linear stability analysis
DESCRIPTION
Linear stability analysis. x. Transcription-translation model. Eigenvectors and eigenvalues. Nullclines and critical points. The cribsheet of linear stability analysis. f. m. Transcription-translation model. m. x. +1. -1. +1. -1. f. Nullclines and critical points. x. 1.0. f. - PowerPoint PPT PresentationTRANSCRIPT
1
Linear stability analysis
Transcription-translation model
Nullclines and critical points
Eigenvectors and eigenvalues
The cribsheet of linear stability analysis
f
m
x
[βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
ππ₯ππ‘
=π π₯
ππ +ΒΏππ +ΒΏ
ππ‘+ π π₯ππ β
π π β
ππ‘ΒΏΒΏ
ππππ‘
=ππ
ππ π+ΒΏππ π+ΒΏ
ππ‘+ ππππ πβ
π π πβ
ππ‘ΒΏΒΏ
2
Transcription-translation model
f
π½π
π½
πΌπ
πΌ
m
x
π½π+1 πΌππ-1 π½π+1 πΌ π₯-1
ππππ‘
=π½πβπΌππππ₯ππ‘
=π½πβπΌ π₯
π π+ΒΏ ΒΏ
π πβ
π +ΒΏΒΏ
π β
π½π=1πΌπ=2
πΌ=1π½=1
3
Nullclines and critical points
ππππ‘
=π½πβπΌππ=0
ππ₯ππ‘
=π½πβπΌ π₯=0
π½π=πΌπππ½π
πΌπ
=π
π½π=πΌ π₯π½πΌπ=π₯
ππΆ=π½ππΌπ
π₯πΆ=π½πΌπ½π
πΌπ
m
x
0
π₯=π½πΌπ
π½π=1πΌπ=2
πΌ=1π½=1
0.5
1.0
1.00.5π=π½π
πΌπ
ππ
/ππ‘=
0
ππ₯/ππ‘=0
f
4
Nullclines and critical points
ππππ‘
=π½πβπΌππ
ππ₯ππ‘
=π½πβπΌ π₯
ππΆ=π½ππΌπ
π₯πΆ=π½πΌπ½π
πΌπ
m
x
0
π₯=π½πΌπ
βπβ ππππ‘
β π‘>0
π½π=1πΌπ=2
πΌ=1π½=1
0.5
1.0
1.00.5π=π½π
πΌπ
ππ
/ππ‘=
0
ππ₯/ππ‘=0
f
5
Nullclines and critical points
ππππ‘
=π½πβπΌππ
ππ₯ππ‘
=π½πβπΌ π₯
ππΆ=π½ππΌπ
π₯πΆ=π½πΌπ½π
πΌπ
m
x
0
π₯=π½πΌπ
βπβ ππππ‘
β π‘>0
β π₯β ππ₯ππ‘β π‘>0
π½π=1πΌπ=2
πΌ=1π½=1
0.5
1.0
1.00.5π=π½π
πΌπ
ππ
/ππ‘=
0
ππ₯/ππ‘=0
f
6
Nullclines and critical points
m
x
0
π₯=π½πΌπππ
ππ‘=π½πβπΌππ
ππ₯ππ‘
=π½πβπΌ π₯
ππΆ=π½ππΌπ
π₯πΆ=π½πΌπ½π
πΌπ
βπβ ππππ‘
β π‘>0
β π₯β ππ₯ππ‘β π‘>0
ππππ‘
=π½πβπΌππππ₯ππ‘
=π½πβπΌ π₯
π½π=1πΌπ=2
πΌ=1π½=1
0.5
1.0
1.00.5π=π½π
πΌπ
ππ
/ππ‘=
0
ππ₯/ππ‘=0
f
7
Nullclines and critical points
m
x
0 ππππ‘
=π½πβπΌππππ₯ππ‘
=π½πβπΌ π₯
π½π=1πΌπ=2
πΌ=1π½=1
1.00.5
ππ/ππ‘=0
ππ₯/ππ‘=0
f
0 1 2 3 4 5t
0.5
0.6
0.7
0.8
0.9
1.0
x or m
mRN
A
Protein
8
Linear stability analysis
Transcription-translation model
Nullclines and critical points
Eigenvectors and eigenvalues
The cribsheet of linear stability analysis
f
m
x
[βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
9
Unbending trajectories
m
x
0 ππππ‘
=π½πβπΌππππ₯ππ‘
=π½πβπΌ π₯
π½π=1πΌπ=2
