Routh-Hurwitz Criterion - MTBI · Routh-Hurwitz Criterion Introduce Routh-Hurwitz Criterion for 2 2...

Routh-Hurwitz Criterion

Baojun Song, Ph.D.

Department of Mathematical SciencesMontclair State University

June 20, 2016

[email protected]

B. Song (Montclair State) Routh-Hurwitz Criterion June 20, 2016 1 / 1

Introduce Routh-Hurwitz Criterion for 2× 2 and 3× 3matrices. Check book for more general formula.

Routh-Hurwitz Criterion for 2 by 2 matrices

|λI −A| is the characteristic polynomial of A. Let λ1 and λ2 be theeigenvalues of A.

∣∣∣∣λ− a11 −a12−a21 λ− a22

∣∣∣∣ = (λ− λ1)(λ− λ2)

(1) Let λ = 0 , detA = λ1λ2, detA = λ1λ2 > 0. First we needdetA > 0.

(2) Compare the coefficient of λ on both sides ,

−(a11 + a22) = LHS=RHS = −(λ1 + λ2).

As λ1 + λ2 < 0, tr(A) < 0.

Conclusion: All eigenvalues of a 2 by 2 matrix have negative real partsif and only if |A| > 0 and tr(A) < 0.

|λI −A| is the characteristic polynomial of A. Let λ1 and λ2 be theeigenvalues of A.∣∣∣∣λ− a11 −a12

−a21 λ− a22

∣∣∣∣ = (λ− λ1)(λ− λ2)

−(a11 + a22) = LHS=RHS = −(λ1 + λ2).

As λ1 + λ2 < 0, tr(A) < 0.

−a21 λ− a22

∣∣∣∣ = (λ− λ1)(λ− λ2)

−(a11 + a22) = LHS=RHS = −(λ1 + λ2).

As λ1 + λ2 < 0, tr(A) < 0.

−a21 λ− a22

∣∣∣∣ = (λ− λ1)(λ− λ2)

−(a11 + a22) = LHS=RHS = −(λ1 + λ2).

As λ1 + λ2 < 0, tr(A) < 0.

−a21 λ− a22

∣∣∣∣ = (λ− λ1)(λ− λ2)

−(a11 + a22) = LHS=RHS = −(λ1 + λ2).

As λ1 + λ2 < 0, tr(A) < 0.

|λI −A| is the characteristic polynomial of A. Let λ1, λ2 and λ3 be theeigenvalues of A.

∣∣∣∣∣∣λ− a11 −a12 −a13−a21 λ− a22 −a23−a31 −a32 λ− a33

∣∣∣∣∣∣ = (λ− λ1)(λ− λ2)(λ− λ3)

(1) Let λ = 0 , −detA = −λ1λ2λ3, detA = λ1λ2λ3 < 0. First we needdetA < 0.

(2) Compare the coefficient of λ2 on both sides ,

−(a11 + a22 + a33) = LHS=RHS = −(λ1 + λ2 + λ3).

As λ1 + λ2 + λ3 < 0, tr(A) < 0.

|λI −A| is the characteristic polynomial of A. Let λ1, λ2 and λ3 be theeigenvalues of A.∣∣∣∣∣∣

λ− a11 −a12 −a13−a21 λ− a22 −a23−a31 −a32 λ− a33

∣∣∣∣∣∣ = (λ− λ1)(λ− λ2)(λ− λ3)

−(a11 + a22 + a33) = LHS=RHS = −(λ1 + λ2 + λ3).

As λ1 + λ2 + λ3 < 0, tr(A) < 0.

λ− a11 −a12 −a13−a21 λ− a22 −a23−a31 −a32 λ− a33

∣∣∣∣∣∣ = (λ− λ1)(λ− λ2)(λ− λ3)

−(a11 + a22 + a33) = LHS=RHS = −(λ1 + λ2 + λ3).

As λ1 + λ2 + λ3 < 0, tr(A) < 0.

