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3.6 : Existence and Uniqueness of Solutions of Nonlinear Equations

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

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Although there are methods for solving some nonlinear equations, it's impossible to find useful formulas for the solutions of most. Whether we're looking for exact solutions or numerical approximations, it's useful to know conditions that imply the existence and uniqueness of solutions of initial value problems for nonlinear equations. In this section we state such a condition and illustrate it with examples.

\vspace*{-15pt}

\begin{figure}[H]

\centering

\scalebox{.9}{

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020301}}

\color{blue}

\vspace*{-35pt}

\caption{An open rectangle}

\label{figure:2.3.1}

\end{figure}

Some terminology:

an {\color{blue}\it open rectangle $R$\/}

is a set of points $(x,y)$ such that

$$a<x<b\mbox{\quad and \quad}c<y<d$$

(Figure~\ref{figure:2.3.1}). We'll denote this set by

$R: \{ a < x < b, c < y < d \}$.

Open'' means that the

boundary rectangle (indicated by the dashed lines in

Figure~\ref{figure:2.3.1}) isn't included in $R$ .

The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for first order nonlinear differential equations. We omit the proof, which is beyond the scope of this book.

\begin{theorem}\color{blue} \label{thmtype:2.3.1} \mbox{}\newline

\vspace*{-5pt}

\begin{alist}

\item %(a)

If $f$ is continuous

on an open rectangle

$$R: \{ a < x < b, c < y < d \}$$

that contains $(x_0,y_0)$

then the initial value problem

\begin{equation} \label{eq:2.3.1}

\end{equation}

has at least one solution on some open subinterval

of $(a,b)$ that contains $x_0.$

\item % (b)

If both $f$ and $f_y$ are

continuous on $R$ then \eqref{eq:2.3.1} has a unique

solution on some open subinterval of $(a,b)$ that contains $x_0$.

\end{alist}

\end{theorem}

It's important to understand exactly what Theorem~\ref{thmtype:2.3.1}

says.

\begin{itemize}

\item \part{a} is an {\color{blue}\it existence theorem\/}. It guarantees that a

solution exists on some open interval that contains $x_0$, but provides

no information on how to find the solution, or to determine the open

interval on which it exists. Moreover, \part{a} provides no

information on the number of solutions that \eqref{eq:2.3.1} may have. It

leaves open the possibility that \eqref{eq:2.3.1} may have two or more

solutions that differ for values of $x$ arbitrarily close to $x_0$. We

will see in Example~\ref{example:2.3.6} that this can happen.

\item \part{b} is a {\color{blue}\it uniqueness theorem\/}. It guarantees that

\eqref{eq:2.3.1} has a unique solution on some open interval (a,b) that

contains $x_0$. However, if $(a,b)\ne(-\infty,\infty)$,

\eqref{eq:2.3.1} may have more than one solution on a larger interval

that contains $(a,b)$. For example, it may happen that $b<\infty$ and all

solutions have the same values on $(a,b)$, but two solutions $y_1$ and

$y_2$ are defined on some interval $(a,b_1)$ with $b_1>b$, and have

different values for $b<x<b_1$; thus, the graphs of the $y_1$ and

$y_2$ branch off'' in different directions at $x=b$. (See

Example~\ref{example:2.3.7} and Figure~\ref{figure:2.3.3}). In this case,

continuity implies that $y_1(b)=y_2(b)$ (call their common value

$\overline y$), and $y_1$ and $y_2$ are both solutions of the initial

value problem

\begin{equation} \label{eq:2.3.2}

\end{equation}

that differ on every open interval that contains $b$. Therefore

$f$ or $f_y$ must have a discontinuity at some point in each open

rectangle that contains $(b,\overline y)$, since if this were not so,

\eqref{eq:2.3.2} would have a unique solution on some open interval

that contains $b$. We leave it to you to give a similar analysis of the

case where $a>-\infty$.

\end{itemize}

\begin{example}\label{example:2.3.1}

\rm Consider the initial value problem

\begin{equation} \label{eq:2.3.3}

y'={x^2-y^2 \over 1+x^2+y^2}, \quad y(x_0)=y_0.

\end{equation}

Since

$$f(x,y) = {x^2-y^2 \over 1+x^2+y^2}\mbox{\quad and \quad} f_y(x,y) = -{2y(1+2x^2)\over (1+x^2+y^2)^2}$$

are continuous for all $(x,y)$, Theorem~\ref{thmtype:2.3.1} implies that

if $(x_0,y_0)$ is arbitrary, then

\eqref{eq:2.3.3} has a unique solution on some open interval that contains

$x_0$.

