Download - Constant-Coefficient Systems(1)

Constant-Coefficient Systems(1)

– Then the corresponding solutions are

– The general solution is

– Wronskian of (2) is

Constant-Coefficient Systems(2)

– We can graph solutions of (1), 𝐲 𝑡 =𝑦1(𝑡)𝑦2(𝑡)

as a single curve in the

𝑦1- 𝑦2 plane.

– That curve is a parametric representation with parameter 𝑡, and

called a trajectory of (1).

– The 𝑦1- 𝑦2 plane is called the phase plane.

– If we fill the phase plane with trajectories, we obtain the so-called

phase portrait of (1)

How to graph solutions in the phase plane

𝐲′ = 𝐀𝐲⇒𝑦1′ = 𝑎11𝑦1 + 𝑎12𝑦2

𝑦2′ = 𝑎21𝑦1 + 𝑎22𝑦2

⋯(1)

– By substituting 𝐲 = 𝐱𝑒𝜆𝑡 and 𝐲′ = 𝜆𝐱𝑒𝜆𝑡, we get 𝐀𝐱 = 𝜆𝐱.

– Characteristic equation

⇒ det(𝐀 − 𝜆𝐈) =−3 − 𝜆 1

1 −3 − 𝜆= 𝜆2 + 6𝜆 + 8 = 0

⇒ 𝜆1 = −2, 𝜆2 = −4

– For 𝜆1 = −2, 𝐱(𝟏) = 𝟏 𝟏 𝑻, and for 𝜆2 = −4, 𝐱(𝟐) = 𝟏 −𝟏 𝑻

– The general solution is

Phase portrait : example (1)

𝐲′ = 𝐀𝐲 =−3 11 −3

𝐲 ⇒𝑦1′ = −3𝑦1 + 𝑦2𝑦2′ = 𝑦1 − 3𝑦2

𝐲 =𝑦1𝑦2

= 𝑐1𝐲(1) + 𝑐2𝐲

(2) = 𝑐111𝑒−2𝑡 + 𝑐2

1−1

𝑒−4𝑡

⋯(1)

– Phase portrait of (1)

– The point 𝐲 = 𝟎 is a common point of all

projectories.

– Unique tangent direction for every point (𝑦1, 𝑦2)

except for the point (0, 0).

– This point (0, 0), at which 𝒅𝒚𝟐/𝒅𝒚𝟏 is undetermined, is called

a critical point.

Phase portrait : example (1)

𝑑𝑦2𝑑𝑦1

=𝑦2′𝑑𝑡

𝑦1′𝑑𝑡

=𝑎21𝑦1 + 𝑎22𝑦2𝑎11𝑦1 + 𝑎12𝑦2

– Improper node (two distinct real eigenvalues of the same sign)

: a critical point 𝑃0 at which all the trajectories, except for two of them,

have the same limiting direction of the tangent. The two exceptional

trajectories also have a limiting direction of the tangent at 𝑃0 which,

however, is different

Five types of critical points

𝐲′ =−3 11 −3

𝐲

⇒ 𝐲 = 𝑐111𝑒−2𝑡 + 𝑐2

1−1

𝑒−4𝑡

– Proper node (two identical eigenvalues and two linearly independent

eigenvectors)

: a critical point 𝑃0 at which every trajectory has a definite limiting

direction and for any given direction 𝒅 at 𝑃0 there is a trajectory

having 𝒅 as its limiting direction.


𝐲′ =1 00 1

𝐲

⇒ 𝐲 = 𝑐110𝑒𝑡 + 𝑐2

01𝑒𝑡 or 𝑐1𝑦2 = 𝑐2𝑦1

– Saddle point (two real eigenvalues of opposite signs)

: a critical point 𝑃0 at which there are two incoming trajectories, two

outgoing trajectories, and all the other trajectories in a

neighborhood of 𝑃0 bypass 𝑃0.


