01. ODEs and Picard's Theorem

1.1 Introduction

ODE

Equation for an unknown function $y(x)$ of one variable $x$ in form

$$G(x,y(x),y^{\prime }(x),y^{\prime \prime }(x),\cdots ,y^{(n)}(x))=0$$

where $y^{(n)}(x)=\frac{d^{n}y}{d{x}^n}(x)$

This can be solved for the highest derivative of $t$ in the form

$$y^{(n)}(x)=F(x,y,y^{\prime },\cdots ,y^{(n-1})$$

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Order of an ODE

Order $n$ of an ODE is order of the highest derivative that appears

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Non Uniqueness of solutions to ODEs with initial values

Consider IVP

$$y^{\prime }(x)=3y(x{)}^{2/3},y(0)=0$$

By separation of variables you get

$$\int \frac{dy}{3y^{2/3}}=\int dx$$

Solving that you get

$$y(x)=(x+A{)}^3\text{ where }A\text{ is a constant}$$

Using initial condition $y(0)=0$ we get

1. Solution $y(x)=x^3$
2. Trivial Solution $y(x)=0$
3. Infinite Solutions for $c,d$ where $c\le 0\le d$ then

$$y_{c,d}(x):=\{\begin{array}{ll}(x-c{)}^3& \text{ for }x<c\\ 0& \text{ for }c\le x\le d\\ (x-d{)}^3& \text{ for }x>d\end{array}$$

All these functions satisfy the initial conditions and are differentiable everywhere

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1.1.1 Setting up for Picard’s Theorem

General form of the IVP

$$y^{\prime }(x)=f(x,y(x))\text{ and }y(a)=b$$

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Rectangular Neighbourhood for the Initial Value Problem

We seek a solution $y(x)$ which is defined on interval $[a-h,a+h]$ for suitable $h>0$
We also require the graph to be contained in rectangle $R$ around point $(x,y)=(a,b)$

$$R=\{(x,y):|x-a|\le h,|y-b|\le k\}$$

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Assumptions and Properties of function $f:R\to \mathbb{R}$

Properties

1. $f$ is a function such that $(x,y)\mapsto f(x,y)$

Assumptions

1. $f$ is continuous in $R$ ($R$ is bounded so this guarantees that $f$ is bounded on $R$) with

$$M:=\underset{R}{\sup }|f|=\underset{R}{\max }|f|$$

Don't get confused with $x\mapsto f(x,y(x))$

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Lipschitz Condition (with constant $L$)

Function $f:[a-h,a+h]\times [b-k,b+k]$ satisfies Lipschitz Condition

If there exists $L>0$ such that

$$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|$$

for all $x\in [a-h,a+h]$ and for all $y_1,y_2\in [b-k,b+k]$

Stronger than being continuous in the second variable
Weaker then continuously differentiable

Way to show a function $f$ satisfies the condition

Suppose on $R$ that $f$ is differentiable with respect to $y$, with

$$|f_y(x,y)|\le K\text{ for some }K\text{ and all }(x,y)\in R$$

For each $x\in [a-h,a+h]$, we can apply MVT to function $[b-k,b+k]\ni y\mapsto f(x,y)$
So that for any $y,\tilde{y}\in [b-k,b+k]$, there exists $\xi$ between $y$ and $\tilde{y}$ such that

$$|f(x,\tilde{y})-f(x,y)|=|f_y(x,\xi )(y-\tilde{y})|\le K|y-\tilde{y}|$$

So $f$ satisfies the Lipschitz Condition on rectangle $R$ with $L=K$

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1.2 Picard’s Theorem

01 - Picard's Existence Theorem

Picard's Existence Theorem

Let $f:R\to \mathbb{R}$ be a function defined on rectangle

$$R:=\{(x,y):|x-a|\le h,|y-b|\le k\}$$

Which satisfies

• $P(i)$ - $(a)$: $f$ is continuous in $R$ with $|f(x,y)|\le M$ for all $(x,y)\in R$
• $P(i)$ - $(b)$: $Mh\le k$
• $P(ii)$: $f$ satisfies Lipschitz Condition in $R$

