An unstable node hasBoth linear modes grow forward in time. Nearby nonconstant orbits move away from the equilibrium, and approach it under reversed time; their local directions come from the two real eigenvectors.

A stable node hasBoth exponential modes decay as $t\to \infty$, so the equilibrium is asymptotically stable. Generic linear orbits approach tangent to the eigenvector for the less negative eigenvalue; the eigenvector directions themselves are invariant straight lines. There is no rotation because both eigenvalues are real.

Suppose, towards a contradiction, that for some $r,w,\epsilon$ the punctured-disk image misses the open disk of radius $\epsilon$ about $w$. Thenis holomorphic and bounded by $1/\epsilon$ on $0<|z|<r$. Part (i) extends it to a holomorphic function $H$ on the full disk. If $H(0)\ne 0$, then $w+1/H$ extends $f$ holomorphically at zero, making zero a removable singularity.

If $H(0)=0$, the function $H$ is not identically zero, since it is nonzero on the punctured disk. The Taylor series and identity theorem give a finite zero order $k\ge 1$, so $H(z)=z^{k}q(z)$ with $q$ holomorphic and $q(0)\ne 0$. Hencehas a pole of order $k$. Both alternatives contradict an essential singularity. ConsequentlyThis is the Casorati-Weierstrass theorem: every complex value can be approximated arbitrarily closely in every punctured neighbourhood of an essential singularity.

Take the Laurent series of $f$ on $0<|z|<r$. For every integer $k\ge 1$, its coefficient of $z^{-k}$ satisfies, on a circle of radius $0<\rho <r$,The coefficient is independent of $\rho$. Letting $\rho \downarrow 0$ therefore gives $c_{-k}=0$ for every $k\ge 1$. The Laurent series is a power series with no negative powers and supplies a holomorphic extension at zero. Thus zero is a removable singularity. This is the Riemann removable singularity theorem, with boundedness used to eliminate the entire principal part.

An essential singularity is an isolated singularity that is neither a removable singularity nor a pole. In its Laurent series, infinitely many coefficients of negative powers are nonzero. The alternatives follow because no negative terms give a removable singularity, while finitely many with a lowest nonzero power give a pole.

A pole of order $n\ge 1$ meansEquivalently, the Laurent series has a finite nonzero principal part, with lowest power exactly $(z-z_0{)}^{-n}$. Thus $(z-z_0{)}^{n}f(z)$ extends holomorphically with nonzero value at $z_0$; using a smaller positive power would not remove the singularity.

The point is a removable singularity if assigning a value to $f(z_0)$ makes $f$ holomorphic on a full neighbourhood of $z_0$. Equivalently, its Laurent series has no nonzero coefficients of negative powers. The assigned value is its constant coefficient, or equivalently the finite limit ${\lim }_{z\to z_0}f(z)$.

An isolated singularity at $z_0$ means that some radius $r>0$ has $f$ holomorphic on the punctured disk $0<|z-z_0|<r$, while $f$ is not initially defined as a holomorphic function at $z_0$. The adjective isolated refers to the absence of other singularities in this punctured neighbourhood; it permits a removable singularity.

Either equality condition says that the holomorphic function $h$ from part (i) has modulus one at an interior point: at zero in the derivative case, or at the given nonzero point in the other case. Since $|h|\le 1$ everywhere, the maximum modulus principle makes $h$ a constant $w$ with $|w|=1$. ThereforeThese maps are rotations of the unit disk and do achieve both equalities, so this also identifies all equality cases of the Schwarz lemma.

Because $f(0)=0$, the Taylor series of $f$ at zero shows thatis holomorphic throughout the unit disk. Fix $0<r<1$. On $|z|=r$ we have $|h(z)|<1/r$, and the maximum modulus principle on the closed disk of radius $r$ gives $|h(z)|\le 1/r$ for $|z|\le r$. For any fixed $z$ inside the unit disk we may let $r$ increase to one, obtaining $|h(z)|\le 1$. Hence the Schwarz lemma inequalities areThe derivative bound is the same estimate at $z=0$, using the holomorphic extension of the quotient.

The closed unit disk is a compact set, and $|f|$ is continuous on it because $f$ is holomorphic on an open neighbourhood. The extreme value theorem therefore supplies a point where its maximum is attained. If such a point is on the boundary, the conclusion follows. If it is in the interior, part (i) makes $f$ constant on the open disk; continuity then gives the same value on the boundary. ConsequentlyThe assumption that the closed disk lies in $U$ is what makes the boundary values and the compactness argument available.

Write $M=|f(z_0)|$. The local maximum property gives a disk around $z_0$ on which $|f|\le M$, contained in the unit disk. If $M=0$, then $f=0$ on that neighbourhood, and the identity theorem makes $f$ zero on the connected unit disk.

Otherwise multiply $f$ by the unit complex number $\overline{f(z_0)}/M$, obtaining a holomorphic function $g$ with $g(z_0)=M$ and $|g|\le M$ nearby. Applying the Cauchy integral formula on any sufficiently small circle of radius $r$ gives the mean value propertyTake real parts. The continuous function $M-\mathrm{Re}g$ is nonnegative on the circle and has integral zero, so it vanishes everywhere there. Since $|g|\le M$, having real part $M$ forces imaginary part zero. Thus $g=M$ throughout the circle. The zeros of $g-M$ have an accumulation point inside its domain, so the identity theorem gives $f$ constant on the entire unit disk. This proves the local maximum modulus principle directly from the Cauchy integral formula.

