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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

Summer School: KPI, 2010 Nonlinear transport phenomena 1 / 38

THE AIM OF THIS LECTURE IS TO PRESENT THESIMPLEST PATTERNS SUPPORTED BY TRANSPORTEQUATIONS

WHAT ARE PATTERNS?


THE AIM OF THIS LECTURE IS TO PRESENT THESIMPLEST PATTERNS SUPPORTED BY TRANSPORTEQUATIONS

WHAT ARE PATTERNS?


Examples of patterns:I Solution to the system of reaction-diffusion equation

ut = Du ∆u + a(1− u)

vt = Dv ∆v + u v2 − k v

u(0, x) = f(x), v(0, x) = g(x),

∆ =∑2

i=1∂ 2

∂x2i

I periodically initiated cordial impulse; formation of thetrain of solitary waves;

I tsunami;I tornadoI etc., etc.



ut = Du ∆u + a(1− u)

vt = Dv ∆v + u v2 − k v

u(0, x) = f(x), v(0, x) = g(x),

∆ =∑2

i=1∂ 2

∂x2i





ut = Du ∆u + a(1− u)

vt = Dv ∆v + u v2 − k v

u(0, x) = f(x), v(0, x) = g(x),

∆ =∑2

i=1∂ 2

∂x2i





ut = Du ∆u + a(1− u)

vt = Dv ∆v + u v2 − k v

u(0, x) = f(x), v(0, x) = g(x),

∆ =∑2

i=1∂ 2

∂x2i





ut = Du ∆u + a(1− u)

vt = Dv ∆v + u v2 − k v

u(0, x) = f(x), v(0, x) = g(x),

∆ =∑2

i=1∂ 2

∂x2i





ut = Du ∆u + a(1− u)

vt = Dv ∆v + u v2 − k v

u(0, x) = f(x), v(0, x) = g(x),

∆ =∑2

i=1∂ 2

∂x2i





ut = Du ∆u + a(1− u)

vt = Dv ∆v + u v2 − k v

u(0, x) = f(x), v(0, x) = g(x),

∆ =∑2

i=1∂ 2

∂x2i




Our aim to discuss the patterns supported by the varioustransport equations in case of one spatial variable.

Advantages:I ability to present and analyze the analytical solutions to

non-linear modelsI the possibility to analyze the problem by means of

qualitative theory methods.

By patterns we temporarily mean:I non-monotonic solution to the modelling system,

maintaining their shape during the evolution;I or the non-monotonic solutions evolving in a self-similar

modeI or the non-monotonic solutions tending to infinity in finite

time (blowing-up patterns)Summer School: KPI, 2010 Nonlinear transport phenomena 4 / 38









































Heat transport: the balance equation

d

d t

∫Ω

ρ u(t, x)d x = −∫

∂ Ω~q d~σ +

∫Ω

f (u(t, x); t, x)d x, (1)

where ~q is the density of the heat flux on the boundary ∂ Ω.On virtue of the Fick law, is as follows:

~q = −κ∇u(t, x);

u(t, x) is the energy per unit mass (the temperature);ρ is the heat capacity per unit volume;

κ(u; t, x) is the heat transport coefficient;f (u(t, x); t, x) is the voluminal heat source.



d

d t

∫Ω

ρ u(t, x)d x = −∫

∂ Ω~q d~σ +

∫Ω

f (u(t, x); t, x)d x, (1)


~q = −κ∇u(t, x);





d

d t

∫Ω

ρ u(t, x)d x = −∫

∂ Ω~q d~σ +

∫Ω

f (u(t, x); t, x)d x, (1)


~q = −κ∇u(t, x);





d

d t

∫Ω

ρ u(t, x)d x = −∫

∂ Ω~q d~σ +

∫Ω

f (u(t, x); t, x)d x, (1)


~q = −κ∇u(t, x);




Using the Green-Gauss-Ostrogradsky theorem, we are able towrite down the following identities:

−∫

∂ Ω~q d~σ ≡

∫∂ Ω

κ∇u(t, x) d~σ =

∫Ω∇ [κ∇u(t, x)] d x.