Then IVP

$$y^{\prime }(x)=f(x,y(x))\text{ with }y(a)=b$$

has a unique solution $y$ on interval $[a-h,a+h]$

Picard’s Theorem doesn’t give the best (largest) interval where existence/uniqueness holds and is a local result

Proof - Existence - Via Successive Approximation

Successive Approximation (Iteration)

Start of with constant function $y_0(x)$

$$y_0(x)=b\forall x\in [a-h,a+h]$$

Then for $n=0,1,\cdots$ then

$$y_{n+1}(x):=b+{\int }_a^{x}f(t,y_n(t))dt$$

Equivalent to asking that $y_{n+1}$ satisfies $y_{n+1}^{\prime }(x)=f(x,y_n(x))$ and $y_{n+1}(a)=b$

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Claim 1: Each $y_n(x)$ is well-defined and continuous on $[a-h,a+h]$ which satisfies

$$|y_n(x)-b|\le k\text{ for all }x\in |a-h,a+h|$$

Claim 1 - By Induction

True for $n=0$ as it is constant
For $t\in [a-h,a+h]$ then $(t,y_n(t))\in R$ so evaluate $f$ at this point

As $y_n:[a-h,a+h]\to [b-k,b+k]$ is continuous then

$$t\mapsto (t,y_n(t))\text{ is a continious function}$$

from $[a-h,a+h]$ to the rectangle $R$
As $f$ is continuous function from $R$ to $\mathbb{R}$ then

$$\text{composition }[a-h,a+h]\ni t\mapsto f(t,y_n(t))\in \mathbb{R}\text{ is continious }$$

Hence it is integrable

So next iterate $y_{n+1}$ is well-defined function from $[a-h,a+h]\to \mathbb{R}$
So by properties of integration it is differentiable hence continuous

Hence we have

$$|y_{n+1}(x)-b|\le \left|{\int }_a^x|f(t,y_n(t))|dt\right|\le \left|{\int }_a^{x}Mdt\right|=M|x-a|\le Mh\le k$$

for every $x\in [a-h,a+h]$
Note that we need modulus outside integrals for cases where $x\le a$

Claim 2: Let $\hat{M}$ be so that $|f(x,b)|\le \hat{M}$ for all $x\in [a-h,a+h]$
Let $L$ be so that the the Lipschitz Condition holds so

Define the differences between consecutive iterations as

$$e_n(x):=y_n(x)-y_{n-1}(x),\text{ for }n=1,2,\cdots$$

which satisfy

$$|e_n(x)|\le \frac{L^{n-1}\hat{M}}{n!}|x-a{|}^n$$

for every $x\in [a-h,a+h]$ and every $n\in \mathbb{N}$
Note that we can choose $\hat{M}=M$ as the bound on $|f|$ (as $\hat{M}\le M$)

Claim 2 - By Induction

For $n=1$ use $y_0(x)=b$ so

$$e_1(x)=y_1(x)-y_0(x)=b+{\int }_a^{x}f(t,y_0(t))dt-b={\int }_a^{x}f(t,b)dt$$

Hence

$$|e_1(x)|\le \left|{\int }_a^x|f(t,b)|dt\right|\le \left|{\int }_a^x\hat{M}dt\right|\le \hat{M}|x-a|$$

Hence case is true for $n=1$

Suppose claim holds for $n\in \mathbb{N}$ then using the definition of Picard iterations

$$\begin{array}{rl}{e}_{n+1}(x)& =y_{n+1}(x)-y_n(x)=\left[b+{\int }_a^{x}f(t,y_n(t))dt\right]-\left[b+{\int }_a^{x}f(t,y_{n-1}(t))dt\right]\\ & ={\int }_a^{x}f(t,y_n)(t)-f(t,y_{n-1}(t))dx\end{array}$$

