Download Differences Between Linear and Nonlinear Equation Theorem 1: If

p1
Differences Between Linear and Nonlinear Equation
Theorem 1:
If the function p and g are continuous on an open interval
I : α < t < β containing the point t = t0 , then there exists a
unique function y = φ(t) that satisfies the differential equation
y 0 + p(t)y = g(t)
for each t ∈ I, and that also satisfies the initial condition
y(t0 ) = y0
where y0 is an arbitrary prescribed initial value.
p2
Differences Between Linear and Nonlinear Equation
Theorem 2:(Containing nonlinear differential equation)
Let the functions f and ∂f
∂y be continuous in some rectangle
α < t < β, γ < y < δ containing the point (t0 , y0 ). Then, in some
interval t0 − h < t < t0 + h contained in α < t < β, there is a
unique solution y = φ(t) of the initial value problem
y 0 = f (t, y),
y(t0 ) = y0
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Example. Use Theorem 1 to find an interval in which the initial
value problem
ty 0 + 2y = 4t2 , y(1) = 2
has a unique solution.
Sol. Rewriting ty 0 + 2y = 4t2 in the form
2
y 0 + y = 4t.
t
Thus, p(t) = 2t is continuous only for t < 0 or t > 0, and g(t) = 4t
is continuous for all t. Since t0 = 1 ∈ (0, ∞); consequently, the
initial value problem
ty 0 + 2y = 4t2 ,
y(1) = 2
p4: continuation p3
has a unique solution on the interval 0 < t < ∞ since
2
ty 0 + 2y = 4t2 ⇔ y 0 + y = 4t,
t
for t > 0.
Quiz:
1. Determine(without solving the problem) an interval in which
the solution of the initial value problem
(4 − t2 )y 0 + 2ty = 3t2 ,
is certain to exist.
y(1) = −3
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Example. Use Theorem 2 to find an interval in which the initial
value problem
3x2 + 4x + 2
dy
=
,
dx
2(y − 1)
has a unique solution.
2 +4x+2
Sol. f (x, y) = 3x2(y−1)
and
∂f
∂y (x, y)
y(0) = −1
2
+4x+2
= − 3x2(y−1)
2 . Thus each of
f and ∂f
∂y functions are continuous everywhere except on the line
y = 1.
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Consequently, the initial value problem
3x2 + 4x + 2
dy
=
,
dx
2(y − 1)
y(0) = −1
has a unique solution in some interval about x = 0.
Note: Though f and ∂f
∂y are continuous on
(−∞, ∞) × (−∞, 1) = {(x, y) | − ∞ < x < ∞, −∞ < y < 1}
containing a point (0, −1). This does not necessarily mean that
the solution exists for all x.
Exercise. Find the unique solution, and verify that the initial value
problem has a unique solution only for x > −2.
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Example. Consider the initial value problem
1
y0 = y 3 ,
y(0) = 0
for t ≥ 0. Apply Theorem 2 to this initial value problem and then
solve the problem.
1 −2/3
Sol. f (x, y) = y 1/3 and ∂f
= 3y12/3 . Thus f is
∂y = 3 y
continuous everywhere, ∂f
∂y does not exist when y = 0 and not
continuous at y = 0. Hence there is not a rectangle R containing
(0, 0) such that f and ∂f
∂y are continuous on R. So Theorem 2
does not apply to this problem.
p8: continuation p7
y 0 = y 1/3 is rewritten as y −1/3 dy
dt = 1 (separable)
3
y −1/3 dy = 1 · dt ⇒ y 2/3 = t + C
2
and
3/2
2
or y = − (t + C)
3
3/2
2
t=0⇒y=
(0 + C)
=0⇒C=0
3
2
y=
(t + C)
3
3/2
so y = φ1 (t) = ( 23 t)3/2 , t ≥ 0 is a solution. On the other hand,
the function y = φ2 (t) = −( 23 t)3/2 , t ≥ 0 is also a solution of the
initial value problem.
p9: continuation p8
Moreover,
y = ψ(t) = 0,
t≥0
is yet another solution. For an arbitrary positive t0 , the function
(
0,
if 0 ≤ t < t0
y = X(t) =
2
3/2
± 3 (t − t0 )
, if t ≥ t0
are continuous, differentiable, and are solutions of the initial value
problem
1
y 0 = y 3 , y(0) = 0.
