2.4E: Existence and Uniqueness of Solutions of Nonlinear Equations (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
Q2.3.1
In Exercises 2.3.1-2.3.13, find all (x0,y0) for which Theorem 2.3.1 implies that the initial value problem y′=f(x,y), y(x0)=y0 has (a) a solution and (b) a unique solution on some open interval that contains x0.
1. y′=x2+y2sinx
2. y′=ex+yx2+y2
3. y′=tanxy
4. y′=x2+y2lnxy
5. y′=(x2+y2)y1/3
6. y′=2xy
7. y′=ln(1+x2+y2)
8. y′=2x+3yx−4y
9. y′=(x2+y2)1/2
10. y′=x(y2−1)2/3
11. y′=(x2+y2)2
12. y′=(x+y)1/2
13. y′=tanyx−1
Q2.3.2
14. Apply Theorem 2.3.1 to the initial value problem y′+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.
- Verify that the function y={(x2−1)5/3,−1<x<1,0,|x|≥1, is a solution of the initial value problem y′=103xy2/5,y(0)=−1 on (−∞,∞). HINT: You'll need the definition y′(¯x)=limx→¯xy(x)−y(¯x)x−¯x to verify that y satisfies the differential equation at ¯x=±1.
- Verify that if ϵi=0 or 1 for i=1, 2 and a, b>1, then the function y={ϵ1(x2−a2)5/3,−∞<x<−a,0,−a≤x≤−1,(x2−1)5/3,−1<x<1,0,1≤x≤b,ϵ2(x2−b2)5/3,b<x<∞, is a solution of the initial value problem of a on (−∞,∞).
16. Use the ideas developed in Exercise 2.3.15 to find infinitely many solutions of the initial value problem y′=y2/5,y(0)=1 on (−∞,∞).
17. Consider the initial value problem y′=3x(y−1)1/3,y(x0)=y0.
- For what points (x0,y0) does Theorem 2.3.1 imply that (A) has a solution?
- For what points (x0,y0) does Theorem 2.3.1 imply that (A) has a unique solution on some open interval that contains x0?
18. Find nine solutions of the initial value problem y′=3x(y−1)1/3,y(0)=1that are all defined on (−∞,∞) and differ from each other for values of x in every open interval that contains x0=0.
19. From Theorem 2.3.1, the initial value problem y′=3x(y−1)1/3,y(0)=9 has a unique solution on an open interval that contains x0=0. Find the solution and determine the largest open interval on which it is unique.
20.
- From Theorem 2.3.1, the initial value problem y′=3x(y−1)1/3,y(3)=−7 has a unique solution on some open interval that contains x0=3. Determine the largest such open interval, and find the solution on this interval.
- Find infinitely many solutions of (A), all defined on (−∞,∞).
21. Prove:
- If f(x,y0)=0,a<x<b, andx0 is in (a,b), then y≡y0 is a solution of y′=f(x,y),y(x0)=y0 on (a,b).
- If f and fy are continuous on an open rectangle that contains (x0,y0) and (A) holds, no solution of y′=f(x,y) other than y≡y0 can equal y0 at any point in (a,b).