Using Lipschitz Condition in $P(ii)$ and that $y_n,y_{n-1}$ are contained in rectangle $R$
So for $|t-a|\le h$ then

$$|f(t,y_n(t))-f(t,y_{n-1}(t))|\le L|y_n(t)-y_{n-1}(t)|=L|e_n(t)|$$

Combining with the induction assumption we get

$$|e_{n+1}(x)|\le L\left|{\int }_a^x|e_n(t)|dt\right|\le L\left|{\int }_a^x\frac{L^{n-1}\hat{M}}{n!}|t-a{|}^{n}dt\right|=\frac{L^n\hat{M}}{(n+1)!}|x-a{|}^{n+1}$$

So it is true by induction

Claim 3: The iterates $y_n(x)=y_0(x)+{\sum }_{j=1}^n{e}_j(x)$ converge uniformly to

$$\text{continuous function }{y}_{\infty }(x)$$

on the interval $[a-h,a+h]$ and $y_{\infty }(x)$ is a solution of the recursive integral equation

Claim 3

Note that from Claim 2 then for every $n\in N$

$$|e_n(x)|\le \frac{L^{n-1}\hat{M}}{n!}{h}^n=:M_n\text{ for all }x\in [a-h,a+h]$$

As ${\sum }_{n=1}^{\infty }{M}_n$ converges by ratio test then by Weierstrass M-test then

$$\sum _{j=1}^n{e}_j(x)\text{ converges uniformly on }[a-h,a+h]\text{ as }n\to \infty$$

As $y_0(x)$ is just a fixed function (independent of $n$) then

$$y_n(x)=y_0(x)+\sum _{j=1}^n{e}_j(x)$$

also converges uniformly to limit function $y_{\infty }(x)$ on $[a-h,a+h]$

Uniform convergence combined with the continuity of the functions $y_n$ means

$$y_{\infty }\text{ is also continuous}$$

Using Lipschitz-Condition and uniform convergence of $y_n(t)$ then

$$\underset{t\in [a-h,a+h]}{\sup }|f(t,y_n(t))-f(t,y_{\infty }(t))|\le L\underset{t\in [a-h,a+h]}{\sup }|y_n(t)-y_{\infty }(t)|\to 0$$

where the last step follows from $y_n$ converging uniformly

Since we can exchange limit and integrals for uniformly convergent sequences

$$\underset{n\to \infty }{\lim }{\int }_a^{x}f(t,y_n(t))dt={\int }_a^{x}f(t,y_n(t))dt={\int }_a^{x}f(t,y_{\infty }(t))dt$$

So by passing the limit to the recursive equation to define $y_{n+1}$ then
For all $x\in [a-h,a+h]$

$$y_{\infty }(x)=\underset{n\to \infty }{\lim }{y}_{n+1}(x)=b+\underset{n\to \infty }{\lim }{\int }_a^{x}f(t,y_n(t))dt=b+{\int }_a^{x}f(t,y_{\infty }(x))dt$$

Hence $y_{\infty }$ is indeed a solution of the integral equation
So as the integral equation is equivalent to the IVP then the existence part of Picard’s Theorem is proven

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1.3 Rewriting an IVP into an equivalent integral equation#

Rewriting the IVP into an equivalent Integral Equation

As $y(x)$ is differentiable and satisfies IVP on interval $[a-h,a+h]$ then

$$y(x)\text{ is certainly continuous so }f\text{ is continuous}$$

As $x\mapsto f(x,y(x))$ is a continuous function on $[a-h,a+h]$ it is integrable

Hence

$$y(x)-y(a)=[y(t){]}_a^x={\int }_a^{x}f(t,y(t))dt\text{ for any }x\in [a-h,a+h]$$

By rearranging then we get any solution of IVP satisfies

$$y(x)=b+{\int }_a^{x}f(t,y(t))dt\text{ for any }x\in [a-h,a+h]$$

Converse Argument $y:[a-h,a+h]\to [b-k,b+k]$ is a continuous function satisfying integral equation Then

If

$$y(a)=b$$

Since the integrand $t\mapsto f(t,y(t))$ is continuous then by
Fundamental Theorem of Calculus that $y$ is differentiable in every $x\in [a-h,a+h]$ with

$$y^{\prime }(x)=f(x,y(x))$$

So $y$ is a solution of the IVP

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1.4 Proof of the Existence Part of Picard’s Theorem via Method of Successive Approximation

