6 Two ODEs: steady states and stability Flashcards
u nullcines
v nullclines
u nullcines
points s.t
du/dt=0
v nullclines
points s.t
dv/dt=0
Act as a boundary showing the change of sign of du/dt and useful to identify SS dynamics
in general 2 ODEs: interacting populations
du/dt=f(u(t),v(t))
dv/dt=g(u(t),v(t))
Steady states
Nullclines
Steady states
Pairs (u_s,v_s) s.t. f(u_s,v_s) = 0 AND g(u_s,v_s)=0
u-nullclines points s.t. f(u(t),v(t))=0
v-nullclines points s.t. g(u(t),v(t))=0
in general 2 ODEs: interacting populations
du/dt=f(u(t),v(t))
dv/dt=g(u(t),v(t))
STABILITY
Linearize the eq. about the steady state: Consider
u(t) = us + U(t) and v(t) = vs + V (t).
Then
f(u(t), v(t)) = f(us, vs) + U(t)
(∂f/∂u)(us, vs) + V (t)(∂f/∂v) (us, vs) + . . .
g(u(t), v(t)) = g(us, vs) + U(t)(∂g/∂u)(us, vs) + V (t)(∂g/∂v) (us, vs) +
close enough to u_s,v_s values U(t) and V(t) are small and dU/dt=du/dt dV/dt=dv/dt
so THE SET OF ODES below IS SAID TO BE LINEARISED ABOUT THE STEADY STATE (u_s,v_s)
dU(t)/dt =
(∂f/∂u)(u_s, v_s)U(t) + (∂f/∂v) (u_s, v_s)V (t),
dV (t)/dt =
∂g/∂u(u_s, v_s)U(t) + (∂g/∂v) (us, vs)V(t)
in general 2 ODEs: interacting populations
du/dt=f(u(t),v(t))
dv/dt=g(u(t),v(t))
The linearised equation
We write the linearised equation as a matrix equation
d/dt
(U(t))
(V (t))
=
[(∂f/∂u)(us, vs) (∂f/∂v) (us, vs) ]
[∂g/∂u(us, vs) (∂g/∂v) (us, vs)]
*[U(t)]
[V (t)]
Jacobian evaluated at (u_s,v_s) s.t.
d/dt (U)
(V) =
J_(u_s,v_s) * (U)
(V)
in general 2 ODEs: interacting populations
du/dt=f(u(t),v(t))
dv/dt=g(u(t),v(t))
The linearised equation GENERAL SOLUTON linking to eigenvalues
The general solution of the linearised equation is
(U(t))
(V(t))
=
c₁(α₁)e^{λ₁t}+c₂(α₂)e^{λ₂t}
(β₁) (β₂)
where above are eigenvalues and eigenvectors
s.t. J(u_s,v_s) v₁=λ₁v etc
general sol approach SS dep on eigenvalues so useful
Matrix equations in general
dx(t)/dt= Ax(t), where A =
[a b]
[c d]
and x =
(U(t))
(V (t))
is shorthand for the pair of differential equations
dU(t)/dt = aU(t) + bV(t)
dV (t)/dt = cU(t) + dV (t),
where a, b, c and d are constants. The origin (the point U = V = 0) is a steady state.
Trajectories started at the origin remain there. Notice that, because we have shifted the origin
when we changed variables from (u(t), v(t)), the new variables U(t) and V (t) may take negative
values
How do the eigenvalues help?
The type of steady state at the origin tells us what happens to trajectories that do not start
at the origin:
Jacobian A: how do i find eigenvalues?
eigenvalues of the matrix A, which satisfy det(A − λI) = 0, where I is the 2 × 2 identity matrix. Explicitly
λ^2 − (a + d)λ + ad − bc = 0 .
If form of matrix:
[a 0]
[0 d ] λ₁=a λ₂=d
[a 0 ]
[c d ] λ₁=a λ₂=d
[a b]
[0 d ] λ₁=a λ₂=d
eigenvalues and stability
- If the eigenvalues are REAL and have the SAME SIGN:
(a) STABLE NODE (both negative)
(b)UNSTABLE NODE (both positive). - If the eigenvalues are REAL and have OPPOSITE SIGNS then
a SADDLE POINT. - If the eigenvalues are COMPLEX numbers with non-zero imaginary part, then
(a)STABLE SPIRAL (real part is negative),
(b) UNSTABLE SPIRAL (if the real part is positive), or
(c) CENTRE (if the real part is zero, neither un\stable).
In order for the steady state to be STABLE both eigenvalues must have negative real part.
TRACE DET method for eigenvalues
(a − λ)(d − λ) − cb =0 (evaluating the determinant)
λ^2 − trace(A)λ + det(A) =0
trace(A) = a + d
det(A) = ad − bc
trace(A)=λ₁+ λ₂
det(A)= =λ₁* λ₂
The two eigenvalues of A are
λ =
(1/2) trace(A)
±(1/2) sqrt(trace(A)^2− 4 det(A))
how to use signs of trace and det tell us about stability
If det(a)>0 and trace(A)<0 then SS is STABLE
If det(a)<0 or trace(A)>0 then SS is UNSTABLE
two ODES interacting populations
Example 2: Viral infections
All plants and animals on earth are infected by parasitical viruses that reproduce inside their
cells. Viruses are not cells; they can only make copies of themselves inside a cell. New virus
particles, produced inside a host cell, may be released from that cell to infect other cells.
A simple model that describes what happens to the populations of virus particles V and infected cells I, when a small amount of virus enters a host, is
dV(t)/dt = pI(t) − cV (t)
dI(t)/dt = aV (t) − δI(t).
The constants p, c, a and δ are positive.
V(t) amount of extracellular virus, measured in the blood of an infected (particles per ml of blood)
I(t) amount of infected cells at time t≥ 0
p, the rate at which infected cells release new virus;
c, the rate at which extracellular virus is cleared from the blood (for example, by cells in
the liver or kidney);
a, the rate at which a released virus is able to infect uninfected host cells;
δ, the rate at which infected cells die
two ODES interacting populations
Example 2: Viral infections
dV(t)/dt = pI(t) − cV (t)
dI(t)/dt = aV (t) − δI(t).
SS
(0,0) stability:
J_(0,0) =
[-c p]
[a −δ]
trace <0
unstable if: infection will grow if det<0
ie ap/cδ > 1
(R_0>1)
two ODES interacting populations
Example 2: Viral infections
R_0
the steady state (0,0) stability depends on R_0
If R_0>1 then viral population grows
If R_0<1 then virus dies out
R_ 0 =(pa)/(cδ)
is called the “basic reproductive number”. It is non-dimensional and can be understood as
follows: a viral particle survives outside host cells for a mean time of 1/c before it is cleared, so
on average it infects a/c host cells before being cleared. On average, an infected cell lives for a
time 1/δ and releases p/δ virus particles during this time. Thus the product
(a/c) × (p/δ)
is the mean number of new virus particles that result from each initial viral particle. If this number is greater than 1 then the viral population will grow, otherwise it will die out.
If the infection does grow and establish itself in the body, we cannot use (6.5) as a model.
We must also consider the population of uninfected cells that the virus is able to infect, often
called target cells. Researchers such as Rob de Boer (Utrecht) and Alan Perelson (Los Alamos
National Laboratory) have developed sophisticated models over the last 30 years.
