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Put $u=a+i\omega t$. ThenThe $G$ equation from part (b) is satisfied whensoWith this value, the $F$ equation follows from ${\tanh }^{2}u=1-{\mathrm{sech}}^{2}u$: both its constant and ${\mathrm{sech}}^{2}u$ coefficients agree. Thus the stated $F,G$ satisfy both shape equations; the remaining scalar equation determines the normalization factor $N(t)$.

Fordirect differentiation givesandSubstitution into the Schrodinger equation and equality of the coefficients of $x^2,x,1$ give the necessary and sufficient Riccati equation system

For the one-dimensional quantum harmonic oscillator,This is the Time-dependent Schrödinger equation with harmonic potential.

The first mode remains the solution from part (b). The applied force excites only the third mode, sowhere, using $\pi c/l=k$,A particular solution isAdding the decaying complementary solution and imposing the initial conditions gives

As $t\to \infty$, all transients decay but the driven third-mode oscillation remains. The string approaches a periodic steady-state response, rather than coming to rest.

With gravity omitted, its mechanical energy isWriting $s=\alpha /(34k^2)$ and letting $t\to \infty$ givesThe limiting energy is periodic because the external force continually supplies the energy dissipated by drag.

With gravity omitted, the initial data contain only the first fixed-end normal mode, so writeThenBecause $k=\pi c/l$, this oscillator is critically damped, andTherefore

Take positive $y$ vertically downward. For a short string element, the transverse tension resultant is $T{y}_{xx}dx$, while gravity and linear drag contribute $\mu gdx$ and $-2k\mu y_{t}dx$. Newton's second law givesSince $T=\mu c^2$, the damped wave equation isChoosing upward displacement reverses the sign of the gravity term.

Because $\cos 0=1$, continuity gives a disc $U$ about zero on which $\cos z\ne 0$. The function $-\tan z=(\log \cos z{)}^{\prime }$ has a holomorphic primitive $G$ on a sufficiently small disc, normalized by $G(0)=0$. Thenis holomorphic and satisfies $f(0)=1$ and $f(z{)}^2=\cos z$ on $U$.

No such square root exists throughout $|z|<2$. The point $z=\pi /2$ lies in that disc and is a simple zero of $\cos z$, since $-\sin (\pi /2)=-1$. Every zero of the square of a holomorphic function has even order of a zero of a holomorphic function, contradicting this simple zero.

The set $D$ is star-shaped with centre $z_0$: if $z\in D$, every point of $[z_0,z]$ has its segment to $z_0$ contained in the same segment and therefore avoids the origin. Applying part (a) and the stated triangular-integral fact to $1/z$ gives a holomorphic functionChoose either square root $c$ of $z_0$ and defineThen $h$ is holomorphic. Moreover,and its value at $z_0$ is one. Hence

The hypothesis makes $\Omega$ star-shaped with centre $z_0$. For small $h$ with the triangle having vertices $z_0,w,w+h$ contained in $\Omega$, the vanishing triangular integral givesParametrizing the final segment,by continuity. Thus $g$ is holomorphic andThis is the triangle-integral construction behind Morera's theorem.

Use an orientation-preserving isometry to send $P$ to $i$ and $\ell$ to the imaginary axis, whose boundary points are $0,\infty$. A geodesic $m$ through $i$ meeting it at angle $\theta$ has boundary pointsIndeed $uv=-1$, so the semicircle with endpoints $u,v$ passes through $i$, and its tangent there makes angle $\theta$ with the vertical.

The cross-ratio is invariant under the normalizing Möbius transformation. With the suitable ordering $(u,v,\infty ,0)$ and the convention used here,

An isometry sends the geodesic through $P,Q$ to the imaginary axis and sends the two points to $ir$ and $i/r$ for some $r>0$. The transformationbelongs to $PSL(2,\mathbb{R})$ and exchanges $ir$ with $i/r$. Conjugating it back gives an orientation-preserving isometry exchanging $P$ and $Q$.

For uniqueness, the quotient of two such isometries fixes both $P$ and $Q$. An orientation-preserving hyperbolic isometry fixing two distinct interior points fixes their connecting geodesic and both tangent directions there, hence is the identity. Therefore the exchanging isometry is unique; geometrically it is the hyperbolic half-turn about the midpoint of the segment $PQ$.

By point transitivity, it suffices to take $P=i$. The conditionis equivalent to $d=a$, $c=-b$, and the determinant condition becomes $a^2+b^2=1$. Thus the stabilizer consists ofmodulo $\pm I$. Writing $a=\cos \phi$, $b=\sin \phi$ identifies it with rotations of the tangent plane, and the doubled matrix angle removes the $\pm I$ quotient. Hence

Every orientation-preserving isometry of the Poincaré half-plane model has the formwith matrices identified up to multiplication by $-I$. Thus $G\cong PSL(2,\mathbb{R})$.

For $z=x+iy\in \mathbb{H}$, the affine map $(z-x)/y$ sends $z$ to $i$, proving transitivity on points. Real translations and positive dilations act transitively on finite boundary points, while inversion $z\mapsto -1/z$ exchanges $0$ and $\infty$; hence the action on $\mathbb{R}\cup \{\infty \}$ is transitive. A hyperbolic line is determined by its unordered pair of boundary endpoints, and a real Möbius transformation can send any such pair to $\{0,\infty \}$. Therefore $G$ is also transitive on hyperbolic lines.

This is always true. Since $f$ is bounded and $f_n\to f$ uniformly, there are $M$ and $N$ such thatfor every $x\in X$ and $n\ge N$. The continuous function $g$ is uniformly continuous on the compact interval $[-M-1,M+1]$. Therefore, given $\epsilon >0$, a sufficiently small uniform bound on $|f_n-f|$ impliesHence $g\circ f_n\to g\circ f$ uniformly.

This is always true. Uniform convergence makes $(f_n)$ a uniformly Cauchy sequence: for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$ and $m,n\ge N$. Letting $x\to x_0$ through points of $A\setminus \{x_0\}$ and using the assumed limits givesThus $(y_n)$ is a Cauchy sequence. Since $Y$ is a complete metric space, $(y_n)$ converges in $Y$.

For every $(x,y)\in [1,2{]}^2$,Moreover,The bound is independent of $(x,y)$ and tends to zero, so

This may be false when the limiting functions are unbounded. On $X=\mathbb{R}$, letBoth sequences converge uniformly to their stated limits, butis unbounded for every $n$. Thus the products do not converge uniformly.

This is always true. Given $\epsilon >0$, choose $N_A,N_B$ such thaton $A$ for $n\ge N_A$ and on $B$ for $n\ge N_B$. For $n\ge \max (N_A,N_B)$ the same inequality holds at every point of $X=A\cup B$. Hence $f_n\to f$ uniformly on $X$.