πΌ=1π½=1
1.00.5
f
10
Finding the βspecialβ direction
m
x
0
Dx
Dm
ππΆ=π½ππΌπ
π₯πΆ=π½πΌπ½π
πΌπ
ππππ‘
=π½πβπΌππππ₯ππ‘
=π½πβπΌ π₯
π½π=1πΌπ=2
πΌ=1π½=1
0.5
1.0
1.00.5
-0.25
0.25
0.25-0.25
f
11m
x Dx
Dm
ππΆ=π½ππΌπ
π₯πΆ=π½πΌπ½π
πΌπ
βπβπβπ½ππΌπ
πβπππ‘
= πππ‘ (πβ π½π
πΌπ)
ππππ‘
=π½πβπΌππππ₯ππ‘
=π½πβπΌ π₯
ππππ‘
=βπΌπ(πβ π½ππΌπ
)
πβπππ‘
=βπΌπβπ+0
β π₯βπ₯β π½πΌπ½π
πΌπ
πβ π₯ππ‘
=ππ₯ππ‘
πβ π₯ππ‘
=π½βπβπΌ β π₯
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ]
ππππ‘
=π½πβπΌππππ₯ππ‘
=π½πβπΌ π₯
Finding the βspecialβ direction
ππΆ=π½ππΌπ
π₯πΆ=π½πΌπ½π
πΌπ
ΒΏππππ‘
12
Finding the βspecialβ direction
m
x Dx
Dm
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ]
0.5
-0.5
0.5-0.5
13
Finding the βspecialβ direction
m
x Dx
Dm
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[βπβπ₯ ]= 1π Ξ π‘
β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=π[βπβ π₯ ][βπΌπ 0
π½ βπΌ] [βπβ π₯ ]Want eigenvectors!
(βπΌπβ π) (βπΌβ π )βπ½ β0=0
π1=βπΌπ π2=βπΌ
π1β[ 1π½ / (πΌβπΌπ ) ]π2β[01 ]
0.5
-0.5
0.5-0.5
m
x
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[βπβπ₯ ]= 1π Ξ π‘
β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=π[βπβ π₯ ][βπΌπ 0
π½ βπΌ] [βπβ π₯ ]Want eigenvectors!
(βπΌπβ π) (βπΌβ π )βπ½ β0=0
π1=βπΌπ
π1β[ 1π½ / (πΌβπΌπ ) ]
π2=βπΌ
π2β[01 ]
Finding the βspecialβ direction
Dx
Dm
14
0.25π2β[ 00.25]
0.5
-0.5
0.5-0.5
m
x
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[βπβπ₯ ]= 1π Ξ π‘
β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=π[βπβ π₯ ][βπΌπ 0
π½ βπΌ] [βπβ π₯ ]Want eigenvectors!
(βπΌπβ π) (βπΌβ π )βπ½ β0=0
π1=βπΌπ
π1β[ 1π½ / (πΌβπΌπ ) ]
π2=βπΌ
π2β[01 ]
Finding the βspecialβ direction
Dm
15
?π2β? [01 ]
0.5
-0.5
0.5-0.5
Dx
m
x
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[βπβπ₯ ]= 1π Ξ π‘
β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=π[βπβ π₯ ][βπΌπ 0
π½ βπΌ] [βπβ π₯ ]Want eigenvectors!
(βπΌπβ π) (βπΌβ π )βπ½ β0=0
π2=βπΌ
π2β[01 ]
Finding the βspecialβ direction
π1=βπΌπ
π1β[ 1π½ / (πΌβπΌπ ) ]
Dx
Dm
16
π½π=1πΌπ=2
πΌ=1π½=1
β1
0.25π1β[ 0.25β0.25]
0.5
-0.5
0.5-0.5
m
x
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[βπβπ₯ ]= 1π Ξ π‘
β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=π[βπβ π₯ ][βπΌπ 0
π½ βπΌ] [βπβ π₯ ]Want eigenvectors!