Successive Approximation (Iteration)

Start of with constant function $y_0(x)$

$$y_0(x)=b\forall x\in [a-h,a+h]$$

Then for $n=0,1,\cdots$ then

$$y_{n+1}(x):=b+{\int }_a^{x}f(t,y_n(t))dt$$

Equivalent to asking that $y_{n+1}$ satisfies $y_{n+1}^{\prime }(x)=f(x,y_n(x))$ and $y_{n+1}(a)=b$

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Refer to proof of Picard's Existence Theorem

Picard's Theorem for Non-Symmetric Rectangles

Replace condition $Mh\le k$ with

$$M\max (h_1,h_2)\le \min (k_1,k_2)$$

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Boundedness Lemma for IVPs lemma

Let IVP refer to $y^{\prime }(x)=f(x,y(x))$ with $y(a)=b$

Let $f:R\to \mathbb{R}$ be so that $P(i)$ holds
Let $y(x)$ be any solution of the IVP where $y(x)$ is defined on interval $[a-h_1,a{+}_{h_2}]$ for some $0<h_{1,2}\le h$ then

$$|y(x)-b|\le k\text{ for all }x\in [a-h_1,a+h_2]$$

and the graph doesn’t leave $R$

Note that this doesn’t require the Lipschitz Condition

Proof - By Contradiction

Suppose the claim is wrong so there exists a solution $y:[a-h_1,a+h_2]\to \mathbb{R}$ of IVP
such that there exists some $x_1\in [a-h_{1}a+h_2]$ with $|y(x_1)-b|>k$

By symmetry WLOG that $x_1>a$
As the graph starts out in rectangle $R=[a-h,a+h]\times [b-k,b+k]$ but contains

$$(x_1,y(x_1))$$

with $x_1\in [a-h,a+h]$ which is outside rectangle then there must be a first point

$$(x_0,y(x_0))$$

where the graph intersects the boundary of the rectangle

Hence there is some $x_0\in (a,x_1)$ such that

$$|y(x_0)-b|=k\text{ while }|y(x)-b|<k\text{ for all }x\in [a,x_0)$$

Since for $t\in [a,x_0]$ the points $(t,y(t))$ are in rectangle where $|f|$ is bounded by $M$
So since $y$ satisfies the IVP and hence integral equation then we get

$$\begin{array}{rl}|y(x_0)-b|& =\left|{\int }_a^{x_0}f(t,y(t))dt\right|\le \left|{\int }_a^{x_0}|f(t,y(t))|dt\right|\le M|x_0-a|\\ & <M(x_1-a)\le M{h}_1\le Mh\le k\end{array}$$

which is a contradiction

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1.5 Gronwall’s Inequality

02 - Gronwall's Inequality

Gronwall's Inequality

Suppose $A\ge 0$ and $B\ge 0$ are constants and $v$ is a non-negative continious function with

$$v(x)\le B+A\left|{\int }_a^{x}v(s)ds\right|$$

for all $x$ in an interval $[a-h_1,a+h_2]$ and $h_{1,2}\ge 0$ then

$$v(x)\le B{e}^{A|x-a|}\text{ for all }x\in [a-h_1,a+h_2]x$$

Note: need modulus for case where $x\le a$

Proof

For $x\ge a$ let $V(x)={\int }_a^{x}v(s)ds$ so $V^{\prime }(x)=v(x)$
As $x\ge a$ and $v\ge 0$ then $V(x)\ge 0$ and $V(a)=0$ so

$$V^{\prime }(x)\le B+AV(x)$$

By multiplying it through by integrating factor $e^{-}Ax$ then

$$\begin{array}{rl}(V^{\prime }(x)-AV(x))e^{-Ax}& \le B{e}^{-Ax}\\ \frac{d}{dx}(V(x)e^{-Ax})& \le B{e}^{-Ax}\\ V(x)e^{-Ax}-V(a)e^{-Aa}& \le {\int }_a^{x}B{e}^{-As}ds=\frac{B}{A}(e^{-Aa}-e^{-Ax})\\ V(x)& \le \frac{B}{A}(e^{A(x-a)}-1)\end{array}$$