\end{example}

\begin{example}\label{example:2.3.2}

\rm Consider the initial value problem

\begin{equation} \label{eq:2.3.4}

y'={x^2-y^2 \over x^2+y^2}, \quad y(x_0)=y_0.

\end{equation}

Here

$$f(x,y) = {x^2-y^2 \over x^2+y^2}\mbox{\quad and \quad} f_y(x,y) = \dst -{4x^2y \over (x^2+y^2)^2}$$

are continuous everywhere except at $(0,0)$. If $(x_0,y_0) \ne(0,0)$, there's an open rectangle $R$ that contains

$(x_0,y_0)$ that does not contain $(0,0)$. Since $f$ and $f_y$ are

continuous on $R$, Theorem~\ref{thmtype:2.3.1} implies that if

$(x_0,y_0)\ne(0,0)$ then

\eqref{eq:2.3.4}

has a unique solution on some open interval that contains $x_0$.

\end{example}

\begin{example}\label{example:2.3.3}

\rm Consider the initial value problem

\begin{equation} \label{eq:2.3.5}

\end{equation}

Here

$$f(x,y) = {x+y\over x-y}\mbox{\quad and \quad} f_y(x,y) = {2x\over (x-y)^2}$$

are continuous everywhere except on the line $y=x$. If $y_0\ne x_0$,

there's an open rectangle $R$ that contains $(x_0,y_0)$ that

does not intersect the line $y=x$. Since $f$ and $f_y$ are continuous

on $R$, Theorem~\ref{thmtype:2.3.1} implies that if $y_0\ne x_0$,

\eqref{eq:2.3.5} has a unique solution on some open interval that contains

$x_0$.

\end{example}

\begin{example}\label{example:2.3.4}

\rm In Example~\ref{example:2.2.4} we saw that the

solutions of

\begin{equation} \label{eq:2.3.6}

y'=2xy^2

\end{equation}

are

$$y\equiv0\mbox{\quad and \quad} y=-{1 \over x^2+c},$$

where $c$ is an arbitrary constant. In particular, this implies that

no solution of \eqref{eq:2.3.6} other than $y\equiv0$ can equal zero for

any value of $x$. Show that Theorem~\ref{thmtype:2.3.1}\part{b} implies

this.

\end{example}

\solution

We'll obtain a contradiction

by assuming that \eqref{eq:2.3.6} has a solution $y_1$ that equals

zero for some value of $x$, but isn't identically zero. If $y_1$

has this property, there's a point $x_0$ such that $y_1(x_0)=0$,

but $y_1(x)\ne0$ for some value of $x$ in every open interval that contains

$x_0$. This means that the initial value problem

\begin{equation} \label{eq:2.3.7}

\end{equation}

has two solutions $y\equiv0$ and $y=y_1$ that differ for some value of

$x$ on every open interval that contains $x_0$. This contradicts

Theorem~\ref{thmtype:2.3.1}(b), since in \eqref{eq:2.3.6} the functions

$$f(x,y)=2xy^2 \mbox{\quad and \quad} f_y(x,y)= 4xy.$$

are both continuous

for all $(x,y)$, which implies that \eqref{eq:2.3.7} has a unique

solution on some open interval that contains $x_0$.

\begin{example}\label{example:2.3.5}

\rm Consider the initial value problem

\begin{equation} \label{eq:2.3.8}

y' = {10\over 3}xy^{2/5}, \quad y(x_0) = y_0.

\end{equation}

\begin{alist}

\item %(a)

For what points $(x_0,y_0)$ does Theorem~\ref{thmtype:2.3.1}\part{a}

imply that

\eqref{eq:2.3.8} has a solution?

\item %(b)

For what points $(x_0,y_0)$ does Theorem~\ref{thmtype:2.3.1}\part{b}

imply that

\eqref{eq:2.3.8} has a unique solution on some open interval that contains

$x_0$?

\end{alist}

\end{example}

\solutionpart{a} Since

$$f(x,y) = {10\over 3}xy^{2/5}$$

is continuous for all $(x,y)$, Theorem~\ref{thmtype:2.3.1}

implies that \eqref{eq:2.3.8} has a solution for every $(x_0,y_0)$.

