Linear stability analysis
35
1 Linear stability analysis Transcription-translation model Nullclines and critical points Eigenvectors and eigenvalues The cribsheet of linear stability ana f m x [ ∆ ( + ∆ ) ∆ ( + ∆ ) ] ≅ [ ∆ ( ) ∆ ( ) ] + ∆ [ ∆ / ∆ / ]
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 Presentation
Transcript of Linear stability analysis
1
Linear stability analysis
Transcription-translation model
Nullclines and critical points
Eigenvectors and eigenvalues
The cribsheet of linear stability analysis
f
m
x
[∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
𝑑𝑥𝑑𝑡
=𝜕 𝑥
𝜕𝑅+¿𝑑𝑅+¿
𝑑𝑡+ 𝜕 𝑥𝜕𝑅−
𝑑 𝑅−
𝑑𝑡¿¿
𝑑𝑚𝑑𝑡
=𝜕𝑚
𝜕𝑅𝑚+¿𝑑𝑅𝑚+¿
𝑑𝑡+ 𝜕𝑚𝜕𝑅𝑚−
𝑑 𝑅𝑚−
𝑑𝑡¿¿
2
Transcription-translation model
f
𝛽𝑚
𝛽
𝛼𝑚
𝛼
m
x
𝛽𝑚+1 𝛼𝑚𝑚-1 𝛽𝑚+1 𝛼 𝑥-1
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝑅𝑚+¿ ¿
𝑅𝑚−
𝑅+¿¿
𝑅−
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
3
Nullclines and critical points
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚=0
𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥=0
𝛽𝑚=𝛼𝑚𝑚𝛽𝑚
𝛼𝑚
=𝑚
𝛽𝑚=𝛼 𝑥𝛽𝛼𝑚=𝑥
𝑚𝐶=𝛽𝑚𝛼𝑚
𝑥𝐶=𝛽𝛼𝛽𝑚
𝛼𝑚
m
x
0
𝑥=𝛽𝛼𝑚
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
0.5
1.0
1.00.5𝑚=𝛽𝑚
𝛼𝑚
𝑑𝑚
/𝑑𝑡=
0
𝑑𝑥/𝑑𝑡=0
f
4
Nullclines and critical points
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚
𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝑚𝐶=𝛽𝑚𝛼𝑚
𝑥𝐶=𝛽𝛼𝛽𝑚
𝛼𝑚
m
x
0
𝑥=𝛽𝛼𝑚
∆𝑚≅𝑑𝑚𝑑𝑡
∆ 𝑡>0
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
0.5
1.0
1.00.5𝑚=𝛽𝑚
𝛼𝑚
𝑑𝑚
/𝑑𝑡=
0
𝑑𝑥/𝑑𝑡=0
f
5
Nullclines and critical points
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚
𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝑚𝐶=𝛽𝑚𝛼𝑚
𝑥𝐶=𝛽𝛼𝛽𝑚
𝛼𝑚
m
x
0
𝑥=𝛽𝛼𝑚
∆𝑚≅𝑑𝑚𝑑𝑡
∆ 𝑡>0
∆ 𝑥≅𝑑𝑥𝑑𝑡∆ 𝑡>0
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
0.5
1.0
1.00.5𝑚=𝛽𝑚
𝛼𝑚
𝑑𝑚
/𝑑𝑡=
0
𝑑𝑥/𝑑𝑡=0
f
6
Nullclines and critical points
m
x
0
𝑥=𝛽𝛼𝑚𝑑𝑚
𝑑𝑡=𝛽𝑚−𝛼𝑚𝑚
𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝑚𝐶=𝛽𝑚𝛼𝑚
𝑥𝐶=𝛽𝛼𝛽𝑚
𝛼𝑚
∆𝑚≅𝑑𝑚𝑑𝑡
∆ 𝑡>0
∆ 𝑥≅𝑑𝑥𝑑𝑡∆ 𝑡>0
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
0.5
1.0
1.00.5𝑚=𝛽𝑚
𝛼𝑚
𝑑𝑚
/𝑑𝑡=
0
𝑑𝑥/𝑑𝑡=0
f
7
Nullclines and critical points
m
x
0 𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
1.00.5
𝑑𝑚/𝑑𝑡=0
𝑑𝑥/𝑑𝑡=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
[∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
9
Unbending trajectories
m
x
0 𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
1.00.5
f
10
Finding the “special” direction
m
x
0
Dx
