2.3 Existence and Uniqueness of Solutions of Nonlinear Equations

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.

Some terminology: an open rectangle \(R\) is a set of points \((x,y)\) such that

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

(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 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.

Theorem 2.3.1

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} y'=f(x,y), \quad y(x_0)=y_0 \tag{2.3.1}\end{equation}

has at least one solution on some open subinterval of \((a,b)\) that contains \(x_0.\)
If both \(f\) and \(f_y\) are continuous on \(R\) then (2.3.1) has a unique solution on some open subinterval of \((a,b)\) that contains \(x_0\).

It’s important to understand exactly what Theorem 2.3.1 says.

(a) is an 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, (a) provides no information on the number of solutions that (2.3.1) may have. It leaves open the possibility that (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 2.3.6 that this can happen.
(b) is a uniqueness theorem. It guarantees that (2.3.1) has a unique solution on some open interval (a,b) that contains \(x_0\). However, if \((a,b)\ne(-\infty,\infty)\), (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 2.3.7 and 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} y'=f(x,y),\quad y(b)=\overline y \tag{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, (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 \(a>-\infty\).

Example 2.3.1

Consider the initial value problem

\begin{equation} y'={x^2-y^2 \over 1+x^2+y^2}, \quad y(x_0)=y_0. \tag{2.3.3}\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 2.3.1 implies that if \((x_0,y_0)\) is arbitrary, then (2.3.3) has a unique solution on some open interval that contains \(x_0\).

Example 2.3.2

Consider the initial value problem

\begin{equation} y'={x^2-y^2 \over x^2+y^2}, \quad y(x_0)=y_0. \tag{2.3.4}\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 2.3.1 implies that if \((x_0,y_0)\ne(0,0)\) then (2.3.4) has a unique solution on some open interval that contains \(x_0\).

Example 2.3.3

Consider the initial value problem

\begin{equation} y'={x+y\over x-y},\quad y(x_0)=y_0. \tag{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 2.3.1 implies that if \(y_0\ne x_0\), (2.3.5) has a unique solution on some open interval that contains \(x_0\).

Example 2.3.4

In Example 2.2.4 we saw that the solutions of

\begin{equation} y'=2xy^2 \tag{2.3.6}\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 (2.3.6) other than \(y\equiv0\) can equal zero for any value of \(x\). Show that Theorem 2.3.1(b) implies this.

Solution We’ll obtain a contradiction by assuming that (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} y'=2xy^2,\quad y(x_0)=0 \tag{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 2.3.1(b), since in (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 (2.3.7) has a unique solution on some open interval that contains \(x_0\).

Example 2.3.5

Consider the initial value problem

\begin{equation} y' = {10\over 3}xy^{2/5}, \quad y(x_0) = y_0. \tag{2.3.8}\end{equation}

For what points \((x_0,y_0)\) does Theorem 2.3.1(a) imply that (2.3.8) has a solution?
For what points \((x_0,y_0)\) does Theorem 2.3.1(b) imply that (2.3.8) has a unique solution on some open interval that contains \(x_0\)?

Solution (a) Since

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

is continuous for all \((x,y)\), Theorem 2.3.1 implies that (2.3.8) has a solution for every \((x_0,y_0)\).

Solution (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 2.3.1 implies that (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 2.3.1 does not apply to (2.3.8) if \(y_0=0\).

Example 2.3.6

Example 2.3.5 leaves open the possibility that the initial value problem

\begin{equation} y'={10 \over 3}xy^{2/5}, \quad y(0)=0 \tag{2.3.9}\end{equation}

has more than one solution on every open interval that contains \(x_0=0\). Show that this is true.

Solution By inspection, \(y\equiv0\) is a solution of the differential equation

\begin{equation} y'={10 \over 3} xy ^{2/5}. \tag{2.3.10}\end{equation}

Since \(y\equiv0\) satisfies the initial condition \(y(0)=0\), it’s a solution of (2.3.9).

Now suppose \(y\) is a solution of (2.3.10) that isn’t identically zero. Separating variables in (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} y = (x^2+c)^{5/3}. \tag{2.3.11}\end{equation}

Since we divided by \(y\) to separate variables in (2.3.10), our derivation of (2.3.11) is legitimate only on open intervals where \(y\) has no zeros. However, (2.3.11) actually defines \(y\) for all \(x\), and differentiating (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 (2.3.11) satisfies (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 (2.3.11) yields

\[ y=x^{10/3} \]

as a second solution of (2.3.9). Both solutions are defined on \((-\infty,\infty)\), and they differ on every open interval that contains \(x_0=0\) (see Figure 2.3.2.) In fact, there are four distinct solutions of (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?