λ− a11 −a12 −a13−a21 λ− a22 −a23−a31 −a32 λ− a33

∣∣∣∣∣∣ = (λ− λ1)(λ− λ2)(λ− λ3)

−(a11 + a22 + a33) = LHS=RHS = −(λ1 + λ2 + λ3).

As λ1 + λ2 + λ3 < 0, tr(A) < 0.

a11 a12 a13a21 a22 a23a31 a32 a33

w1 = −det(A) = −|A|, w2 = −tr(A) = −(a11 + a22 + a33)

w3 = det

(a11 a12a21 a22

)+ det

(a11 a13a31 a33

)+ det

(a22 a23a32 a33

Then the real parts of all eigenvalues of A are negative if and only ifwi > 0, i = 1, 2, 3 and w2w3 > w1.

a11 a12 a13a21 a22 a23a31 a32 a33

w1 = −det(A) = −|A|, w2 = −tr(A) = −(a11 + a22 + a33)

w3 = det

(a11 a12a21 a22

)+ det

(a11 a13a31 a33

)+ det

(a22 a23a32 a33

a11 a12 a13a21 a22 a23a31 a32 a33

w1 = −det(A) = −|A|, w2 = −tr(A) = −(a11 + a22 + a33)

w3 = det

(a11 a12a21 a22

)+ det

(a11 a13a31 a33

)+ det

(a22 a23a32 a33

Lorenz Equation

dt= σ(y − x)

dt= rx− y − xz

dt= −bz + xy

(0, 0, 0)

z = r − 1, x2 = bz, x2 = b(r − 1), x = ±√b(r − 1)

If r > 1, there are two more equilibria:C+ = (

√b(r − 1),

√b(r − 1), r − 1)

C− = (−√b(r − 1),−

√b(r − 1), r − 1)

Lorenz Equation

dt= σ(y − x)

dt= rx− y − xz

dt= −bz + xy

(0, 0, 0)z = r − 1,

x2 = bz, x2 = b(r − 1), x = ±√b(r − 1)

√b(r − 1),

√b(r − 1), r − 1)

C− = (−√b(r − 1),−

√b(r − 1), r − 1)

Lorenz Equation

dt= σ(y − x)

dt= rx− y − xz

dt= −bz + xy

(0, 0, 0)z = r − 1, x2 = bz,

x2 = b(r − 1), x = ±√b(r − 1)

√b(r − 1),

√b(r − 1), r − 1)

C− = (−√b(r − 1),−

√b(r − 1), r − 1)

Lorenz Equation

dt= σ(y − x)

dt= rx− y − xz

dt= −bz + xy

(0, 0, 0)z = r − 1, x2 = bz, x2 = b(r − 1),

x = ±√b(r − 1)

√b(r − 1),

√b(r − 1), r − 1)

C− = (−√b(r − 1),−

√b(r − 1), r − 1)

Lorenz Equation

dt= σ(y − x)

dt= rx− y − xz

dt= −bz + xy

(0, 0, 0)z = r − 1, x2 = bz, x2 = b(r − 1), x = ±

√b(r − 1)

√b(r − 1),

√b(r − 1), r − 1)

C− = (−√b(r − 1),−

√b(r − 1), r − 1)

Lorenz Equation

dt= σ(y − x)

dt= rx− y − xz

dt= −bz + xy

(0, 0, 0)z = r − 1, x2 = bz, x2 = b(r − 1), x = ±

√b(r − 1)

√b(r − 1),

√b(r − 1), r − 1)

C− = (−√b(r − 1),−

√b(r − 1), r − 1)

Lorenz Equation

dt= σ(y − x)

dt= rx− y − xz

dt= −bz + xy

(0, 0, 0)z = r − 1, x2 = bz, x2 = b(r − 1), x = ±

√b(r − 1)

√b(r − 1),

√b(r − 1), r − 1)

