Hence the ODE of the orthogonal trajectories is
. By separation, .
Integration and taking exponents gives
ln .
This shows that the ratio has substantial influence on the form of the trajectories.
For the given curves are circles, and we obtain straight lines as trajectories.
gives quadratic parabolas. For higher integer values of we obtain parabolas of higher order. Intuitively, the “flatter” the ellipses are, the more rapidly the trajectories must increase to have orthogonality.
Note that our discussion also covers families of parabolas; simply interchange the roles of the curves and their trajectories.
Note further that, in the light of the present answer, our example in the text turns out to be typical.
16. . Since c is just an additive constant, the statement about the curves follows; these curves are obtained from any one of them by translation in the y-direction. Similarly for the OTs, whose ODE is with the function on the right independent of .
SECTION 1.7. Existence and Uniqueness of Solutions for Initial Value Problems, page 38
Purpose. To give the student at least some impression of the theory that would occupy a central position in a more theoretical course on a higher level.
Short Courses. This section can be omitted.
Comment on Iteration Methods
Iteration methods were used rather early in history, but it was Picard who made them popular. Proofs of the theorems in this section (given in books of higher level, e.g., [A11]) are based on the Picard iteration (see CAS Project 6).
Iterations are well suited for the computer because of their modest storage demand and usually short programs in which the same loop or loops are used many times, with different data. Because integration is generally not difficult for a CAS, Picard’s method has gained some popularity during the past few decades.
y~
y~
r
⫽ ⫺1>f(x)y⫽ 兰f(x) dx⫹ c
a2>b2 a2>ba22⫽ 2⫽ b2 a2>b2
ƒy~ ƒ ⫽ a2
b2 ln ƒ x ƒ ⫹ c**, y~⫽ c*xa2>b2 dy~
y~ ⫽ a2 b2
dx y~
r
⫽ a x2y~
b2x 2x a2⫹ 2yy
r
b2 ⫽ 0, y
r
⫽ ⫺2x>a2
2y>b2 ⫽ ⫺b2x a2y y~2⫹ (x ⫺ c*)2⫽ c*2⫺ 1
y~2⫹ x2⫹ 1 ⫽ 2c*x y~2
x ⫹ x ⫹1 x⫽ 2c*
Example 1 is simple, involving only , and is typical inasmuch as it illustrates that the actual interval of existence is much larger than the interval guaranteed by Existence Theorem 1.
Example 2 shows that IVPs violating uniqueness can be constructed relatively easily.
Lipschitz and Hölder conditions play a basic role in the theory of PDEs on a level substantially higher than that of our Chap.12.
SOLUTIONS TO PROBLEM SET 1.7, page 42
2. The initial condition is given at the point . The coefficient of is 0 at that point, so from the ODE we already see that something is likely to go wrong. Separating variables, integrating, and taking exponents gives
.
This last expression is the general solution. It shows that for any c. Hence the initial condition cannot be satisfied. This does not contradict the theorems because we first have to write the ODE in standard form:
.
This shows that f is not defined when (to which the initial condition refers).
4. For we still get no solution, violating the existence as in Prob. 2. For we obtain infinitely many solutions, because c remains unspecified. Thus in this case the uniqueness is violated. Neither of the two theorems is violated in either case.
6. CAS Project. (b)
(c)
(d) . It approximates . General solution .
(e) would be a good candidate to begin with. Perhaps you write the initial choice as ; then corresponds to the choice in the text, and you see how the expressions in a are involved in the approximations. The conjecture is true for any choice of a constant (or even of a continuous function of x).
It was mentioned in footnote 10 that Picard used his iteration for proving his existence and uniqueness theorems. Since the integrations involved in the method can be handled on the computer quite efficiently, the method has gained in importance in numerics.
8. The student should get an understanding of the “intermediate” position of a Lipschitz condition between continuity and (partial) differentiability.
The student should also realize that the linear equation is basically simpler than the nonlinear one. The calculation is straightforward because we have
f(x, y)⫽ r(x) ⫺ p(x)y a⫽ 0
y0⫹ a y
r
⫽ yy⫽ (x ⫹ c)2 y⫽ 0
y⫽ (x ⫺ 1)2, y⫽ 0 y(x)⫽ 1
1⫺ 2x ⫽ 1 ⫹ 2x ⫹ 4x2⫹ 8x3⫹ . . . y0⫽ 1, y1⫽ 1 ⫹ 2x, y2⫽ 1 ⫹ 2x ⫹ 4x2⫹ 8x3
3 , . . . yn⫽x2
2! ⫹ x3
3! ⫹ . . . ⫹ xnⴙ1
(n⫹ 1)!, y⫽ ex⫺ x ⫺ 1
k⫽ 0 k⫽ 0
x⫽ 2 y
r
⫽ f(x, y)⫽ 2yx⫺ 2
y(1)⫽ 1 y(2)⫽ 0
dy
y ⫽ 2 dx
x⫺ 2, ln ƒ y ƒ ⫽ 2 ln ƒ x ⫺ 2 ƒ ⫹ c*, y⫽ c(x ⫺ 2)2 y
r
x⫽ 2 y⫽ tan x
and this implies that
(A) .
This becomes a Lipschitz condition if we note that the continuity of for implies that is bounded, say for all these x. Taking absolute values on both sides of (A) now gives
. 10. By separation and integration,
. Taking exponents gives the general solution
. From this we can see the answers:
No solution if .