Dm
𝑚𝐶=𝛽𝑚𝛼𝑚
𝑥𝐶=𝛽𝛼𝛽𝑚
𝛼𝑚
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
0.5
1.0
1.00.5
-0.25
0.25
0.25-0.25
f
11m
x Dx
Dm
𝑚𝐶=𝛽𝑚𝛼𝑚
𝑥𝐶=𝛽𝛼𝛽𝑚
𝛼𝑚
∆𝑚≔𝑚−𝛽𝑚𝛼𝑚
𝑑∆𝑚𝑑𝑡
= 𝑑𝑑𝑡 (𝑚− 𝛽𝑚
𝛼𝑚)
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
𝑑𝑚𝑑𝑡
=−𝛼𝑚(𝑚− 𝛽𝑚𝛼𝑚
)
𝑑∆𝑚𝑑𝑡
=−𝛼𝑚∆𝑚+0
∆ 𝑥≔𝑥− 𝛽𝛼𝛽𝑚
𝛼𝑚
𝑑∆ 𝑥𝑑𝑡
=𝑑𝑥𝑑𝑡
𝑑∆ 𝑥𝑑𝑡
=𝛽∆𝑚−𝛼 ∆ 𝑥
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]
𝑑𝑚𝑑𝑡
=𝛽𝑚−𝛼𝑚𝑚𝑑𝑥𝑑𝑡
=𝛽𝑚−𝛼 𝑥
Finding the “special” direction
𝑚𝐶=𝛽𝑚𝛼𝑚
𝑥𝐶=𝛽𝛼𝛽𝑚
𝛼𝑚
¿𝑑𝑚𝑑𝑡
12
Finding the “special” direction
m
x Dx
Dm
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]
0.5
-0.5
0.5-0.5
13
Finding the “special” direction
m
x Dx
Dm
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[∆𝑚∆𝑥 ]= 1𝜆 Δ 𝑡
∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=𝜆[∆𝑚∆ 𝑥 ][−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]Want eigenvectors!
(−𝛼𝑚− 𝜆) (−𝛼− 𝜆 )−𝛽 ∙0=0
𝜆1=−𝛼𝑚 𝜆2=−𝛼
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]𝑏2→[01 ]
0.5
-0.5
0.5-0.5
m
x
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[∆𝑚∆𝑥 ]= 1𝜆 Δ 𝑡
∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=𝜆[∆𝑚∆ 𝑥 ][−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]Want eigenvectors!
(−𝛼𝑚− 𝜆) (−𝛼− 𝜆 )−𝛽 ∙0=0
𝜆1=−𝛼𝑚
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]
𝜆2=−𝛼
𝑏2→[01 ]
Finding the “special” direction
Dx
Dm
14
0.25𝑏2→[ 00.25]
0.5
-0.5
0.5-0.5
m
x
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[∆𝑚∆𝑥 ]= 1𝜆 Δ 𝑡
∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=𝜆[∆𝑚∆ 𝑥 ][−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]Want eigenvectors!
(−𝛼𝑚− 𝜆) (−𝛼− 𝜆 )−𝛽 ∙0=0
𝜆1=−𝛼𝑚
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]
𝜆2=−𝛼
𝑏2→[01 ]
Finding the “special” direction
Dm
15
?𝑏2→? [01 ]
0.5
-0.5
0.5-0.5
Dx
m
x
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[∆𝑚∆𝑥 ]= 1𝜆 Δ 𝑡
∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=𝜆[∆𝑚∆ 𝑥 ][−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]Want eigenvectors!
(−𝛼𝑚− 𝜆) (−𝛼− 𝜆 )−𝛽 ∙0=0
𝜆2=−𝛼
𝑏2→[01 ]
Finding the “special” direction
𝜆1=−𝛼𝑚
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]
Dx
Dm
16
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
−1
0.25𝑏1→[ 0.25−0.25]
0.5
-0.5
0.5-0.5
m
x
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[∆𝑚∆𝑥 ]= 1𝜆 Δ 𝑡
∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=𝜆[∆𝑚∆ 𝑥 ][−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]Want eigenvectors!
