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 $\boxed{T\text{ is compact}}$.

The Brouwer fixed-point theorem says that every continuous map $T:{\overline{B}}_R(0)\to {\overline{B}}_R(0)$ has a fixed point. To use the supplied outward vector field zero theorem, first extend the construction to all of ${\mathbb{R}}^n$. LetThe radial retraction onto a closed ball $P_R$ is continuous, so $g$ is a continuous vector field on the required whole space. On $|x|=R$, the Cauchy-Schwarz inequality givesThe supplied theorem gives $g(x_0)=0$ with $|x_0|\le R$. There $P_R(x_0)=x_0$, hence $\boxed{T(x_0)=x_0}$. This explicit extension avoids presuming that the original map $T$ was defined outside the ball.

The contraction mapping theorem states that a self-map $T$ of a nonempty complete metric space with a uniform contraction constant $q<1$ has a unique fixed point $p$. For every starting point, the iteration $x_{n+1}=T{x}_n$ converges to $p$, with $d(x_n,p)\le q^{n}d(x_1,x_0)/(1-q)$.

For the cosine fixed-point iteration, any real $x_0$ gives $x_1\in [-1,1]$. This closed interval is complete, and cosine maps it into $[\cos 1,1]\subseteq [-1,1]$. The mean value theorem giveswith $\sin (1)<1$. Apply the contraction mapping theorem to the tail starting at $x_1$. Therefore $\boxed{x_n\to p,p=\cos p}$ for every real initial value. Numerically $p\approx 0.739085$. The global real-line derivative bound is only one; entering the invariant interval supplies the strict uniform bound needed by the theorem.

Here “contractive” means strictly distance-decreasing: $\boxed{d(Tx,Ty)<d(x,y)\text{ whenever }x\ne y}$. “Strongly contractive” means a contraction mapping with one uniform constant $q\in [0,1)$ satisfying $\boxed{d(Tx,Ty)\le qd(x,y)\text{ for all }x,y\in M}$.

If distinct $p,q$ were both fixed points, strict decrease would give $d(p,q)=d(Tp,Tq)<d(p,q)$, a contradiction. Thus there is at most one fixed point. This proves uniqueness if a fixed point exists; the pointwise strict inequality alone does not assert existence.

Take the clamp at $s=0$ with $x(0)=y(0)=\theta (0)=0$. An end force without an applied end moment gives the free-end condition ${\theta }^{\prime }(1)=0$. With $F=0$, the Euler elastica equation is $EI{\theta }^{\prime \prime }+G\cos \theta =0$. For $|G|/(EI)\ll 1$, $\theta$ is small, so the leading equation isIntegrating with the two boundary conditions givesTo first order, $x^{\prime }=\cos \theta =1+O(G^2)$ and $y^{\prime }=\sin \theta =\theta +O(G^3)$. HenceEquivalently $y(x)=G{x}^2(3-x)/(6EI)$ at leading order, the familiar cantilever response. Because $G<0$, the deflection is downward. The free-end displacement field and slope are $y(1)=G/(3EI)$ and $\theta (1)=G/(2EI)$. The remainder notation holds with fixed positive $EI$ and nondimensional unit rod length.

Set $\ell =\sqrt{EI/F}$. The Euler elastica equation becomes ${\theta }^{\prime \prime }={\ell }^{-2}\sin \theta$. Multiplying by ${\theta }^{\prime }$ gives the first integralStraight asymptotic ends require ${\theta }^{\prime }\to 0$ and a horizontal tangent, so $C=F$. This is the separatrix of a simple pendulum with its angle shifted by $\pi$. A nonconstant solution moves from $\theta =0$ to $2\pi$, with the same physical tangent at both ends, taking infinite arc length to approach either limit. One explicit representative isUsing $\cos \theta =1-2{\mathrm{sech}}^2(s/\ell )$ and $\sin \theta =-2\mathrm{sech}(s/\ell )\tanh (s/\ell )$ and integrating $r^{\prime }=(\cos \theta ,\sin \theta )$ gives the tensile elastica loopThis solution is nontrivial, tends to horizontal straight tails, and has finite excess bending energy. It is also necessarily self-intersecting. Indeed $y$ is even and $x$ is odd. For $z>0$, let $f(z)=z-2\tanh z$. Near zero $f(z)<0$, while $f(2)>0$, so there is a root $z_{\ast }\in (0,2)$. At the two distinct parameters $s=\pm \ell z_{\ast }$, both points have $x=0$ and the same $y>0$. Numerically $z_{\ast }\simeq 1.91501$.

