apde4 3 - PKUdsec.pku.edu.cn/~tlu/lecture notes/apde4_3.pdf · We derived the conservation laws of...

Reivew of Last lecture5 zero-order approximation

UUUþþþÅÅÅððð111gggYYY

A^ ©§

©

®ÆêÆÆÆ

2007-4-3

©©© AAA^ ©©©§§§



ÌSN

1 Reivew of Last lecture

2 5 zero-order approximation

3 UþÅð

4 1gY

©©© AAA^ ©©©§§§



Review of last lecture

We derived the conservation laws of mass, momentum andenergy from the Botzmann equation. In the equations, there aresome quantities that depends on the solution of the Boltzmannequation. For exmaple, the pressure tensor

pij =

∫

R3

M(vi − Vi)(vj − Vj)f(t, ~x,~v)dv

and thermodensity flux vector

qi =1

2nM

⟨

(vi − Vi)∣

∣

∣~v − ~V∣

∣

∣

2⟩

=M

2

∫

R3

(vi − Vi)∣

∣

∣~v − ~V

∣

∣

∣

2f(t, ~x,~v)dv

It is hard to solve f from Boltzmann equation, so we will use anapproximation of f to get the expression of the pressure tensor andthe theromdensity flux vector. Then we can solve the conservationlaws. ©©© AAA^ ©©©§§§



zero-order approximation I

Consider an nonequilibrium state that is not far from theequilibrium state. Assume that the distribution can be written intoa local equibrium Maxwell-Boltzmann distribution at aneighbourhood of a point, and the density of mass, temperatureand the mean velocity vary slowly with respect to the position andthe time. It is natural that we can get an approxiamtion as

f(t, ~x,~v) ≈ f (0) (t, ~x,~v)

where f (0) (t, ~x,~v) is

f (0) (t, ~x,~v) = n

(

M

2πkT

)3/2

exp

−M

∣

∣

∣~v − ~V

∣

∣

∣

2kT

©©© AAA^ ©©©§§§



zero-order approximation II

Here n, T, and ~V are slowly-varying functions of t and ~x.This is NOT a solution of Boltzmann equation Eq (2.54), but

makes the scattering term be zero.

J =

∫

R3

dw

∫

dΩσ (Ω) |~w − ~v|(

f (0)(t, ~x,~v′)

f (0)(

t, ~x, ~w′)

− f (0)(t, ~x,~v)f (0) (t, ~x, ~w))

= 0

But we still consider f (0) (t, ~x,~v) is an good approximationand will substitute it into the expressions of the pressure tensor ~Pand ~Q.

©©© AAA^ ©©©§§§



zero-order approximation of P and Q I

Use the notations

C(t, ~x) = n

(

M

2πkT

)3/2

, A (t, ~x) =M

2kT

f (0)(t, ~x,~v) = C(t, ~x) exp

(

−A(t, ~x)∣

∣

∣~v − ~V∣

∣

∣

2)

Then we have

q(0)i =

M

2

∫

R3

(vi − Vi)∣

∣

∣~v − ~V

∣

∣

∣C (t, ~x) · exp

(

−A (t, ~x)∣

∣

∣~v − ~V

∣

∣

∣

2)

dv

= 0

©©© AAA^ ©©©§§§



zero-order approximation of P and Q II

and

p(0)ij = M

∫

R3

(vi − Vi) (vi − Vj) C (t, ~x) · exp(

−A (t, ~x)∣

∣

∣~v − ~V

∣

∣

∣

2)

dv

= MC(t, ~x

∫

R3

vivj exp(

−A (t, ~x) |~v|2)

dv

= pδij

where

p = p(t, ~x) =1

3MC(t, ~x)

∫

R3

|~v|2 exp(

−A (t, ~x) |~v|2)

dv

=1

3MC(t, ~x) · 3π3/2

2A5/2

= nkT

©©© AAA^ ©©©§§§



zero-order approximation of P and Q III

is the local static pressure.

