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Continuity on the compact rectangle gives finite constants $M={\max }_R|F|$ and $L={\max }_R|F_y|$. The real mean value theorem in the second coordinate gives $|F(x,y)-F(x,z)|\le L|y-z|$ within the rectangle. Define the Picard iterates on the entire interval byIf the previous iterate stays in the vertical range of $R$, then $|y_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Induction therefore makes all these iterates well defined on $|x-a|\le h$.

Repeated use of the Lipschitz bound gives the factorial estimateThe case $n=0$ is the direct integral bound; integrating the estimate for $n-1$ proves the next one, on either side of $a$. The series of uniform bounds converges, so the iterates converge uniformly to a continuous function $y$ taking values within the closed vertical range. The Lipschitz estimate permits passage through the integral to give $y(x)=b+{\int }_a^{x}F(t,y(t))dt$. The fundamental theorem of calculus now gives $y^{\prime }=F(x,y)$ and $y(a)=b$. Thus the solution exists on the full stated interval; no extra restriction $Lh<1$ was used.

For the specified equation take $k=1/4$, so $3/4\le y\le 5/4$. On $|x|\le h$,The condition $Mh\le k$ holds at $h^3=9/64$, namely $\boxed{h=3^{2/3}/4}$. The rectangle avoids $y=0$, so both continuity hypotheses hold there.

Direct separation gives $4y^3{y}^{\prime }=3x^2$, hence $(y^4{)}^{\prime }=(x^3{)}^{\prime }$ and the initial value yields $y^4=1+x^3$. The branch through $y(0)=1$ isIt remains positive on this interval. At $x=-1$ it reaches zero, where the differential equation is undefined, so this is its maximal real interval containing zero. The rectangle theorem supplies a sufficient interval, smaller than that actual interval.

For $t>0$, integrate $e^{itz}/(1+z^2)$ over the upper semicircle of radius $R$. The arc tends to zero, since $|e^{itz}|\le 1$ there and the denominator is of order $R^2$. The sole pole is $i$, of residue $e^{-t}/(2i)$, so the real part gives $I(t)=\pi e^{-t}$. The integrand defining $I$ is even in $t$, and $I(0)=\pi$ by the arctangent integral. Hence

For $J$ use $e^{itz}/[z(1+z^2)]$ with a small clockwise semicircular indentation above zero. Its residue at zero is one, so the small arc tends to $-i\pi$. The pole $i$ has residue $-e^{-t}/2$, and the large arc again vanishes. ThusThe imaginary part is the ordinary convergent integral $J(t)$: its apparent singularity at zero is removable, with limit $t$, and its tails are absolutely integrable. Therefore $J(t)=\pi (1-e^{-t})$ for $t>0$. Oddness in $t$ and the value zero at $t=0$ give

Differentiation is valid at every real $t$, including zero. Indeed the mean value theorem bounds the absolute difference quotient of $\sin (tx)/[x(1+x^2)]$ with respect to $t$ by $1/(1+x^2)$, an integrable majorant independent of $t$. Dominated convergence therefore gives $\boxed{J^{\prime }(t)=I(t)\text{ for all }t\in \mathbb{R}}$. In particular both one-sided derivatives at zero equal $\pi$.

Laurent's theorem states that a holomorphic function on $R<|z-a|<S$ has a unique expansion ${\sum }_{n\in \mathbb{Z}}{a}_n(z-a{)}^n$, converging absolutely there and uniformly on compact subannuli, withHere $\sin z=z(1-z^2/6+\cdots )$ has a simple zero at zero and no other zero in $|z|<\pi$. Hence $1/\sin z$ has only a simple pole there, giving the stated series with no powers below $z^{-1}$. Multiplying by $\sin z$ and comparing the constant, linear and quadratic coefficients yields

Use the coefficient integrals on the two circles. The annular region between them contains exactly the simple poles $\pi ,-\pi$, and the residue of $1/\sin z$ at each is $1/\cos (\pm \pi )=-1$. Apply the residue theorem to $1/(z^{n+1}\sin z)$ between the circles:For lower indices, set $c_n=0$ because the inner Laurent series has no such terms; the same contour argument then still applies. Since $f$ is odd, its even-index coefficients vanish. A complete explicit list for the requested range isThe subtraction accounts for the two poles crossed; using the inner coefficients unchanged outside their annulus would miss precisely these terms.

