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