# 3.5: Existence and Uniqueness of Solutions of Nonlinear Equations - Mathematics

Although there are methods for solving some nonlinear equations, it is impossible to find useful formulas for the solutions of most. 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

(Figure (PageIndex{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 (PageIndex{1})) is not included in (R).

The next theorem gives sufficient conditions for the 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 (PageIndex{1}): existence and uniqueness

1. If (f) is continuous on an open rectangle [R: { a < x < b, c < y < d } onumber] that contains ((x_0,y_0)) then the initial value problem [label{eq:3.5.1} y'=f(x,y), quad y(x_0)=y_0] has at least one solution on some open subinterval of ((a,b)) that contains (x_0.)
2. If both (f) and (f_y) are continuous on (R) then Equation ef{eq:3.5.1} has a unique solution on some open subinterval of ((a,b)) that contains (x_0)

It’s important to understand exactly what Theorem (PageIndex{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 Equation ef{eq:3.5.1} may have. It leaves open the possibility that Equation ef{eq:3.5.1} may have two or more solutions that differ for values of (x) arbitrarily close to (x_0). We will see in Example (PageIndex{6}) that this can happen.

• (b) is a uniqueness theorem. It guarantees that Equation ef{eq:3.5.1} has a unique solution on some open interval (a,b) that contains (x_0). However, if ((a,b) e(-infty,infty)), Equation ef{eq:3.5.1} may have more than one solution on a larger interval that contains ((a,b)). For example, it may happen that (bb), and have different values for (b

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, y)), since if this were not so, ef{eq:3.5.2} 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 > −∞).

Example (PageIndex{1})

Consider the initial value problem

[label{eq:3.5.3} y'={x^2-y^2 over 1+x^2+y^2}, quad y(x_0)=y_0.]

Since

[f(x,y) = {x^2-y^2 over 1+x^2+y^2} quad ext{and} quad f_y(x,y) = -{2y(1+2x^2)over (1+x^2+y^2)^2} onumber]

are continuous for all ((x,y)), Theorem (PageIndex{1}) implies that if ((x_0,y_0)) is arbitrary, then Equation ef{eq:3.5.3} has a unique solution on some open interval that contains (x_0).

Example (PageIndex{2})

Consider the initial value problem

[label{eq:3.5.4} y'={x^2-y^2 over x^2+y^2}, quad y(x_0)=y_0.]

Here

[f(x,y) = {x^2-y^2 over x^2+y^2}quad ext{and} quad f_y(x,y) = -{4x^2y over (x^2+y^2)^2} onumber]

are continuous everywhere except at ((0,0)). If ((x_0,y_0) e(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 (PageIndex{1}) implies that if ((x_0,y_0) e(0,0)) then Equation ef{eq:3.5.4} has a unique solution on some open interval that contains (x_0).

Example (PageIndex{3})

Consider the initial value problem

Here

are continuous everywhere except on the line (y=x). If (y_0 e 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 (PageIndex{1}) implies that if (y_0 e x_0), Equation ef{eq:3.5.5} has a unique solution on some open interval that contains (x_0).

Example (PageIndex{4})

In Example 2.2.4, we saw that the solutions of

[label{eq:3.5.6} y'=2xy^2]

are

where (c) is an arbitrary constant. In particular, this implies that no solution of Equation ef{eq:3.5.6} other than (yequiv0) can equal zero for any value of (x). Show that Theorem (PageIndex{1b}) implies this.

We’ll obtain a contradiction by assuming that Equation ef{eq:3.5.6} has a solution (y_1) that equals zero for some value of (x), but is not 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) e0) for some value of (x) in every open interval that contains (x_0). This means that the initial value problem

has two solutions (yequiv0) and (y=y_1) that differ for some value of (x) on every open interval that contains (x_0). This contradicts Theorem (PageIndex{1})(b), since in Equation ef{eq:3.5.6} the functions

are both continuous for all ((x,y)), which implies that Equation ef{eq:3.5.7} has a unique solution on some open interval that contains (x_0).

Example (PageIndex{5})

Consider the initial value problem

[label{eq:3.5.8} y' = {10over 3}xy^{2/5}, quad y(x_0) = y_0.]

1. For what points ((x_0,y_0)) does Theorem (PageIndex{1a}) imply that Equation ef{eq:3.5.8} has a solution?
2. For what points ((x_0,y_0)) does Theorem (PageIndex{1b}) imply that Equation ef{eq:3.5.8} has a unique solution on some open interval that contains (x_0)?

Solution a

Since

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

is continuous for all ((x,y)), Theorem (PageIndex{1}) implies that Equation ef{eq:3.5.8} has a solution for every ((x_0,y_0)).

