Problem 5 comes with a worked problem as a hint. Please only solve if you know what you're doing.
Posted: Mon May 23, 2022 10:28 am
Problem 5 comes with a worked problem as a hint. Please only
solve if you know what you're doing.
Problem 5 (20pts) Consider the toggle switch model given by a a = X, Ý y. 1+ y2 1 + x2 Determine the number and stability of all equilibria as a function of a > 0. = Hint Show that equilibrium points satisfy either y = x or y = 1/x. For both of these, find the number of positive equilibria (you might find Descartes' rule of sign helpful in the former) and how their existence depends on the value of a > 0. Following this, sketch the nullclines in each case separately, linearize the dynamics, and use the implicit function theorem to conclude stability properties of the equilibria by analyzing the sign of the trace and determinant of the system matrix obtained via the linearization. See Example 1 under BFS_Practice_ Problems/Modules.
CHAPTER #5 Show that the toggle switch can be either monorable or bistable, and derive parameter conditious ensuring these stability profiles, focus . the symmetric realization with monomerization . ou and Since this example is rather long, I included the outline of the solution below with the main steps and tasks within. 1. Dynamics re-scale concentrations 2. De-dimensionalization re-scale time collapse 3 parameters into 1 Toggle switch localize equilibria 3. Monomerization + stability analysis via linearization graphical stability analysis localize equilibria 3. Dimerization stability analysis via linearization graphical stability analysis
Dynamics in case of symmetric realization for wanawerization I dimerization ) Crol and u=2 žk dX dt kh tua - 8x + 3 parameters ă, ki8 dY à kh dt 8Y k+xn 2 De-demensionalize dynamics to reduce the - muce ber of parameter Х X = X ༢ ° Y K ī=tr all 3 are dimensionless よど 'd (x K) d (=/;) -8kx / Ligh : K + "K Jag a I parameter dx dł 85 1 + Х -X ye ityn similarly : 2 = st y tx"
© Mocomerization (n=1) 3a] Localize equilibrium points In case of amerization wowo : n=1 a = x (1+y) = y(1+x) xay at equilibrium Substituting this back Х 1+x x (1+) - a x?7 x-a = 0 - 1 - -Sitha 2. zo X I Į I tha + 2 X eq -l+Vitha o 2 Huique positive equilibriu : +/Itha Хэ y a There is a single positive equilibrium in case of noraueritation point
3] Stability analysis via linearization Linearize the dynamics around the equilibria x = f(xey? ty Ý ý Itt my g(xry) ola la capace g (1+x)² -1 ay (itys" ( 용 ag 2 ។ We are interested in the dynamics at the unique positive equilibriu -lt Sitha *- qe ។ 2 Trace . 2 x² Determinant. 1 so -1+itha 2 :)4
1 % Monomer alpha = linspace(0,1e3, 1e2); x = (-1+sqrt(1+4*alpha))/2; d = 1 - alpha.^2./(1+x).^4; 0.9 0.8 0.7 0.6 figure() plot(alpha, d) grid on axis square xlabel('$\alpha$') ylabel('determinant') determinant 0.5 0.4 0.3 0.2 0.1 0 0 100 200 300 400 500 600 700 800 900 1000 0 The determinant is positive and the trace is negative independent of the value of unique positive equilibrium equilibrium is stable a>o al x=5 5 5 4.5 4.5 3.5 3.5 3 3 $ 2.5 9 2.5 2N 2 1.5 1.5 1 1 0.5 0.5 0 0 0.5 1 1 1.5 2 2 2.5 3 3.5 4 4.5 5 0 0 0.5 1 1.5 2 3 3.5 4 4.5 5 21 2.5 01
a 3c] Let's do 3b] in simpler way using the implicit function theorem via graphical analysis if(x,y) = 0 3.5 dy dx 3 f(x,y) = 0 2.5 2 that alla costeslapen 1.5 1 7689) dy 았 dx 0.5 360p) 0 0 0.5 1 1.5 2.5 3 3.5 4 x At the equilibrium de filtracupen dy . < dx co dx gorgeo co afico ag 1 < 0 of lola co og 84Jco of da > determinant 警英 ay of og of de ay dx [trace co the unique positive equilibrium is stable af ax
solve if you know what you're doing.
