\mathbf{n}}_B$ point outward from the body into the fluid. The outward normal of the exterior fluid domain at the body is $-{\mathbf{n}}_B$. The planar divergence theorem, equivalently Green's theorem, therefore givesOrient $\partial B$ counterclockwise around the body. Then ${\mathbf{n}}_{B}ds=(dy,-dx)$ andFor the finite-energy exterior rigid-body flow intended here, the outer boundary term vanishes as $R\to \infty$. In particular, zero net volume flux and zero circulation eliminate the $1/z$ term in the exterior Laurent series of the complex velocity. Thus $\mathbf{u}=O(R^{-2})$ and, after choosing the additive potential constant, $\varphi =O(R^{-1})$. The outer integral is then $O(R^{-2})$.

Multiplying by half the constant mass density gives the boundary integral for kinetic energy of planar potential flow:The minus sign comes from the fluid's inner-boundary orientation, rather than a negative physical kinetic energy.

There is a necessary qualification to the printed hypotheses. Decaying velocity and zero circulation alone do not ensure a finite kinetic energy: $\varphi =A\log r$ gives $\mathbf{u}=A{\mathbf{e}}_r/r$, zero circulation, and logarithmically divergent exterior kinetic energy. It carries a nonzero volume flux from the body and is not the flow of an impermeable rigid body. The finite-energy condition for planar exterior potential flow makes explicit the source-free assumption needed to discard the outer boundary term.

Use instantaneous polar coordinates centered on the moving circular cylinder, but measure the fluid velocity in the laboratory frame. Impermeability equates the normal components of body and fluid velocity:Since ${\mathbf{e}}_x\cdot {\mathbf{e}}_r=\cos \theta$ and $u_r={\varphi }_r$, the no-penetration boundary condition isThe velocity potential obeys the exterior Laplace equation, by incompressibility. The boundary data have only the first cosine harmonic. A decaying harmonic potential of this form is $\varphi =C\cos \theta /r$. Matching its radial derivative gives $-C/a^2=U$, so $C=-U{a}^2$. The zero circulation condition rules out an angular potential term, and the integrated boundary condition gives zero net volume flux, ruling out a radial logarithmic source term. The remaining decaying harmonics have zero coefficients by the boundary data. HenceUsing the streamfunction in polar coordinates, $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$, givesOne may also obtain both functions as the real and imaginary parts of $w(z)=-U{a}^2/z$, a two-dimensional potential doublet. The velocity decays as $r^{-2}$ and the source of this singular representation lies inside the excluded body. These are the fields used to calculate the added mass of a circular cylinder.

Choose the planar streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. For an irrotational flow, a velocity potential gives $\mathbf{u}=\nabla \varphi$. ThusThe streamfunction enforces incompressibility, since $u_x+v_y={\psi }_{yx}-{\psi }_{xy}=0$. Conversely, incompressibility permits a local streamfunction, while zero curl permits a local velocity potential. These are the Cauchy-Riemann equations for the complex potential of a two-dimensional potential flow $w=\varphi +i\psi$.

The gradients of the two functions satisfyEach gradient is normal to its regular level curves. Therefore regular curves of constant velocity potential and constant streamfunction meet orthogonally. At a stagnation point, both gradients vanish and this tangent-direction argument needs a limiting interpretation. The conclusion concerns regular portions of the two families of curves.

Let $u,w\in M$ both solve the variational inequality. Test the inequality for $u$ with $w$ and the one for $w$ with $u$, then add. Writing $d=u-w$ givessince cosine is Lipschitz with constant one. Since $d(0)=0$, the fundamental theorem of calculus and Cauchy-Schwarz inequality give $|d(x)|\le \sqrt{l}\Vert d^{\prime }{\Vert }_2$; integrating gives $\Vert d{\Vert }_2^2\le l^2\Vert d^{\prime }{\Vert }_2^2$. ThereforeFor $\boxed{0<\gamma <2c_0/l^2}$, the coefficient is positive, so $d^{\prime }=0$ and the zero boundary value gives $d=0$. This proves uniqueness with an explicit sufficient smallness condition. More generally any Poincaré inequality constant $C_P$ gives the condition $\gamma C_P^2<2c_0$; optimizing that constant is unnecessary for the requested conclusion. The underlying principle is uniqueness for a strongly monotone variational inequality.

The noncontact set $D=\{x\in (0,l):u(x)>g(x)\}$ is open. For $\phi \in C_0^{\infty }(D)$, the positive gap above the obstacle has a positive minimum on the support of $\phi$. Thus both $u+t\phi$ and $u-t\phi$ are admissible for sufficiently small $t>0$. Applying the variational inequality to both signs yieldsHence the Euler-Lagrange equation on the noncontact set isIt holds classically under the assumed smoothness. On the contact set the obstacle prevents the negative variations used to obtain this equality, so a variational inequality remains the appropriate global description.