𝐲′ =1 00 −1

𝐲

⇒ 𝐲 = 𝑐110𝑒𝑡 + 𝑐2

01𝑒−𝑡 or 𝑦1𝑦2 = 𝑐𝑜𝑛𝑠𝑡

– Center (two purely imaginary conjugate eigenvalues)

: a critical point 𝑃0 that is enclosed by infinitely many closed

trajectories.


𝐲′ =0 1

−4 0𝐲

⇒ 𝐲 = 𝑐112𝑖

𝑒2𝑖𝑡 + 𝑐20−2𝑖

𝑒−2𝑖𝑡

or 2𝑦12 +

1

2𝑦22 = 𝑐𝑜𝑛𝑠𝑡

– Spiral point (two complex conjugate eigenvalues)

: a critical point 𝑃0 about which the trajectories spiral, approaching 𝑃0as 𝑡 → ∞ (or tracing these spirals in the opposite sense, away from

𝑃0).


𝐲′ =−1 1−1 −1

𝐲

⇒ 𝐲 = 𝑐11𝑖𝑒(−1+𝑖)𝑡 + 𝑐2

1−𝑖

𝑒(−1−𝑖)𝑡

or 𝑟 = 𝑐𝑒−𝑡 where 𝑟2 = 𝑦12 + 𝑦2

2

– Two identical eigenvalues and one eigenvector

– 𝐲(1) =1

−1𝑒3𝑡, and let 𝐲(2) = 𝐱𝑡𝑒3𝑡 + 𝐮𝑒3𝑡

– Then 𝐲 2 ′= 𝐱𝑒3𝑡 + 3𝐱𝑡𝑒3𝑡 + 3𝐮𝑒3𝑡 - (*)

𝐀𝐲 2 = 𝐀𝐱𝑡𝑒3𝑡 + 𝐀𝐮𝑒3𝑡 - (**)

⇒ 𝐀 − 𝟑𝐈 𝐮 = 𝐱, 𝐮 = 0 1 𝑇 (linearly independent of 𝐱)

⇒ 𝐲 = 𝑐11

−1𝑒3𝑡 + 𝑐2

1−1

𝑡 +01

𝑒3𝑡

Degenerate node

𝐲′ =4 1

−1 2𝐲 ⇒ 𝜆1 = 𝜆𝟐 = 𝜆 = 3, 𝐱1 = 𝐱2 = 𝐱 = 1 − 1 𝑇

(*)=(**) where 𝐀 =4 1

−1 2

Stability

– A critical point 𝑃0 is called stable if all trajectories that at some

instant are close to 𝑃0 remain close to 𝑃0 at all future times

– For every disk 𝐷 of radius 휀 > 0 with center 𝑃0 there is a disk 𝐷𝛿 of

radius 𝛿 > 0 with center 𝑃0 such that every trajectory that has a point

𝑃1(corresponding to 𝑡 = 𝑡1 say) in 𝐷 has all its points corresponding

to 𝑡 ≥ 𝑡1 in 𝐷 .

Criteria for critical points & stability

Stable critical point 𝑃0(The trajectory initiating at 𝑃1stays in the disk of radius 휀.)

Stable and attractive

critical point 𝑃0

𝐲′ = 𝐀𝐲 =𝑎11 𝑎12𝑎21 𝑎22

𝐲Ch. eqn. det 𝐀 − 𝜆𝐈 = 𝜆2 − 𝑎11 + 𝑎22 𝜆 + det𝐀

⇒ 𝑝 = 𝑎11 + 𝑎22, 𝑞 = det𝐀, Δ = 𝑝2 − 4𝑞

𝑝 = 𝜆1 + 𝜆2, 𝑞 = 𝜆1𝜆2, Δ = 𝜆1 − 𝜆22

Stability criteria for critical points

Eigenvalue criteria for critical points

Stability chart

Criteria for critical points & stability

Stability of critical points : Example

– Free motions of a mass on a spring

𝑚𝑦′′ + 𝑐𝑦′ + 𝑘𝑦 = 0

𝐲′ =0 1

−𝑘/𝑚 −𝑐/𝑚𝐲

𝑦1 = 𝑦𝑦2 = 𝑦′

det 𝐀 − 𝜆𝐈 = 𝜆2 + 𝑐/𝑚 𝜆 + 𝑘/𝑚 = 0𝑝 = −𝑐/𝑚, 𝑞 = 𝑘/𝑚, Δ = 𝑐/𝑚 2 − 4𝑘/𝑚

No damping (𝑐 = 0)

