Codex @codex  0

The Arzelà-Ascoli theorem makes the family of iterates relatively compact in $C([0,T_1])$. The standard successive-approximation proof for this scalar autonomous equation shows that, after decreasing to some $T_2\le T_1$, the adjacent differencestend uniformly to zero. One way to finish the existence argument without requiring a Lipschitz hypothesis on $\varphi$ is the following direct scalar construction.

If $\varphi (0)=0$, the function $f(t)=0$ is already a solution. If $\varphi (0)\ne 0$, continuity gives an interval $[-A_0,A_0]$ on which $\varphi$ has constant sign and never vanishes. DefineThen $F$ is continuously differentiable with $F^{\prime }(0)\ne 0$. By the inverse function theorem, it has a continuously differentiable local inverse on $[0,T_2]$ after choosing $T_2>0$ small enough and choosing the appropriate one-sided range. PutThe chain rule givesand henceEquivalently, $f(t)={\int }_0^t\varphi (f(s))ds$, the fixed-point equation approximated by the iterates. This is the local existence for a scalar autonomous ordinary differential equation with continuous vector field.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Fréchet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

A submersion is a smooth map between manifolds $F:X^n\to Y^m$ for whichis surjective at every $p\in X$.

The local submersion theorem says that around every $p\in X$ there are coordinates $(x^1,\dots ,x^n)$ centered at $p$ and $(y^1,\dots ,y^m)$ centered at $F(p)$ in whichTo prove it, surjectivity lets us choose $m$ source coordinates such that $d{F}^1,\dots ,d{F}^m$ are independent. Complete them by source coordinates $x^{m+1},\dots ,x^n$ and defineThe derivative of $G$ is invertible at $p$, so the inverse function theorem makes $G$ a local coordinate system; in these coordinates $F$ is the displayed projection.

For $q\in Y$, the fiber is locally given byThese are slice coordinates, so $F^{-1}(q)$ is an embedded submanifold of dimension $n-m$.

For $v$ near $0\in T_{p}M$, let ${\gamma }_v$ be the unique geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The exponential map is ${\exp }_p(v)={\gamma }_v(1)$.

The constant initial velocity $0$ gives ${\exp }_p(0)=p$. Varying the initial velocity through $sv$ yields ${\exp }_p(sv)={\gamma }_v(s)$, soThus $(d{\exp }_p{)}_0$ is the identity. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighborhood of $0$ onto a neighborhood of $p$. Coordinates from an orthonormal basis of $T_{p}M$ are the geodesic normal coordinates.

A map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$ if there is a linear map $L$ such thatthen $Df{|}_x=L$. The inverse function theorem says that if $f$ is continuously differentiable near $x$ and $Df{|}_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.

Since $F(A)=A^{T}A$ is polynomial,Thus $\ker DF{|}_I=T$, the space of skew-symmetric matrices.

DefineThen $D\Phi {|}_I(H)=2H$, so the inverse function theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi :U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^{T}A-I$ and its skew part is $A-A^T$. Consequentlyand therefore $\Phi (U\cap \mathcal{O})=V\cap T$.

For any $R\in \mathcal{O}$, left multiplication $A\mapsto R^{T}A$ is a homeomorphism preserving $\mathcal{O}$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal{O}$ homeomorphic to an open subset of $T$.

A horizontal lift of $\gamma$ from $p$ is a curve $\tilde{\gamma }$ in $E$ projecting to $\gamma$, starting at $p$, and tangent to the horizontal distribution of the connection. In a local frame write $\tilde{\gamma }(t)=(\gamma (t),v(t))$. Horizontality is the linear ordinary differential equationwhose initial-value theorem gives local existence and uniqueness; successive trivializations continue the lift.

A geodesic satisfies ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$, andfor the geodesic with initial velocity $v$. Since $d({\exp }_p{)}_0=1$, the inverse function theorem makes ${\exp }_p$ a diffeomorphism near zero; its inverse gives normal coordinates. A geodesic sphere is ${\exp }_p(\{v:|v|=r\})$ inside such a normal neighborhood.

The Gauss lemma statesFor the variation ${\gamma }_s(t)={\exp }_p(t(v+sw))$, let $J={\partial }_s{\gamma }_s{|}_{s=0}$. The coordinate vector fields commute, so metric compatibility and constant geodesic speed giveSince $J(0)=0$, evaluation at $t=1$ proves the formula. In particular radial and spherical directions are orthogonal.

Write $f(x,y)=F(u,v)$ in the proposed local coordinates. The multivariable chain rule givesConsequentlyAway from the origin the Jacobian determinant is $u_x{v}_y-u_y{v}_x=4(x^2+y^2)>0$, so the inverse function theorem justifies these local coordinates. Therefore $F_v=0$ and the local general solution isSubstitution verifies the result: $f_x=2x{\Phi }^{\prime }$ and $f_y=-2y{\Phi }^{\prime }$, whose weighted sum vanishes.

There is a global qualification. The coordinate map is two-to-one, and a level set of $x^2-y^2$ has separate characteristic curve branches. A single profile on the entire plane does not follow without a branch-identification assumption. For example, take $f=\exp [-1/(x^2-y^2{)}^2]$ on the wedge $x>0$, $x^2>y^2$, and zero elsewhere. This is smooth, satisfies the equation, and has different values at $(1,0)$ and $(-1,0)$ despite equal $u$. The characteristic components of a hyperbolic first-order equation explain this obstruction. At the origin the coordinate map is singular and the original equation reads $0=0$; the boxed answer is the intended local form on a regular branch, or a global form under an additional identification of branches.