\solutionpart{b} Here

$$f_y(x,y) = {4 \over 3}xy^{-3/5}$$

is continuous for all $(x,y)$ with $y\ne 0$. Therefore, if $y_0\ne0$

there's an open rectangle on which both $f$ and $f_y$ are

continuous, and Theorem~\ref{thmtype:2.3.1} implies that \eqref{eq:2.3.8} has

a unique solution on some open interval that contains $x_0$.

If $y=0$ then $f_y(x,y)$ is undefined, and therefore discontinuous;

hence, Theorem~\ref{thmtype:2.3.1} does not apply to \eqref{eq:2.3.8} if

$y_0=0$.

\begin{example}\label{example:2.3.6}

\rm Example~\ref{example:2.3.5} leaves open the possibility that the

initial value problem

\begin{equation} \label{eq:2.3.9}

y'={10 \over 3}xy^{2/5}, \quad y(0)=0

\end{equation}

has more than one solution on every open interval that contains $x_0=0$. Show

that this is true.

\end{example}

\solution

By inspection, $y\equiv0$ is a solution of the

differential equation

\begin{equation} \label{eq:2.3.10}

y'={10 \over 3} xy ^{2/5}.

\end{equation}

Since $y\equiv0$ satisfies the initial condition $y(0)=0$,

it's a solution of \eqref{eq:2.3.9}.

Now suppose $y$ is a solution of \eqref{eq:2.3.10} that isn't

identically zero.

Separating variables in \eqref{eq:2.3.10} yields

$$y^{-2/5}y'={10 \over 3}x$$

on any open interval where $y$ has no zeros.

Integrating this and rewriting the arbitrary constant as $5c/3$ yields

$${5\over 3}y^{3/5} = {5\over 3}(x^2+c).$$

Therefore

\begin{equation} \label{eq:2.3.11}

y = (x^2+c)^{5/3}.

\end{equation}

Since we divided by $y$ to separate variables in \eqref{eq:2.3.10}, our

derivation of \eqref{eq:2.3.11} is legitimate only on open intervals where $y$

has no zeros. However, \eqref{eq:2.3.11} actually defines $y$ for all $x$,

and differentiating \eqref{eq:2.3.11} shows that

$$y'={10 \over 3}x(x^2+c)^{2/3}={10 \over 3}xy^{2/5},\,-\infty<x<\infty.$$

Therefore \eqref{eq:2.3.11} satisfies \eqref{eq:2.3.10} on

$(-\infty,\infty)$

even if $c\le 0$, so that $y(\sqrt{|c|})=y(-\sqrt{|c|})=0$. In

particular, taking $c=0$ in \eqref{eq:2.3.11} yields

$$y=x^{10/3}$$

as a second solution of \eqref{eq:2.3.9}.

Both solutions are defined on

$(-\infty,\infty)$, and they differ on every

open interval that contains $x_0=0$ (see Figure~\ref{figure:2.3.2}.)

In fact, there are {\color{blue}\it four\/} distinct solutions of

\eqref{eq:2.3.9} defined on $(-\infty,\infty)$ that differ

from each other on every open interval that contains $x_0=0$.

Can you identify the other two?

\begin{figure}[tbp]

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020302}

\color{blue}

\caption{Two solutions ($y=0$ and $y=x^{1/2}$) of \eqref{eq:2.3.9} that

differ on every interval containing $x_{0}=0$}

\label{figure:2.3.2}

\end{figure}

\begin{example} \label{example:2.3.7}

\rm From Example~\ref{example:2.3.5}, the initial value

problem

\begin{equation} \label{eq:2.3.12}

y'={10 \over 3}xy^{2/5}, \quad y(0)=-1

\end{equation}

has a unique solution on some open interval that contains $x_0=0$.

Find a solution and determine the largest open interval $(a,b)$ on

which it's unique.

\end{example}

\solution

Let $y$ be any solution of \eqref{eq:2.3.12}. Because of the initial

condition $y(0)=-1$ and the continuity of $y$, there's an open interval

$I$ that contains $x_0=0$ on which $y$ has no zeros, and is consequently

of the form \eqref{eq:2.3.11}. Setting $x=0$ and $y=-1$ in \eqref{eq:2.3.11}

yields $c=-1$, so

\begin{equation} \label{eq:2.3.13}

y=(x^2-1)^{5/3}

\end{equation}

for $x$ in $I$.

