EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUATIONS

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First Order Equations

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.

y

x

a b

c d

Some terminology: anopen rectangleRis a set of points(x, y)such that a < x < b and c < y < d

(Figure2.3.1). We’ll denote this set byR :{a < x < b, c < y < d}. “Open” means that the boundary rectangle (indicated by the dashed lines in Figure2.3.1) isn’t included inR.

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

(a) Iff is continuous on an open rectangle

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

that contains(x0, y0)then the initial value problem

y0=f(x, y), y(x0) =y0 (2.3.1) has at least one solution on some open subinterval of(a, b)that containsx0.

(b) If bothf andfyare continuous onRthen(2.3.1)has a unique solution on some open subinterval

of(a, b)that containsx0.

It’s important to understand exactly what Theorem2.3.1says.

• (a)is anexistence theorem. It guarantees that a solution exists on some open interval that contains x0, 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 ofxarbitrarily close tox0. We will see in Example2.3.6that this can happen.

• (b)is auniqueness theorem. It guarantees that (2.3.1) has a unique solution on some open interval (a,b) that containsx0. However, if(a, b)6= (−∞,∞), (2.3.1) may have more than one solution on a larger interval that contains(a, b). For example, it may happen thatb <∞and all solutions have the same values on(a, b), but two solutionsy1 andy2 are defined on some interval(a, b1)with b1> b, and have different values forb < x < b1; thus, the graphs of they1andy2“branch off” in different directions atx=b. (See Example2.3.7and Figure2.3.3). In this case, continuity implies thaty1(b) =y2(b)(call their common valuey), andy1andy2are both solutions of the initial value problem

y0=f(x, y), y(b) =y (2.3.2)

that differ on every open interval that containsb. Thereforef orfy must have a discontinuity at some point in each open rectangle that contains(b, y), since if this were not so, (2.3.2) would have a unique solution on some open interval that containsb. We leave it to you to give a similar analysis of the case wherea >−∞.

Example 2.3.1 Consider the initial value problem y0 = x2−y2 1 +x2+y2, y(x0) =y0. (2.3.3) Since f(x, y) = x 2y2 1 +x2+y2 and fy(x, y) =− 2y(1 + 2x2) (1 +x2+y2)2

Section 2.3Existence and Uniqueness of Solutions of Nonlinear Equations 57

are continuous for all(x, y), Theorem2.3.1implies that if(x0, y0)is arbitrary, then (2.3.3) has a unique solution on some open interval that containsx0.

Example 2.3.2 Consider the initial value problem

y0 =x2−y2 x2+y2, y(x0) =y0. (2.3.4) Here f(x, y) = x 2y2 x2+y2 and fy(x, y) =− 4x2y (x2+y2)2

are continuous everywhere except at(0,0). If(x0, y0)6= (0,0), there’s an open rectangleRthat contains

(x0, y0)that does not contain(0,0). Sincef andfy are continuous onR, Theorem2.3.1implies that if

(x0, y0)6= (0,0)then (2.3.4) has a unique solution on some open interval that containsx0. Example 2.3.3 Consider the initial value problem

y0 =x+y x−y, y(x0) =y0. (2.3.5) Here f(x, y) = x+y x−y and fy(x, y) = 2x (x−y)2

are continuous everywhere except on the liney=x. Ify06=x0, there’s an open rectangleRthat contains

(x0, y0)that does not intersect the liney=x. Sincefandfyare continuous onR, Theorem2.3.1implies that ify06=x0, (2.3.5) has a unique solution on some open interval that containsx0.

Example 2.3.4 In Example2.2.4we saw that the solutions of y0 = 2xy2

(2.3.6)

are

y≡0 and y=−x21+c,

wherecis an arbitrary constant. In particular, this implies that no solution of (2.3.6) other thany≡0can equal zero for any value ofx. Show that Theorem2.3.1(b)implies this.