Two solutions (y=0 and y=x^1/2) of eq:2.3.9 that differ on every interval containing x_0=0 — Figure 2.3.2. Two solutions (\(y=0\) and \(y=x^{1/2}\)) of (2.3.9) that differ on every interval containing \(x_{0}=0\)

Example 2.3.7

From Example 2.3.5, the initial value problem

\begin{equation} y'={10 \over 3}xy^{2/5}, \quad y(0)=-1 \tag{2.3.12}\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.

Solution Let \(y\) be any solution of (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 (2.3.11). Setting \(x=0\) and \(y=-1\) in (2.3.11) yields \(c=-1\), so

\begin{equation} y=(x^2-1)^{5/3} \tag{2.3.13}\end{equation}

for \(x\) in \(I\). Therefore every solution of (2.3.12) differs from zero and is given by (2.3.13) on \((-1,1)\); that is, (2.3.13) is the unique solution of (2.3.12) on \((-1,1)\). This is the largest open interval on which (2.3.12) has a unique solution. To see this, note that (2.3.13) is a solution of (2.3.12) on \((-\infty,\infty)\). From Exercise 2.2. 15, there are infinitely many other solutions of (2.3.12) that differ from (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 2.3.3).

Two solutions of eq:2.3.12 on (-,) that coincide on (-1,1), but on no larger open interval — Figure 2.3.3. Two solutions of (2.3.12) on \((-\infty,\infty)\) that coincide on \((-1,1)\), but on no larger open interval

The unique solution of eq:2.3.14 — Figure 2.3.4. The unique solution of (2.3.14)

Example 2.3.8

From Example 2.3.5, the initial value problem

\begin{equation} y'={10 \over 3}xy^{2/5}, \quad y(0)=1 \tag{2.3.14}\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.

Solution Let \(y\) be any solution of (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 (2.3.11). Setting \(x=0\) and \(y=1\) in (2.3.11) yields \(c=1\), so

\begin{equation} y=(x^2+1)^{5/3} \tag{2.3.15}\end{equation}

for \(x\) in \(I\). Therefore every solution of (2.3.14) differs from zero and is given by (2.3.15) on \((-\infty,\infty)\); that is, (2.3.15) is the unique solution of (2.3.14) on \((-\infty,\infty)\). Figure 2.3.4 shows the graph of this solution.

2.3 Exercises

In Exercises 1-13 find all \((x_0,y_0)\) for which Theorem 2.3.1 implies that the initial value problem \(y'=f(x,y),\ y(x_0)=y_0\) has (a) a solution (b) a unique solution on some open interval that contains \(x_0\).

\(\dst {y'={x^2+y^2 \over \sin x}}\)

Show answer

(a), (b) \(x_0\ne k \pi\) (\(k=\) integer)
\(\dst {y'={e^x+y \over x^2+y^2}}\)

Show answer

(a), (b) \((x_0,y_0)\ne(0,0)\)
\(y'= \tan xy\)

Show answer

(a), (b) \(x_0y_0\ne(2k+1){\pi\over2}\) (\(k\)= integer)
\(\dst {y'={x^2+y^2 \over \ln xy}}\)

Show answer

(a), (b) \(x_0y_0>0\) and \(x_0y_0\ne1\)
\(y'= (x^2+y^2)y^{1/3}\)

Show answer

(a) all \((x_0,y_0)\) (b) \((x_0,y_0)\) with \(y_0\ne 0\)
\(y'=2xy\)

Show answer

(a), (b) all \((x_0,y_0)\)
\(\dst {y'=\ln(1+x^2+y^2)}\)

Show answer

(a), (b) all \((x_0,y_0)\)
\(\dst {y'={2x+3y \over x-4y}}\)

Show answer

(a), (b) \((x_0,y_0)\) such that \(x_0\ne 4y_0\)
\(\dst {y'=(x^2+y^2)^{1/2}}\)

Show answer

(a) all \((x_0,y_0)\) (b) all \((x_0,y_0)\ne(0,0)\)
\(y' = x(y^2-1)^{2/3}\)

Show answer

(a) all \((x_0,y_0)\)
(b) all \((x_0,y_0)\) with \(y_0\ne \pm 1\)
\(y'=(x^2+y^2)^2\)

Show answer

(a), (b) all \((x_0,y_0)\)
\(y'=(x+y)^{1/2}\)

Show answer

(a), (b) all \((x_0,y_0)\) such that \(x_0+y_0>0\)
\(\dst {y'={\tan y \over x-1}}\)

Show answer

(a), (b) all \((x_0,y_0)\) with \(x_0\ne 1,\quad y_0\ne(2k+1)\dst{{\pi\over2}}\) (\(k=\) integer)
Apply Theorem 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.
1. 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 \(\overline{x} = \pm 1\).
2. 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 (a) on \((-\infty,\infty)\).
Use the ideas developed in Exercise 15 to find infinitely many solutions of the initial value problem

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

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

Show answer

\(\dst{y = \left( {3\over 5} x+1 \right)^{5/3},\; - \infty < x < \infty,}\) is a solution.

