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A function $f:{\mathbb{R}}^m\to {\mathbb{R}}^r$ is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^r$ such thatThe map $L$, which is unique, is the derivative $f^{\prime }(x)$.

Expanding in a matrix increment $H$ givesTherefore $p$ is differentiable everywhere and

The inverse function theorem states that if $f$ is continuously differentiable near $a$ and $f^{\prime }(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivativeIn the normalized case $f(0)=0$ and $f^{\prime }(0)=I$, continuity of $f^{\prime }$ gives a closed ball ${\overline{B}}_r(0)$ on whichFor small $y$, define $T_y(x)=x-f(x)+y$. The mean value inequality makes $T_y$ a contraction mapping with constant $q$. If $\Vert y\Vert \le (1-q)r$, thenso $T_y$ maps the ball into itself. The contraction mapping theorem gives a unique fixed point $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,so $\Vert x_y-x_z\Vert \le \Vert y-z\Vert /(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.

For the polynomial map in the question,The derivative is invertible, so the inverse function theorem supplies some $ϵ>0$ and a continuously differentiable $q:D_{ϵ}(I)\to {\mathcal{M}}_n$ such thatThere can be no inverse on all of ${\mathcal{M}}_n$, becauseso $p$ is not injective.

The preceding estimate gives constants $A,B>0$ such that, whenever $\Vert w{\Vert }_{H^2}\le R$,Choose $R>0$ so that $B{R}^2\le 1/2$, and then choose $r>0$ so that $Ar\le R/2$. If $\Vert f{\Vert }_2\le r$, the closed ball $B_R(0)$ is mapped into itself.

For $w,z\in B_R(0)$, factor the cubic gradient nonlinearity asThe same Sobolev embedding theorem and the Holder inequality implyThe elliptic estimate therefore yieldsShrinking $R$ further makes $C^{\prime }{R}^2<1$, so $\Phi$ is a contraction mapping of the closed ball. This ball is complete because $H^2(U)\cap H_0^1(U)$ is a Banach space. The contraction mapping theorem gives a fixed point $u=\Phi (u)$, and its defining equation isThus the nonlinear elliptic boundary value problem has a solution for sufficiently small $\Vert f{\Vert }_2$.

A map $T:X\to Y$ between metric spaces is a contraction mapping if some $q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

The Banach fixed-point theorem states that a contraction from a nonempty complete metric space to itself has a unique fixed point. Indeed, for $x_{n+1}=T{x}_n$,so the geometric-series estimate makes $(x_n)$ Cauchy. Completeness gives a limit $x_{\ast }$, continuity gives $T{x}_{\ast }=x_{\ast }$, andproves uniqueness.

Every solution of $x=\cos x$ lies in $[-1,1]$. The cosine maps this complete interval into itself and, by the mean value theorem, has Lipschitz constant at most $\sin 1<1$ there. It therefore has exactly one real fixed point.

The mean value inequality says that on a convex domain, a uniform derivative bound $\Vert Df\Vert \le M$ implies $\Vert f(x)-f(y)\Vert \le M\Vert x-y\Vert$. Equip ${\mathbb{R}}^2$ with the maximum norm and takeFor $(x,y)\in D$, elementary cosine bounds giveso $f(D)\subseteq D$. The maximum absolute row sum ofis at most $2\sin (1/2)<1$. The mean value inequality makes $f$ a contraction on the complete set $D$, so it has a fixed point.

A Cauchy sequence $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. A complete metric space is one in which every Cauchy sequence converges to a point of the space.

Every Cauchy sequence is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\ge N$, and putThen every term lies in the ball $B(x_N,R)$.

Now suppose $M$ is complete and $(F_n)$ is a decreasing sequence of nonempty closed sets with $\mathrm{diam}{F}_n\to 0$. Choose $x_n\in F_n$. Given $\epsilon >0$, choose $N$ with $\mathrm{diam}{F}_N<\epsilon$. For $m,n\ge N$, both points lie in $F_N$, so $d(x_m,x_n)<\epsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore

Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. DefineThese sets are nonempty, closed, and decreasing. The Cauchy property implies $\mathrm{diam}{F}_n\to 0$; taking a closure does not change the diameter. Choose $x\in {\bigcap }_n{F}_n$. Since $x_n,x\in F_n$,Thus every Cauchy sequence converges and $M$ is complete. This proves the Cantor intersection theorem characterization.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point.