𝑝 = 0, 𝑞 > 0 ⇒ a center

Underdamping (𝑐2 < 4𝑚𝑘)

𝑝 < 0, 𝑞 > 0, Δ < 0⇒ a stable and attractive spiral point

Critical damping (𝑐2 = 4𝑚𝑘)

𝑝 < 0, 𝑞 > 0, Δ = 0⇒ a stable and attractive node

Overdamping (𝑐2 > 4𝑚𝑘)

𝑝 < 0, 𝑞 > 0, Δ > 0⇒ a stable and attractive node

Linearization of nonlinear systems

– If 𝑓1 and 𝑓2 in (1) are continuous and have continuous partial

derivatives in a neighborhood of the critical point 𝑃0: (0, 0), and if

det 𝐀 ≠ 0 in (2), then the kind and stability of the critical point of (1)

are the same as those of the linearized system

– Exception : 𝐀 has equal and pure imaginary eigenvalues

⇒ same kind of critical points of (3) or a spiral point

𝐲′ = 𝐟(𝐲) ⇒𝑦1′ = 𝑓1(𝑦1, 𝑦2)

𝑦2′ = 𝑓2(𝑦1, 𝑦2)

⋯ (1)Nonlinear

system

𝐲′ = 𝐀𝐲 + 𝐡(𝐲) ⇒𝑦1′ = 𝑎11𝑦1 + 𝑎12𝑦2 + ℎ1(𝑦1, 𝑦2)

𝑦2′ = 𝑎21𝑦1 + 𝑎22𝑦2 + ℎ2(𝑦1, 𝑦2)

⋯(3)𝐲′ = 𝐀𝐲 ⇒𝑦1′ = 𝑎11𝑦1 + 𝑎12𝑦2

𝑦2′ = 𝑎21𝑦1 + 𝑎22𝑦2

Linearization : Example

Free undamped pendulum

Step 1 : modeling

Step 2 : critical points ±2𝑘𝜋, 0 𝑘 ∶ 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

– Consider linearization near 0, 0 , and set 𝜃 = 𝑦1, 𝜃′ = 𝑦2

– Calculate 𝑝 = 0, 𝑞 = 𝑘, ∆= −4𝑘 ⇒ a center

– Due to periodicity, critical points ±2𝑘𝜋, 0 (𝑘 ∶ 𝑖𝑛𝑡𝑒𝑔𝑒𝑟) are all

centers.

𝑚𝐿𝜃′′ + 𝑚𝑔𝑠𝑖𝑛𝜃 = 0 ⇒ 𝜃′′ + 𝑘𝑠𝑖𝑛𝜃 = 0 (𝑘 = 𝑔/𝐿)

𝜃′′ + 𝑘𝑠𝑖𝑛𝜃 = 0⇒𝑦1′ = 𝑦2

𝑦2′ = −𝑘𝑦1 (𝑠𝑖𝑛𝑦1 ≈ 𝑦1)

Linearization : Example

Step 3 : critical points ±(2𝑘 − 1)𝜋, 0 𝑘 ∶ 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

– Consider linearization near 𝜋, 0 , and set 𝜃 − 𝜋 = 𝑦1, (𝜃 − 𝜋)′= 𝑦2

– Calculate 𝑝 = 0, 𝑞 = −𝑘, ∆= 4𝑘 ⇒ a saddle point

– Due to periodicity, critical points ±(2𝑘 − 1)𝜋, 0 (𝑘 ∶ 𝑖𝑛𝑡𝑒𝑔𝑒𝑟) are

all saddle points.