Using the assumption again then

$$v(x)\le B+A{\int }_a^{x}v(s)ds=B+AV(x)\le B+A\cdot \frac{B}{A}(e^{A(x-a)}-1)=B{e}^{A(x-a)}$$

Similar case for $x\le a$

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1.6 Uniqueness and Continuous Dependence on the Initial Data

Uniqueness of the Picard's Theorem

Let $y(x)$ and $\tilde{y}(x)$ be two solutions of the ODE to the same initial data then

$$y(x)=\tilde{y}(x)$$

Proof

Suppose $f$ is a continuous function which satisfies Lipschitz condition on rectangle

$$R=[c,d]\times [b-k,b+k]$$

Let $a\in [c,d]$ and $y(x),\tilde{y}(x)$b be solutions of same ODE

$$y^{\prime }(x)=f(x,y(x))\text{ and }y(a)=b,\tilde{y}(a)=\tilde{b}$$

where $b$ may not necessarily be the same as $\tilde{b}$ whose graph is in $R$

Consider the difference function

$$v(x)=|\tilde{y}(x)-y(x)|\ge 0$$

And we have that $y(x)$ and $\tilde{y}(x)$ are solutions of corresponding integral equations

$$y(x)=b+{\int }_a^{x}f(t,y(t))dt\text{ and }\tilde{y}(x)=\tilde{b}+{\int }_a^{x}f(t,\tilde{y}(t))dt$$

We also have Lipschitz condition

$$|f(t,y(t))-f(t,\tilde{y}(t))|\le L|y(t)-\tilde{y}(t)|\le Lv(t)$$

Hence we get

$$\begin{array}{rl}v(x)& =\left|b-\tilde{b}+{\int }_a^{x}f(t,y(t))-f(t,\tilde{y}(t))dt\right|\\ & \le |b-\tilde{b}|+\left|{\int }_a^x|f(t,y(t))-f(t,\tilde{y}(t))|dt\right|\\ & \le |b-\tilde{b}|+L\left|{\int }_a^{x}v(t)dt\right|\end{array}$$

So by using Gronwall’s Inequality then

$$|y(x)-\tilde{y}(x)|=v(x)\le |b-\tilde{b}|e^{L|x-a|}\text{ for all }x\in [c,d]$$

When $b=\tilde{b}$ then we get uniqueness of solutions on the whole interval

In addition to proving uniqueness part of Picard’s Theorem then
For close $b,\tilde{b}$ then

$$||y-\tilde{y}|{|}_{\sup }:=\underset{x\in [c,d]}{\sup }|y(x)-\tilde{y}(x)|$$

If $y(x),\tilde{y}(x)$ satisfy ODE with initial values $b,\tilde{b}$ with $|b-\tilde{b}|<\delta$ then

$$|y(x)-\tilde{y}(x)|\le |b-\tilde{b}|e^{L|x-a|}\le \delta e^{Lh}=ϵ$$

for any $x\in [c,d]$ with $h:=\max \{a-c,d-a\}$

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1.7 Extension of Solutions and Characterisation of Maximal Existence Interval

Transclude of 02---Setup-for-Picard's-Theorem#^006aa0

Upper Existence Bound

Let $f:{\mathbb{R}}^2\to \mathbb{R}$ be a function then define

$$T_{+}:=\sup \{t>a:\text{ there exists a solution of IVP on }[a,t)\}$$

Note that $T_{+}:=+\infty$ if for every $t>a$ then there is a solution on $[a,t)$

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Lower Existence Bound

Let $f:{\mathbb{R}}^2\to \mathbb{R}$ be a function then define

$$T_{-}:=\inf \{t<a:\text{ there exists a solution of IVP on }(t,a]\}$$

Note that $T_{-}:=-\infty$ if for every $t<a$ then there is a solution on $(t,a]$

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03 - Maximal Existence Theorem