(−𝛼𝑚− 𝜆) (−𝛼− 𝜆 )−𝛽 ∙0=0
𝜆2=−𝛼
𝑏2→[01 ]
Finding the “special” direction
𝜆1=−𝛼𝑚
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]
Dx
Dm
17
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=1
−1
?𝑏1→? [ 1−1]
0.5
-0.5
0.5-0.5
Dx
m
x
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]
𝜆1=−𝛼𝑚
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]
𝜆2=−𝛼
𝑏2→[01 ]
Finding the “special” direction
Dm
18
Trajectories along these directions do not bend0.5
-0.5
0.5-0.5
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]Eigenvectors and eigenvalues provide analytic solution
Dx
m
x
Dm
19
[∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑠1 (𝑡 )[ 1
𝛽 / (𝛼−𝛼𝑚 )]
𝜆1=−𝛼𝑚
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]
𝜆2=−𝛼
𝑏2→[01 ]
𝑑∆𝑚𝑑𝑡
=𝑑𝑠1𝑑𝑡
𝑑∆ 𝑥𝑑𝑡
=𝑑𝑑𝑡 [𝑠1 (𝑡 ) 𝛽
𝛼−𝛼𝑚 ][−𝛼𝑚 0𝛽 −𝛼]
−𝛼𝑚
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]= 𝑑𝑠1
𝑑𝑡 [ 1𝛽 /(𝛼−𝛼𝑚 )]=−𝛼𝑚𝑠1 (𝑡 )[ 1
𝛽 / (𝛼−𝛼𝑚 )]
𝛽𝛼−𝛼𝑚
𝑑 𝑠1𝑑𝑡
𝑑𝑠1𝑑𝑡
=−𝛼𝑚 𝑠1 ⟹𝑠1 (𝑡 )=𝑤1𝑒−𝛼𝑚 𝑡
[∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑤1𝑒
−𝛼𝑚 𝑡 [ 1𝛽 /(𝛼−𝛼𝑚 )]
Trajectories along these directions do not bend
[−𝛼𝑚 0𝛽 −𝛼]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]Eigenvectors and eigenvalues provide analytic solution
Dx
m
x
Dm
20
[∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑠2 (𝑡 )[01]
𝜆1=−𝛼𝑚
𝑏1→[ 1𝛽 / (𝛼−𝛼𝑚 ) ]
𝜆2=−𝛼
𝑏2→[01 ]
𝑑∆𝑚𝑑𝑡
=0 𝑑∆ 𝑥𝑑𝑡
=𝑑𝑠2𝑑𝑡
[−𝛼𝑚 0𝛽 −𝛼] −𝛼
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]= 𝑑𝑠2
𝑑𝑡 [01]=−𝛼𝑠2 (𝑡 )[01 ]𝑑𝑠2𝑑𝑡
=−𝛼 𝑠2 ⟹𝑠2 (𝑡 )=𝑤2𝑒−𝛼𝑡
[∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑤2𝑒
−𝛼 𝑡[01 ]
Trajectories along these directions do not bend
21
Eigenvectors and eigenvalues provide analytic solution
[∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑤1𝑒
−𝛼𝑚 𝑡 [ 1𝛽 /(𝛼−𝛼𝑚 )]+𝑤2𝑒
−𝛼𝑡[01 ][𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ][−𝛼𝑚 0𝛽 −𝛼] −𝛼−𝛼𝑚
[−𝛼𝑚 0𝛽 −𝛼] [∆𝑚∆ 𝑥 ]=−𝛼𝑚𝑤1𝑒
−𝛼𝑚 𝑡[ 1𝛽/ (𝛼−𝛼𝑚 )]−𝛼𝑤2𝑒
−𝛼𝑡 [01 ][−𝛼𝑚 0𝛽 −𝛼]
[−𝛼𝑚 0𝛽 −𝛼]
22
Eigenvectors and eigenvalues provide analytic solution
[∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑤1𝑒
−𝛼𝑚 𝑡 [ 1𝛽 /(𝛼−𝛼𝑚 )]+𝑤2𝑒
−𝛼𝑡[01 ][𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ]
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[ 𝑑
𝑑𝑡(𝑤1𝑒
−𝛼𝑚𝑡 ∙1+𝑤2𝑒−𝛼𝑡 ∙0 )
𝑑𝑑𝑡 (𝑤1𝑒
−𝛼𝑚𝑡 ∙𝛽
𝛼−𝛼𝑚
+𝑤2𝑒−𝛼𝑡 ∙1) ]
−𝛼𝑚
−𝛼−𝛼𝑚
¿ [ −𝛼𝑚 𝑤1𝑒−𝛼𝑚𝑡 ∙1
−𝛼𝑚 𝑤1𝑒−𝛼𝑚 𝑡 ∙
𝛽𝛼−𝛼𝑚
−𝛼 𝑤2𝑒−𝛼𝑡 ∙1]
[−𝛼𝑚 0𝛽 −𝛼] [∆𝑚∆ 𝑥 ]=−𝛼𝑚𝑤1𝑒
−𝛼𝑚 𝑡[ 1𝛽/ (𝛼−𝛼𝑚 )]−𝛼𝑤2𝑒
−𝛼𝑡 [01 ]
23