𝜃′′ + 𝑘𝑠𝑖𝑛𝜃 = 0⇒𝑦1′ = 𝑦2

𝑦2′ = 𝑘𝑦1 (𝑠𝑖𝑛𝑦1 ≈ −𝑦1)

Transformation to a 1st order eq. in the phase plane

– 2nd order autonomous ODE(an ODE in which 𝑡 does not occur explicitly)

– Take 𝑦 = 𝑦1 as the independent variable, set 𝑦′ = 𝑦2 and transform 𝑦′′

by the chain rule.

– Then, the ODE becomes of first order.

𝐹 𝑦, 𝑦′, 𝑦′′ = 0

𝑦′′ = 𝑦2′ =

𝑑𝑦2𝑑𝑡

=𝑑𝑦2𝑑𝑦1

𝑑𝑦1𝑑𝑡

=𝑑𝑦2𝑑𝑦1

𝑦2

𝐹 𝑦1, 𝑦2,𝑑𝑦2𝑑𝑦1

𝑦2 = 0

Example

Free undamped pendulum

– set 𝜃 = 𝑦1, 𝜃′ = 𝑦2

– Using separation of variables, we get

𝑚𝐿𝜃′′ + 𝑚𝑔𝑠𝑖𝑛𝜃 = 0 ⇒ 𝜃′′ + 𝑘𝑠𝑖𝑛𝜃 = 0 (𝑘 = 𝑔/𝐿)

𝜃′′ + 𝑘𝑠𝑖𝑛𝜃 = 0 ⇒𝑑𝑦2

𝑑𝑦1𝑦2 = −𝑘𝑠𝑖𝑛𝑦1

1

2𝑦22 = 𝑘𝑐𝑜𝑠𝑦1 + 𝐶 ⇒

1

2𝑚 𝐿𝑦2

2 −𝑚𝐿2𝑘𝑐𝑜𝑠𝑦1 = 𝑚𝐿2𝐶

Nonhomogeneous linear systems of ODEs

Example

Method 1 : method of undetermined coefficients

– 𝐲(ℎ) = 𝑐111𝑒−2𝑡 + 𝑐2

1−1

𝑒−4𝑡

– Set 𝐲(𝑝) = 𝐮𝑡𝑒−2𝑡 + 𝐯𝑒−2𝑡 (by modification rule) where 𝐮 = 𝑎 1 1 𝑇

– Substitute 𝑦(𝑝) into (1), then we get

𝐮 − 2𝐯 = 𝐀𝐯 +−62

⇒𝑎𝑎

−2𝑣12𝑣2

=−3𝑣1 + 𝑣2

𝑣1 − 3𝑣2+

−62

⇒ 𝑎 = −2, 𝑣1 = 𝑘, 𝑣2 = 𝑘 + 4

– We can choose any real number 𝑘, like 𝑘 = 0

𝐲′ = 𝐀𝐲 + 𝐠 =−3 11 −3

𝐲 +−62

𝑒−2𝑡 ⋯(1)