Maximal Existence Theorem

Let $f:{\mathbb{R}}^2\to \mathbb{R}$ be continuous and locally Lipschitz with respect to $y$
Let $a,b\in \mathbb{R}$ and let $(T_{-},T_{+})$ be the maximal existence interval of solution $y(x)$ for IVP

$$y^{\prime }(x)=f(x,y(x))\text{ and }y(a)=b$$

If $T_{+}<\infty$ then the solutions blows up as we approach $T_{+}$ ($|y(x)|\to \infty$ as $t\to T_{+}$)
If $T_{-}>-\infty$ then $|y(x)|\to \infty$ as $t\to T_{-}$

By thinking of $x$ is a time parameter then we have the following cases

1. Global Existence - Solution of IVP exist for all $x\in \mathbb{R}$
2. Solution exists for all times in future but blows up at some finite time $T_{-}<a$ in the past
3. Solution exists for all times in past but blows up at some finite time $T_{+}>a$ in the future
4. Solution blows up at some finite time in the past and future

Note that can have solutions that exist for all times but tend to infinity as $x\to \pm \infty$

Proof

We only need to prove the case where $T_{+}<\infty$ as we can reverse time direction by

$$\tilde{y}(x)=y(-x)$$

to handle $T_{-}>-\infty$

Suppose that $T_{+}<\infty$ then we need to show that

$$|y(x)|\to \infty \text{ as }x\to T_{+}$$

That is, for every $K>0$ there exists $\delta >0$ such that

$$|y(x)|>K\text{ for all }x\in (T_{+}-\delta ,T_{+})$$

Suppose by contradiction that this isn’t true so there is $K>0$ so that for every $\delta >0$

$$\exists x_{\delta }\in (T_{+}-\delta ,T_{+})\text{ with }|y(\delta )|\le K$$

For some compact rectangle $R_0$ containing set $[-K,K]\times \{T_{+}\}$ then
As $f$ is continuous and $R_0$ is compact then $f$ is bounded on $R_0$ so suppose

$$|f|\le M_0\text{ on }{R}_0$$

Note that the bound still applies for any rectangle $R\subseteq R_0$

So by setting $h:=\min \left\{\frac{1}{M_0},\frac{1}{2}\right\}$ and $k:=1$ so$M_{0}h\le k$ still holds
Hence $P(i)$ holds on

$$R=[\tilde{a}-h,\tilde{a}+h]\times [\tilde{b}-1,\tilde{b}+1]\text{ for all }\tilde{a}\in \left[T_{+}-\frac{1}{2},T_{+}+\frac{1}{2}\right],\tilde{b}\in [-K,K]$$

Since $f$ is locally Lipschitz wrt $y$ on all of ${\mathbb{R}}^2$ then Lipschitz Condition holds on all $R$
Hence we can apply Picard’s Theorem with same $h$ for all initial conditions $y(\tilde{a})=\tilde{b}$

Choosing $\delta <h$ and letting $x_{\delta }\in (T_{+}-\delta ,T_{+})$ so that $|y(x_{\delta })|\le K$
So use $\tilde{a}=x_{\delta }$ and $\tilde{b}=y(x_{\delta })$ as initial data to get solution $\tilde{y}$ of the ODE which satisfies

$$\tilde{y}(x_{\delta })=y(x_{\delta })$$

which is defined on $[x_{\delta }-h,x_{\delta }+h]$

As $x_{\delta }+h>T_{+}-\delta +h>T_{+}$ which allows us to extend the original solution $y$
beyond the maximal existence time $T_{+}$ which is a contradiction

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1.8 Comparison Principe and a Priori Estimates

