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Elements $g,h\in G$ are conjugate whenfor some $x\in G$. Thenfor every integer $n$, so $h^n=e$ exactly when $g^n=e$. They therefore have the same order.

The Möbius group consists of the Möbius transformationsof the Riemann sphere, under composition. If $h=xg{x}^{-1}$, thenThus $p$ is fixed by $g$ exactly when $x(p)$ is fixed by $h$. Conjugation gives a bijection between their fixed-point sets, so conjugate elements have the same number of fixed points.

A nonidentity Möbius transformation has at most two fixed points because its fixed-point equation is quadratic on the Riemann sphere. If it has only one repeated fixed point, conjugate that point to infinity; the transformation becomes a nontrivial translation $z\mapsto z+c$, which has infinite order. Consequently every nontrivial finite-order element has two distinct fixed points:This is fixed points of a finite-order Möbius transformation.

Termwise differentiation of the matrix exponential givesThus $y(t)=e^{tA}{y}_0$ satisfies $y^{\prime }=Ay$ and $y(0)=y_0$. If two solutions existed, multiplying their difference by $e^{-tA}$ would give a vector with zero derivative and zero initial value, proving uniqueness.

For the inhomogeneous equation, multiply by the integrating factor $e^{-tA}$:Integration and multiplication by $e^{tA}$ give the variation-of-constants formula

For the stated matrix,soWriting $f(t)=v\sin t$ with $v=(1,3,0{)}^T$, one findsThereforeor explicitly

Put $\omega =\dot{\theta }$. The first-order system isWriteOn $[0,\pi ]$, $h$ is strictly increasing. Moreover,when $3\lambda <4\mu$. Thus there is one ${\theta }_{\ast }\in (0,\pi )$ with $h({\theta }_{\ast })=0$, in addition to the fixed points at $\theta =0,\pi$. Explicitly,

Let $g(\theta )=\sin \theta h(\theta )$. Linearization at $({\theta }_0,0)$ has eigenvalues satisfyingAt zero, $g^{\prime }(0)=h(0)<0$, and at $\pi$, $g^{\prime }(\pi )=-h(\pi )<0$, so both endpoints are saddles. At the interior point,so $({\theta }_{\ast },0)$ is a center. On the full angular interval there is a second center at $(-{\theta }_{\ast },0)$.

The system has the conserved energy of a conservative planar phase portrait,For $\lambda =1$, $\mu =3/2$,The phase portrait on the cylinder consists of centers at $\theta =\pm \arccos (-11/16)$ surrounded by closed periodic-energy curves, with saddles at $\theta =0$ and the identified point $\theta =\pi =-\pi$. The saddle-energy contours form the separatrices between librations in the potential wells and trajectories crossing the lower barrier.

If $3\lambda >4\mu$, then $h(0)>0$ and monotonicity gives no interior zero on $(0,\pi )$. The point $(0,0)$ becomes a center, while $(\pi ,0)$ remains a saddle. At $3\lambda =4\mu$, the two off-axis centers coalesce with the origin and the linearization there is degenerate.

With $\eta =x-t$ and $\xi =x+t$, the chain rule givesHenceso the wave equation is equivalent to $U_{\eta \xi }=0$. Integrating in each variable givesThe initial conditions implySolving and integrating yields the D'Alembert formula with initial velocity

If $|x-x_0|>r+t$, then both $x-t$ and $x+t$ lie outside the support interval $[x_0-r,x_0+r]$, and every point between them does as well. All three terms therefore vanish. Thuswhich is finite propagation speed.

For the finite string, differentiate its energy and use $y_{tt}=y_{xx}$:The fixed-end conditions hold for every $t$, so differentiation gives $y_t(0,t)=y_t(L,t)=0$. The boundary term vanishes and

For$x_0$ is an ordinary point when $P$ and $Q$ are analytic there, and a singular point otherwise. A singular point is regular singular when $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. These are the ordinary-point and regular-singular criteria.

For Kummer's equation, substituteEquating the coefficient of $x^m$ givesTaking $c_0=1$,where $(a{)}_m=a(a+1)\cdots (a+m-1)$ is the rising factorial. Thus

Putting $y=x^{1-b}u$ and simplifying givesThereforeFor nonintegral $b$, the powers $x^0$ and $x^{1-b}$ at zero are distinct, so these solutions are linearly independent.

When $b\to 1$, both solutions tend to $M(x,a,1)$. Differentiate their difference with respect to $b$. Writing $M_a,M_b$ for derivatives with respect to the second and third arguments,Hence at $b=1$ one may takeas two linearly independent solutions. Their independence follows from the logarithmic term in the second solution.