C− = (−√b(r − 1),−

√b(r − 1), r − 1)

Stability of (0, 0, 0) of Lorenz Equation

−σ σ 0r − z −1 −xy x −b

J(0,0,0) =

−σ σ 0r −1 00 0 −b

w1 = bσ(1− r) > 0 if r < 1. w2 = σ + 1 + b > 0.w3 = σ(1− r) + b(σ + 1) > 0w2w3 = (σ + 1 + b)(σ(1− r) + b(σ + 1)) =bσ(1− r) plus positive terms > bσ(1− r) = w1

(0, 0, 0) is locally asymptotically stable if r < 1.

−σ σ 0r − z −1 −xy x −b

J(0,0,0) =

−σ σ 0r −1 00 0 −b

−σ σ 0r − z −1 −xy x −b

J(0,0,0) =

−σ σ 0r −1 00 0 −b

w1 = bσ(1− r) > 0 if r < 1.

w2 = σ + 1 + b > 0.w3 = σ(1− r) + b(σ + 1) > 0w2w3 = (σ + 1 + b)(σ(1− r) + b(σ + 1)) =bσ(1− r) plus positive terms > bσ(1− r) = w1

−σ σ 0r − z −1 −xy x −b

J(0,0,0) =

−σ σ 0r −1 00 0 −b

w1 = bσ(1− r) > 0 if r < 1. w2 = σ + 1 + b > 0.

w3 = σ(1− r) + b(σ + 1) > 0w2w3 = (σ + 1 + b)(σ(1− r) + b(σ + 1)) =bσ(1− r) plus positive terms > bσ(1− r) = w1

−σ σ 0r − z −1 −xy x −b

J(0,0,0) =

−σ σ 0r −1 00 0 −b

w1 = bσ(1− r) > 0 if r < 1. w2 = σ + 1 + b > 0.w3 = σ(1− r) + b(σ + 1) > 0

w2w3 = (σ + 1 + b)(σ(1− r) + b(σ + 1)) =bσ(1− r) plus positive terms > bσ(1− r) = w1

−σ σ 0r − z −1 −xy x −b

J(0,0,0) =

−σ σ 0r −1 00 0 −b

−σ σ 0r − z −1 −xy x −b

J(0,0,0) =

−σ σ 0r −1 00 0 −b

Stability of C+ of Lorenz Equation

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

(σ + r)(σ + 1 + b) > 2σ(r − 1)σ(σ + 3 + b) > (σ − 1− b)rC+ is locally asymptotically stable if r > 1 andσ(σ + 3 + b) > (σ − 1− b)r.

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1.

w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,

w3 = b(σ + r) > 0w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

(σ + r)(σ + 1 + b) > 2σ(r − 1)

σ(σ + 3 + b) > (σ − 1− b)rC+ is locally asymptotically stable if r > 1 andσ(σ + 3 + b) > (σ − 1− b)r.

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

(σ + r)(σ + 1 + b) > 2σ(r − 1)σ(σ + 3 + b) > (σ − 1− b)r

C+ is locally asymptotically stable if r > 1 andσ(σ + 3 + b) > (σ − 1− b)r.

C+ = (√b(r − 1),

√b(r − 1), r − 1)

−σ σ 0r − z −1 −xy x −b

−σ σ 01 −1 −xx x −b

w1 = 2bσ(r − 1) > 0 if r > 1. w2 = σ + 1 + b > 0,w3 = b(σ + r) > 0

w2w3 = b(σ + r)(σ + 1 + b) > 2bσ(r − 1) = w1

A Model for TB

dt= γN(1− N

K)− βS I

N− µS + γ1L+ γ2I

dt= βS

N− (µ+ k + γ1)L

dt= kL− (µ+ d+ γ2)I

where N = S + L+ I.

