Nonlinear transport phenomena: models, method of solving and unusual features (2)

Nonlinear transport phenomena:models, method of solving and unusual

features

Vsevolod Vladimirov

AGH University of Science and technology, Faculty of AppliedMathematics

Krakow, August 10, 2010

KPI, 2010 Nonlinear transport phenomena, Burgers Eqn. 1 / 29

Burgers equation

Consider the second law of Newton for viscous incompressiblefluid:

∂ ui

∂ t+ uj ∂ u

i

∂ xj+

1

ρ

∂ P

∂ xi= ν∆ui, i = 1, ...n, n = 1, 2 or 3,

~u(t, x) is the velocity field,∂∂ t + uj ∂

∂ xj is the time ( substantial) derivative;ρ is the constant density ;P is the pressure ;ν is the viscosity coefficient;∆ =

∑ni=1

∂2

∂ x2i

is the Laplace operator.

For P = const, n = 1, we get the Burgers equation

ut + u ux = ν ux x. (1)


Burgers equation


∂ ui

∂ t+ uj ∂ u

i

∂ xj+

1

ρ

∂ P

∂ xi= ν∆ui, i = 1, ...n, n = 1, 2 or 3,



∑ni=1

∂2

∂ x2i



ut + u ux = ν ux x. (1)


Burgers equation


∂ ui

∂ t+ uj ∂ u

i

∂ xj+

1

ρ

∂ P

∂ xi= ν∆ui, i = 1, ...n, n = 1, 2 or 3,



∑ni=1

∂2

∂ x2i



ut + u ux = ν ux x. (1)


Hyperbolic generalization to Burgers equation

Let us consider delayed equation

∂ u(t+ τ, x)

∂ t+ u(t, x)ux(t, x) = ν ux x(t, x).

Applying to the term ∂ u(t+τ, x)∂ t the Taylor formula, we get, up

to O(τ2) the equation called the hyperbolic generalization ofthe Burgers equation (GBE to abbreviate):

τ utt + ut + u ux = ν ux x. (2)

GBE appears when modeling transport phenomena in mediapossessing internal structure: granular media,polymers, cellularstructures in biology.




∂ u(t+ τ, x)

∂ t+ u(t, x)ux(t, x) = ν ux x(t, x).








∂ u(t+ τ, x)

∂ t+ u(t, x)ux(t, x) = ν ux x(t, x).








∂ u(t+ τ, x)

∂ t+ u(t, x)ux(t, x) = ν ux x(t, x).






Various generalizations of Burgers equation

Convection-reaction diffusion equation

ut + u ux = ν [un ux]x + f(u), (3)

and its hyperbolic generalization

τ ut t + ut + u ux = ν [un ux]x + f(u) (4)


Various generalizations of Burgers equation

Convection-reaction diffusion equation

ut + u ux = ν [un ux]x + f(u), (3)

and its hyperbolic generalization

τ ut t + ut + u ux = ν [un ux]x + f(u) (4)


Solution to BELemma 1. BE is connected with the equation

ψt +1

2ψ2

x = ν ψx x (5)

by means of the transformation

ψx = u, ψt = ν ux −u2

2. (6)

Lemma 2. The equation (5) is connected with the heattransport equation

Φt = ν Φx x


ψ = −2 ν log Φ.



ψt +1

2ψ2

x = ν ψx x (5)



2. (6)


Φt = ν Φx x





ψt +1

2ψ2

x = ν ψx x (5)



2. (6)


Φt = ν Φx x




Corollary. Solution to the initial value problem

ut + uux = ν ux x, (7)u(0, x) = F (x)

is connected with the solution to the initial value problem

Φt = ν Φx x, (8)

Φ(0, x) = exp

[− 1

2 ν

∫ x

0F (z) d z

]:= θ(x)

via the transformation

u(t, x) = −2 ν {log[Φ(t, x)]}x .



ut + uux = ν ux x, (7)u(0, x) = F (x)


Φt = ν Φx x, (8)

Φ(0, x) = exp

[− 1

2 ν

∫ x

0F (z) d z

]:= θ(x)


u(t, x) = −2 ν {log[Φ(t, x)]}x .



ut + uux = ν ux x, (7)u(0, x) = F (x)


Φt = ν Φx x, (8)

Φ(0, x) = exp

[− 1

2 ν

∫ x

0F (z) d z

]:= θ(x)


u(t, x) = −2 ν {log[Φ(t, x)]}x .