©©© AAA^ ©©©§§§



conservation laws I

Plug ~P and ~Q into the conservation laws of mass, momentumand energy, we have

∂ρ

∂t+ ∇ ·

(

ρ~V)

= 0

∂

∂t

(

ρ~V)

+ ∇ ·(

ρ~V⊗

~V + p~I)

= ρ~g

∂

∂t

(

ρe +1

2ρ

∣

∣

∣

~V∣

∣

∣

2)

+ ∇ ·((

ρe +1

2ρ

∣

∣

∣

~V∣

∣

∣

2+ p

)

~V

)

= ρ~g · ~V

where the equation of state is

p = nkT = RρT Here R =k

M

©©© AAA^ ©©©§§§



conservation laws II

Recall the equation (2.11), we have

e =3

2

kT

M= CV T Ù¥(CV =

3

2R)

This shows that what we are considering is not only an idealor perfect gas but also poly

Def. A gas is called perfect if the followng equation o stateholds:

p = ρRT

with T the temperature measured in K (Kelvin).Def. A gas is called polytropic (õíN¤if the following

equation holds:e = CV T

©©© AAA^ ©©©§§§



conservation laws III

where CV is a positive constant and is called volumetric specificheat capacity.

So we have got the fluid dynamic system for the perfact gasexcluding the viscosity and the thermal conduction.

we can simplify the above equations by using the continuumequation and get

d~V

dt+

1

ρ∇p = ρ~g

de

dt+ p∇ · ~V = 0

AlternativelydT

dt+

2

3T∇ · ~V = 0

©©© AAA^ ©©©§§§



UþÅð I

Last quantity χ = 12M |~v|2 . Eq. (4.22) is

1

2

∂

∂t

⟨

nM |~v|2⟩

+1

2

3∑

i=1

⟨

nMvi |v|2⟩

− n

3∑

i=1

〈Mgivi〉 = 0

From Eq. (2.10)-(2.11), we have

1

2

⟨

nM |~v|2⟩

=3

2nkT +

1

2ρ

∣

∣

∣

~V∣

∣

∣

2= ρe +

1

2ρ

∣

∣

∣

~V∣

∣

∣

2

where e is the internal energy of a unit of mass. And we also have

n

3∑

i=1

〈Mgivi〉 = ρ

3∑

i=1

gi 〈vi〉 = ρ~g · ~V

©©© AAA^ ©©©§§§



UþÅð II

Now it is time to have a careful look at the second term of the lefthand side of equation (4.22).

1

2

⟨

nMvi |v|2⟩

=1

2

⟨

nM(vi − Vi) |~v|2⟩

+1

2

⟨

nMVi |~v|2⟩

1

2

⟨

nMVi |~v|2⟩

=1

2Vi

⟨

nM |~v|2⟩

=

(

ρe +1

2ρ

∣

∣

∣

~V∣

∣

∣

2)

Vi

1

2

⟨

nM(vi − Vi) |~v|2⟩

=1

2

⟨

nM(vi − Vi)

(

∣

∣

∣~v − ~V∣

∣

∣

2+ 2(~v − ~V ) · ~V +

∣

∣

∣

~V∣

∣

∣

2)⟩

©©© AAA^ ©©©§§§



UþÅð III

=1

2nM

⟨

(vi − Vi)∣

∣

∣~v − ~V

∣

∣

∣

2⟩

+ nM⟨

(vi − Vi)(~v − ~V )⟩

· ~V +1

2nM 〈vi

=1

2nM

⟨

(vi − Vi)∣

∣

∣~v − ~V∣

∣

∣

2⟩

+ nM

3∑

j=1

〈(vi − Vi)(vj − Vj)〉Vj

= qi +3

∑

j=1

pijV j

Now we can write the equation of the conservation of energyinto

∂

∂t

(

ρe +1

2ρ

∣

∣

∣

~V∣

∣

∣

2)

+ ∇ ·((

ρe +1

2ρ

∣

∣

∣

~V∣

∣

∣

2)

~V + ~P ~V

)

(4.29)

= −∇ · ~Q + ρ~g · ~V

©©© AAA^ ©©©§§§



UþÅð IV

By using of the continuum equation (4.16), we can rewrite (4.29)into

ρ∂

∂t

(

e +1

2

∣

∣

∣

~V∣

∣

∣

2)

+ ρ~V · ∇(

e +1

2

∣

∣

∣

~V∣

∣

∣

2)

+ ∇ ·(

~P ~V)

(4.29a)

= −∇ · ~Q + ρ~g · ~V

Note that we use the vector identity

∇ · (a~V ) = a∇ · ~V + (∇a) · ~V

where a is a scaler and ~V is a vector.

