Handbook of Differential Equations: Ordinary Differential Equations, Volume 1
Math 2C03 - Differential...
Transcript of Math 2C03 - Differential...
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Math 2C03 - Differential EquationsSlides shown in class - Winter 2015
March 25 - April 8, . . . 2015
Series Solutions
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Review of Power Series
Definition
(a) A power series in x − x0 (or “centred at x0”) has the form:
∞∑n=0
cn (x − x0)n = c0 + c1 (x − x0) + c2 (x − x0)
2 + . . .
(b) A power series converges at x if a finite limit.
limN→∞
N∑n=0
cn (x − x0)n exists.
(c) A power series is said to converge absolutely at x if
∞∑n=0
|cn| |x − x0|n converges.
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Radius of Convergence
Theorem (Radius of convergence, R)
Given a power series∑∞
n=0 cn (x − x0)n centred at x0, there
exists a number R with 0 ≤ R ≤ ∞, called the radius ofconvergence of the series such that
the series converges absolutely if |x − x0| < R.the series diverges if |x − x0| > R.
|x−x0| < R ⇔ x0−R < x < x0+R ⇔ x ∈ (x0−R, x0+R)
A power series centred at x0 always converges at x = x0.Interval of convergence can be:(x0 − R, x0 + R), [x0 − R, x0 + R), (x0 − R, x0 + R],[x0 − R, x0 + R], (−∞,∞), or [x0, x0].Convergence at end points must be tested separately.“Absolute Convergence”⇒ “Convergence”.
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Test for Convergence
Theorem (The ratio test)
Consider the power series∑∞
n=0 cn (x − x0)n and assume that
there exists an integer N ≥ 0 such that cn 6= 0 for all n ≥ N.
If limn→∞
∣∣∣∣cn+1
cn
∣∣∣∣ = L, then R =1L.
If L = 0, then R =∞.
If L =∞, then R = 0.
It the limit does not exist other methods, e.g. Root Test can beused.
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Examples: Maclaurin Series
11−x =
∑∞n=0 xn = 1 + x + x2 + x3 + . . . if |x | < 1 : R = 1.
(Geometric series.)ex =
∑∞n=0
xn
n! = 1 + x + x2
2! +x3
3! + . . . : R =∞.sin(x) =
∑∞n=0 (−1)n x2n+1
(2n+1)! = x − x3
3! +x5
5! . . . : R =∞cos(x) =
∑∞n=0 (−1)n x2n
(2n)! = 1− x2
2! +x4
4! . . . : R =∞.
Since a power series converges absolutely, within its openinterval of convergence, (x0 − R, x0 + R)
it represents a continuous function,has derivatives of all orders,can be differentiated term-by-term and integratedterm-by-term & the resulting series has radius ofconvergence at least R,its terms can be rearranged.
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Theorem
If f (x) =∑∞
n=0 cn (x − x0)n is a power series in x − x0 with a
radius of convergence R > 0, then,(a) f (x) is infinitely differentiable for |x − x0| < R.Within its open interval of convergence:(b) a power series can be differentiated term by term
f ′(x) =∞∑
n=1
n cn (x − x0)n−1, |x − x0| < R;
(c) a power series can be integrated term by term∫ x
x0
f (t)dt =∞∑
n=0
cn(x − x0)
n+1
n + 1, |x − x0| < R;
(d)
cn =f (n)(x0)
n!, n ≥ 0.
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More Properties of Power Series:within the Radius of Convergence: can add, subtract, multiply & divide
If f (x) =∑∞
n=0 an (x − x0)n is a power series in x − x0 with a
radius of convergence Rf > 0, &g(x) =
∑∞n=0 bn (x − x0)
n is a power series in x − x0 with aradius of convergence Rg > 0, thenf (x)± g(x) =
∑∞n=0 (an ± bn) (x − x0)
n
with radius of convergence at least min(Rf ,Rg).
f (x)g(x) =∑∞
n=0 an (x − x0)n∑∞
n=0 bn (x − x0)n =
(a0+a1(x−x0)+a2(x−x0)2+. . .)(b0+b1(x−x0)+b2(x−x0)
2+. . .)= a0b0 + (a0b1 + a1b0)(x − x0) + (a0b2 + a1b1 + a2b0)(x − x0)
2
+(. . . )(x − x0)3 + . . .
with radius of convergence at least min(Rf ,Rg).
Can also do f (x)g(x) provided g(x0) 6= 0, BUT here radius of
convergence can be smaller.
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Key Observation
Theorem
(a) If∞∑
n=0
an (x − x0)n =
∞∑n=0
bn (x − x0)n
for all x in an open interval that contains x0,then an = bn for all n = 0,1,2, . . ..
(b) If∞∑
n=0
an (x − x0)n = 0
for all x in an open interval that contains x0,then an = 0 for all n = 0,1,2, . . ..