A unique solution if equals any .
Infinitely many solutions if .
This does not contradict the theorems because
is not defined when .
SOLUTIONS TO CHAP. 1 REVIEW QUESTIONS AND PROBLEMS, page 43 12. . Note that the solution curves are congruent.
14. . The figure also shows the solution curves through .
y(⫺1) ⫽ 1], (1, 0.1), (1, 1), and (1, 2)
(⫺1, 1) [thus, y⫽ x2⫹ cx
y⫽ tanh (x ⫹ c)
x⫽ 0 or 1
f(x, y)⫽ 2x⫺ 1 x2⫺ x y(0)⫽ 0 or y(1) ⫽ 0
y0 and x0⫽ 0 or x0⫽ 1 y(x0)
y(0)⫽ k ⫽ 0 or y(1) ⫽ k ⫽ 0 y⫽ c(x2⫺ x) dy
y ⫽2x⫺ 1
x2⫺ x dx, ln ƒ y ƒ ⫽ ln ƒ x2⫺ x ƒ ⫹ c*
ƒf(x, y2)⫺ f(x, y1) ƒ ⬉ M ƒ y2⫺ y1ƒ ƒp(x) ƒ ⬉ M p(x)
ƒx⫺ x0ƒ ⬉ a p(x)
f(x, y2)⫺ f(x, y1)⫽ ⫺p(x)(y2⫺ y1)
1
–1
–2 2
1
–1 0
–2 2
y
x y(x)
Problem 14. Direction field of xyr⫽y⫹x˛
2
16. Solution . y⫽ 1>(1 ⫹ 4eⴚx) Computations:
Error 0 0.2000 0 0.1 0.2160 0.0005 0.2 0.2329 0.0010 0.3 0.2509 0.0015 0.4 0.2696 0.0021 0.5 0.2893 0.0026 0.6 0.3098 0.0032 0.7 0.3312 0.0037 0.8 0.3534 0.0041 0.9 0.3762 0.0046 1.0 0.3997 0.0048
yn
xn
18. .
20. This Bernoulli equation (a Verhulst equation if ) can be reduced to linear form, as shown in Example 4 of Sec. 1.5 (except for the notation). The general solution is (see (12) in Sec. 1.5)
22. The general solution of this linear differential equation is obtained as explained in Sec. 1.6,
From this and the initial condition we have . Answer:
24. To solve this Bernoulli equation we set . Then , Substitution into the given ODE gives
. We now multiply by , obtaining
. General solution: u⫽ cex⫹ 2. u
r
⫺ u ⫽ ⫺2⫺2u3>2
⫺12uⴚ3>2u
r
⫹12uⴚ1>2⫽ uⴚ3>2y
r
⫽ ⫺12uⴚ3>2 ur
.y⫽ uⴚ1>2 u⫽ yⴚ2
y⫽ (x ⫺ 4.3)eⴚ2x2.
c⫽ ⫺4.3 y(0)⫽ ⫺4.3
y⫽ eⴚ2x2a冮e2x2eⴚ2x2dx⫹ cb ⫽ 1x ⫹ c2eⴚ2x2
y⫽ 1
ceⴚax⫺ b>a. b ⬍ 0 y⫽ ce0.4x⫺ 25 cos x ⫺ 10 sin x
Problem 16. Solution curve and computed values
0.4
0.35
0.3
0.25
0.2
0.2
0 0.4 0.6 0.8 1 x
y
Hence
From this and the initial condition y(0) we get Answer:
26. Theorem 1 in Sec. 1.4 gives the integrating factor . We thus obtain the exact equation
By inspection or systematically by integration (as explained in Sec. 1.4), we obtain
; thus, .
From this and the initial condition we get . Answer:
28. We proceed as in Sec. 1.3. The time rate of change equals the inflow of salt minus the outflow per minute.
y
r
⫽ 20 ⫺ 20500 y.
y
r
⫽ dy>dtcosh y⫽13x.
1 3 #
1⫽ c 1
x cosh y⫽ c d a1
x cosh yb ⫽ 0 1
x sinh y dy⫺ 1
x2 cosh y dx⫽ 0.
F⫽ 1>x2
y⫽ 1
22 ⫹ 7ex. c⫽ 7.
1
⫽3
y⫽ u⫺1>2⫽ 1 2cex⫹ 2.
The initial condition is . This gives the particular solution
The limiting value is 500 lb; are 475 lb, so that we get the condition
from which we can determine
so it will take a little over an hour.
30. By Newton’s law of cooling, since the surrounding temperature is and the initial temperature of the metal is , we first obtain
k can be determined from the condition that ; that is,
so that . With this value of k we can now find the time at which the metal has the temperature .
Answer: The temperature of the metal has practically reached that of the boiling water after 13.4 min.
99.9⫽ 100 ⫺ 80eⴚ0.5t, 0.1⫽ 80eⴚ0.5t, t⫽ ln 800
0.5 ⫽ 13.4.
99.9°C k⫽ ln (48.5>80) ⫽ ⫺0.500
T(1)⫽ 100 ⫺ 80ek⫽ 51.5, T(1)⫽ 51.5 T(t)⫽ 100 ⫺ 80ekt.
T(0)⫽ 20 100°C
t⫽ 25 ln 420
25 ⫽ 70.5 [min];
500⫺ 420eⴚ0.04t⫽ 475, 95%
y⫽ 500 ⫺ 420e⫺0.04t. y(0)⫽ 80