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Separation $y=T(t)X(x)$ and the fixed-end conditions givePutThe initial displacement and zero initial velocity then give the separated solution of the damped string equation

For the triangular initial displacement,Thus the even coefficients vanish and the odd ones alternate in sign.

LetThenOrthogonality of the sine and cosine modes, equivalently the Parseval identity, gives

The sign of its derivative is clearest directly from the equation. Integration by parts, using $y_t=0$ at the fixed endpoints, yieldsThis is the energy dissipation identity for a linearly damped string.

For a closed piecewise smooth curve $\gamma$ avoiding $w$, its winding number about $w$ is

The argument principle says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic function $F$ has no zeros or poles on $\gamma$, thenwhere zeros and poles are counted with multiplicity.

Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline{U}$ andFor $0\le t\le 1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus Rouché's theorem states that

Now let $w_0=f(z_0)$. Since $|w_0|<1\le |f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. ThereforeThis is the unit-disc image from a boundary modulus lower bound.

Finally, take a sufficiently small positively oriented circle $C$ around zero. The residueis the winding number of $g(C)$ around zero, so $k\in \mathbb{Z}$. The pole is simple, hence its residue is nonzero and $k\ne 0$. Forwe haveThe second term cancels the complete principal part at zero, so the integer residue of a logarithmic derivative proves that $h^{\prime }/h$ has a removable singularity there.

In polar coordinates the metric isAlong a radius, the hyperbolic distance from the origin to Euclidean radius $r$ is thereforeso $r=\tanh (R/2)$. The Riemannian area element isConsequently a hyperbolic disc of radius $R$ has areaThis proves the area of a hyperbolic disc formula from the metric.

For area $\pi /2$ we get $\cosh R=5/4$. Hence $\sinh R=3/4$ and $e^R=2$, so $R=\log 2$. Tangent equal discs have centers at distance $2R$. Since the centers lie successively on the same radial geodesic,If $r_n$ is the Euclidean coordinate of $c_n$, the radial distance formula givesTherefore the radial chain of equal hyperbolic discs has

No such isometry to the stated upper-half-plane configuration exists for $n\ge 3$. The centers $c_0,\dots ,c_n$ lie on one hyperbolic geodesic, so their images under an isometry would also lie on one geodesic. In the upper-half-plane model, geodesics are vertical lines or semicircles orthogonal to the real axis. Neither type can contain three distinct points of the horizontal line $y=1$, while the centers of the distinct discs $D_0^{\prime },\dots ,D_n^{\prime }$ all lie on that line. This is the geodesic obstruction for a horizontal chain of hyperbolic discs.

Writing the velocity as $(u,v)=(y,ax)$ givesso the flow is incompressible. With the conventiona stream function isThe streamlines are its level sets. For $a>0$, they are the hyperbolaswith separatrices $y=\pm \sqrt{a}x$ and a saddle at the origin. For $a<0$, they are concentric ellipsestraversed clockwise. This is the streamline classification of a planar linear saddle or centre.

The scalar vorticity isHence the flow is irrotational exactly whenThen $(u,v)=(y,x)=\nabla \varphi$, and a velocity potential is

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is