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- Problem 5 Comes With A Worked Problem As A Hint Please Only Solve If You Know What You Re Doing 3 (98.43 KiB) Viewed 11 times
CHAPTER #5 Show that the toggle switch can be either monorable or bistable, and derive parameter conditious ensuring these stability profiles, focus . the symmetric realization with monomerization . ou and Since this example is rather long, I included the outline of the solution below with the main steps and tasks within. 1. Dynamics re-scale concentrations 2. De-dimensionalization re-scale time collapse 3 parameters into 1 Toggle switch localize equilibria 3. Monomerization + stability analysis via linearization graphical stability analysis localize equilibria 3. Dimerization stability analysis via linearization graphical stability analysis
Dynamics in case of symmetric realization for wanawerization I dimerization ) Crol and u=2 žk dX dt kh tua - 8x + 3 parameters ă, ki8 dY à kh dt 8Y k+xn 2 De-demensionalize dynamics to reduce the - muce ber of parameter Х X = X ༢ ° Y K ī=tr all 3 are dimensionless よど 'd (x K) d (=/;) -8kx / Ligh : K + "K Jag a I parameter dx dł 85 1 + Х -X ye ityn similarly : 2 = st y tx"
© Mocomerization (n=1) 3a] Localize equilibrium points In case of amerization wowo : n=1 a = x (1+y) = y(1+x) xay at equilibrium Substituting this back Х 1+x x (1+) - a x?7 x-a = 0 - 1 - -Sitha 2. zo X I Į I tha + 2 X eq -l+Vitha o 2 Huique positive equilibriu : +/Itha Хэ y a There is a single positive equilibrium in case of noraueritation point
3] Stability analysis via linearization Linearize the dynamics around the equilibria x = f(xey? ty Ý ý Itt my g(xry) ola la capace g (1+x)² -1 ay (itys" ( 용 ag 2 ។ We are interested in the dynamics at the unique positive equilibriu -lt Sitha *- qe ។ 2 Trace . 2 x² Determinant. 1 so -1+itha 2 :)4
1 % Monomer alpha = linspace(0,1e3, 1e2); x = (-1+sqrt(1+4*alpha))/2; d = 1 - alpha.^2./(1+x).^4; 0.9 0.8 0.7 0.6 figure() plot(alpha, d) grid on axis square xlabel('$\alpha$') ylabel('determinant') determinant 0.5 0.4 0.3 0.2 0.1 0 0 100 200 300 400 500 600 700 800 900 1000 0 The determinant is positive and the trace is negative independent of the value of unique positive equilibrium equilibrium is stable a>o al x=5 5 5 4.5 4.5 3.5 3.5 3 3 $ 2.5 9 2.5 2N 2 1.5 1.5 1 1 0.5 0.5 0 0 0.5 1 1 1.5 2 2 2.5 3 3.5 4 4.5 5 0 0 0.5 1 1.5 2 3 3.5 4 4.5 5 21 2.5 01
a 3c] Let's do 3b] in simpler way using the implicit function theorem via graphical analysis if(x,y) = 0 3.5 dy dx 3 f(x,y) = 0 2.5 2 that alla costeslapen 1.5 1 7689) dy 았 dx 0.5 360p) 0 0 0.5 1 1.5 2.5 3 3.5 4 x At the equilibrium de filtracupen dy . < dx co dx gorgeo co afico ag 1 < 0 of lola co og 84Jco of da > determinant 警英 ay of og of de ay dx [trace co the unique positive equilibrium is stable af ax