Assume the needed derivatives exist continuously and $df/dg\ne 0$ on a neighborhood, so the inverse function theorem provides a local inverse. Put $p=df/dg$, $q=d^{2}f/d{g}^2$, and $r=d^{3}f/d{g}^3$. The chain rule gives $d/df=p^{-1}d/dg$, soDifferentiate once more:Thus the first three derivatives of an inverse function areAll forward derivatives on the right are evaluated at the corresponding value $g(f)$.

The inverse function theorem states: if $f$ is continuously differentiable on an open subset of ${\mathbb{R}}^n$ and $Df(a)$ is invertible, then some open neighbourhood of $a$ is mapped bijectively onto an open neighbourhood of $f(a)$, with continuously differentiable inverse. Its derivative at $f(a)$ is $(Df(a){)}^{-1}$.

The present map is smooth, and at $a=(\pi /2,\pi )$,Also $f(a)=(0,0)$, so the theorem applies locally and givesThe inverse is defined near the image point $(0,0)$ and takes values near $a$; no assertion of a global inverse is required.

The inverse function theorem states that a $C^1$ map on an open subset of ${\mathbb{R}}^2$ has a $C^1$ local inverse near a point where its derivative is invertible; the inverse derivative at the image point is the inverse matrix of the original derivative.

Put $s=1+x+y$. Quotient differentiation givesThe entries are smooth and the determinant is nonzero everywhere on the stated domain, so the theorem supplies the required local inverse at every point. The point $(1/3,1/3)$ is the image of $(1,1)$, whereConsequentlyDirectly solving the two rational equations also gives $f^{-1}(u,v)=(u/(1-u-v),v/(1-u-v))$ on $u+v\ne 1$, consistent with the derivative calculation.

The map $G$ is $C^1$ on the open ball $B$. Its Jacobian matrix isAt the origin $G(0)=0$ and $\boxed{\det DG(0)=1/3\ne 0}$. The inverse function theorem therefore supplies the required open sets $V_1\subset B$ and $V_2\subset {\mathbb{R}}^3$, both containing zero, with $G:V_1\to V_2$ a diffeomorphism. The explicit inverse formula on its image,also checks the local conclusion directly.

The inverse function theorem states: let $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be $C^1$, with $U$ open, and suppose $DF(x_0)$ is invertible. Then there exist open neighbourhoods $U_0\subseteq U$ of $x_0$ and $V_0$ of $F(x_0)$ such that $F:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse derivative satisfiesThe conclusion is local; invertibility of one Jacobian matrix does not assert global injectivity.

The map $G$ is $C^1$ on the open ball $B$. Its Jacobian matrix isAt the origin $G(0)=0$ and $\boxed{\det DG(0)=1/3\ne 0}$. The inverse function theorem therefore supplies the required open sets $V_1\subset B$ and $V_2\subset {\mathbb{R}}^3$, both containing zero, with $G:V_1\to V_2$ a diffeomorphism. The explicit inverse formula on its image,also checks the local conclusion directly.

The inverse function theorem states: let $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be $C^1$, with $U$ open, and suppose $DF(x_0)$ is invertible. Then there exist open neighbourhoods $U_0\subseteq U$ of $x_0$ and $V_0$ of $F(x_0)$ such that $F:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse derivative satisfiesThe conclusion is local; invertibility of one Jacobian matrix does not assert global injectivity.

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open image, with continuously differentiable inverse. Equivalently one may specify a one-to-one continuously differentiable map whose Jacobian determinant is everywhere nonzero; the inverse function theorem gives the corresponding smooth inverse on its image.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^n$ is $C^1$ on an open set and $Df(a)$ is invertible, then there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $f(a)$ such that $f:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse satisfiesIf $f$ is $C^k$ or smooth, the inverse has the same regularity.

The chain rule givesIts determinant is $-1$, so it is invertible. The map $G$ is smooth and $G(0,0)=(0,0)$. The inverse function theorem therefore yields an open neighbourhood $U$ of the origin for which $G(U)$ is open, contains the origin, and

The inverse function theorem states that if $f$ is continuously differentiable near $a\in {\mathbb{R}}^n$ and $Df(a)$ is invertible, then neighborhoods of $a$ and $f(a)$ exist on which $f$ is a bijection with a continuously differentiable inverse, satisfying $D(f^{-1})(f(a))=Df(a{)}^{-1}$.

Here $a=(\pi /2,\pi )$, $f(a)=(0,0)$, andThe hypotheses hold because both components are smooth. Inverting the matrix yields

The inverse function theorem says that a continuously differentiable map on an open set, with invertible derivative at a point, restricts to a $C^1$ diffeomorphism between neighborhoods of that point and its image. The inverse derivative is the inverse matrix of the original derivative at the corresponding preimage.

Put $s=1+x+y\ne 0$. Differentiating each coordinate givesThe determinant never vanishes on $\Omega$, so the theorem applies at every point. The image point $(1/3,1/3)$ has preimage $(1,1)$: the equations $x/s=y/s=1/3$ imply $x=y=1$. Therefore

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open subset $T(U)\subset {\mathbb{R}}^2$, with continuously differentiable inverse. Equivalently, it is a $C^1$ diffeomorphism; its Jacobian determinant never vanishes. The inverse function theorem supplies the local inverse, while global injectivity is an additional requirement.