Let us remind, that solution to the initial value problem (8) canbe presented by the formula

Φ(t, x) =1√

4π ν t

∫ ∞

−∞θ(ξ) e−

(x−ξ)2

4 ν t d ξ.

Corollary. Solution to the initial value problem (7) is given bythe formula

u(t, x) =

∫∞−∞

x−ξt e−

f(ξ;t, x)2 ν d ξ∫∞

−∞ e− f(ξ;t, x)

2 ν d ξ, (9)

where

f(ξ; t, x) =

∫ ξ

0F (z) d z +

(x− ξ)2

2 t.

So, the formula (9)completely defines the solution to Cauchyproblem to BE!



Φ(t, x) =1√

4π ν t

∫ ∞

−∞θ(ξ) e−

(x−ξ)2

4 ν t d ξ.


u(t, x) =

∫∞−∞

x−ξt e−

f(ξ;t, x)2 ν d ξ∫∞

−∞ e− f(ξ;t, x)

2 ν d ξ, (9)

where

f(ξ; t, x) =

∫ ξ

0F (z) d z +

(x− ξ)2

2 t.




Φ(t, x) =1√

4π ν t

∫ ∞

−∞θ(ξ) e−

(x−ξ)2

4 ν t d ξ.


u(t, x) =

∫∞−∞

x−ξt e−

f(ξ;t, x)2 ν d ξ∫∞

−∞ e− f(ξ;t, x)

2 ν d ξ, (9)

where

f(ξ; t, x) =

∫ ξ

0F (z) d z +

(x− ξ)2

2 t.




Φ(t, x) =1√

4π ν t

∫ ∞

−∞θ(ξ) e−

(x−ξ)2

4 ν t d ξ.


u(t, x) =

∫∞−∞

x−ξt e−

f(ξ;t, x)2 ν d ξ∫∞

−∞ e− f(ξ;t, x)

2 ν d ξ, (9)

where

f(ξ; t, x) =

∫ ξ

0F (z) d z +

(x− ξ)2

2 t.



Example: solution of the ”point explosion” problem

Letu(0, x) = F (x) = Aδ(x)H(x),

δ(x) = limt→+0

1√4π ν t

e−(x−ξ)2

4 ν t , H(x) =

{1 if x ≥ 0,

0 if x < 0.

Figure:KPI, 2010 Nonlinear transport phenomena, Burgers Eqn. 8 / 29

Example: solution of the ”point explosion” problem

Letu(0, x) = F (x) = Aδ(x)H(x),

δ(x) = limt→+0

1√4π ν t

e−(x−ξ)2

4 ν t , H(x) =

{1 if x ≥ 0,

0 if x < 0.

Figure:KPI, 2010 Nonlinear transport phenomena, Burgers Eqn. 8 / 29

Performing simple but tedious calculations, we finally get thefollowing solution to the point explosion problem:

u(t, x) =

√ν

t

(eR − 1

)e−

x2

4 ν t

√π

2

[(eR + 1) + erf( x√

4 ν t) (1− eR)

] ,where

erf(z) =2√π

∫ z

0e−x2

d x,

R = A2 ν plays the role of the Reynolds number!



u(t, x) =

√ν

t

(eR − 1

)e−

x2

4 ν t

√π

2

[(eR + 1) + erf( x√

4 ν t) (1− eR)

] ,where

erf(z) =2√π

∫ z

0e−x2

d x,




u(t, x) =

√ν

t

(eR − 1

)e−

x2

4 ν t

√π

2

[(eR + 1) + erf( x√

4 ν t) (1− eR)

] ,where

erf(z) =2√π

∫ z

0e−x2

d x,




u(t, x) =

√ν

t

(eR − 1

)e−

x2

4 ν t

√π

2

[(eR + 1) + erf( x√

4 ν t) (1− eR)

] ,where

erf(z) =2√π

∫ z

0e−x2

d x,



Suppose now, that ν becomes very large. Then

R→ 0 eR ≈ 1 +R, erf(

x√4 ν t

)≈ 0,

and

u(t, x) =

√ν

t

A2 ν

e−x2

4 ν t

√π

+ O(R2) ≈ A√4 π ν t

e−x2

4 ν t .

Corollary.Solution to the ”point explosion” problem for the BEapproaches solution to the ”heat explosion” problem for thelinear heat transport equation, when ν becomes large.



R→ 0 eR ≈ 1 +R, erf(

x√4 ν t

)≈ 0,

and

u(t, x) =

√ν

t

A2 ν

e−x2

4 ν t

√π

+ O(R2) ≈ A√4 π ν t

e−x2

4 ν t .