Solution b

Here

[f_y(x,y) = {4 over 3}xy^{-3/5} onumber]

is continuous for all ((x,y)) with (y e 0). Therefore, if (y_0 e0) there’s an open rectangle on which both (f) and (f_y) are continuous, and Theorem (PageIndex{1}) implies that Equation ef{eq:3.5.8} 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 (PageIndex{1}) does not apply to Equation ef{eq:3.5.8} if (y_0=0).

Example (PageIndex{6})

Example (PageIndex{5}) leaves open the possibility that the initial value problem

[label{eq:3.5.9} y'={10 over 3}xy^{2/5}, quad y(0)=0]

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

Solution

By inspection, (yequiv0) is a solution of the differential equation

[label{eq:3.5.10} y'={10 over 3} xy ^{2/5}.]

Since (yequiv0) satisfies the initial condition (y(0)=0), it is a solution of Equation ef{eq:3.5.9}.

Now suppose (y) is a solution of Equation ef{eq:3.5.10} that is not identically zero. Separating variables in Equation ef{eq:3.5.10} yields

[y^{-2/5}y'={10 over 3}x onumber]

on any open interval where (y) has no zeros. Integrating this and rewriting the arbitrary constant as (5c/3) yields

[{5over 3}y^{3/5} = {5over 3}(x^2+c) . onumber]

Therefore

[label{eq:3.5.11} y = (x^2+c)^{5/3}. ]

Since we divided by (y) to separate variables in Equation ef{eq:3.5.10}, our derivation of Equation ef{eq:3.5.11} is legitimate only on open intervals where (y) has no zeros. However, Equation ef{eq:3.5.11} actually defines (y) for all (x), and differentiating Equation ef{eq:3.5.11} shows that

[egin{aligned}y'={10 over 3}x(x^2+c)^{2/3}={10 over 3}xy^{2/5},,-inftyend{aligned} ]

Therefore Equation ef{eq:3.5.11} satisfies Equation ef{eq:3.5.10} on ((-infty,infty)) even if (cle 0), so that (y(sqrt{|c|})=y(-sqrt{|c|})=0). In particular, taking (c=0) in Equation ef{eq:3.5.11} yields

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

as a second solution of Equation ef{eq:3.5.9}. Both solutions are defined on ((-infty,infty)), and they differ on every open interval that contains (x_0=0) (Figure (PageIndex{2})). In fact, there are four distinct solutions of Equation ef{eq:3.5.9} defined on ((-infty,infty)) that differ from each other on every open interval that contains (x_0=0). Can you identify the other two?

Example (PageIndex{7})

From Example (PageIndex{5}), the initial value problem

[label{eq:3.5.12} y'={10 over 3}xy^{2/5}, quad y(0)=-1]

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 is unique.

Solution

Let (y) be any solution of Equation ef{eq:3.5.12}. 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 Equation ef{eq:3.5.11}. Setting (x=0) and (y=-1) in Equation ef{eq:3.5.11} yields (c=-1), so

[label{eq:3.5.13} y=(x^2-1)^{5/3}]

for (x) in (I). Therefore every solution of Equation ef{eq:3.5.12} differs from zero and is given by Equation ef{eq:3.5.13} on ((-1,1)); that is, Equation ef{eq:3.5.13} is the unique solution of Equation ef{eq:3.5.12} on ((-1,1)). This is the largest open interval on which Equation ef{eq:3.5.12} has a unique solution. To see this, note that Equation ef{eq:3.5.13} is a solution of Equation ef{eq:3.5.12} on ((-infty,infty)). From Exercise 2.2.15, there are infinitely many other solutions of Equation ef{eq:3.5.12} that differ from Equation ef{eq:3.5.13} on every open interval larger than ((-1,1)). One such solution is

[y = left{ egin{array}{cl} (x^2-1)^{5/3}, & -1 le x le 1, [6pt] 0, & |x|>1. end{array} ight. onumber]

Example (PageIndex{8})

From Example (PageIndex{5})), the initial value problem

[label{eq:3.5.14} y'={10 over 3}xy^{2/5}, quad y(0)=1]

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 is unique.

Solution

Let (y) be any solution of Equation ef{eq:3.5.14}. 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 Equation ef{eq:3.5.11}. Setting (x=0) and (y=1) in Equation ef{eq:3.5.11} yields (c=1), so

[label{eq:3.5.15} y=(x^2+1)^{5/3}]

for (x) in (I). Therefore every solution of Equation ef{eq:3.5.14} differs from zero and is given by Equation ef{eq:3.5.15} on ((-infty,infty)); that is, Equation ef{eq:3.5.15} is the unique solution of Equation ef{eq:3.5.14} on ((-infty,infty)). Figure (PageIndex{4})) shows the graph of this solution.