Eigenvectors and eigenvalues provide analytic solution
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑤1𝑒
−𝛼𝑚 𝑡 [ 1𝛽 /(𝛼−𝛼𝑚 )]+𝑤2𝑒
−𝛼𝑡[01 ]General solution
Dx
m
x
Dm
[∆𝑚(0)∆𝑥 (0) ]=𝑤1[ 1
𝛽 / (𝛼−𝛼𝑚) ]+𝑤2[01]Initial conditions
Differential equations
0.5
-0.5
0.5-0.5
Dx
m
x
Dm
24
Eigenvectors and eigenvalues provide analytic solution
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑤1𝑒
−𝛼𝑚 𝑡 [ 1𝛽 /(𝛼−𝛼𝑚 )]+𝑤2𝑒
−𝛼𝑡[01 ]General solution
[∆𝑚(0)∆𝑥 (0) ]=𝑤1[ 1
𝛽 / (𝛼−𝛼𝑚) ]+𝑤2[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
[𝑑∆𝑚/𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]=[−𝛼𝑚 0
𝛽 −𝛼] [∆𝑚∆ 𝑥 ] [∆𝑚(𝑡)∆𝑥 (𝑡) ]=𝑤1𝑒
−𝛼𝑚 𝑡 [ 1𝛽 /(𝛼−𝛼𝑚 )]+𝑤2𝑒
−𝛼𝑡[01 ]General solution
[∆𝑚(0)∆𝑥 (0) ]=𝑤1[ 1
𝛽 / (𝛼−𝛼𝑚) ]+𝑤2[01]Initial conditions
Differential equations
𝛽𝑚=1𝛼𝑚=2
𝛼=1𝛽=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
[∆𝑚 (𝑡+∆ 𝑡 )∆𝑥 (𝑡+∆𝑡 ) ]≅ [∆𝑚 (𝑡 )
∆ 𝑥 (𝑡 ) ]+∆ 𝑡 [𝑑∆𝑚 /𝑑𝑡𝑑∆𝑥 /𝑑𝑡 ]
27
Distinct positive eigenvalues
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0
28
Distinct positive eigenvalues
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0
29
Distinct positive eigenvalues
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0Node
30
Distinct negative eigenvalues
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0
𝜆1<𝜆2<0
Node
Node
31
Eigenvalues of opposite signs
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0
𝜆1<𝜆2<0
𝜆1<0<𝜆2
Node
Node
Saddle
32
Equal eigenvalues
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0
𝜆1<𝜆2<0
𝜆1<0<𝜆2
𝜆1=𝜆2>0
𝜆1=𝜆2<0
Node
Node
Saddle
Star
Star
Degenerate node
Degenerate node
33
Complex eigenvalues
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0
𝜆1<𝜆2<0
𝜆1<0<𝜆2
𝜆1=𝜆2>0
𝜆1=𝜆2<0
𝜆±=𝜎 ± 𝑖𝜔Node
Node
Saddle
Star
Star
Degenerate node
Degenerate node
34
Complex eigenvalues: Oscillatory and spiral solutions
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆±=𝜎 ± 𝑖𝜔
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤+¿ 𝑒 (𝜎 + 𝑖 𝜔 ) 𝑡¿ ¿
[𝑥 (𝑡)𝑦 (𝑡)]=𝑒𝜎𝑡 ¿
[𝑥 (𝑡)𝑦 (𝑡)]=𝑒𝜎𝑡 ¿
Scaling Rotation
35
The big cribsheet of linear stability analysis
[𝑥 (𝑡)𝑦 (𝑡)]=𝑤1𝑒
𝜆1𝑡 [𝑏𝑥1
𝑏 𝑦1 ]+𝑤2𝑒
𝜆2𝑡 [𝑏𝑥2
𝑏 𝑦2 ][𝑑𝑥 /𝑑𝑡
𝑑 𝑦 /𝑑𝑡 ]=[𝑎 𝑏𝑐 𝑑] [𝑥𝑦 ]
General solution Initial conditions
[𝑥 (0)𝑦 (0)]=𝑤1[𝑏𝑥
1
𝑏𝑦1 ]+𝑤2[𝑏𝑥
2
𝑏 𝑦2 ]
Differential equations
𝜆1>𝜆2>0
𝜆1<𝜆2<0
𝜆1<0<𝜆2
𝜆1=𝜆2>0
𝜆1=𝜆2<0
𝜆±=𝜎 ± 𝑖𝜔
𝜎 <0
𝜎=0
𝜎 >0Node
Node
Saddle
Star
Star
Degenerate node
Degenerate node
Center
Spiral
Spiral
