p0
教學方式:
1. 課程內容講解
2. 隨堂練習 (可互相討論與提問)
3. 自主學習包含部分課程內容與演練題目 (課堂會給)
學期評量:
1. 課堂參與: 10%
2. 隨堂練習: 30% (每天)
3. 作業: 20% (每週一題)
4. 期中考: 15% / 期末考: 25%
參考教材:
William E. Boyce and Richard C. DiPrima
Elementary Differential Equations and Boundary Value Problems /
8e
p1
The behavior of the natural world, such as
(1) the motion of fluids (流體的運動)
(2) the flow of current(電流的流動) in electric circuits (電路)
(3) the dissipation(耗散) of heat in solid objects
(4) the propagation(傳播) and detection(檢測) of seismic(地震)
waves
(5) the increase or decrease of populations
Q: How to be expressed for these behaviors?
A: Many of the principles, or laws, underlying the behavior of the
natural world are statements or relations involving rates at
which things happen. When expressed in mathematical terms(數學
術語), the relations are equations and the rates are derivatives.
p2
Example: Suppose that an object is falling in the atmosphere(大
氣層) near sea level. Formulate(規劃) a differential equation that
describes the motion.
Argument :
t : time (seconds)
v : velocity (meters/second)
The velocity will presumably change with time,
we think of v as a function of t.
Newton’s second law:
F = ma
m : the mass of the object (kilograms)
a : acceleration (meters/second2 )
F : the net force (newtons)
p3: Quiz (隨堂練習)
Recall(definition of derivative ) : the derivative of v, written as
dv
dt ,
dv
v(t + h) − v(t)
= lim
.
h→0
dt
h
Quiz:
1. Let k and r be an constant number, functions f (x) and g(x)
be differentiable. If f 0 (x) and g 0 (x) are expressed as the
derivatives of f (x) and g(x), repsectively. Then filling right
answer in the following blank:
d
d r
k=
x =
(1) dx
(2) dx
d
d x
(3) dx ln x =
(4) dx e =
d
d
(5) dx [f (x) + g(x)] =
(6) dx
[f (x)g(x)] =
d f (x)
d
(7) dx g(x) =
(8) dx f (g(x)) =
p4 : continuation p2 (延續 p2)
Since a =
dv
dt ,
F = ma = m
dv
dt
and F = mg − γv;
(a) Gravity : mg
g ≈ 9.8m/sec2 is the acceleration due to gravity.
(b) Air resistance(阻力) or drag(摩擦力): γv
It is often assumed that the drag is proportional to velocity,
the drag force has the magnitude
γv
where γ is a constant, called the drag coefficient.
p5: continuation p4 (延續 p4)
Therefore,
dv
= mg − γv
dt
is a mathematical model of an object falling in the atmosphere
near sea level.
Example. Let m = 10 kg and γ = 2 kg/sec. Then
m·
10 ·
dv
dv
v
= 10 · 9.8 − 2 · v ⇔
= 9.8 − .
dt
dt
5
Investigate the behavior of solutions of dv
dt = 9.8 −
actually finding the solutions in question.
v
5
without
p6: continuation p5 (延續 p5)
40
Argument:If v = 40, then dv
dt = 9.8 − 5 = 1.8. This means that
the slope of a solution v = v(t) has the value 1.8 at any point
where v = 40. We can display this information graphically in the
tv-plane by drawing short line segment or arrows.
direction field or slope field
p7: continuation p6 (延續 p6)
In fact, the constant function v(t) = 49 is a solution of
dv
v
dt = 9.8 − 5 .
Verify: the left side of the equation:
dv
d
= (49) = 0
dt
dt
and the right side of the equation:
v
49
= 9.8 −
=0
5
5
Thus, v(t) = 49 satisfy the differential equation
9.8 −
dv
v
= 9.8 − .
dt
5
Because it does not change with time, the solution v(t) = 49 is
called an equilibrium solution.
p8: quiz
Quiz:
2. Which of the following functions are solutions of the
differential equation y 00 + y = sin x?
