Write $N=T-I$. Cayley-Hamilton theorem gives $N^3=0$. Its minimal polynomial is therefore $X^k$ for exactly one $k\in \{1,2,3\}$, so the possibilities for that of $T$ areLet $d=\dim \ker N$, which is the dimension of the only eigenspace. If $d=3$, then $N=0$ and $T=I$.

If $d=2$, rank-nullity theorem gives $\dim \mathrm{im}N=1$. The restriction of the nilpotent map $N$ to this invariant line is zero, so $N^2=0$. Choose $v_3$ with $v_2=N{v}_3\ne 0$, and choose $v_1\in \ker N$ independent of $v_2$. Then $v_3\notin \ker N$, so these three vectors form a basis; its matrix for $T$ is the printed $J_1(1)\oplus J_2(1)$.

If $d=1$, then $\mathrm{im}N$ has dimension two, so $N^2$ cannot vanish: otherwise $\mathrm{im}N\subseteq \ker N$. Choose $v_3$ with $N^2{v}_3\ne 0$ and set $v_2=N{v}_3$, $v_1=N^2{v}_3$. Applying $N^2$ and then $N$ to a putative linear relation proves these three vectors independent. The matrix is the printed $J_3(1)$.

These cases exhaust the possible eigenspace dimensions. They are pairwise dissimilar because similarity preserves the eigenspace dimension, equivalently the three different minimal polynomials. Hence every matrix is similar to exactly one of the three displayed forms.

Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb{C}$, the fundamental theorem of algebra supplies a root $\lambda$ of the characteristic polynomial. Since $\det (T-\lambda I)=0$, there is a nonzero vector $v_1$ with $T{v}_1=\lambda v_1$. The line $W=\mathbb{C}{v}_1$ is invariant, so $T$ induces a linear map $\overline{T}$ on $V/W$. By induction, choose an upper triangular basis for that quotient and lift its vectors to $v_2,\dots ,v_n$. Then $T{v}_j\in \mathrm{span}(v_1,\dots ,v_j)$ for every $j$. Thus $v_1,\dots ,v_n$ is the required basis and $\boxed{T\text{ is triangularizable}}$.

The same conclusion holds over $F$ when the characteristic polynomial splits over $F$. Indeed an eigenvalue then lies in $F$, and the block matrix arising from the invariant line gives ${\chi }_T(X)=(X-\lambda ){\chi }_{\overline{T}}(X)$. The quotient characteristic polynomial therefore also splits, so the induction works over $F$ without requiring algebraic closure.

Integrating the exponential directly givesThe inequality is the absolute-convergence domain of this Laplace transform.

For $\mathrm{Re}p>0$, the starting integral is ${\int }_0^{\infty }{e}^{-pt}dt=1/p$. Repeated integration by parts, with vanishing boundary terms, givesThis also follows from the gamma function. The stated open half-plane ensures absolute convergence.

For $\mathrm{Re}p>a$, differentiation under the integral is justified by the dominated convergence theorem: on a smaller closed half-plane $\mathrm{Re}p\ge a+\delta$, the additional factor $t$ is controlled by $t{e}^{-\delta t}$. HenceThis is the multiplication rule for the Laplace transform.

Translation in $x$ leaves this autonomous differential equation unchanged. For every $c\ge 1$, the same formulasatisfies $u_c(1)=0$, and the zero function does too. Thus solutions exist and are not unique, exactly as for the initial point $(0,0)$. The relevant failure is the Lipschitz condition at $y=0$; moving the initial value of $x$ does not remove it.

Near $y=1$, the right side is continuously differentiable in $y$, with derivative $1/(4y^{3/4})$ bounded on, for example, $1/2\le y\le 3/2$. It therefore satisfies a local Lipschitz condition; it is also bounded on an appropriate bounded neighborhood. The existence and uniqueness result applies there. Separation of variables givesThe unique extension to the whole real axis is $u(x)=[\max \{1+3x/4,0\}{]}^{4/3}$. It joins zero with zero derivative at $x=-4/3$. To see uniqueness of this extension, $u^{\prime }\ge 0$ makes every solution nondecreasing. Before it reaches zero it cannot be positive, while a negative interval would have $u^{\prime }=0$ and could not continuously reach zero. Thus it must be zero for $x\le -4/3$; the positive part is fixed by the separated formula.

At zero the right side is continuous but fails every local Lipschitz condition in its second variable, since ${\delta }^{1/4}/\delta ={\delta }^{-3/4}\to \infty$. For every $c\ge 0$, the functionis continuously differentiable and satisfies the differential equation, including at the joining point where both one-sided derivatives are zero. Each has $u_c(0)=0$. The identically zero function is another possibility. Thus solutions exist but are not unique, even locally: choosing $c=0$ and taking the zero function already gives two different solutions on every neighborhood of zero. These are instances of waiting-time solutions of a sublinear Bernoulli equation.

Choose the branch $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi /2<\arg z<3\pi /2$. It is holomorphic off the downward imaginary ray and takes values $\log x$ on the positive real axis and $\log x+i\pi$ at $-x$ for $x>0$.

Take $R>2$ and $0<\epsilon <2$. The positively oriented contour consists of the real segments $[-R,-\epsilon ]$ and $[\epsilon ,R]$, the small upper semicircle from $-\epsilon$ to $\epsilon$ traversed clockwise, and the large upper semicircle from $R$ to $-R$ traversed counterclockwise. For $F(z)=\mathrm{Log}z/(4+z^2)$, the only enclosed pole is $2i$, withThe integral over the large arc is $O(\log R/R)$, since its length is $\pi R$ and $|4+z^2|\ge R^2-4$. The small arc contributes $O(\epsilon |\log \epsilon |)$, since its length is $\pi \epsilon$ and $|4+z^2|\ge 4-{\epsilon }^2$. Both tend to zero.