R→ 0 eR ≈ 1 +R, erf(

x√4 ν t

)≈ 0,

and

u(t, x) =

√ν

t

A2 ν

e−x2

4 ν t

√π

+ O(R2) ≈ A√4 π ν t

e−x2

4 ν t .




R→ 0 eR ≈ 1 +R, erf(

x√4 ν t

)≈ 0,

and

u(t, x) =

√ν

t

A2 ν

e−x2

4 ν t

√π

+ O(R2) ≈ A√4 π ν t

e−x2

4 ν t .



For large R the way of getting the approximating formula is less clear, sowe restore to the results of the numerical simulation. Below it is shown thesolution to ”point explosion” problem obtained for ν = 0.05 and R = 35:

Figure:

It reminds the shock wave profile

u(t, x) =

xt

if t > 0, 0 < x <√

2 A t,

0 if t > 0, x < 0 or x >√

2 A t,

which the BE ”shares” with the hyperbolic-type equation

ut + u ux = 0,



Figure:


u(t, x) =

xt

if t > 0, 0 < x <√

2 A t,

0 if t > 0, x < 0 or x >√

2 A t,


ut + u ux = 0,



Figure:


u(t, x) =

xt

if t > 0, 0 < x <√

2 A t,

0 if t > 0, x < 0 or x >√

2 A t,


ut + u ux = 0,


Figure:

A common solution

u(t, x) =

{xt if t > 0, 0 < x <

√2A t,

0 if t > 0, x < 0 or x >√

2A t,

to the Burgers and the Euler equations


So the solutions to the point explosion problem for BE arecompletely different in the limiting cases: whenR = A/(2 ν) → 0 it coincides with the solution of the heatexplosion problem,while for large R it reminds the shock wave solution!


So the solutions to the point explosion problem for BE arecompletely different in the limiting cases: whenR = A/(2 ν) → 0 it coincides with the solution of the heatexplosion problem,while for large R it reminds the shock wave solution!


The hyperbolic generalization of BE

Let us consider the Cauchy problem for the hyperbolicgeneralization of BE:

τ utt + ut + uux = ν ux x, (10)u(0, x) = ϕ(x).

Considering the linearization of (10)

τ utt + ut + u0 ux = ν ux x,

we can conclude, that the parameter C =√ν/τ is equal to the

velocity of small (acoustic) perturbations.

If the initial perturbation ϕ(x) is a smooth compactly supportedfunction, and D = max ϕ(x), then the number M = D/C (the”Mach number”) characterizes the evolution of nonlinear wave.




















Results of the numerical simulation: M = 0.3


Figure: M = 0.3


Figure: M = 0.3


Figure: M = 0.3


Figure: M = 0.3

The solution of the initial perturbation reminds the evolution ofthe point explosion problem for BE in the case whenR = A/(2 ν) is large.




Figure: M = 1.45


Figure: M = 1.45

For M = 1 + ε a formation of the blow-up regime is observed atthe beginning of evolution,


Figure: M = 1.45


Figure: M = 1.45

but for larger t it is suppressed by viscosity and returns to theshape of the BE solution!




Figure: M = 1.8


Figure: M = 1.8


Figure: M = 1.8

For M = 1.8 (and larger ones) a blow-up regime is formed atthe wave front in finite time!


Appendix 1. Calculation of point explosion problemfor BE

Since,∫ ξ

0+F (x) d x = −A lim

B→+0

∫ ∞

−∞δ(x)φB(x)H(x) d x =

{−A, if ξ < 0,0, if ξ > 0,

where φB(x) is any C∞0 function such that φ(x)|<B, ξ> ≡ 1, andsuppφ ⊂< B/2, ξ +B/2 > then

f(ξ; t, x) =

{(x−ξ)2

2 t −A if ξ < 0,(x−ξ)2

2 t , if ξ > 0.


Appendix 1. Calculation of point explosion problemfor BE

Since,∫ ξ

0+F (x) d x = −A lim

B→+0

∫ ∞

−∞δ(x)φB(x)H(x) d x =

{−A, if ξ < 0,0, if ξ > 0,

where φB(x) is any C∞0 function such that φ(x)|<B, ξ> ≡ 1, andsuppφ ⊂< B/2, ξ +B/2 > then

f(ξ; t, x) =

{(x−ξ)2

2 t −A if ξ < 0,(x−ξ)2

2 t , if ξ > 0.


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Nonlinear transport phenomena: models, method of solving and unusual features (2)