Solution We’ll obtain a contradiction by assuming that (2.3.6) has a solutiony1that equals zero for some value ofx, but isn’t identically zero. Ify1has this property, there’s a pointx0such thaty1(x0) = 0, but y1(x)6= 0for some value ofxin every open interval that containsx0. This means that the initial value problem

y0 = 2xy2, y(x0) = 0 (2.3.7)

has two solutionsy ≡0andy =y1 that differ for some value ofxon every open interval that contains x0. This contradicts Theorem2.3.1(b), since in (2.3.6) the functions

f(x, y) = 2xy2 and fy(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 containsx0.

Example 2.3.5 Consider the initial value problem y0 =10

3 xy

2/5, y(x0) =y0. (2.3.8)

(a) For what points(x0, y0)does Theorem2.3.1(a)imply that (2.3.8) has a solution?

(b) For what points(x0, y0)does Theorem2.3.1(b)imply that (2.3.8) has a unique solution on some open interval that containsx0?

SOLUTION(a) Since

f(x, y) =10 3 xy

2/5

is continuous for all(x, y), Theorem2.3.1implies that (2.3.8) has a solution for every(x0, y0).

SOLUTION(b) Here

fy(x, y) = 4

3xy −3/5

is continuous for all(x, y)withy 6= 0. Therefore, ify0 6= 0there’s an open rectangle on which bothf andfyare continuous, and Theorem2.3.1implies that (2.3.8) has a unique solution on some open interval that containsx0.

Ify= 0thenfy(x, y)is undefined, and therefore discontinuous; hence, Theorem2.3.1does not apply to (2.3.8) ify0= 0.

Example 2.3.6 Example2.3.5leaves open the possibility that the initial value problem y0 = 10

3 xy

2/5, y(0) = 0 (2.3.9)

has more than one solution on every open interval that containsx0= 0. Show that this is true.

Solution By inspection,y≡0is a solution of the differential equation y0 =10

3 xy

2/5. (2.3.10)

Sincey≡0satisfies the initial conditiony(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−2/5y0= 10 3 x

on any open interval whereyhas no zeros. Integrating this and rewriting the arbitrary constant as5c/3

yields 5 3y 3/5= 5 3(x 2+c). Therefore y = (x2+c)5/3. (2.3.11)

Since we divided byyto separate variables in (2.3.10), our derivation of (2.3.11) is legitimate only on open intervals whereyhas no zeros. However, (2.3.11) actually definesy for allx, and differentiating (2.3.11) shows that y0 =10 3 x(x 2+c)2/3= 10 3 xy 2/5, −∞< x <∞.

Section 2.3Existence and Uniqueness of Solutions of Nonlinear Equations 59

x y

Figure 2.3.2 Two solutions (y= 0andy =x1/2) of (2.3.9) that differ on every interval containing

x0= 0

Therefore (2.3.11) satisfies (2.3.10) on(−∞,∞)even ifc ≤0, so thaty(p

|c|) = y(−p

|c|) = 0. In particular, takingc= 0in (2.3.11) yields

y=x10/3

as a second solution of (2.3.9). Both solutions are defined on(−∞,∞), and they differ on every open interval that containsx0= 0(see Figure2.3.2.) In fact, there arefourdistinct solutions of (2.3.9) defined on(−∞,∞)that differ from each other on every open interval that containsx0 = 0. Can you identify the other two?

Example 2.3.7 From Example2.3.5, the initial value problem y0= 10

3 xy

2/5

, y(0) =−1 (2.3.12)

has a unique solution on some open interval that containsx0 = 0. Find a solution and determine the largest open interval(a, b)on which it’s unique.

Solution Letybe any solution of (2.3.12). Because of the initial conditiony(0) =−1and the continuity ofy, there’s an open intervalIthat containsx0= 0on whichyhas no zeros, and is consequently of the form (2.3.11). Settingx= 0andy=−1in (2.3.11) yieldsc=−1, so

y= (x2−1)5/3 (2.3.13)

forxinI. 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(−∞,∞). 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 = ( (x21)5/3, 1x1, 0, |x|>1. (Figure2.3.3). 1 −1 x y (0, −1)

Figure 2.3.3 Two solutions of (2.3.12) on(−∞,∞)

that coincide on(−1,1), but on no larger open interval

x y

(0,1)

Figure 2.3.4 The unique solution of (2.3.14)

Example 2.3.8 From Example2.3.5, the initial value problem y0 = 10

3 xy

2/5, y(0) = 1 (2.3.14)

has a unique solution on some open interval that containsx0 = 0. Find the solution and determine the largest open interval on which it’s unique.