Let $J={\int }_0^{\infty }\log x/(4+x^2)dx$; it converges because $|\log x|$ is integrable near zero and $\log x/x^2$ is integrable at infinity. The two real segments, after substituting $z=-x$ on the negative one, tend toThe residue theorem therefore givesTaking real parts yields

On the unit circle, put $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\cos \theta =(z+z^{-1})/2$, soWrite $r=2-\sqrt{3}$ and $s=2+\sqrt{3}$. Only $r$ lies inside the unit circle; the rational function $h(z)=z/[(z-r{)}^2(z-s{)}^2]$ has a double pole there. Its residue isBy the residue theorem,

With parameter $t$ along a characteristic curve, the equations areTheir solutions are $y=t$, $x=s+t^2/2$ and $u=s{e}^t$. Thus the projected curve satisfies $y^2=2(x-s)$, as required. Eliminating the parameters givesIndeed $u_x=e^y$ and $u_y=e^y(x-y^2/2-y)$, so $y{u}_x+u_y=u$, and the initial values hold.

The parameter map $(s,t)\mapsto (s+t^2/2,t)$ has Jacobian one and is globally invertible. Every point whose characteristic curve meets the initial segment therefore lies in the stripThe uniquely determined open region is its interior, with the boundary values determined on the two endpoint characteristic curves. Both signs of $y$ are included. Outside the strip, a general solution is $u=e^{y}F(x-y^2/2)$ with $F(s)=s$ only required for $1\le s\le 2$. Smooth perturbations of $F$ supported outside that interval produce different solutions with the same prescribed data, proving the stated region is exact.

Assume the vector field and multiplier are continuously differentiable, as required for Green's theorem and uniqueness of trajectories. A nonconstant periodic trajectory of a planar autonomous system is a simple closed orbit: uniqueness prevents crossings, and one traversal can be taken over its least period. Let $\Omega$ be its bounded interior. Along the orbit $dx=fdt$, $dy=gdt$, soGreen's theorem identifies this with ${\iint }_{\Omega }[{\partial }_x(\varphi f)+{\partial }_y(\varphi g)]dxdy$, which is strictly positive by hypothesis, a contradiction. Thus there is no nonconstant periodic orbit. This standard periodic-orbit criterion does not by itself exclude stationary solutions.

For the first concrete system choose $\varphi =-1$. ThenThere are also no equilibria: $f=0$ requires $y=1$, but then $g=-e^{-x}-x^2<0$. Consequently it has no periodic solution, including constant ones.

For the second system the origin is outside the domain of the printed vector field. For $r>0$,Thus for an admissible initial point with $\rho =\sqrt{a^2+b^2}>0$,where $a=\rho \cos \alpha$, $b=\rho \sin \alpha$. These formulas are valid while $r>0$, in particular for every $t\ge 0$. Trajectories rotate counterclockwise, spiral outward when $0<r<1$, and inward when $r>1$, approaching the unit circle. On $r=1$ they give $x(t)=\cos (\alpha +t)$, $y(t)=\sin (\alpha +t)$, a periodic orbit with least period $2\pi$. It is an attracting limit cycle. There is no contradiction with the preceding criterion: this vector field is singular at the origin and has the exhibited closed orbit.

At a zero $a$, Taylor's theorem gives $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ near $a$. If some coefficient is nonzero, let $m$ be the least such index. Then $f(z)=(z-a{)}^{m}q(z)$ with $q(a)\ne 0$, and continuity makes $q$ nonzero on a smaller disc. Thus $a$ is an isolated zero. A limit point of zeros is also a zero by continuity and cannot be isolated, so all Taylor coefficients there vanish and $f$ vanishes on a nonempty open disc.

Let $A$ be the set of points having a neighborhood on which $f$ vanishes. It is nonempty and open. A point in its relative closure either already belongs to $A$ or is a limit point of distinct zeros; the preceding argument puts it in $A$ in either case. Thus $A$ is also relatively closed. Connectedness of $D$ gives $A=D$, proving $f$ is identically zero.

For conjugate reflection, the disc is invariant under conjugation, andHence $h$ is holomorphic, with derivative $\overline{g^{\prime }(\bar{z})}$. If $g$ is real on the real diameter, then for real $x\in (-1,1)$, $h(x)=\overline{g(x)}=g(x)$. The holomorphic difference $g-h$ therefore has zeros accumulating at zero, and the identity theorem just proved gives $g=h$ on the entire disc.

Choose a small positively oriented circle $C_{\epsilon }$ about $z_0$ inside the given region. Cauchy's theorem on the region between the two contours givesContinuity makes the last integrand converge uniformly to $f(z_0)$ as $\epsilon \downarrow 0$. Its limit is $2\pi if(z_0)$, proving the Cauchy integral formula.

The same formula holds with an interior point $w$ near $z_0$. Subtract its value at $z_0$ and divide by $w-z_0$:The contour is a positive distance from $z_0$, so the integrands converge uniformly as $w\to z_0$. Passing to the limit gives the differentiated formulaOn the unit circle, $\bar{z}=1/z$ and $\mathrm{Im}z=(z-\bar{z})/(2i)$, proving the requested identity. Hence the requested integral isThe first integral is $2\pi i/3$. For the second useThe differentiated formula applied to the constant function gives $\int z^{-2}dz=0$; the remaining two terms integrate to $-6\pi i+6\pi i=0$. Therefore