(a) y = sin x
(b) y = cos x
1
(c) y = 2 x sin x (d) y = − 21 x cos x
3. For what values of r does the function y = erx satisfy the
differential equation 2y 00 + y 0 − y = 0?
p9:
Direction Fields :
Direction fields are valuable tools in studying the solutions of
differential equation of the form
dy
= f (t, y)
dt
where f is given function of the two variables t and y, sometimes
referred to as the rate function.
Example. Field Mice(田鼠) and owls(貓頭鷹):
Consider a population of field mice who inhabit(居住) a certain
rural(田園) area. In the absence of predators(捕食者) we assume
that the mouse population increases at a rate proportional to the
current population, that is,
p10: continuation p9
dp
= rp
dt
t =time(month), p(t) = mouse population where the proportionally
factor r is called the rate constant or growth rate. Let r = 0.5 and
suppose that several owls live in the same neighborhood and that
they kill 15 field mice per day, per month 15 × 30 = 450.
Let pe(t) represent no predators population and p(t) = pe(t) at time
t. Then
p(t + ∆t) − p(t) pe(t + ∆t) − 450 · ∆t − pe(t)
=
∆t
∆t
pe(t + ∆t) − pe(t)
=
− 450
∆t
p11: continuation p10
dp
p(t + ∆t) − p(t)
pe(t + ∆t) − pe(t)
= lim
= lim
− 450
∆t→0
dt ∆t→0
∆t
∆t
de
p
p(t) − 450 = 0.5p(t) − 450
= (t) − 450 = 0.5e
dt
Therefore,
d
(1) P (t) = 900 is a equilibrium solution since dp
dt = dt (900) = 0
and 0.5p(t) − 450 = 0.5(900) − 450 = 0.
(2) When p(t0 ) > 900, dp
= 0.5p(t0 ) − 450 > 0 implies p(t)
dt t=t0
increase for t > t0 .(Why?)
(3) When p(t0 ) < 900, dp
= 0.5p(t0 ) − 450 < 0 implies p(t)
dt t=t0
decrease for t > t0 .(Why?)
p12: continuation p11
Direction field
Field mice and owls : A more general version
dp
= rp − k
dt
where the growth rate r and the predation rate(捕食率) k are
unspecified.
p13: continuation p12
In this case, the equilibrium solution of
p(t) =
dp
dt
= rp − k is
k
.
r
Example. Find a function p(t) (not constant function) to satisfy
the differential equation
dp
= 0.5p − 450.
dt
Review: Anti-derivative (Indefinite Integral)
d
dx (y)
0
−−
−−
−−
−
−−
−−
−−
−−
−−
−→
y ←
R−
−− y
0
y dx
y 0 is also represented as
dy
dx .
p14:
Such as,
dy
dx
R
= k ⇒ kdx = kx + C. Thus,
y = kx or y = kx − 1 or · · ·
R
r+1
dy
(2) dx
= xr ⇒ xr dx = xr+1 + C. Thus
r+1
r+1
y = xr+1 , or y = xr+1 + 1, or · · ·
R
dy
(3) dx
= x1 ⇒ x1 dx = ln |x| + C. Thus
y = ln |x| or y = ln |x| + 2 or · · ·
R
dy
(4) dx
= ex ⇒ ex dx = ex + C. Thus
y = ex or y = ex + π, or · · ·
(1)
where C is an arbitrary constant
p15: continuation p13
Sol.
dp
dt
= 0.5p − 450 =
dp
dt
p−900
2
⇒
dp
dt
p−900
= 12 ,
1
dp
dt (dp =
dt)
p − 900
2
dt
Z
Z
1
1
⇒
dp =
dt
p − 900
2
1
⇒ ln |p − 900| = t + C1 (C1 arbitrary constant)
2
t
⇒|p − 900| = e 2 +C1 ⇒ |p − 900| = eC1 et/2 ⇒ p − 900 = ±eC1 et/2
Z
Z
dt =
⇒p = 900 + Cet/2
where C = ±eC1 is an arbitrary nonzero constant(since eC1 > 0).