Any nonconstant smooth solution with the same straight asymptotic conditions has this first integral. Between consecutive horizontal orientations, separating variables gives the displayed solution up to translation, reflection, and reversal. It cannot pass through a straight equilibrium at a finite parameter, by uniqueness of the ordinary differential equation. Thus all such nontrivial branches have the same loop crossing. This is an ideal zero-thickness rod solution; a physical rod with self-contact excluded cannot realize the crossing without adding contact forces.

Put $P=-F>0$. With $G=0$, the Euler elastica equation is $EI{\theta }^{\prime \prime }+P\sin \theta =0$. Linearization about the straight configuration gives $EI{\theta }^{\prime \prime }+P\theta =0$. Under the boundary conditions explicitly stated, $\theta (0)=\theta (L)=0$, a nonzero solution requiresThus with only the stated boundary conditions, the first critical compressive force is $\boxed{F_c=-{\pi }^{2}EI/L^2}$. The corresponding linear transverse displacement field, with $y(0)=0$, isand $x(s)=s$ to first order. These are the successive small-amplitude buckling modes. The nonlinear branches retain the simple pendulum equation; if the angular amplitude is ${\theta }_m$ and $m=\sin ({\theta }_m/2)$, fitting $n$ half-periods into the rod gives $P=4EI{n}^{2}K(m{)}^2/L^2$, which tends to $P_n$ as $m\downarrow 0$. Here $K$ is the complete elliptic integral of the first kind, with $m$ its modulus, not its squared parameter.

Endpoint position compatibility for the Euler elastica adds a condition when fully fixed clamps have equal transverse height. If the word “clamped” also imposes $y(L)=y(0)$, then ${\int }_0^L{\theta }_{n}ds=0$. With $G=0$, this retains only even $n=2j$, yieldingThe same exclusion is exact on the nonlinear branches: integrating the elastica equation gives $P\int \sin \theta ds=-EI[{\theta }^{\prime }(L)-{\theta }^{\prime }(0)]$, and an odd number of half-periods reverses the nonzero endpoint slope. Hence odd $n$ changes the endpoint height. The conventional fixed-clamp first Euler buckling load is the second boxed value, whereas the source's explicitly listed boundary conditions alone give the first. Additional clamped modes with a nonzero transverse reaction require releasing the imposed $G=0$. Load-controlled buckling also permits axial shortening; an exactly inextensible rod whose endpoint separation is fixed equal to its full length has no room to buckle.

Let $s$ be arc length, so inextensibility makes $r^{\prime }=d_3$ a unit tangent vector. The curvature is $\kappa ={\theta }^{\prime }$, hence $m^{\prime }=EI{\theta }^{\prime \prime }{e}_z$ for uniform bending stiffness. Balance of linear momentum in static equilibrium gives a constant resultant $n=F{e}_x+G{e}_y$.

Taking the cross product directly,The scalar component of conservation of angular momentum therefore becomesTogether with $x^{\prime }=\cos \theta$ and $y^{\prime }=\sin \theta$, this Euler elastica equation determines the centerline once the appropriate boundary conditions are supplied.

Substituting both profiles into the shear traction condition givesFor the normal traction, use $M{b}_1^2=M{k}^2-\rho k^2{c}^2$ and $b_2^2+k^2=k^2(2-c^2/c_T^2)$. Removing the common nonzero factors yields the second equation. ThusA nonzero displacement field requires the determinant to vanish:Let $s=c^2/c_T^2$ and $\eta =c_T^2/c_L^2$. Cancelling $k^4$ gives the Rayleigh secular equationSquaring and excluding the zero root gives $s^3-8s^2+(24-16\eta )s-16(1-\eta )=0$; any candidate must still satisfy the unsquared equation and the decay inequalities. For stable isotropic elasticity $0<\eta <3/4$, this cubic has one root in $(0,1)$: its values at zero and one have opposite signs, and its strictly negative second derivative on this interval prevents a second crossing back to a positive final value. Both sides of the unsquared equation are positive there, so this root is valid.

The root $s=0$ of the squared equation is not a propagating surface wave: at $c=0$, the two roots coincide, and the null combination $A_2=-A_1$ in this representation produces zero displacement field. As a numerical example, $\lambda =\mu$ gives $\eta =1/3$ and $c\simeq 0.9194c_T$. The surface-wave speed is independent of $k$ in this homogeneous half-space model.

Write the in-plane displacement field as $(u,v)$. The infinitesimal strain tensor gives $2e_{XY}=u_Y+v_X$, $e_{YY}=v_Y$, and $\mathrm{tr}e=u_X+v_Y$. The two nontrivial zero-traction conditions becomeEquivalently for the complex amplitudes they are $U^{\prime }(0)+ikV(0)=0$ and $ik\lambda U(0)+(\lambda +2\mu )V^{\prime }(0)=0$. Because the constitutive equation is real and linear, vanishing of these complex amplitudes enforces the conditions at every position and time.