So, the balance equation can be rewritten as∫Ω

ρ

∂ u(t, x)

∂ t−∇ [κ∇u(t, x)]− f [u(t, x); t, x]

d x = 0.

Since the volume Ω is arbitrary, fulfillment of the balanceequation is possible providing that the integrand is equal to zero

In the case of one spatial variable, we get this way the equation

∂ u(t, x)

∂ t=

∂

∂ x

[κ (u(t, x), t, x)

∂ u(t, x)

∂ x

]+f (u(t, x), t, x) ,

(2)

κ = κ/ρ, f = f/ρ.



−∫

∂ Ω~q d~σ ≡

∫∂ Ω

κ∇u(t, x) d~σ =

∫Ω∇ [κ∇u(t, x)] d x.


ρ

∂ u(t, x)

∂ t−∇ [κ∇u(t, x)]− f [u(t, x); t, x]

d x = 0.



∂ u(t, x)

∂ t=

∂

∂ x

[κ (u(t, x), t, x)

∂ u(t, x)

∂ x

]+f (u(t, x), t, x) ,

(2)

κ = κ/ρ, f = f/ρ.



−∫

∂ Ω~q d~σ ≡

∫∂ Ω

κ∇u(t, x) d~σ =

∫Ω∇ [κ∇u(t, x)] d x.


ρ

∂ u(t, x)

∂ t−∇ [κ∇u(t, x)]− f [u(t, x); t, x]

d x = 0.



∂ u(t, x)

∂ t=

∂

∂ x

[κ (u(t, x), t, x)

∂ u(t, x)

∂ x

]+f (u(t, x), t, x) ,

(2)

κ = κ/ρ, f = f/ρ.



−∫

∂ Ω~q d~σ ≡

∫∂ Ω

κ∇u(t, x) d~σ =

∫Ω∇ [κ∇u(t, x)] d x.


ρ

∂ u(t, x)

∂ t−∇ [κ∇u(t, x)]− f [u(t, x); t, x]

d x = 0.



∂ u(t, x)

∂ t=

∂

∂ x

[κ (u(t, x), t, x)

∂ u(t, x)

∂ x

]+f (u(t, x), t, x) ,

(2)

κ = κ/ρ, f = f/ρ.


Self-similar solutions to the heat equation.

Statement 1. Any physical law can be written down in theform

a = ϕ (a1, a2, ..., an; an+1, ...an+m) , where

a1, a2, ..., an

are the physical quantities expressed in the basic units,

a, an+1, ...an+m

are the physical quantities expressed in derived units.Example of basic units:

the length[L], the time [T ], the mass [M ]

Examples of physical quantities expressed in derivedunits:

V =d x

d t

[L

T

], F = m

d2 x

d t2

[M · LT 2

], etc., etc.





a1, a2, ..., an


a, an+1, ...an+m




V =d x

d t

[L

T

], F = m

d2 x

d t2

[M · LT 2

], etc., etc.





a1, a2, ..., an


a, an+1, ...an+m




V =d x

d t

[L

T

], F = m

d2 x

d t2

[M · LT 2

], etc., etc.





a1, a2, ..., an


a, an+1, ...an+m




V =d x

d t

[L

T

], F = m

d2 x

d t2

[M · LT 2

], etc., etc.





a1, a2, ..., an


a, an+1, ...an+m




V =d x

d t

[L

T

], F = m

d2 x

d t2

[M · LT 2

], etc., etc.