Equivalent Model Equations

dt= γN(1− N

K)− µN − dI

dt= β(N − L− I)

N− (µ+ k + γ1)L

A Model for TB

dt= γN(1− N

K)− βS I

N− µS + γ1L+ γ2I

dt= βS

N− (µ+ k + γ1)L

where N = S + L+ I.Equivalent Model Equations

dt= γN(1− N

K)− µN − dI

N− (µ+ k + γ1)L

A Model for TB (Cont’d)

Disease-free equilibrium (N0, 0, 0), N0 = (1− µγ )K and for N0, γ > µ is

demographically necessary.

dt= γN(1− N

K)− µN − dI

N− (µ+ k + γ1)L

Jacobian µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

Disease-free equilibrium (N0, 0, 0), N0 = (1− µγ )K and for N0, γ > µ is

demographically necessary.

dt= γN(1− N

K)− µN − dI

N− (µ+ k + γ1)L

Jacobian µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

w1 = −detA = −(µ− γ)[(µ+ d+ γ2)(µ+ k + γ1)− kβ]

= (γ − µ)(µ+ d+ γ2)(µ+ k + γ1)(1−R0)

R0 =kβ

(µ+ d+ γ2)(µ+ k + γ1)

w1 > 0 requires R0 < 1,

µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

w1 = −detA = −(µ− γ)[(µ+ d+ γ2)(µ+ k + γ1)− kβ]

= (γ − µ)(µ+ d+ γ2)(µ+ k + γ1)(1−R0)

R0 =kβ

(µ+ d+ γ2)(µ+ k + γ1)

w1 > 0 requires R0 < 1,

µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2 > 0

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

(µ+ d+ γ2)(µ+ k + γ1)− kβ = (µ+ d+ γ2)(µ+ k + γ1)(1−R0).Because R0 < 1, w3 > 0.

µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2 > 0

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2 > 0

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2 > 0

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

(µ+ d+ γ2)(µ+ k + γ1)− kβ = (µ+ d+ γ2)(µ+ k + γ1)(1−R0).

Because R0 < 1, w3 > 0.

µ− γ 0 −d0 −(µ+ k + γ1) β0 k −(µ+ d+ γ2)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2 > 0

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

w1 = (γ − µ)(µ+ d+ γ2)(µ+ k + γ1)(1−R0)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

w2w3 = (µ+ d+ γ2 + µ+ k + γ1 + γ − µ)×[(µ+ d+ γ2)(γ − µ) + (µ+ k + γ1)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)(1−R0)]

= (γ − µ)(µ+ d+ γ2)(µ+ k + γ1)(1−R0) + positive terms

By Routh-Hurwitz Criterion, disease-free equilibrium is locallyasymptotically stable whenever R0 < 1.

w1 = (γ − µ)(µ+ d+ γ2)(µ+ k + γ1)(1−R0)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

+ (µ+ d+ γ2)(µ+ k + γ1)(1−R0)]

w1 = (γ − µ)(µ+ d+ γ2)(µ+ k + γ1)(1−R0)

w2 = γ − µ+ µ+ k + γ1µ+ d+ γ2

w3 = (µ+ k + γ1)(γ − µ) + (µ+ d+ γ2)(γ − µ)

+ (µ+ d+ γ2)(µ+ k + γ1)− kβ

+ (µ+ d+ γ2)(µ+ k + γ1)(1−R0)]

Routh-Hurwitz Criterion for Polynomials of Degree Four

Consider f(λ) = λ4 + a1λ3 + a2λ

2 + a3λ+ a4. Solutions to f(λ) = 0 allhave negative real parts iff

∆1 = a1 > 0

∆2 =

∣∣∣∣a1 1a3 a2

∣∣∣∣ > 0

∆3 =

∣∣∣∣∣∣a1 1 0a3 a2 a1a5 a4 a3

∣∣∣∣∣∣ > 0

a4 > 0

Routh-Hurwitz Criterion - MTBI · Routh-Hurwitz Criterion Introduce Routh-Hurwitz Criterion for 2 2...

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