Traveling Waves in a bistable Reaction-Diffusion System
Wave phenomena in RD systems
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∂u∂t
= f (u) + D∂2u∂x 2
, ∂u∂x ±∞
= 0
Keener & Sneyd (1998) Mathematical Physiology; p 270-275
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f (u) = au(u −1)(α − u) 0 <α <1
f (u) = −u + H(u −α) 0 <α <1
With f cubic:
or
Piecewise linear:
1 α
McKean, 1970
f(u)
f(u)
U is the shape of the wave, z is the position along the wave front
Bistable well-mixed system
1 α Steady states:
u = 0, α, 1
With diffusion
1 α
u=1
u=0 x
Moving Wave
u=1
u=0 x
Wave profile
z
Traveling wave coordinates: z=x-ct, U(z)=u(x,t)
u=1
u=0
What is the wave profile? Traveling wave coordinates: z=x-ct, U(z)=u(x,t) Scaling and transformation of coordinates:
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∂∂t→−c d
dz∂∂x
→ddz
∂u∂t
= f (u) +∂ 2u∂x 2
→ − c dUdz
= f (U) +d2Udz2PDE ODE
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Uzz − cUz + f (U) = 0
Uz =WWz = cW − f (U)
2nd order (nonlin) ODE
Equivalent ODE system: We can study this qualitatively in the UW phase plane to get insight into the shape of the wave.
Example: f(u)=u(1-u)(u- α)
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∂u∂t
= f (u) +∂ 2u∂x 2
W
U U
W
W’=0
U’=0
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Uzz − cUz + f (U) = 0
Uz =WWz = cW − f (U)
(Rescaled)
Xpp file # bistable.ode # classic example of a wave joining 0 and 1 # u_t = u_xx + u(1-u)(u-a) # # -cu'=u''+u(1-u)(u-a) f(u)=u*(1-u)*(u-a) par c=0,a=.25 u'=up up'=-c*up-f(u) init u=1 @ xp=u,yp=up,xlo=-.5,xhi=1.5,ylo=-.5,yhi=.5 done
Bard Ermentrout’s XPP file
Bistable.ode
Interpretation: Each trajectory describes how U (and W) vary as z increases over range -∞< z < ∞
In bio applications: u= density>0, so we want: a positive bounded wave: Look for a positive bounded trajectory connecting two steady states.
W
U U
Here is a heteroclinic trajectory:
W
U
Here it is on its own:
U
W
And here is the shape of the wave it represents:
U
z
What is the speed of the wave? • A reaction-diffusion equation with “bistable” kinetics will admit a traveling wave that looks like a moving “front”, with high level at back, and low level ahead.
• But how fast does the wave move and does it ever stop?
There is a cute trick that allows us to answer this question for arbitrary function f(u).
Determining wave speed
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∂u∂t
= f (u) +∂ 2u∂x 2
, z = x − ct
−c dUdz
− f (U) =∂ 2Udz2
−c dUdz
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
− f (U) dUdz
⎛
⎝ ⎜
⎞
⎠ ⎟ =12ddz
dUdz
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
−c dUdz
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
−∞
∞
∫ dz − f (U)uback
ufront
∫ dU =12
ddz
dUdz
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
= 0−∞
∞
∫
−c dUdz
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
−∞
∞
∫ dz = f (U)uback
ufront
∫ dU
Multiply by dU/dz,
uback ufront
z
c
integrate,
The wave speed:
Finally..
z
u=1
u=0
f(u)=u(1-u)(u- α)
Result!
But what does this tell us????
Which way does it move?
1 α
f(u)
1 α
f(u)
Always positive
Which way does it move?
1 α
f(u)
1 α
f(u)
> 0 < 0
c > 0 c < 0
Which way does it move?
1
f(u)
1
f(u)
c > 0 c < 0
Which way does it move?
1
f(u)
1
f(u)
c > 0 c < 0
We can now answer the question:
Under what conditions would the wave stop?
Wave stops:
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0 = c = − f (U)0
1
∫ dU dUdz
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
−∞
∞
∫ dz
f (U)0
1
∫ dU = 0
Speed is zero if:
i.e.: if :
Geometry of stalled wave:
1 α
f(u)
Maxwell condition:
1 α
f(u)
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f (U)0
1
∫ dU = 0
“Equal areas”
“Maxwell condition”:
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f (U)0
α
∫ dU = − f (U)α
1
∫ dU
Can we get an explicit solution for the wave speed?
Explicit solution? In some special cases, e.g. f(u) cubic or piecewise linear, can calculate wave speed fully by this method.
(See Keener & Sneyd p 274, Murray p 305)
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f (u) = au(u −1)(α − u) 0 <α <1For
Speed of the wave is:
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c =a2(1− 2α)
Implications: Wave moves right (c>0) if α<1/2 Moves left (c<0) if α>1/2
Wave stops (c=0) precisely for one value of the parameter, α=1/2
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f (u) = au(u −1)(α − u) 0 <α <1