Statement 2. Derived physical quantities a, an+1, ...an+m canbe expressed through the basic ones in the following form:

a = ar11 ar2

2 ...arnn Π,

an+j = arj1

1 arj2

2 ...arjn

n Πj , j = 1, 2, ....,m,

where Π, Π1, ...., Πm are dimensionless parameters.So, the physical law

a = F (a1, ...an; ...an+j , ....)

is equivalent to

Π =1

ar11 ar2

2 ...arnn

F

[a1, a2, ..., an;...a

rj1

1 arj2

2 ...arjn

n Πj ...

]. (3)


Statement 2. Derived physical quantities a, an+1, ...an+m canbe expressed through the basic ones in the following form:

a = ar11 ar2

2 ...arnn Π,

an+j = arj1

1 arj2

2 ...arjn

n Πj , j = 1, 2, ....,m,

where Π, Π1, ...., Πm are dimensionless parameters.So, the physical law

a = F (a1, ...an; ...an+j , ....)

is equivalent to

Π =1

ar11 ar2

2 ...arnn

F

[a1, a2, ..., an;...a

rj1

1 arj2

2 ...arjn

n Πj ...

]. (3)


The essence of Π theorem

Theorem. The RHS of the formula (3) depends, at most,on the m dimensionless parameters Π1, ...., Πm, and doesnot depend on the dimension quantities a1, a2, ..., an.

1

So, the physical law

a = F (a1, ...an; ...an+j , ....)

in dimensionless variables takes the form to

Π = Ψ(Π1, Π2, ...Πm ) . (4)

1A. Samarskij, A. Mikhailov, Mathematical Modelling: Ideas, Methods,Examples, Mosclw, 1997, Ch. V.




1


a = F (a1, ...an; ...an+j , ....)


Π = Ψ(Π1, Π2, ...Πm ) . (4)





1


a = F (a1, ...an; ...an+j , ....)


Π = Ψ(Π1, Π2, ...Πm ) . (4)



Applications. Point explosion (linear case)

ut = κ ux x, (5)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞

−∞u(t, x) d x = Q = const.

The solution to (5) can be expressed as u = f(t, κ, Q; x). Basicunits:

t[T ], κ

[E · L2

T

], Q[E · L].

Derived units:

u =Q√κ t

Π, x =√

κ t Π1.

Thus, passing to dimensionless variables we get

Π =

√κ t

Qf(t, κ, Q;

√κ t Π1) = ϕ (Π1) .



ut = κ ux x, (5)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[E · L2

T

], Q[E · L].

Derived units:

u =Q√κ t

Π, x =√

κ t Π1.


Π =

√κ t

Qf(t, κ, Q;

√κ t Π1) = ϕ (Π1) .



ut = κ ux x, (5)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[E · L2

T

], Q[E · L].

Derived units:

u =Q√κ t

Π, x =√

κ t Π1.


Π =

√κ t

Qf(t, κ, Q;

√κ t Π1) = ϕ (Π1) .



ut = κ ux x, (5)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[E · L2

T

], Q[E · L].

Derived units:

u =Q√κ t

Π, x =√

κ t Π1.


Π =

√κ t

Qf(t, κ, Q;

√κ t Π1) = ϕ (Π1) .



ut = κ ux x, (5)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[E · L2

T

], Q[E · L].

Derived units:

u =Q√κ t

Π, x =√

κ t Π1.


Π =

√κ t

Qf(t, κ, Q;

√κ t Π1) = ϕ (Π1) .


Corollary.Ansatz

u =Q√κ t

ϕ(ξ), ξ = Π1

should strongly simplify the point explosion problem (5).In fact, inserting this ansatz into the heat transport equation,we get

2 ϕ[ξ] + ξ ϕ[ξ] + ϕ[ξ] ≡ d

d ξ(ξϕ[ξ] + 2 ϕ[ξ]) = 0, (6)

with the additional condition (as well expressed in thedimensionless variables):∫ ∞

−∞ϕ[ξ] dξ = 1. (7)