Differentiating the transform with respect to $p$ givesApplying this to $\mathcal{L}[1]=1/p$ also gives $\boxed{\mathcal{L}[t]=1/p^2}$ for $\mathrm{Re}p>0$.

Now put $Y=\overline{y}$, $y_0=y(0)$ and $v_0=\dot{y}(0)$. Transforming each term of the differential equation, and using the derivative identities, givesThe initial-value terms cancel, leavingIntegrating once yields $(p+1)Y=C+1/(2p^2)$. Partial fractions giveThe transform pairs therefore imply, on writing $A=C+1/2$,For the final parameter count, the differential equation is the derivative of $t(\dot{y}+y)$: it says $[t(\dot{y}+y){]}^{\prime }=t$. Integrating gives $t(\dot{y}+y)=t^2/2+B$. A regular solution at $t=0$ has finite $y$ and $\dot{y}$, forcing $B=0$. Thus $\dot{y}+y=t/2$ is first order and leaves only one constant. Equivalently, the original equation at zero requires $\dot{y}(0)+y(0)=0$, so the two usual initial values cannot be chosen independently. The displayed solution satisfies this constraint and verifies the original equation directly.

Integration by parts, where the transform boundary term at infinity vanishes, givesFor a sufficiently regular function with the required exponential growth bounds, apply this again to $\dot{y}$ to deduceThese identities hold in a right half-plane where the displayed transforms converge; the regular solutions obtained below satisfy those requirements.

Combine the exponentials and integrate:The parameter $a>0$ makes this convergence half-plane larger than the one for the constant function.

The Laplace transform isA sufficient condition is that $y$ be piecewise continuous on each bounded interval and of exponential order $c$, meaning $|y(t)|\le C{e}^{ct}$ for all sufficiently large $t$. Then the integral is absolutely convergent for $\mathrm{Re}p>c$, by comparison with $e^{-(\mathrm{Re}p-c)t}$.

For the constant function, direct integration gives

The same quotient $|x||y{|}^{-1/3}$ is unbounded near $(1,0)$, so the uniform Lipschitz hypothesis again fails. Two explicit solutions through the initial point areFor the second, put $u=(x^2-1)/6$. Then $y_2^{\prime }=3u^2(x/3)=x{u}^2=x(y_2{)}^{2/3}$, and $y_2(1)=0$. The real power here means the square of the real cube root, so the identity also holds when $y_2$ is negative. Both functions are differentiable through their zeros. Thus uniqueness fails in this case as well.

No rectangle of positive width around $(0,0)$ has a uniform Lipschitz constant in $y$. For any nonzero $x$ in such a rectangle,The vanishing factor $x$ at the single initial point does not control neighbouring $x$-values. The failure of the hypothesis is accompanied by actual nonuniqueness: bothsatisfy the equation and have value zero at $x=0$. Direct differentiation gives $y_2^{\prime }=x^5/36=x(y_2{)}^{2/3}$, including the zero point by continuity.

First prove the general uniqueness claim. The bound $Mh\le k$ prevents a solution from leaving the dependent-variable interval before $|x-a|=h$: up to any first exit, $|y(x)-b|\le M|x-a|\le k$. Let $y_1,y_2$ be two solutions with the same initial value. Their integral equations and the Lipschitz condition give, for $x\ge a$,Since their difference is bounded by $2k$, repeated substitution into this inequality proves, for every integer $n\ge 0$,The same argument with reversed limits works for $x\le a$. Letting $n\to \infty$ forces the difference to zero on the whole interval. This proves at most one solution without assuming that $Ah<1$.

For case (i), choose a rectangle around $(0,1)$ that stays in $y>0$. There $f_y=(2/3)x{y}^{-1/3}$ is bounded, giving a uniform Lipschitz constant by the mean value theorem. For example $h=k=1/2$ works: $M\le \frac{1}{2}(3/2{)}^{2/3}$ and $Mh<1/2$. Separate variables using $u=y^{1/3}$, obtaining $3u^{\prime }=x$. The initial condition gives $u=1+x^2/6$, henceIt remains positive for all real $x$, so the same local uniqueness argument continues along this solution.

For each triangle compactly contained in $U$, part (i) permits the interchange of the series and its boundary integral. Cauchy's integral theorem makes every term zero. The uniform limit is continuous, so Morera's theorem yields that $f$ is holomorphic on $U$.

For the final application, each function $(z-k{)}^{-2}$ is holomorphic away from its pole. Fix a positive integer $n$. For $z\in U_n$ and $k>2n$,The numerical series ${\sum }_{k>2n}4/k^2$ converges, so the Weierstrass M-test gives uniform convergence of the tail on $U_n$. Adding the finitely many initial terms preserves uniform convergence. Those initial terms need not be bounded near their excluded poles; only the tail needs the uniform bound. By the first conclusion, the full sum is holomorphic on each $U_n$. Their union is $\mathbb{C}\setminus \mathbb{N}$, hence