p16: continuation p15
Since p(t) = 900 is also solution of the differential equation
dp
dt = 0.5p − 450 and that it is contained in the expression
p = 900 + Cet/2 if we allow C to take the value zero. That is,
solutions of dp
dt = 0.5p − 450 are
p = 900 + Cet/2
where C is an arbitrary constant. If p(0) = 850, then
p(0) = 900 + Ce0/2 ⇒ 850 = 900 + C ⇒ C = −50
We obtain the desired solution
p = 900 − 50et/2
to satisfy an initial condition : p(0) = 850.
p17
Note: Both m ·
the general form
dv
dt
= mg − γv and
dp
dt
dy
= ay − b
dt
where a and b are given constants.
Initial value problem :
dy
= ay − b
dt
and the initial condition
y(0) = y0
where y0 is an arbitrary initial value.
= rp − k equations are of
p18
Quiz:
4. Solve the initial value problem
dy
= −2y + 10,
dt
y(0) = y0
and plot the solutions for several values of
y0 (y0 = 3, 4, 5, 6, 7). Then describe in a few words how the
solution resemble(類似), and differ(不同) from, each other.
p19
Classification of Differential Equations:
• Ordinary differential equation (ODE)
L
d2 Q(t)
1
dQ(t)
+ Q(t) = E(t)
+R
2
dt
dt
C
• Partial differential equation (PDE)
the heat conduction(傳導) equation
α2
∂ 2 u(x, t)
∂u(x, t)
=
2
∂x
∂t
the wave equation
a2
∂ 2 u(x, t)
∂ 2 u(x, t)
=
∂x2
∂t2
p20
• Systems of Differential Equations:
For example,
(
dx
= ax − αxy
dt
dy
= −cy − γxy
dt
where x(t) and y(t) are the respective populations of the
prey(獵物) and predator(捕食者) species.
• Order
y=u(t)
F [t, u(t), u0 (t), · · · , u(n) (t)] = 0 =⇒ F (t, y, y 0 , · · · , y (n) ) = 0
is an ODE of the nth order. For example,
y 000 + 2et y 00 + yy 0 = t4
is a third order ODE for y = u(t).
p21
• Linear and Nonlinear Equations
The ODE
F (t, y, y 0 , y 00 , · · · , y (n) ) = 0
is said to be linear if F is a linear function of the variables
y, y 0 , y 00 , · · · , y (n) ; a similar definition applies to partial
differential equations. The general linear ODE of order n is
an (t)y (n) + an−1 (t)y (n−1) + · · · + a1 (t)y 0 + a0 (t)y = g(t).
An equation that is not a linear equation is a nonlinear
equation. For example,
g
d2 θ
+ sin θ = 0
dt2
L
p22
A solution of the ODE
y (n) = f (t, y, y 0 , · · · , y (n−1) )
on the interval α < t < β is a function φ such that φ0 , φ00 , · · · , φ(n)
exists and satisfy
φ(n) (t) = f (t, φ(t), φ0 (t), · · · , φ(n−1) (t))
for every t in α < t < β.
p23
Exercise. Consider an electric circuit containing a capcitor(電容),
resistor(電阻), and battery
The charge Q(t) on the capacitor satisfies the equation
R
dQ Q
+
=V
dt
C
p24 : continuation p23
where R is the resistance, C is the capacitance, and V is the
constant voltage supplied by the battery.
(a) If Q(0) = 0, find Q(t) at any time t, and sketch the graph of
Q versus t. (Ans: Q(t) = CV (1 − e−t/RC ) )
(b) Find the limiting value QL that Q(t) approaches after a long
time. (Ans: QL = CV )
(c) Suppose that Q(t1 ) = QL and that the battery is removed
from the circuit at t = t1 . Find Q(t) for t > t1 , and sketch its
graph. (Ans: Q(t) = CV exp[−(t − t1 )/RC] )