Corollary.Ansatz

u =Q√κ t

ϕ(ξ), ξ = Π1


2 ϕ[ξ] + ξ ϕ[ξ] + ϕ[ξ] ≡ d

d ξ(ξϕ[ξ] + 2 ϕ[ξ]) = 0, (6)


−∞ϕ[ξ] dξ = 1. (7)


Corollary.Ansatz

u =Q√κ t

ϕ(ξ), ξ = Π1


2 ϕ[ξ] + ξ ϕ[ξ] + ϕ[ξ] ≡ d

d ξ(ξϕ[ξ] + 2 ϕ[ξ]) = 0, (6)


−∞ϕ[ξ] dξ = 1. (7)


Statement 3. Solution to the problem

d

d ξ(ξϕ[ξ] + 2 ϕ[ξ]) = 0,

∫ ∞

−∞ϕ[ξ] dξ = 1

is given by the formula

ϕ[ξ] =1√4 π

e−ξ2

4 .

Corollary. The only solution to the linear point explosionproblem (5) is given by the formula

u =Q√

4 κπ te−

x2

4 κ t . (8)


Statement 3. Solution to the problem

d

d ξ(ξϕ[ξ] + 2 ϕ[ξ]) = 0,

∫ ∞

−∞ϕ[ξ] dξ = 1


ϕ[ξ] =1√4 π

e−ξ2

4 .

Corollary. The only solution to the linear point explosionproblem (5) is given by the formula

u =Q√

4 κπ te−

x2

4 κ t . (8)


Figure:


Figure:


Figure:


Remark 1. For Q = 1 the previous formula written as

E =1√

4 π κ te−

x2

4 κ t

defines so called fundamental solution of the operator

L =∂

∂ t− κ

∂2

∂ x2.

Knowledge of this solution enables to solve the initial valueproblem.

Theorem Solution to the initial value problem

ut − κux x, u(0, x) = F (x)


u(t, x) = E ∗ F (t, x) ≡∫ +∞

−∞E (t, x− y) · F (y) d y

provided that the integral in the RHS does exist.


Remark 1. For Q = 1 the previous formula written as

E =1√

4 π κ te−

x2

4 κ t

defines so called fundamental solution of the operator

L =∂

∂ t− κ

∂2

∂ x2.

Knowledge of this solution enables to solve the initial valueproblem.

Theorem Solution to the initial value problem

ut − κux x, u(0, x) = F (x)


u(t, x) = E ∗ F (t, x) ≡∫ +∞

−∞E (t, x− y) · F (y) d y

provided that the integral in the RHS does exist.


Peculiarity of the solution

u =Q√

4 κ π te−

x2

4 κ t

to the linear point explosion:

At any moment of time all the space ”knows ” that the pointexplosion took place. This is physically incorrect!!!


Peculiarity of the solution

u =Q√

4 κ π te−

x2

4 κ t

to the linear point explosion:

At any moment of time all the space ”knows ” that the pointexplosion took place. This is physically incorrect!!!


Point explosion (nonlinear case)

ut = κ [u ux]x , (9)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[L2

E · T

], Q[E · L].

Derived units:

u =Q2/3

[κ t]1/3Π, x = [κ Q t]1/3 Π1.


Π =[κ t]1/3

Q2/3f

(t, κ, Q; Π1 [κ Q t]1/3

)= ϕ (Π1) .



ut = κ [u ux]x , (9)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[L2

E · T

], Q[E · L].

Derived units:

u =Q2/3

[κ t]1/3Π, x = [κ Q t]1/3 Π1.


Π =[κ t]1/3

Q2/3f

(t, κ, Q; Π1 [κ Q t]1/3

)= ϕ (Π1) .



ut = κ [u ux]x , (9)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[L2

E · T

], Q[E · L].

Derived units:

u =Q2/3

[κ t]1/3Π, x = [κ Q t]1/3 Π1.