©©© AAA^ ©©©§§§



UþÅð V

(4.30a) can be written into by using the substantial derivative(Nê)

ρd

dt

(

e +1

2

∣

∣

∣

~V∣

∣

∣

2)

+ ∇ ·(

~P ~V)

(4.30)

= −∇ · ~Q + ρ~g · ~V

Then we use the equation of the conservation law for momentum(4.21), we can rewrite (4.30) into

ρde

dt+

3∑

i,j=1

pij∂Vj

∂xi= −∇ · ~Q (4.31)

©©© AAA^ ©©©§§§



UþÅð VI

The process to derive (4.31) from (4.30) needs to know thefollowing identities

d

dt

∣

∣

∣

~V∣

∣

∣

2

=d

dt(~V · ~V )

= 2

(

d

dt~V

)

· ~V

∇ ·(

~P ~V)

=

3∑

i=1

∂

∂xi

3∑

j=1

pijVj

©©© AAA^ ©©©§§§



UþÅð VII

=

3∑

i=1

∂

∂xi

3∑

j=1

pijVj

=

3∑

i=1

3∑

j=1

∂pij

∂xiVj +

3∑

i=1

3∑

j=1

pij∂Vj

∂xi

We exchange the dumb variables i and j and can rewrite thefirst term into

3∑

i=1

3∑

j=1

∂pij

∂xiVj

=

3∑

i,j=1

∂pji

∂xjVi make use of pij = pji

©©© AAA^ ©©©§§§



UþÅð VIII

=

3∑

i=1

3∑

j=1

∂pij

∂xjVi

=(

∇ · ~P)

~V

Then we have

∇ ·(

~P ~V)

=(

∇ · ~P)

~V + +

3∑

i=1

3∑

j=1

pij∂Vj

∂xi

Make another expression of the conservation law of energy byintroducing

Sij =1

2

(

∂Vi

∂xj+

∂Vj

∂xi

)

©©© AAA^ ©©©§§§



UþÅð IX

We have

ρde

dt+

3∑

i,j=1

pijSij = −∇ · ~Q

Note that we have used

3∑

i=1

3∑

j=1

pij∂Vj

∂xiexchange i and j

=

3∑

j=1

3∑

i=1

pji∂Vi

∂xjfor pij = pji

=1

2

3∑

i=1

3∑

j=1

pij

(

∂Vj

∂xi+

∂Vi

∂xj

)

©©© AAA^ ©©©§§§



UþÅð X

Now we have derived the conservation laws in fluid dynamics,but we have know the solution of the Boltzmann equation (2.54)in order to determine the pressure tensor ~P and ~Q. It is a task fullof challenge and we will introduce some approximate methods togive the approximations of various order for the fluid dynamicalequations.

©©© AAA^ ©©©§§§



1gY I

1. (a) Find the general solution of u(x, y) of the PDE

ux + 2x(

e−x2 − y)

uy = x2

(b)Describe (with the help of a sketch) the region R of thexy-plane in which u(x, y) is determined by prescribed values of ualong the line segment that connects the points (0, 0) and (0, 1). Ifu = 1 + y on this line segment, find u(x, y) in R.

Solution: CHAR:

dx

1=

dy

2x(

e−x2 − y) =

du

x2

©©© AAA^ ©©©§§§



1gY II

dx1 = du

x2 gives u = 13x3 + K1 along the characteristic line

determined bydx

1=

dy

2x(

e−x2 − y)

with the solution

y = K2e−x2

+ x2e−x2

The hint to solve this ode is to rewrite the equation into byintroducing t = x2

dy

dt= −y + e−t

This is a nonhomegeneous ode with constant coefficients and canbe solved easily.