Also,

\[ y = \left\{ \begin{array}{cl} 0, & -\infty< x \le -{5\over 3} \\[3pt] \left({3\over 5}x + 1\right)^{5/3}, & -{5\over3} < x < \infty \end{array}\right. \]

is a solution, For every \(\dst{a \ge {5\over 3}}\), the following function is also a solution:

\[ y = \left\{ \begin{array}{cl} \left( {3\over 5} (x+a) \right)^{5/3}, & - \infty < x < -a, \\[3pt] 0, & -a \le x \le -{5\over 3} \\[3pt] \left({3\over 5}x + 1\right)^{5/3}, & -{5\over3} < x < \infty. \end{array}\right. \]
Consider the initial value problem

\[ y' = 3x(y-1)^{1/3}, \quad y(x_0) = y_0. \tag*{\rm (A)} \]
1. For what points \((x_0,y_0)\) does Theorem 2.3.1 imply that (A) has a solution?
2. For what points \((x_0,y_0)\) does Theorem 2.3.1 imply that (A) has a unique solution on some open interval that contains \(x_0\)?
Show answer

(a) all \((x_0,y_0)\) (b) all \((x_0,y_0)\) with \(y_0\ne1\)
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 \(x_0=0\).

Show answer

\(y_1\equiv1\); \(y_2=1+|x|^3\); \(y_3=1-|x|^3\); \(y_4=1+x^3\); \(y_5=1-x^3\)

\(y_6=\dst{\left\{\begin{array}{ccl}1+x^3,&x\ge 0,\\ 1,&x<0\end{array}\right.}\); \(y_7=\dst{\left\{\begin{array}{ccl}1-x^3,&x\ge 0,\\ 1,&x<0\end{array}\right.}\);

\(y_8=\dst{\left\{\begin{array}{ccl}1,&x\ge 0,\\ 1+x^3,&x<0\end{array}\right.}\); \(y_9=\dst{\left\{\begin{array}{ccl}1,&x\ge 0,\\ 1-x^3,&x<0\end{array}\right.}\)
From Theorem 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.

Show answer

\(\dst{y=1+(x^2+4)^{3/2}, \quad - \infty < x < \infty}\)
1. From Theorem 2.3.1, the initial value problem
  
  \[ y'=3x(y-1)^{1/3}, \quad y(3)=-7 \tag*{\rm (A)} \]
  
  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.
2. Find infinitely many solutions of (A), all defined on \((-\infty,\infty)\).
Show answer

(a) The solution is unique on \((0,\infty)\). It is given by

\[ y= \left\{ \begin{array}{cl} 1, & 0< x\le\sqrt5,\\ 1-(x^2-5)^{3/2}, & \sqrt5<x<\infty \end{array}\right. \]

(b)

\[ \dst y= \left\{ \begin{array}{cl} 1, &-\infty< x\le\sqrt5,\\ 1-(x^2-5)^{3/2},& \sqrt5<x<\infty \end{array} \right. \]

is a solution of (A) on \((-\infty,\infty)\). If \(\alpha\ge0\), then

\[ y= \left\{ \begin{array}{cl} 1+(x^2-\alpha^2)^{3/2},& -\infty<x<-\alpha, \\ 1, & -\alpha\le x\le\sqrt5,\\ 1-(x^2-5)^{3/2},& \sqrt5<x<\infty, \end{array}\right. \]

and

\[ y= \left\{ \begin{array}{cl} 1-(x^2-\alpha^2)^{3/2},& -\infty<x<-\alpha, \\ 1, & -\alpha\le x\le\sqrt5,\\ 1-(x^2-5)^{3/2},& \sqrt5<x<\infty, \end{array} \right. \]

are also solutions of (A) on \((-\infty,\infty)\).
Prove:
1. If
  
  \[ f(x,y_0) = 0,\quad a<x<b, \tag*{\rm (A)} \]
  
  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 \((a,b)\).
2. 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 \((a,b)\).