Π =[κ t]1/3

Q2/3f

(t, κ, Q; Π1 [κ Q t]1/3

)= ϕ (Π1) .



ut = κ [u ux]x , (9)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[L2

E · T

], Q[E · L].

Derived units:

u =Q2/3

[κ t]1/3Π, x = [κ Q t]1/3 Π1.


Π =[κ t]1/3

Q2/3f

(t, κ, Q; Π1 [κ Q t]1/3

)= ϕ (Π1) .



ut = κ [u ux]x , (9)

u(0, x) = 0, x ∈ R, x 6= 0,

∫ ∞



t[T ], κ

[L2

E · T

], Q[E · L].

Derived units:

u =Q2/3

[κ t]1/3Π, x = [κ Q t]1/3 Π1.


Π =[κ t]1/3

Q2/3f

(t, κ, Q; Π1 [κ Q t]1/3

)= ϕ (Π1) .


Corollary. Ansatz

u =Q

[κ Q t]1/3ϕ(ξ), ξ = Π1 =

x

[κ Q t]1/3


d

d ξ(ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0, (10)


−∞ϕ[ξ] dξ = 1. (11)


Corollary. Ansatz

u =Q

[κ Q t]1/3ϕ(ξ), ξ = Π1 =

x

[κ Q t]1/3


d

d ξ(ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0, (10)


−∞ϕ[ξ] dξ = 1. (11)


Corollary. Ansatz

u =Q

[κ Q t]1/3ϕ(ξ), ξ = Π1 =

x

[κ Q t]1/3


d

d ξ(ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0, (10)


−∞ϕ[ξ] dξ = 1. (11)


Statement 4. Solution to the problemd

d ξ(ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0,∫ ∞

−∞ϕ[ξ] dξ = 1


ϕ[ξ] =

16

(ξ2Φ − ξ2

)if |ξ| < ξΦ,

0 otherwise,

where ξΦ = (9/2)1/3.

Corollary. The only solution to the nonlinear pointexplosion problem (9) is given by the formula

u =Q

6 [κQ t]1/3

(9/2)1/3 − x2

[κ Q t]2/3 if |x| <[

9 (κ t Q)2

2

]1/6

,

0 otherwise(12)



d ξ(ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0,∫ ∞

−∞ϕ[ξ] dξ = 1


ϕ[ξ] =

16

(ξ2Φ − ξ2

)if |ξ| < ξΦ,

0 otherwise,

where ξΦ = (9/2)1/3.


u =Q

6 [κQ t]1/3

(9/2)1/3 − x2

[κ Q t]2/3 if |x| <[

9 (κ t Q)2

2

]1/6

,

0 otherwise(12)



d ξ(ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0,∫ ∞

−∞ϕ[ξ] dξ = 1


ϕ[ξ] =

16

(ξ2Φ − ξ2

)if |ξ| < ξΦ,

0 otherwise,

where ξΦ = (9/2)1/3.


u =Q

6 [κQ t]1/3

(9/2)1/3 − x2

[κ Q t]2/3 if |x| <[

9 (κ t Q)2

2

]1/6

,

0 otherwise(12)


Figure:


Figure:


Figure:


Figure:


Conclusion. Nonlinearity of the heat transport coefficientcauses the localization of the heat wave within the finitedomain

< −L(t), L(t) >,

where

L(t) =

[9 (κ t Q)2

2

]1/6

= C t1/3.


Unlimited growth of temperature and the completelocalization of the heat energy

ut = κ [uσ ux]x , (13)u(t, 0) = A (−t)−n , t < 0, n > 0, x ∈ R+, lim

t→−∞u(t, x) = 0(14)

In general, the problem is self-similar and the ansatz

u =A

(−t)nϕ(ξ), ξ =

x√Aσ κ t1−n σ

reduces the initial problem to corresponding ODE.