04 - Comparison Principle

Definition

Let $g:{\mathbb{R}}^2\to \mathbb{R}$ be continuously differentiable
Let $u(x)$ and $v(x)$ be differentiable functions which satisfy

$$u^{\prime }(x)\le g(x,u(x))\text{ and }{v}^{\prime }(x)\ge g(x,v(x))$$

on some interval $I$

If $u(x_0)\le v(x_0)$ for some $x_0\in I$ then

$$u(x)\le v(x)\text{ for all }x\in I,x\ge x_0$$

If $u(x_0)\ge v(x_0)$ for some $x_0\in I$ then

$$u(x)\ge v(x)\text{ for all }x\in I,x\le x_0$$

Proof

Since $g$ is continuously differentiable use MVT to check that $h\to {\mathbb{R}}^3\to \mathbb{R}$ of 3 variables
which is defined by difference quotient

$$h(x,y_1,y_2):=\{\begin{array}{ll}\frac{g(x,y_1)-g(x,y_2)}{y_1-y_2}& \text{ if }{y}_1\ne y_2\\ {\partial }_{y}g(x,y_1)& \text{ if }{y}_1=y_2\end{array}$$

is continuous

As both functions $u(x)$ and $v(x)$ are continuous on interval $I$ then composition

$$a(x):=h(x,u(x),v(x))\text{ is also continuous}$$

Crucially the function is chosen so that we can write

$$g(x,v(x))-g(x,u(x))=a(x)(v(x)-u(x))$$

From assumptions of theorem then we get $w(x):=v(x)-u(x)$ satisfying

$$w^{\prime }(x)=v^{\prime }(x)-u^{\prime }(x)\ge g(x,v(x))-g(x,u(x))=a(x)w(x)$$

Setting $A(x)={\int }_{x_0}^{x}a(t)dt$ to get function $A^{\prime }(x)=a(x)$ hence

$$(e^{-A(x)}w(x){)}^{\prime }=e^{-A(x)}(w^{\prime }(x)-A^{\prime }(x)w(x))$$

Hence

$$e^{-A(x)}w(x)\text{ is non-decreasing}$$

If $u(x_0)\le v(x_0)$ then

$$e^{-A(x_0)}w(x_0)=w(x_0)\ge 0$$

Hence for $x\ge x_0$ we have $e^{-A(x)}w(x)\ge 0$ and $w(x)\ge 0$ so

$$u(x)\le v(x)$$

On the other hand if $u(x_0)\ge v(x_0)$ for $x\le x_0$ then

$$e^{-A(x)}w(x)\le w(x_0)\le 0$$

Hence

$$u(x)\ge v(x)$$

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1.9 Picard’s Theorem for Systems and Higher Order ODEs

Pair System of First Order ODEs

Consider a pair first so for functions $y_1,y_2$ then

$$\begin{array}{rl}{y}_1^{\prime }(x)& =f_1(x,y_1(x),y_2(x))\\ y_2^{\prime }(x)& =f_2(x,y_1(x),y_2(x))\end{array}$$

with initial data

$$y_1(a)=b_1,y_2(a)=b_2$$

Then we can use vector notation

$$\underline{y}=\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right),\underline{f}=\left(\begin{array}{c}{f}_1\\ f_2\end{array}\right),\underline{b}=\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)$$

So we can write the system of equations as

$$\begin{array}{rl}{\underline{y}}^{\prime }(x)& =\underline{f}(x,\underline{y}(x))\\ \underline{y}(a)& =\underline{b}\end{array}$$

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Ball of Radius $k$

$$B_k(\underline{b})=\{\underline{y}\in {\mathbb{R}}^2:||\underline{y}-\underline{b}|{|}_1\le k\}=\{(y_1,y_2)\in {\mathbb{R}}^2:|y_1-b_1|+|y_2-b_2|\le k\}$$

Note that we are using the $l^1$ norm so that $||\underline{y}||=|y_1|+|y_2|$

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05 - Picard's Existence Theorem for Systems

Picard's Existence Theorem for Systems

Let $f:S\to {\mathbb{R}}^2$ be a function defined on the set $S=[a-h,a+h]\times B_k(\underline{b})$ which satisfies

• $H(i)$: $\underline{f}(x,\underline{y})$ is continuous on $S$, $||\underline{f}(x,\underline{y})|{|}_1\le M$ and $Mh\le k$
• $H(ii)$: $f$ is Lipschitz with respect to $\underline{y}$ on $S$ so
  There exists $L$ such that for all $x\in [a-h,a+h]$ and $\underline{y},\tilde{\underline{y}}\in B_k(b)$ then

$$||\underline{f}(x,\underline{y})-\underline{f}(x,\tilde{\underline{y}})|{|}_1\le L||\underline{y}-\tilde{\underline{y}}|{|}_1$$