Solution Letybe any solution of (2.3.14). Because of the initial conditiony(0) = 1and the continuity ofy, there’s an open intervalIthat containsx0= 0on whichyhas no zeros, and is consequently of the form (2.3.11). Settingx= 0andy= 1in (2.3.11) yieldsc= 1, so

y= (x2+ 1)5/3 (2.3.15)

forxinI. Therefore every solution of (2.3.14) differs from zero and is given by (2.3.15) on(−∞,∞); that is, (2.3.15) is the unique solution of (2.3.14) on(−∞,∞). Figure 2.3.4shows the graph of this solution.

2.3 Exercises

In Exercises1-13find all(x0, y0)for which Theorem2.3.1implies that the initial value problemy0 =

Section 2.3Existence and Uniqueness of Solutions of Nonlinear Equations 61 1. y0= x 2+y2 sinx 2. y 0= ex+y x2+y2 3. y0= tanxy 4. y0= x2+y2 lnxy 5. y0= (x2+y2)y1/3 6. y0 = 2xy 7. y0= ln(1 +x2+y2) 8. y0= 2x+ 3y x−4y 9. y0= (x2+y2)1/2 10. y0 =x(y21)2/3 11. y0= (x2+y2)2 12. y0= (x+y)1/2 13. y0= tany x−1

14. Apply Theorem2.3.1to the initial value problem

y0+p(x)y=q(x), y(x0) =y0

for a linear equation, and compare the conclusions that can be drawn from it to those that follow from Theorem 2.1.2.

15. (a) Verify that the function

y=

(

(x21)5/3, 1< x <1,

0, |x| ≥1, is a solution of the initial value problem

y0 =10 3 xy

2/5, y(0) =

−1

on(−∞,∞). HINT:You’ll need the definition

y0(x) = lim

x→x

y(x)−y(x)

x−x

to verify thatysatisfies the differential equation atx=±1.

(b) Verify that ifi= 0or1fori= 1,2anda,b >1, then the function

y=                    1(x2a2)5/3, −∞< x <a, 0, −a≤x≤ −1, (x21)5/3, 1< x <1, 0, 1≤x≤b, 2(x2b2)5/3, b < x <, is a solution of the initial value problem of(a)on(−∞,∞).

16. Use the ideas developed in Exercise15to find infinitely many solutions of the initial value problem y0 =y2/5, y(0) = 1

on(−∞,∞).

17. Consider the initial value problem y0 = 3x(y

−1)1/3, y(x0) =y0. (A)

(a) For what points(x0, y0)does Theorem2.3.1imply that (A) has a solution?

(b) For what points(x0, y0)does Theorem2.3.1imply that (A) has a unique solution on some open interval that containsx0?

18. Find nine solutions of the initial value problem y0 = 3x(y

−1)1/3, y(0) = 1

that are all defined on(−∞,∞)and differ from each other for values ofxin every open interval that containsx0= 0.

19. From Theorem2.3.1, the initial value problem y0 = 3x(y

−1)1/3, y(0) = 9

has a unique solution on an open interval that containsx0= 0. Find the solution and determine the largest open interval on which it’s unique.

20. (a) From Theorem2.3.1, the initial value problem y0= 3x(y

−1)1/3, y(3) =−7 (A)

has a unique solution on some open interval that containsx0= 3. Determine the largest such open interval, and find the solution on this interval.

(b) Find infinitely many solutions of (A), all defined on(−∞,∞). 21. Prove:

(a) If

f(x, y0) = 0, a < x < b, (A)

andx0is in(a, b), theny≡y0is a solution of

y0=f(x, y), y(x0) =y0

on(a, b).

(b) Iff andfy are continuous on an open rectangle that contains (x0, y0) and (A) holds, no solution ofy0=f(x, y)other thanyy0can equaly0at any point in(a, b).

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