If we assume that n = σ = 1, than ξ will not depend on t:

u =A

(−t)ϕ(ξ), ξ =

x√A κ

.




t→−∞u(t, x) = 0(14)


u =A

(−t)nϕ(ξ), ξ =




u =A

(−t)ϕ(ξ), ξ =

x√A κ

.




t→−∞u(t, x) = 0(14)


u =A

(−t)nϕ(ξ), ξ =




u =A

(−t)ϕ(ξ), ξ =

x√A κ

.




t→−∞u(t, x) = 0(14)


u =A

(−t)nϕ(ξ), ξ =




u =A

(−t)ϕ(ξ), ξ =

x√A κ

.


The reduced equation

ϕϕ + ϕ2 − ϕ = 0,

together with the initial condition ϕ(0) = 1 (arising from thecondition u(t, 0) = A

(−t)), has the solution

ϕ(ξ) =

(1− ξ/

√6)2, if 0 < ξ <

√6,

0 if ξ ≥√

6,

The corresponding solution to the BV problem (13) is asfollows:

u(t, x) =

A−t

(1− x√

6 κ A

)2, if 0 < x <

√6 κ A,

0 if x ≥√

6 κ A.

Moral of the tale: [1]. Heat energy is completely localized inthe segment 0 < x <

√6 κ A0;

[2]. The temperature u(t, x) tends to +∞ in every pointx ∈ [0,

√6 κ A0) as t → 0−.



ϕϕ + ϕ2 − ϕ = 0,



ϕ(ξ) =

(1− ξ/

√6)2, if 0 < ξ <

√6,

0 if ξ ≥√

6,


u(t, x) =

A−t

(1− x√

6 κ A

)2, if 0 < x <

√6 κ A,

0 if x ≥√

6 κ A.


√6 κ A0;


√6 κ A0) as t → 0−.



ϕϕ + ϕ2 − ϕ = 0,



ϕ(ξ) =

(1− ξ/

√6)2, if 0 < ξ <

√6,

0 if ξ ≥√

6,


u(t, x) =

A−t

(1− x√

6 κ A

)2, if 0 < x <

√6 κ A,

0 if x ≥√

6 κ A.


√6 κ A0;


√6 κ A0) as t → 0−.



ϕϕ + ϕ2 − ϕ = 0,



ϕ(ξ) =

(1− ξ/

√6)2, if 0 < ξ <

√6,

0 if ξ ≥√

6,


u(t, x) =

A−t

(1− x√

6 κ A

)2, if 0 < x <

√6 κ A,

0 if x ≥√

6 κ A.


√6 κ A0;


√6 κ A0) as t → 0−.


Figure:


Figure:


Figure:


Blow-up regime caused by the source term incorporation

Model equation:ut = κ [u ux]x + q0 u2. (15)

Ansatz u(t, x) = P (t) Q(x) leads, after some algebraicmanipulation, to the equations

P−2(t)P (t) =[κ(Q(x) Q(x)′)′ + q0 Q(x)2

]Q−1(x) = C1. (16)

Theorem. The equation (15) has the following solution:

u(t, x) =4

3 q0 (tf − t)

cos2

[√q0

8 κx], if |x| <

√2 π2 κ

q0,

0 otherwise.(17)

The above solution

I is localized within the interval [−L, L], L =√

2 π2 κq0

;

I tends to +∞ ∀ x ∈ (−L, L), as t → tf − 0.





P−2(t)P (t) =[κ(Q(x) Q(x)′)′ + q0 Q(x)2

]Q−1(x) = C1. (16)


u(t, x) =4

3 q0 (tf − t)

cos2

[√q0

8 κx], if |x| <

√2 π2 κ

q0,

0 otherwise.(17)

The above solution


2 π2 κq0

;






P−2(t)P (t) =[κ(Q(x) Q(x)′)′ + q0 Q(x)2

]Q−1(x) = C1. (16)


u(t, x) =4

3 q0 (tf − t)

cos2

[√q0

8 κx], if |x| <

√2 π2 κ

q0,

0 otherwise.(17)