Hence this problem has an infinite family of solutions.
p10: continuation p9
Remark. The continuity of f does assure the existence of
solutions, but not their uniqueness.
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Quiz
2. Solve the initial value problem
y0 =
t2
,
y(1 + t3 )
y(0) = y0
and determine how the interval in which the solution exists
depends on the initial value y0 .
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Compare the first linear equation y 0 + p(t)y = g(t) with nonlinear
differential equation, such as y 0 = y 2 :
Interval of Definition
Remark. The solution of a linear equation
y 0 + p(t)y = g(t)
subject to the initial condition y(t0 ) = y0 , exists throughout any
interval about t = t0 in which the function p and g are continuous.
Thus, vertical asymptotes or other discontinuous in the solution
can occur only at points of discontinuity of p or g.
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On the other hand, for a nonlinear initial value problem satisfying
the hypotheses of Theorem 2, the interval in which a solution
exists may be difficult to determine.
Example. Solve the initial value problem
y0 = y2,
y(0) = 1
and determine the interval in which the solution exists.
Sol. f (t, y) = y 2 and ∂f
∂y (t, y) = 2y are continuous everywhere.
So, by Theorem 2, there exists a h > 0 such that the initial value
problem has a unique solution in interval for t in (0 − h, 0 + h).
p14: continuation p13
y0 = y2 ⇒
dy
= y 2 ⇒ y −2 dy = dt
dt
(separable)
and
Z
y
−2
t=0⇒y=
Z
dy =
1
0+C
dt ⇒ −y −1 = t + C ⇒ y = −
1
t+C
= 1 ⇒ C = −1, thus
y=
1
1−t
is the solution of the given initial value problem. Clearly,
1
lim 1−t
= ∞, the solution exists only in the interval
t→1−
−∞ < t < 1. However, there is no indication from the differential
equation itself.
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General Solution
1
is solutions of the nonlinear equation y 0 = y 2
Example. y = t+C
for arbitrary number C. But y(t) = 0 is also a solution, implies
1
y = t+C
is not general solution since there is no a vlue of C such
1
that y(t) = t+C
= 0 for all t.
Implicit Solution
The situation for nonlinear equations is much less satisfactory(令人
滿意的). Usually, the best that we can hope for is to find an
equation
F (t, y) = 0
involving t and y that is satisfied by the solution y = φ(t).
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Graphical or Numerical Construction of Integral Curves
Because of the difficulty in obtaining exact analytical solutions of
nonlinear differential equations, methods that yield approximate
solutions or other qualitative information about solutions are of
correspondingly greater importance.
Summary. The linear equation y 0 + p(t)y = g(t) has several nice
properties:
1. Assuming that the coefficients are continuous, there is a
general solution, containing an arbitrary constant, that
includes all solutions of the differential equation. A particular
solution that satisfies a given initial condition can be picked
out by choosing the proper value for the arbitrary constant.
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R
u(t)g(t)dt
2. There is an expression for the solution, y =
or
u(t)
hR
i
R
t
1
p(t)dt .
y = u(t)
t0 u(s)g(s)ds + c where u(t) = e
Moreover, although it involves two integrations, the expression
is an explicit one for the solution y = φ(t) rather than an
equation that defines φ implicitly.
3. The possible points of discontinuity, or singularities, of the
solution can be identified (without solving the problem)
merely by finding the points of discontinuity of the
coefficients. Thus, if the coefficients are continuous for all t,
then the soloution also exists and is continuous for all t.
None of these statements is true, in general, of nonlinear equations.
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Exercise.
1. It is possible to solve a nonlinear equation by making a change
of the dependent variable that converts it into a linear
equation. The most important such equations has the form
y 0 + p(t)y = q(t)y n
and is called a Bernoulli equation after Jakob Bernoulli.
(a) Solve Bernoulli’s equation when n = 0, when n = 1
(b) Show that if n 6= 0, 1, then the substitution v = y n−1 reduces
Bernoulli’s equation to a linear equation.
(Note: This method of solution was found by Leibniz in 1696 )
p19: continuation p18
2. Solve the following equations by using the substitution
(a) y 0 = ry − ky 2 , r > 0 and k > 0
(This equation is important in population dynamics) Ans:
r
y = k+cre
−rt
(b) y 0 = y − σy 3 , > 0 and σ > 0.
(This equation occurs in the study of the stability of fluid flow)
h
i1/2
Ans: y = ± σ+ce −2t