Therefore every solution of \eqref{eq:2.3.12} differs from zero

and is given by \eqref{eq:2.3.13} on $(-1,1)$; that is,

\eqref{eq:2.3.13} is the unique solution of \eqref{eq:2.3.12} on

$(-1,1)$.

This is the largest open interval on which

\eqref{eq:2.3.12} has a unique solution. To see this, note that

\eqref{eq:2.3.13} is a solution of \eqref{eq:2.3.12} on $(-\infty,\infty)$.

From Exercise~2.2.~\hspace*{-3pt}\ref{exer:2.2.15}, there are infinitely many

other solutions

of \eqref{eq:2.3.12} that differ from \eqref{eq:2.3.13} on every open interval

larger than

$(-1,1)$. One such solution is

$$y = \left\{ \begin{array}{cl} (x^2-1)^{5/3}, & -1 \le x \le 1, \\[6pt] 0, & |x|>1. \end{array} \right.$$

(Figure \ref{figure:2.3.3}).

\begin{figure}[htbp]

\color{blue}

\begin{minipage}[b]{0.5\linewidth}

\centering

\scalebox{.65}{

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020303}}

\caption{ Two solutions of

\eqref{eq:2.3.12} on

$(-\infty,\infty)$ that coincide on $(-1,1)$, but on no larger open

interval}

\label{figure:2.3.3}

\end{minipage}

\hspace{0.6cm}

\begin{minipage}[b]{0.5\linewidth}

\centering

\scalebox{.65}{

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig020304} }

\caption{The unique solution of \eqref{eq:2.3.14}}

\label{figure:2.3.4}

\end{minipage}

\end{figure}

\begin{example}\label{example:2.3.8}

\rm From Example~\ref{example:2.3.5}, the initial value

problem

\begin{equation} \label{eq:2.3.14}

y'={10 \over 3}xy^{2/5}, \quad y(0)=1

\end{equation}

has a unique solution on some open interval that contains $x_0=0$.

Find the solution and determine the largest open interval on which it's

unique.

\end{example}

\solution

Let $y$ be any solution of \eqref{eq:2.3.14}. Because of the initial

condition $y(0)=1$ and the continuity of $y$, there's an open interval

$I$ that contains $x_0=0$ on which $y$ has no zeros, and is consequently

of the form \eqref{eq:2.3.11}. Setting $x=0$ and $y=1$ in \eqref{eq:2.3.11}

yields $c=1$, so

\begin{equation} \label{eq:2.3.15}

y=(x^2+1)^{5/3}

\end{equation}

for $x$ in $I$. Therefore every solution of \eqref{eq:2.3.14}

differs from zero and is given by \eqref{eq:2.3.15} on $(-\infty,\infty)$;

that is, \eqref{eq:2.3.15} is the unique solution of \eqref{eq:2.3.14} on

$(-\infty,\infty)$.

Figure~\ref{figure:2.3.4} shows the graph of this solution.

\exercises

In Exercises \ref{exer:2.3.1}-\ref{exer:2.3.13} find all $(x_0,y_0)$ for

which Theorem~\ref{thmtype:2.3.1}

implies that the initial value problem $y'=f(x,y),\ y(x_0)=y_0$ has

\part{a}

a solution \part{b} a unique solution on some open interval that

contains

$x_0$.

\begin{exerciselist}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.3.1} $\dst {y'={x^2+y^2 \over \sin x}}$ & \item\label{exer:2.3.2}\vspace*{10pt} $\dst {y'={e^x+y \over x^2+y^2}}$ \end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.3.3} $y'= \tan xy$

& \item\label{exer:2.3.4} $\dst {y'={x^2+y^2 \over \ln xy}}$ \end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.3.5} $y'= (x^2+y^2)y^{1/3}$ &\ \item\label{exer:2.3.6}

\ $y'=2xy$ \end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\vspace*{10pt} \item\label{exer:2.3.7} $\dst {y'=\ln(1+x^2+y^2)}$

& \vspace*{4pt} \item\label{exer:2.3.8} $\dst {y'={2x+3y \over x-4y}}$ \end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.3.9} $\dst {y'=(x^2+y^2)^{1/2}}$ &\item\label{exer:2.3.10}\ $y' = x(y^2-1)^{2/3}$ \end{tabular}

\begin{tabular}[t]{@{}p{168pt}@{}p{168pt}}

\item\label{exer:2.3.11} $y'=(x^2+y^2)^2$

& \item\label{exer:2.3.12} $y'=(x+y)^{1/2}$

\end{tabular}

\item\label{exer:2.3.13}

$\dst {y'={\tan y \over x-1}}$

\item\label{exer:2.3.14}

Apply Theorem~\ref{thmtype:2.3.1} to the initial value problem

$$y'+p(x)y = q(x), \quad y(x_0)=y_0$$

for a linear equation, and compare the conclusions that can be drawn

from it to those that follow from Theorem 2.1.2.