For each $\lambda \in \Lambda$, the map $T_{\lambda }(x)=T(\lambda ,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^{\ast }(\lambda )$. This defines the required unique function. Fix ${\lambda }_0$. The fixed-point identities and the triangle inequality giveConsequentlyThe numerator tends to zero as $\lambda \to {\lambda }_0$ by the assumed continuity for the fixed point $x^{\ast }({\lambda }_0)$. Hence $x^{\ast }$ is continuous, an instance of continuous dependence of the fixed point of a uniform contraction.

There is a sign issue in the printed problem. Part (c) constructs the inverse of $\Delta -1$, whereas the second equation printed in part (e) contains $\Delta +1$. The latter operator is invertible with homogeneous Dirichlet data only when $1$ is not a Dirichlet Laplacian eigenvalue. Thus the assertion as printed needs this nonresonance hypothesis; with a minus sign it follows directly from parts (c) and (d).

Under either the intended minus sign or the stated nonresonance condition, let $G_0$ and $G_1$ be the bounded Dirichlet solution operators for the two linear equations. Choose $s>n/2$ and work with $u,v\in H^{s+2}(\Omega )\cap H_0^1(\Omega )$. Since $H^s(\Omega )$ is a Sobolev algebra,DefineElliptic regularity gives, on a ball of radius $R$,and the difference estimate has Lipschitz constant at most $CR$. Choose $R$ small and then ${\epsilon }_0$ so that $C(R^2+{\epsilon }_0\Vert f{\Vert }_{H^s})\le R$. The contraction mapping theorem gives a solution for $0\le \epsilon <{\epsilon }_0$. Repeated elliptic regularity and smoothness of $f$ bootstrap the solution to $C^{\infty }$.

Write the nonlinearity asIn one dimension, the Sobolev embedding theorem gives $H^1(\mathbb{T})\hookrightarrow L^{\infty }(\mathbb{T})$. HenceThus $N:\mathcal{H}\to \mathcal{H}$ is locally Lipschitz.

On $C([0,T];\mathcal{H})$ defineOn a ball of radius $R$, unitarity givesChoose $R>\Vert Z_0{\Vert }_{\mathcal{H}}$ and then $T>0$ small enough that the first bound preserves the ball and $CTR<1$. The contraction mapping theorem gives a unique fixed point. Precisely, the local mild solution is

For general energy data $Z_0\in \mathcal{H}$, this solution need not be differentiable in $\mathcal{H}$. If $Z_0\in D(A)=H^2\times H^1$, standard semilinear evolution theory and the smoothness of $N$ give a local classical solution.

The blow-up alternative for a semilinear evolution equation says that the solution continues while its $\mathcal{H}$ norm stays finite, but global existence does not hold for every datum. Spatially constant solutions obeyFor sufficiently large $y(0)>1$ with $y^{\prime }(0)\ge 0$, the solution grows until $y^{\prime \prime }\ge y^2/2$ and blows up in finite time. These constant functions are periodic and belong to every Sobolev space, so they provide finite-time blow-up examples for the original equation.

A map $h:X\to X$ is a contraction if there is a constant $q<1$ such thatfor every $x,y\in X$.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point, and that the iterates from every starting point converge to it. To prove this, choose $x_0\in X$ and put $x_{n+1}=h(x_n)$. Thenso, for $m>n$,Thus $(x_n)$ is Cauchy and converges, by completeness, to some $x_{\ast }$. A contraction is continuous, soIf $y_{\ast }$ is another fixed point, thenforcing $x_{\ast }=y_{\ast }$.

For the Newton mapone has $g(r)=r$ andOn the given neighbourhood,Since $f(r)=0$, choose a closed interval $U^{\prime }$ centred at $r$ and contained in $U$ so small that $M|f(x)|/{\delta }^2\le 1/2$ there. Thenso $g(U^{\prime })\subseteq U^{\prime }$ and $g$ is a contraction on the complete interval $U^{\prime }$. The local contraction proof for Newton iteration therefore shows that $r$ is the unique fixed point of $g$ on $U^{\prime }$.

By the contraction mapping theorem, the contraction $A:X_{b,\tau }\to X_{b,\tau }$ has a unique fixed point $u$. The identity $Au=u$ says exactly that $u$ is the weak solution of the linear initial boundary value problem whose forcing is $u^2\sin u$. Consequently $u\in X_{\tau }$ satisfieswith the prescribed initial and homogeneous Dirichlet data. Thus a sufficiently small $\tau >0$ gives a local weak solution of the semilinear wave equation, as summarized by the local weak solution by contraction for a semilinear wave equation.