𝐲 = 𝑐111𝑒−2𝑡 + 𝑐2

1−1

𝑒−4𝑡 − 211𝑡𝑒−2𝑡 +

04𝑒−2𝑡

Nonhomogeneous linear systems of ODEs

Example

Method 2 : method of variation of parameters

– 𝐲(ℎ) = 𝑐111𝑒−2𝑡 + 𝑐2

1−1

𝑒−4𝑡 = 𝑒−2𝑡 𝑒−4𝑡

𝑒−2𝑡 −𝑒−4𝑡𝑐1𝑐2

= 𝐘 𝑡 𝐜

– Set 𝐲(𝑝) = 𝐘 𝑡 𝐮 𝑡 and substitute into (1) then we get

𝐘𝐮′ = 𝐠 ∵ 𝐘′ = 𝐀𝐘 , 𝐮′ = 𝐘−𝟏𝐠 =1

2

𝑒2𝑡 𝑒2𝑡

𝑒4𝑡 −𝑒4𝑡−6𝑒−2𝑡

2𝑒−2𝑡=

−2−4𝑒2𝑡

𝐮 = 0𝑡 −2−4𝑒2𝑡

∗ 𝑑𝑡∗ =−2𝑡

−2𝑒2𝑡 + 2∴ 𝐲(𝑝) = 𝐘𝐮 =

−2𝑡 − 2−2𝑡 + 2

𝑒−2𝑡 +2−2

𝑒−4𝑡

𝐲′ = 𝐀𝐲 + 𝐠 =−3 11 −3

𝐲 +−62

𝑒−2𝑡 ⋯(1)

𝐲 = 𝑐111𝑒−2𝑡 + 𝑐2

1−1

𝑒−4𝑡 − 211𝑡𝑒−2𝑡 +

−22

𝑒−2𝑡

• Homogeneous linear DE w/ variable coeff.

– Power series method

Frobenius method: extension of power series

– Power series

--- infinite series (in powers of )

– If , power series in powers of x

Series Solution of DE

0x x

2

0 0 1 0 2 0

0

m

m

m

a x x a a x x a x x

Center of the series

0 0x

2 3

0 1 2 3

0

m

m

m

a x a a x a x a x

Maclaurin Series

2

0

2 3

0

2 2 4

0

2 1 3 5

0

11

1

1! 2! 3!

1cos 1

2 ! 2! 4!

1sin

2 1 ! 3! 5!

m

m

mx

m

m m

m

m m

m

x x xx

x x xe x

m

x x xx

m

x x xx x

m

– Represent p(x) and q(x) as a power series of x (or )

– Assume a sol. in the form of power series

– Collect like powers of x, and solve for the coefficients.

Power Series Method

0y p x y q x y

0x x

2 3

0 1 2 3

0

m

m

m

y a x a a x a x a x

1 2

1 2 3

0

2 3

m

m

m

y ma x a a x a x

2 2

2 3 4

0

1 2 3 2 4 3

m

m

m

y m m a x a a x a x

– Insert the power series

– Collect the like powers of x

– Solve for each coefficient

– General sol.

Example: Power Series Method

0y y

2 2

1 2 3 0 1 22 3 0 a a x a x a a x a x

2

1 0 2 1 3 22 3 0 a a a a x a a x

0 01 21 0 2 3, , ,

2 2! 3 3!

a aa aa a a a

2 3

0 012! 3!

xx xy a x a e

– Nth partial sum

– Remainder

– If at , the sequence s0(x), s1(x), s2(x), … converges to a value

– In case of convergence

--- all sn(x1) with n > N lie between s (x1) - and s (x1) +

Theory of Power Series Method

2

0 1 0 2 0 0 n

n ns x a a x x a x x a x x

1 2

1 0 2 0

n n

n n nR x a x x a x x

1x x

1 1 1 0

0

lim

m

n mn

m

s x s x a x x Value, sum

1 1 1 n nR x s x s x for all n > N

Convergence Interval (1)

– Always converges at . If it is the only point of convergence, all

other terms except a0 are zeros.

– Further values of x for convergence

0x x

Practically no interest

Convergence interval

0 x x R

Radius of convergence1

lim

m

mm

Ra

Convergence Interval (2)

If the limit is infinite, the power series only converges at .

– If the limit is 0, the power series converges for all x.

– Example: geometric series

--- geometric series converges and represents when

1

1

lim

m

mm

Ra

a

0x x

R

2

0

11

1

m

m

x x xx

1, 1 ma R

1x 1 1 x

Operations on Power Series (1)

– Term-wise differentiation

– Term-wise addition

0

0

m

m

m

y x a x x

1

0

0

m

m

m

y x ma x x

2

0

0

1

m

m

m

y x m m a x x

0 x x R

0 0

0 0

,

m m

m m

m m

a x x b x x

0

0

m

m m

m

a b x x


– Term-wise multiplication

– Vanishing of all coefficients

--- a positive radius of convergence, a sum is identically zero.