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More Manipulations of Power Series
Rearranging into even and odd terms:
∞∑n=0
cn (x − x0)n =
∞∑n=0
c2n (x − x0)2n +
∞∑n=0
c2n+1 (x − x0)2n+1
Rearranging into every 3rd term:∑∞n=0 cn (x − x0)
n
=∞∑
n=0
c3n (x−x0)3n+
∞∑n=0
c3n+1 (x−x0)3n+1+
∞∑n=0
c3n+2 (x−x0)3n+2
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Shifting the Index
Shifting the dummy variable:
∞∑n=0
cn (x − x0)n =
∞∑n=k
cn−k (x − x0)n−k
for any integer k .
∞∑n=k
cn (x − x0)n =
∞∑n=0
cn+k (x − x0)n+k
Omitting zero terms:∑∞n=0 n(n − 1)cn (x − x0)
n−2
=∑∞
n=1 n(n − 1)cn (x − x0)n−2
=∑∞
n=2 n(n − 1)cn (x − x0)n−2
6=∑∞
n=3 n(n − 1)cn (x − x0)n−2
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Definition (Analytic Functions)
A function f (x) is analytic at x0 if, for some R > 0,
f (x) =∞∑
n=0
cn (x − x0)n, |x − x0| < R,
i.e. f (x) admits a power series expansion in x − x0 near x = x0with a positive radius of convergence.
sin(x), cos(x),& ex are all analytic.A polynomial P(x) = a0 + a1 x + · · ·+ am xm is analytic atany point x0 ∈ R.
A rational function f (x) = P(x)Q(x) (where P(x),Q(x) are
polynomials with no common factor), is analytic at any x0where Q(x0) 6= 0. The corresponding power series hasradius of convergence equal to the distance from x0 to thenearest zero of Q(x) in the complex plane.
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Ordinary Points and Singular Points2nd Order Homogeneous linear ODE
Definition
Consider a 2nd -order homogeneous linear DE
a2(x) y ′′ + a1(x) y ′ + a0(x) y = 0, (∗)
which we can write in standard form as
y ′′ + P(x) y ′ + Q(x) y = 0.
(a) The number x0 is an ordinary point of (∗) if P(x) and Q(x)are both analytic at x0.
(b) If x0 is not an ordinary point, it is called a singular point of(∗).
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Theorem (Existence of power series solutions near ordinarypoints)
If x0 is an ordinary point of the differential equation
a2(x) y ′′ + a1(x) y ′ + a0(x) y = 0, (∗)
then (∗) has two linearly independent solutions y1(x) and y2(x)that are both analytic at x0:
y1(x) =∞∑
n=0
bn (x − x0)n; y2(x) =
∞∑n=0
cn (x − x0)n,
& each solution has radius of convergence at least R, whereR is the distance to the nearest singular point to x0 in thecomplex plane.
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Useful Notation & Simplifications
[(2)(4)(6) . . . (2n)]︸ ︷︷ ︸n factors
= [(1 · 2)(2 · 2)(3 · 2) · · · (n · 2)] = n!2n
[(3)(6)(9) . . . (3n)]︸ ︷︷ ︸n factors
= [(1 · 3)(2 · 3)(3 · 3) · · · (n · 3)] = n!3n
[f (k)f (k + 1)f (k + 2) · · · f (m)] =m∏
n=k
f (n)
3∏n=1
nn + 1
=
(12
)(23
)(34
)m∏
k=1
1(2k + 1)
=1
[3 · 5 · 7 · · · (2m + 1)]
=[2 · 4 · · · (2m)]
[(2 · 3)(4 · 5) · · · (2m · (2m + 1))]=
2mm!
(2m + 1)!
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Regular (RSP) vs Irregular (ISP) Singular Points
Definition
If x0 is a singular point of (∗)
y ′′ + P(x) y ′ + Q(x) y = 0, (∗)
a 2nd -order homogeneous linear ODE in standard form, then(a) x0 is a regular singular point (RSP) of (∗), if the functions
(x−x0)P(x) & (x−x0)2 Q(x) are BOTH analytic at x = x0.
(b) If x0 is not a regular singular point, it is called an irregularsingular point ( ISP).
(a)⇒ limx→x0
(x − x0)P(x) & limx→x0
(x − x0)2Q(x) exist and
are finite.x0 = 0 is a (RSP) of the 2nd order homogeneousCauchy-Euler ODE: ax2y ′′ + bxy ′ + cy = 0.
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The Method of FrobeniusPower Series Solutions about RSPs
Theorem
If x0 is a regular singular point of
a2(x) y ′′ + a1(x) y ′ + a0(x) y = 0, (∗)
then there exists at least one solution (∗) of the form
y = (x − x0)r∞∑
n=0
cn (x − x0)n =
∞∑n=0
cn (x − x0)n+r , (∗∗)
with c0 6= 0, & cn, n ≥ 1, defined in terms of c0.The series solution in (∗∗) converges at least for0 < |x − x0| < R, where R is the distance from x0 to the nearestsingular point (real or complex) of (∗).