\item\label{exer:2.3.15}

\begin{alist}

\item %(a)

Verify that the function

$$y = \left\{ \begin{array}{cl} (x^2-1)^{5/3}, & -1 < x < 1, \\[6pt] 0, & |x| \ge 1, \end{array} \right.$$

is a solution of the initial value problem

$$y'={10\over 3}xy^{2/5}, \quad y(0)=-1$$

on $(-\infty,\infty)$. \hint{You'll need the definition $$y'(\overline{x}) = \lim_{x \to \overline{x}} {y(x)-y(\overline{x}) \over x-\overline{x}}$$ to verify that $y$ satisfies the differential

equation at $\overline{x} = \pm 1$.}

\item %(b)

Verify that if $\epsilon_i=0$ or $1$ for $i=1$, $2$ and $a$, $b>1$, then

the function

$$y = \left\{ \begin{array}{cl} \epsilon_1(x^2-a^2)^{5/3}, & - \infty < x < -a, \\[6pt] 0, & -a \le x \le -1, \\[6pt] (x^2-1)^{5/3}, & -1 < x < 1, \\[6pt] 0, & 1 \le x \le b, \\[6pt] \epsilon_2(x^2-b^2)^{5/3}, & b < x < \infty, \end{array} \right.$$

is a solution of the initial value problem of \part{a} on

$(-\infty,\infty)$.

\end{alist}

\item\label{exer:2.3.16}

Use the ideas developed in Exercise~\ref{exer:2.3.15} to find

infinitely

many solutions of the initial value problem

$$y'=y^{2/5}, \quad y(0)=1$$

on $(-\infty,\infty)$.

\item\label{exer:2.3.17}

Consider the initial value

problem

$$y' = 3x(y-1)^{1/3}, \quad y(x_0) = y_0. \eqno{\rm (A)}$$

\begin{alist}

\item %(a)

For what points $(x_0,y_0)$ does Theorem~\ref{thmtype:2.3.1} imply that

(A) has a solution?

\item %(b)

For what points $(x_0,y_0)$ does Theorem~\ref{thmtype:2.3.1} imply that

(A) has a unique solution on some open interval

that contains $x_0$?

\end{alist}

\item\label{exer:2.3.18}

Find nine solutions of the initial value problem

$$y'=3x(y-1)^{1/3}, \quad y(0)=1$$

that are all defined on $(-\infty,\infty)$ and differ from each other

for values of $x$ in every open interval that contains $x_0=0$.

\item\label{exer:2.3.19} From Theorem~\ref{thmtype:2.3.1}, the initial

value

problem

$$y'=3x(y-1)^{1/3}, \quad y(0)=9$$

has a unique solution on an open interval that contains $x_0=0$. Find the

solution and determine the largest open interval on which it's unique.

\item\label{exer:2.3.20}

\begin{alist}

\item %(a)

From Theorem~\ref{thmtype:2.3.1},

the initial value problem

$$y'=3x(y-1)^{1/3}, \quad y(3)=-7 \eqno{\rm (A)}$$

has a unique solution on some open interval that contains $x_0=3$.

Determine

the largest such open interval, and find the solution on this interval.

\item %(b)

Find infinitely many solutions of (A), all defined

on

$(-\infty,\infty)$.

\end{alist}

\item\label{exer:2.3.21}

Prove:

\begin{alist}

\item % ()

If

$$f(x,y_0) = 0,\quad a<x<b, \eqno{\rm (A)}$$

and $x_0$ is in $(a,b)$,

then $y\equiv y_0$ is a solution of

$$y'=f(x,y), \quad y(x_0)=y_0$$

on $(a,b)$.

\item % (b)

If $f$ and $f_y$ are continuous on an open rectangle

that contains $(x_0,y_0)$ and (A) holds, no

solution of $y'=f(x,y)$ other than $y\equiv y_0$ can equal $y_0$ at

any point in $(a,b)$.

\end{alist}

\end{exerciselist}