(βπΌπβ π) (βπΌβ π )βπ½ β0=0
π2=βπΌ
π2β[01 ]
Finding the βspecialβ direction
π1=βπΌπ
π1β[ 1π½ / (πΌβπΌπ ) ]
Dx
Dm
17
π½π=1πΌπ=2
πΌ=1π½=1
β1
?π1β? [ 1β1]
0.5
-0.5
0.5-0.5
Dx
m
x
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ]
π1=βπΌπ
π1β[ 1π½ / (πΌβπΌπ ) ]
π2=βπΌ
π2β[01 ]
Finding the βspecialβ direction
Dm
18
Trajectories along these directions do not bend0.5
-0.5
0.5-0.5
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ]Eigenvectors and eigenvalues provide analytic solution
Dx
m
x
Dm
19
[βπ(π‘)βπ₯ (π‘) ]=π 1 (π‘ )[ 1
π½ / (πΌβπΌπ )]
π1=βπΌπ
π1β[ 1π½ / (πΌβπΌπ ) ]
π2=βπΌ
π2β[01 ]
πβπππ‘
=ππ 1ππ‘
πβ π₯ππ‘
=πππ‘ [π 1 (π‘ ) π½
πΌβπΌπ ][βπΌπ 0π½ βπΌ]
βπΌπ
[πβπ/ππ‘πβπ₯ /ππ‘ ]= ππ 1
ππ‘ [ 1π½ /(πΌβπΌπ )]=βπΌππ 1 (π‘ )[ 1
π½ / (πΌβπΌπ )]
π½πΌβπΌπ
π π 1ππ‘
ππ 1ππ‘
=βπΌπ π 1 βΉπ 1 (π‘ )=π€1πβπΌπ π‘
[βπ(π‘)βπ₯ (π‘) ]=π€1π
βπΌπ π‘ [ 1π½ /(πΌβπΌπ )]
Trajectories along these directions do not bend
[βπΌπ 0π½ βπΌ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ]Eigenvectors and eigenvalues provide analytic solution
Dx
m
x
Dm
20
[βπ(π‘)βπ₯ (π‘) ]=π 2 (π‘ )[01]
π1=βπΌπ
π1β[ 1π½ / (πΌβπΌπ ) ]
π2=βπΌ
π2β[01 ]
πβπππ‘
=0 πβ π₯ππ‘
=ππ 2ππ‘
[βπΌπ 0π½ βπΌ] βπΌ
[πβπ/ππ‘πβπ₯ /ππ‘ ]= ππ 2
ππ‘ [01]=βπΌπ 2 (π‘ )[01 ]ππ 2ππ‘
=βπΌ π 2 βΉπ 2 (π‘ )=π€2πβπΌπ‘
[βπ(π‘)βπ₯ (π‘) ]=π€2π
βπΌ π‘[01 ]
Trajectories along these directions do not bend
21
Eigenvectors and eigenvalues provide analytic solution
[βπ(π‘)βπ₯ (π‘) ]=π€1π
βπΌπ π‘ [ 1π½ /(πΌβπΌπ )]+π€2π
βπΌπ‘[01 ][πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ][βπΌπ 0π½ βπΌ] βπΌβπΌπ
[βπΌπ 0π½ βπΌ] [βπβ π₯ ]=βπΌππ€1π
βπΌπ π‘[ 1π½/ (πΌβπΌπ )]βπΌπ€2π
βπΌπ‘ [01 ][βπΌπ 0π½ βπΌ]
[βπΌπ 0π½ βπΌ]
22
Eigenvectors and eigenvalues provide analytic solution
[βπ(π‘)βπ₯ (π‘) ]=π€1π
βπΌπ π‘ [ 1π½ /(πΌβπΌπ )]+π€2π
βπΌπ‘[01 ][πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ]
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[ π
ππ‘(π€1π
βπΌππ‘ β1+π€2πβπΌπ‘ β0 )
πππ‘ (π€1π
βπΌππ‘ βπ½
πΌβπΌπ
+π€2πβπΌπ‘ β1) ]
βπΌπ
βπΌβπΌπ
ΒΏ [ βπΌπ π€1πβπΌππ‘ β1
βπΌπ π€1πβπΌπ π‘ β
π½πΌβπΌπ
βπΌ π€2πβπΌπ‘ β1]
[βπΌπ 0π½ βπΌ] [βπβ π₯ ]=βπΌππ€1π
βπΌπ π‘[ 1π½/ (πΌβπΌπ )]βπΌπ€2π
βπΌπ‘ [01 ]
23
Eigenvectors and eigenvalues provide analytic solution
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ(π‘)βπ₯ (π‘) ]=π€1π
βπΌπ π‘ [ 1π½ /(πΌβπΌπ )]+π€2π
βπΌπ‘[01 ]General solution
Dx
m
x
Dm
[βπ(0)βπ₯ (0) ]=π€1[ 1
π½ / (πΌβπΌπ) ]+π€2[01]Initial conditions
Differential equations
0.5
-0.5
