Orient the body boundary counterclockwise. The outward normal from the body satisfies $nds=(dy,-dx)$, so pressure givesFor steady irrotational flow, Bernoulli's equation makes $p+\rho |u{|}^2/2$ constant in the connected fluid region. Its constant contribution integrates to zero. Write $q=w^{\prime }=u-iv$. Impermeability says the velocity is tangent to the boundary, so $qdz=\bar{q}d\bar{z}$ there. Consequentlywhich proves Blasius' theorem with this contour convention.

The background stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. Since its coefficients are real, the circle theorem givesIts derivative isIn $q^2$ the coefficient of $z^{-1}$ comes from the exponent pairs $(2,-3)$ and $(1,-2)$, including both orders. It is $36{\alpha }^{2}c{a}^4+36{\alpha }^2{c}^3{a}^2$. The residue theorem then givesThe vanishing transverse force also agrees with reflection symmetry about the real axis.

In vacuum, the source-free Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-{\partial }_{t}B$ and $\nabla \times B=c^{-2}{\partial }_{t}E$. Put $\vartheta =\omega (t-z/c)$ and denote the field amplitude by $E_0$. The corresponding travelling-wave magnetic field isBoth divergences vanish because the fields have no longitudinal component and no dependence on the transverse coordinates. Direct differentiation gives $\nabla \times E=(\omega E_0/c)(\cos \vartheta ,\sin \vartheta ,0)=-{\partial }_{t}B$ and $\nabla \times B=(\omega E_0/c^2)(-\sin \vartheta ,\cos \vartheta ,0)=c^{-2}{\partial }_{t}E$.

Introducing potentials by $B=\nabla \times A$ and $E=-\nabla \varphi -{\partial }_{t}A$ automatically satisfies the magnetic-divergence and Faraday equations. The remaining Maxwell equations become equations for these potentials, with the usual gauge freedom. For $\omega \ne 0$ take $\varphi =0$ andDifferentiation verifies both potential relations.

The electromagnetic force is $Q(E+v\times B)$. Confinement also supplies a normal reaction force. For motion in a plane of constant $z$, both $v$ and $B$ are tangent to the plane, so their cross product is normal and is balanced by that reaction. It does not enter the in-plane equation $m\ddot{r}=Q{E}_0(\cos \vartheta ,\sin \vartheta )$.

At the fixed plane, the solution with no uniform drift isIt is circular, withOther initial data can add a constant drift velocity; the question only asks for the possibility of circular motion.

Write $q=(x,X{)}^T$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The linear equations are $M\ddot{q}+Kq=0$. A normal mode has all coordinates oscillating with one frequency and fixed relative amplitudes, $q=v\cos (\omega t+\phi )$. Its squared frequency solvesand its amplitude vector is a nonzero null vector of $K-{\omega }^{2}M$. The assumed distinct positive roots give two normal frequencies, their positive square roots.

For the rods let ${\theta }_1,{\theta }_2$ be their small inclinations from the downward vertical. To first order, ${\theta }_1=x/a$ and ${\theta }_2=(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second has centre-of-mass horizontal velocity $(\dot{x}+\dot{X})/2$ and angular velocity $(\dot{X}-\dot{x})/(2a)$, givingTogether these give $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The centre-of-mass height changes giveproving the displayed quadratic Lagrangian.

Here $M=(m/3)\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)$ and $K=(mg/(2a))\left(\begin{array}{cc}6& -1\\ -1& 1\end{array}\right)$. Set $\lambda =2a{\omega }^2/(3g)$. The determinant equation reduces to $5{\lambda }^2-17\lambda +5=0$, soThe first row of the mode equation gives $X/x=(6-3\lambda )/(1+\lambda )$. The actual rod-angle ratio is thereforeIt is positive for the smaller root and negative for the larger root. Thus the low-frequency mode has the rods in phase, and the high-frequency mode has them out of phase. Using the rod-angle ratio is necessary because $X$ is the displacement of the end of the second rod, not that rod's inclination.

Differentiability at $a$ means that a linear map $L:V\to V_1$ satisfiesThis map is unique and is the total derivative $D{f}_a$. In finite dimensions linear maps are continuous.

The Frobenius norm is submultiplicative. Expanding the noncommuting cube givesThe last four terms have norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. ThusIf $|C|<1$, the Neumann series ${\sum }_{j\ge 0}(-C{)}^j$ converges and is the inverse of $I+C$, as follows by multiplying its partial sums and letting the final power tend to zero. Its terms from degree two onwards have norm at most $|C{|}^2/(1-|C|)$. Hence $(I+C{)}^{-1}=I-C+O(|C{|}^2)$.

For invertible $A$, write $A+H=A(I+A^{-1}H)$. Whenever $|A^{-1}||H|<1$, this is invertible, proving the general linear group is open. The same expansion givesCompose this inverse derivative with the cube derivative, setting $B=A^{-1}$. The result isThe three terms cannot generally be combined into a scalar-style $-3A^{-4}H$, because matrix factors need not commute.

This is the cubic case of the Fréchet derivative of a negative matrix power.

Compactness means every open cover has a finite subcover. Connectedness means the space cannot be written as the union of two disjoint nonempty open sets. The Hausdorff property means every two distinct points have disjoint open neighbourhoods.