GOAL: To find r & to compute the coefficientscn, n ≥ 1, in terms of c0, in the solution (∗∗).
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To find the Indicial Equationif x0 is a RSP of y ′′ + P(x)y ′ + Q(x)y = 0
Since (x − x0)P(x) & (x − x0)2 Q(x) are both analytic at
x = x0 there exists R > 0 such that for |x − x0| < R
(x − x0)P(x) = a0 + a1 (x − x0) + · · · =∞∑
n=0
an (x − x0)n,
(x − x0)2 Q(x) = b0 + b1 (x − x0) + · · · =
∞∑n=0
bn (x − x0)n.
WLOG, assume x0 = 0. If not, substitute v = (x − x0).Therefore, for |x | < R
x P(x) = a0 + a1 x + · · · =∞∑
n=0
an xn,
x2 Q(x) = b0 + b1 x + · · · =∞∑
n=0
bn xn.
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Assume a solution of the form:
y =∞∑
n=0
cn xn+r = c0 x r + c1 x r+1 + . . .
y ′ =∞∑
n=0
(n + r) cn xn+r−1 = c0 r x r−1 + c1 (r + 1) x r + . . .
y ′′ =∞∑
n=0
(n + r) (n + r − 1) cn xn+r−2
= c0 r (r − 1)x r−2 + c1 (r + 1) r x r−1 + . . .
Substitute into the ODE.
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y ′′ + P(x)y ′ + Q(x)y = 0 ⇔ y ′′ + (xP(x))y ′
x+ (x2Q(x))
yx2 = 0
∞∑n=0
(n + r) (n + r − 1) cn xn+r−2
︸ ︷︷ ︸y ′′
+ (a0 + a1 x + a2 x2 + . . . )︸ ︷︷ ︸x P(x)
( ∞∑n=0
(n + r) cn xn+r−2
)︸ ︷︷ ︸
y′x
+ (b0 + b1 x + b2 x2 + . . . )︸ ︷︷ ︸x2 Q(x)
( ∞∑n=0
cn xn+r−2
)= 0︸ ︷︷ ︸
yx2
,
c0 [r (r − 1) + a0 r + b0] x r−2 + (. . . ) x r−1 + (. . . ) x r + · · · = 0.
Since c0 6= 0, [r (r − 1) + a0 r + b0] = 0, the indicial equation.
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The Indicial Equation
[r (r − 1) + a0 r + b0] = 0 OR r2 + (a0 − 1)r + b0 = 0
x P(x) = a0 + a1 x + · · · =∞∑
n=0
an xn,
x2 Q(x) = b0 + b1 x + · · · =∞∑
n=0
bn xn.
a0 = limx→0
xP(x),
b0 = limx→0
x2Q(x).
If ALSO wish to find the solution in the form of a powerseries, y(x) =
∑∞n=0 cnxn+r , can instead find the indicial
equation by substituting into the ODE and setting the factorof c0x r−2 equal to 0.
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Theorem (Form of two linearly indep solns)
Assume that x0 is a RSP of y ′′ + p(x)y ′ + q(x)y = 0 and thatr1 & r2 are roots of the indicial equation with Re(r1) ≥ Re(r2).(a) If r1 − r2 is not an integer, then there exist two linearly
independent solutions of the form:
y1(x)=∞∑
n=0
an(x − x0)n+r1 , a0 6= 0,
y2(x) =∞∑
n=0
bn(x − x0)n+r2 , b0 6= 0,
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Theorem (Form of two linearly indep solns)
Assume that x0 is a RSP of y ′′ + p(x)y ′ + q(x)y = 0 and thatr1 & r2 are roots of the indicial equation with Re(r1) ≥ Re(r2).(b) If r1 = r2, then there exist two linearly independent
solutions of the form:
y1(x)=∞∑
n=0
an(x − x0)n+r1 , a0 6= 0,
y2(x) = y1(x) ln(x − x0) +∞∑
n=1
bn(x − x0)n+r2 ,
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Theorem (Form of two linearly indep solns)
Assume that x0 is a RSP of y ′′ + p(x)y ′ + q(x)y = 0 and thatr1 & r2 are roots of the indicial equation with Re(r1) ≥ Re(r2).(c) If r1 − r2 is a positive integer, then there exist two linearly
independent solutions of the form:
y1(x)=∞∑
n=0
an(x − x0)n+r1 , a0 6= 0,
y2(x) = Cy1(x) ln(x − x0) +∞∑
n=0
bn(x − x0)n+r2 , b0 6= 0,
where C is a constant that could be zero.