Then IVP

$${\underline{y}}^{\prime }(x)=\underline{f}(x,\underline{y}(x))$$

Has a unique solution on interval $[a-h,a+h]$

Tips for Showing Conditions

Note that conditions on functions $\underline{f}:S\to {\mathbb{R}}^2$ is a function of 3 variables and NOT $x\mapsto \underline{f}(x,\underline{y}(x))$

Can extend for systems of $n$ ODEs for any $n\in \mathbb{N}$ but we focus on $n=2$

When checking for the Lipschitz Condition for vector valued function $\underline{f}$
Simply check for the components as we are using the $l^1$ norm i.e. there exists $L_1,L_2$

$$\begin{array}{rl}|f_1(x,y_1,y_2)-f_1(x,{\tilde{y}}_1,{\tilde{y}}_2)|& \le L_1(|y_1-{\tilde{y}}_1|+|y_2-{\tilde{y}}_2|)\\ |f_2(x,y_1,y_2)-f_2(x,{\tilde{y}}_1,{\tilde{y}}_2)|& \le L_2(|y_1-{\tilde{y}}_1|+|y_2-{\tilde{y}}_2|)\end{array}$$

then take $L=L_1+L_2$

May also be handy to add in “smart 0s” e.g.

$$f_1(x,y_1,y_2)-f_1(x,{\tilde{y}}_1,{\tilde{y}}_2)=(f_1(x,y_1,y_2)-f_1(x,{\tilde{y}}_1,y_2))+(f_1(x,{\tilde{y}}_1,y_2)-f_1(x,{\tilde{y}}_1,{\tilde{y}}_2))$$

So we can then use MVT so there is some $\xi$ between $y_1$ and ${\tilde{y}}_1$ so that

$$f_1(x,y_1,y_2)-f_1(x,{\tilde{y}}_1,y_2)={\partial }_{y_1}{f}_1(x,\xi ,y_2)(y_1-{\tilde{y}}_1)$$

and similarily for some ${\xi }^{\prime }$ between $y_2$ and ${\tilde{y}}_2$

$$f_1(x,{\tilde{y}}_1,y_2)-f_1(x,{\tilde{y}}_1,{\tilde{y}}_2)={\partial }_{y_2}{f}_1(x,{\tilde{y}}_1,{\xi }^{\prime })(y_2-{\tilde{y}}_2)$$

Solutions will not only exist on interval $[a-h,a+h]$ but on maximal existence interval

$$(T_{-},T_{+})$$

which for continious function $\underline{f}:{\mathbb{R}}^3\to {\mathbb{R}}^2$ which satisfy Lipschitz condition on
every compact subset $S$ of ${\mathbb{R}}^3$ will be given by all of $\mathbb{R}$ (assuming no blow-up)

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1.9.1 Picard for Higher Order ODEs

Solving Second Order ODEs using Picard

Suppose we have IVP for second-order ODE

$$y^{\prime \prime }=F(x,y(x),y^{\prime }(x))\text{ with }y(a)=b,y^{\prime }(a)=c$$

Note that this can also be extended to $n$-th order linear ODEs

Solution

Reduce it to a first-order paired system by

$$y_1(x)=y(x)\text{ and }{y}_2(x)=y^{\prime }(x)$$

Note that $y(s)$ solves above IVP if and only if $\underline{y}(x)=(y_1(x),y_2(x))$ satisfies

$${\underline{y}}^{\prime }(x)=\underline{f}(x,\underline{y}(x))$$

where $\underline{f}$ is defined as

$$\underline{f}(x,y_1,y_2):=(y_2,F(x,y_1,y_2){)}^T$$

So $\underline{f}$ is continuous if and only if $F(x,y_1,y_2)$ is continuous
Similarly if $F$ satisfies the Lipschitz condition then $\underline{f}$ also does with $L=L_2+1$

Hence we can apply Picard’s Theorem

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