The above solution


2 π2 κq0

;






P−2(t)P (t) =[κ(Q(x) Q(x)′)′ + q0 Q(x)2

]Q−1(x) = C1. (16)


u(t, x) =4

3 q0 (tf − t)

cos2

[√q0

8 κx], if |x| <

√2 π2 κ

q0,

0 otherwise.(17)

The above solution


2 π2 κq0

;






P−2(t)P (t) =[κ(Q(x) Q(x)′)′ + q0 Q(x)2

]Q−1(x) = C1. (16)


u(t, x) =4

3 q0 (tf − t)

cos2

[√q0

8 κx], if |x| <

√2 π2 κ

q0,

0 otherwise.(17)

The above solution


2 π2 κq0

;






P−2(t)P (t) =[κ(Q(x) Q(x)′)′ + q0 Q(x)2

]Q−1(x) = C1. (16)


u(t, x) =4

3 q0 (tf − t)

cos2

[√q0

8 κx], if |x| <

√2 π2 κ

q0,

0 otherwise.(17)

The above solution


2 π2 κq0

;



Figure:


Figure:


Figure:


Figure:


THE ROLE OF SELF-SIMILAR SOLUTIONS


Maximum principle. Comparison theorems ( initialvalue problem).

Let us consider the Cauchy problem

ut = [κ(u) ux]x , (18)u(0, x) = u0(x) ≥ 0, x ∈ R, κ(u) > 0.

Theorem 1. Maximum of the solution u(t, x) at any timet > 0 does not exceed the maximum of the initial data:

maxt>0, −∞<x<∞u(t, x) ≤ max−∞<x<∞u0(x).

2

2A. Samarskij, A. Mikhailov, Mathematical Modelling: Ideas, Methods,Examples, Moscow, 1997, A. Samarskij, V. Galaktionov et al, Blow-upregimes in quasilinear parabolic equations, Berlin: Walter de Gruyter, 1995.






maxt>0, −∞<x<∞u(t, x) ≤ max−∞<x<∞u0(x).

2







maxt>0, −∞<x<∞u(t, x) ≤ max−∞<x<∞u0(x).

2







maxt>0, −∞<x<∞u(t, x) ≤ max−∞<x<∞u0(x).

2



Corollary (comparison theorem). Let u1(t, x),u2(t, x), u3(t, x), are the solutions to the equation

ut = [κ(u) ux]x ,

u(0, x) = u0(x) ≥ 0, x ∈ R, κ(u) > 0,

corresponding to the initial data u10(x), u2

0(x), and u30(x)

such thatu1

0(x) ≤ u20(x) ≤ u3

0(x)

for −∞ < x < ∞. Then

u1(t, x) ≤ u2(t, x) ≤ u3(t, x)

for all −∞ < x < ∞, and t > 0.



ut = [κ(u) ux]x ,

u(0, x) = u0(x) ≥ 0, x ∈ R, κ(u) > 0,


0(x), and u30(x)

such thatu1

0(x) ≤ u20(x) ≤ u3

0(x)


u1(t, x) ≤ u2(t, x) ≤ u3(t, x)

for all −∞ < x < ∞, and t > 0.



ut = [κ(u) ux]x ,

u(0, x) = u0(x) ≥ 0, x ∈ R, κ(u) > 0,


0(x), and u30(x)

such thatu1

0(x) ≤ u20(x) ≤ u3

0(x)


u1(t, x) ≤ u2(t, x) ≤ u3(t, x)

for all −∞ < x < ∞, and t > 0.



ut = [κ(u) ux]x ,

u(0, x) = u0(x) ≥ 0, x ∈ R, κ(u) > 0,


0(x), and u30(x)

such thatu1

0(x) ≤ u20(x) ≤ u3

0(x)


u1(t, x) ≤ u2(t, x) ≤ u3(t, x)

for all −∞ < x < ∞, and t > 0.