A contraction satisfies $d(Tx,Ty)\le qd(x,y)$ for some $0\le q<1$. Starting from $x_0$, define $x_{k+1}=T{x}_k$. The geometric bound on successive distances makes $(x_k)$ Cauchy, so completeness gives a limit $x$. Continuity of $T$ gives $Tx=x$. If $Ty=y$, then $d(x,y)\le qd(x,y)$, hence $x=y$. This is the contraction mapping theorem.

If $T^n$ is a contraction, it has a unique fixed point $x$. Since $T^n(Tx)=T(T^{n}x)=Tx$, the point $Tx$ is also fixed by $T^n$, so uniqueness gives $Tx=x$. Every fixed point of $T$ is fixed by $T^n$, so it too must equal $x$. Thus $T$ has exactly one fixed point.

The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

For every real $t$, $f(t)-t=1/(1+e^t)>0$, so there is no fixed point. Direct differentiation givesWriting $u=e^t>0$, the inequality $(1+u{)}^2\ge 4u$ showsIn particular $0<f^{\prime }(t)<1$, but $f^{\prime }(t)\to 1$ as $t\to \pm \infty$. There is no common contraction constant smaller than one. Strict pointwise bounds on the derivative therefore do not suffice for the contraction mapping theorem.

Let $f$ be continuous in its first variable and uniformly Lipschitz continuous in its second variable, with constant $L\ge 0$. Retain the finite-interval smoothness assumptions and the bound $\Vert {\tau }_n\Vert \le K{h}^2$ from the preceding subpart. We first establish solvability of each implicit step. For a proposed starting state $u$, the maphas Lipschitz constant $hL/2$. If $hL/2<1$, the contraction mapping theorem on the complete state space gives a unique next state $v={\Psi }_{n,h}(u)$.

To compute a useful Lipschitz constant for the method, introduce its increment function byFor two starting states $u,w$,Rearranging gives the implicit midpoint Lipschitz estimateThus, for example, $hL\le 1$ gives an increment Lipschitz constant at most $2L$, uniformly as $h\to 0$.

For an explicit global error estimate, set $e_n=Y_n-y_n$. Subtract the numerical formula from the exact-value identity containing $h{\tau }_n$:The Lipschitz condition and the triangle inequality implyand thereforeIterating this inequality is the elementary discrete Gronwall inequality argument. For $L>0$, summing the geometric series and using $R_h-1=hL/(1-hL/2)$ givesFor $hL\le 1$, $R_h=1+hL/(1-hL/2)\le 1+2hL\le e^{2hL}$. If $nh\le T$, this provesWith the exact initial value, $e_0=0$, the global error is $O(h^2)$ and tends uniformly to zero on every such finite interval. If $L=0$, the recurrence instead gives $\Vert e_n\Vert \le \Vert e_0\Vert +KT{h}^2$, with the same conclusion. Starting values merely converging to the exact initial value also suffice for numerical convergence; second-order accuracy requires their errors to be $O(h^2)$. The result concerns exact implicit solves and convergence on fixed finite intervals, not uniform accuracy over an unbounded time interval.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Suppose $f$ is continuous and Lipschitz in its state variable on a closed cylinder $|t-a|\le h$, $\Vert y-b\Vert \le k$, with bounds $\Vert f\Vert \le M$ and Lipschitz constant $L_f$. On a smaller interval of radius $\eta$, the displayed Picard operator maps continuous curves in that closed ball into itself when $\eta M\le k$, and it is a contraction mapping in the supremum norm when $\eta L_f<1$. This function space is complete because it is a closed subset of a Banach space of continuous vector-valued functions. The contraction mapping theorem gives a unique fixed point, and the fundamental theorem of calculus makes it a solution of the ordinary differential equation. Choosing $\eta M<k$ also prevents any solution with the same initial data from first exiting the cylinder during that interval.

For a vector field uniformly Lipschitz continuous in the state with constant $L$, the stage fixed-point map of an implicit Runge-Kutta method is a contraction in the maximum stage norm if $hL{\max }_i{\sum }_j|a_{ij}|<1$. The contraction mapping theorem then gives unique stages. This is a sufficient small-step condition; a dissipative vector field may permit solvability for larger steps. An algebraic stability or B-stability estimate compares existing stage solutions and should not be mistaken for an unqualified stage-existence theorem.