– Shifting summation indices

0 0

0 0

,

m m

m m

m m

a x x b x x

0 1 1 0 0

0

m

m m m

m

a b a b a b x x

2

0 0 0 1 1 0 0 0 2 1 1 2 0 0 a b a b a b x x a b a b a b x x

2 2 1

2 1

1 2

m m

m m

m m

x m m a x ma x


– Shifting summation indices (cont’d)

– Set

– Existence of power series sol.

--- if p(x), q(x), r(x) can be represented as a power series at ,

then every sol. is analytic.

1

2 1

1 2

m m

m m

m m

m m a x ma x

1, 1 s m m s

1

2 0

1 2 1

s s

s s

s s

s s a x s a x

1

2

1 2 1

s

s s

s

s s a s a x

0x x

Legendre’s Eqn. & Polynomials (1)

--- boundary value problems for spheres

– Sol.: Legendre function

– Since the coeff.’s are analytic, power series method can be applied.

Apply

21 2 1 0 x y xy n n y

Theory of special functions

0

m

m

m

y a x

2 2 1

0 0 0

1 1 2 0

m m m

m m m

m m m

x m m a x x ma x k a x


– Arranging each power

– Coeff. of each power must be zero.

2

0 0

1 1 0

m m

m m

m m

m m a x m m a x

2

2 3 4 22 1 3 2 4 3 2 1 s

sa a x a x s s a x2

22 1 a x

2

1 22 1 2 2 1 s

sa x a x s s a x2

0 1 2 2 0 s

ska ka x ka x sa x

22 1 0 a n n

3 16 2 1 0 a n n a

0x

1x


– Recurrence relation (formula)

– a0, a1: arbitrary constants

22 1 1 2 1 0 s ss s a s s s n n a 2,3,s

2

1

2 1

s s

n s n sa a

s s

2 0

3 1

1,

2!

1 2,

3!

n na a

n na a

0,1,s


– General sol.

– Converges for

– y1, y2: linearly independent

0 1 1 2 y a y x a y x

2 4

1

1 2 1 31

2! 4!

n n n n n ny x x

3 5

2

1 2 3 1 2 4

3! 5!

n n n n n ny x x x

1x

Legendre Polynomials (1)

– n is a nonnegative integer in many applications. Then, when s = n,

an+2 = 0, an+4 = 0, an+6 = 0, · · ·.

– If n is even, y1 reduces to a polynomial of degree n

– If n is odd, y2 reduces to a polynomial of degree n

– All the nonvanishing coeff. can be represented in terms of an

– an is arbitrary, and choose an = 1 when n = 1

Legendre polynomials

2

2 1

1s s

s sa a

n s n s

2s n

2

2 ! 1 3 5 2 1,

!2 !n n

n na

nn

1,2,n


– Then,

– In general, when n – 2m ≥ 0

– Pn(x): Legendre polynomial of degree n

2

2 2 !,

2 1 ! 2 !n n

na

n n

4

2 4 !

2 2! 2 ! 4 !n n

na

n n

2

2 2 !1

2 ! ! 2 !

m

n m n

n ma

m n m n m

2

0

2

2

2 2 !1

2 ! ! 2 !

2 ! 2 2 !

2 1! 1 ! 2 !2 !

Mm n m

n nm

n n

nn

n mP x x

m n m n m

n nx x

n nn


0

2

2

4 2

4

1

3

3

5 3

5

1,

13 1 ,

2

135 30 3 ,

8

1,

15 3 ,

2

163 70 15 ,

8

P x

P x x

P x x x

P x

P x x x

P x x x x

– First few of the functions

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