For any $v\in M$, the segment $u+t(v-u)$ stays in $M$ for $0\le t\le 1$. Since $u$ is a minimizer, the right derivative of its energy at $t=0$ is nonnegative. The first variation therefore gives the variational inequalityThe factor two comes from the quadratic gradient term. Only a one-sided inequality follows because a variation pointing below the obstacle need not be admissible.

One-dimensional Sobolev embedding gives a continuous representative for each $v\in H_0^1(0,l)$, with zero endpoint values, so the pointwise obstacle condition is meaningful. The positive part $g_{+}=\max \{g,0\}$ is Lipschitz and vanishes near both endpoints because $g(0),g(l)<0$. It therefore belongs to $H_0^1$ and satisfies $g_{+}\ge g$, proving that $M$ is nonempty.

If $v,w\in M$ and $t\in [0,1]$, then $(1-t)v+tw\ge g$, so $M$ is a convex set. If $v_n\to v$ in $H_0^1$, the estimate $\Vert v_n-v{\Vert }_{\infty }\le \sqrt{l}\Vert v_n^{\prime }-v^{\prime }{\Vert }_2$ gives uniform convergence. Passing to the limit in $v_n(x)\ge g(x)$ shows $v\in M$. Thus the admissible set is nonempty, closed and convex.

Take the zero-boundary Sobolev space $X=H_0^1(\Omega )$ with norm $\Vert u{\Vert }_X=\Vert \nabla u{\Vert }_2$. The Poincaré inequality makes this equivalent to the usual Sobolev norm. DefineThe weak formulation is $u\in X$ and $⟨A(u),v⟩=⟨f,v⟩$ for every $v\in X$.

In dimension three the Sobolev embedding theorem gives $\Vert u{\Vert }_6\le C_6\Vert u{\Vert }_X$, so $u^3\in L^2$. The Cauchy-Schwarz inequality and Poincaré inequality yieldThus $A(u)\in X^{\ast }$ for every $u$, and $f\in X^{\ast }$. Finallysince $u^2+uw+w^2\ge 0$. Hence $A$ is strongly monotone with constant one in the chosen norm.

Using the dual pairing, a monotone operator satisfies $⟨A(u)-A(v),u-v⟩\ge 0$ for all $u,v\in M$. It is a strongly monotone operator if some $m>0$ givesA nonlinear coercive operator with respect to $u_0\in M$ satisfies $⟨A(u),u-u_0⟩/\Vert u-u_0\Vert \to +\infty$ as $\Vert u-u_0\Vert \to \infty$ in $M$.

Strong monotonicity, with $v=u_0$, givesAfter division, the lower bound is $m\Vert u-u_0\Vert -\Vert A(u_0){\Vert }_{X^{\ast }}\to +\infty$, proving coercivity. Reflexivity and separability are not needed for this particular estimate.

Work in the Banach space $C([0,1])$ with its supremum norm, and take the nonempty closed bounded convex set $M=\{u:100\le u(t)\le 121\text{ for all }t\}$. DefineFor $u\in M$ the absolute value is unnecessary, and $\boxed{101\le Tu(t)\le 101+12=113\le 121}$. Thus $T:M\to M$. Also, for $a,b\ge 100$,so $\Vert Tu-Tv{\Vert }_{\infty }\le \frac{1}{20}\Vert u-v{\Vert }_{\infty }$ and in particular $T$ is continuous.

The image is uniformly bounded, and $|Tu(t)-Tu(s)|\le 12|t-s|$, so it is equicontinuous. The Arzelà-Ascoli theorem says that a uniformly bounded equicontinuous family in $C([0,1])$ has relatively compact closure in the uniform norm. Hence $T$ is compact, and the Schauder fixed-point theorem supplies $\boxed{u=Tu}$, the required solution. In fact the displayed contraction estimate also lets the contraction mapping theorem give uniqueness within $M$ and convergence of the corresponding Picard iteration. The initial constant here is the printed integer $101$.

The Schauder fixed-point theorem states that a continuous compact self-map of a nonempty closed bounded convex set in a Banach space has a fixed point. Equivalently in this bounded-domain version, it suffices for the image of the map to be relatively compact. The theorem guarantees existence, without requiring a contraction estimate or promising uniqueness.

A nonlinear compact operator is a continuous map that sends bounded sets of its domain to relatively compact subsets of the target space; linearity is not part of this general definition.

For the specified diagonal operator on sequence space on l2 sequence space,so it is well-defined, linear and bounded. Let $T_N$ retain only its first $N$ output coordinates. This is a finite-rank operator, andThe upper bound follows by estimating each tail multiplier by $1/(N+1)$; equality follows on the unit coordinate vector $e_{N+1}$. Finite-rank operators between Banach spaces are compact, and a limit of compact linear operators in operator norm is compact, the norm-closedness of compact operators. Therefore $\boxed{T\text{ is compact}}$.