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Mathematics LibreTexts

3.6 : Existence and Uniqueness of Solutions of Nonlinear Equations

  • Page ID
    17141
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

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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}

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

    \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}

    y'=f(x,y),\quad y(b)=\overline y

    \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}

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

    \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}

    y'=2xy^2,\quad y(x_0)=0

    \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}