If $c(\lambda )=\det (\lambda I-T)=0$, then $\lambda I-T$ has nonzero kernel of a linear map. A nonzero vector $v$ in that kernel of a linear map satisfies $Tv=\lambda v$, proving that $\lambda$ is an eigenvalue. Extend $v$ to a basis. Its first column is $(\lambda ,0,\dots ,0{)}^T$, so the matrix has block formHere $B$ represents the induced operator on $V/⟨v⟩$.

If the algebraic multiplicity is $r$, repeat this argument in that quotient: its characteristic polynomial still has $\lambda$ with multiplicity $r-1$. Induction and lifting a quotient basis gives a basis whose first $r$ columns have the prescribed upper-triangular form. With $U$ the span of those $r$ vectors, the full matrix isThus $U$ is invariant and is contained in $W_{\lambda }$. Conversely, if $(T-\lambda I{)}^{s}v=0$, project to $V/U$. The induced map $C-\lambda I$ is invertible, so its $s$th power cannot kill a nonzero class. Hence $v\in U$. Consequently $\boxed{W_{\lambda }=U,\dim W_{\lambda }=r}$; this proves simultaneously that it is a subspace and $T$-invariant. It also proves $(T-\lambda I{)}^r=0$ on this space, without using Cayley-Hamilton theorem in advance.

To prove independence of the different spaces, take a relation ${\sum }_{\lambda }{v}_{\lambda }=0$ with $v_{\lambda }\in W_{\lambda }$. Fix $\lambda$ and apply $P_{\lambda }(T)={\prod }_{\mu \ne \lambda }(T-\mu I{)}^{r_{\mu }}$. It kills all terms except $v_{\lambda }$. On $W_{\lambda }$, each factor $T-\mu I=(\lambda -\mu )I+N$ is invertible: its inverse is a finite geometric series in the nilpotent $N$. Hence $P_{\lambda }(T)v_{\lambda }=0$ implies $v_{\lambda }=0$. The sum is direct, and its dimension is $\sum r_{\lambda }=\dim V$, soFinally $c(T)={\prod }_{\lambda }(T-\lambda I{)}^{r_{\lambda }}$ kills each summand, hence every vector. This proves $\boxed{c(T)=0}$ without circularly assuming the theorem during the decomposition.

Suppose a nonconstant periodic orbit exists for the continuously differentiable planar vector field. Uniqueness implies that its image over a minimal period is a simple closed curve, enclosing a region $\Omega$ of positive area. Along the orbit, the differential form $fdy-gdx$ vanishes because $dx=fdt$ and $dy=gdt$. Green's theorem, with positive orientation, therefore givesa contradiction. This is the Bendixson-Dulac criterion in the whole plane; the fact that the enclosed region also lies in the domain is essential.

For the second-order equation set $y=\dot{x}$. The equivalent system is $\dot{x}=y$, $\dot{y}=-(1+x^2)y-\cos x$. Its divergence is $\boxed{-(1+x^2)<0}$ throughout the plane. Hence there are no nonconstant periodic solutions.

The nonconstant qualification is necessary: the equilibrium solutions $x(t)\equiv \pi /2+k\pi$, $k\in \mathbb{Z}$, do exist and are periodic in the literal sense of satisfying every positive period. The source's usual dynamical interpretation of “periodic solution” here is a nonconstant periodic orbit.

The equilibrium equations are $x(1+x-2y)=0$ and $(x-1)y=0$. If $x=0$, the second equation gives $y=0$. Otherwise either $y=0$, giving $x=-1$, or $x=1$, giving $y=1$. Thus there are exactly three equilibria. The Jacobian isAt $(0,0)$ it is $\mathrm{diag}(1,-1)$: a saddle. At $(-1,0)$ it is $\left(\begin{array}{cc}-1& 2\\ 0& -2\end{array}\right)$, with eigenvalues $-1,-2$: a stable node. At $(1,1)$ it is $\left(\begin{array}{cc}1& -2\\ 1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2-\lambda +2$ and eigenvalues $(1\pm i\sqrt{7})/2$: an unstable spiral. All three are hyperbolic, so the nonlinear local classifications follow from linearization.