0.5-0.5
Dx
m
x
Dm
24
Eigenvectors and eigenvalues provide analytic solution
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ(π‘)βπ₯ (π‘) ]=π€1π
βπΌπ π‘ [ 1π½ /(πΌβπΌπ )]+π€2π
βπΌπ‘[01 ]General solution
[βπ(0)βπ₯ (0) ]=π€1[ 1
π½ / (πΌβπΌπ) ]+π€2[01]Initial conditions
Differential equations
0.5
-0.5
0.5-0.5
f
Dx
m
x
Dm
0.5
-0.5
0.5-0.5
25
Eigenvectors and eigenvalues provide analytic solution
[πβπ/ππ‘πβπ₯ /ππ‘ ]=[βπΌπ 0
π½ βπΌ] [βπβ π₯ ] [βπ(π‘)βπ₯ (π‘) ]=π€1π
βπΌπ π‘ [ 1π½ /(πΌβπΌπ )]+π€2π
βπΌπ‘[01 ]General solution
[βπ(0)βπ₯ (0) ]=π€1[ 1
π½ / (πΌβπΌπ) ]+π€2[01]Initial conditions
Differential equations
π½π=1πΌπ=2
πΌ=1π½=1
0 1 2 3 4 5t
0.0
0.1
0.2
0.3
0.4
0.5
Dxor
Dm
mRN
AProtein
26
Linear stability analysis
Transcription-translation model
Nullclines and critical points
Eigenvectors and eigenvalues
The cribsheet of linear stability analysis
f
m
x
[βπ (π‘+β π‘ )βπ₯ (π‘+βπ‘ ) ]β [βπ (π‘ )
β π₯ (π‘ ) ]+β π‘ [πβπ /ππ‘πβπ₯ /ππ‘ ]
27
Distinct positive eigenvalues
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0
28
Distinct positive eigenvalues
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0
29
Distinct positive eigenvalues
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0Node
30
Distinct negative eigenvalues
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0
π1<π2<0
Node
Node
31
Eigenvalues of opposite signs
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0
π1<π2<0
π1<0<π2
Node
Node
Saddle
32
Equal eigenvalues
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0
π1<π2<0
π1<0<π2
π1=π2>0
π1=π2<0
Node
Node
Saddle
Star
Star
Degenerate node
Degenerate node
33
Complex eigenvalues
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0
π1<π2<0
π1<0<π2
π1=π2>0
π1=π2<0
πΒ±=π Β± ππNode
Node
Saddle
Star
Star
Degenerate node
Degenerate node
34
Complex eigenvalues: Oscillatory and spiral solutions
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
πΒ±=π Β± ππ
[π₯ (π‘)π¦ (π‘)]=π€+ΒΏ π (π + π π ) π‘ΒΏ ΒΏ
[π₯ (π‘)π¦ (π‘)]=πππ‘ ΒΏ
[π₯ (π‘)π¦ (π‘)]=πππ‘ ΒΏ
Scaling Rotation
35
The big cribsheet of linear stability analysis
[π₯ (π‘)π¦ (π‘)]=π€1π
π1π‘ [ππ₯1
π π¦1 ]+π€2π
π2π‘ [ππ₯2
π π¦2 ][ππ₯ /ππ‘
π π¦ /ππ‘ ]=[π ππ π] [π₯π¦ ]
General solution Initial conditions
[π₯ (0)π¦ (0)]=π€1[ππ₯
1
ππ¦1 ]+π€2[ππ₯
2
π π¦2 ]
Differential equations
π1>π2>0
π1<π2<0
π1<0<π2
π1=π2>0
π1=π2<0
πΒ±=π Β± ππ
π <0
π=0
π >0Node
Node
Saddle
Star
Star
Degenerate node
Degenerate node
Center
Spiral
Spiral