©©© AAA^ ©©©§§§



1gY III

To every K1 we map K2 vice versa: K1 = F (K2)

u =1

3x3 + F

(

ex2y − x2

)

where F is determined by the initial values.(b) By substituting the initial condition, you can find the

solution.2. trffac problem. This problem is an amusing application of

PDE to traffic flow. Assume that the car speed V is given interms of the car density ρ by the quation

V = V (ρ) = V0

(

1 − ρ

ρm

)

©©© AAA^ ©©©§§§



1gY IV

where ρm is known and 0 ≤ ρ ≤ ρm. After a long red light atx = 0, the road is empty for x > 0, and is full of stationary,bumper-to-bumper cars for x < 0. At time t = 0, the light turnsgreen.

(a)Sketch the characteristics. Hint: You may fill the xt-planewith characteristics by considering first a smooth function ρ(x, 0)that is very steep near x = 0, and then let the slope at x = 0become infinite.

(b)How long after the light turns green will a car initially atx = −a (a > 0) start moving?

(c)Where will this car be when its speed becomes V0/2?(d)Where will this car be when its speed becomes

λV0 (0 < λ < 1) , how long after the light turns green will it getthere? Hint: Denote the position of the car as a function of time

©©© AAA^ ©©©§§§



1gY V

by xa(t), and note that xa(t) satisfies the differential equationdxa/dt = V (xa, t).

Solution: The differentioan equation is given by theconservation law

ρt + (ρV )x = 0

i.e

ρt + V0

(

1 − 2ρ

ρm

)

ρx = 0

with the i.c.

ρ(x, 0) =

ρm, x < 00, x > 0

©©© AAA^ ©©©§§§



1gY VI

The solution is a rarefaction shock (DÕÅ¤

ρ(x, t) =

ρm, x < −V0tρm

2

(

1 − xV0t

)

−V0t < x < V0t

0, x > V0t

The above solution can be obtained by using the kown solution theburger’s equation

ut + uux = 0

u =

ul, x < 0ur, x > 0

with u = V0

(

1 − 2ρρm

)

©©© AAA^ ©©©§§§



1gY VII

(b) −a = −V0t ⇒ t = aV0

(c)

V (x, t) =

0, x < −V0tV02 + x

2t −V0t < x < V0tV0, x > V0t

x = 0 is the position of the car when its velocity is zero.(d) Solve the initial value problem

dxa

dt=

V0

2+

xa

2t

xa

(

V0

a

)

= −a

©©© AAA^ ©©©§§§



1gY VIII

This is linear first-oder differential equation. The integral factor is

eR

−12t

dt = t−12

Multiplying the integral factor on both sides of the ode

dxa

dt− xa

2t=

V0

2

yeilds

t−12dxa

dt− t−

12xa

2t=

V0

2t−

12

And we reconize that the left hand side is ddt

(

t−12 xa

)

and have

xa (t) =√

t(

C + V0

√t)

©©© AAA^ ©©©§§§



1gY IX

Applying the intial condition yields C = −2√

aV0. So the solutionof the IVP is

xa (t) = V0t − 2√

aV0t

The position of this car when the velocity is λV0 is obtained bysolving

V0

2+

x

2t= λV0

So

x = (2λ − 1)V0t

After we substitute this into the solution of the IVP, we have thetime

t =1

V0 (1 − λ)2

©©© AAA^ ©©©§§§



1gY X

and the position

x =2λ − 1

(1 − λ)2

3. Describe the solution u = u(x, y, z) of the PDExux + yuy + uuz = 0, where u satisfies the conditionu(x, y, 0) = xy for x > 0, y > 0.

Solution: CHAR:

dx

x=

dy

y=

dz

u=

du

0

u = K1 is a constant along the characteric curve. Thecharacteristic curve is

z = K1 lnx + K2

©©© AAA^ ©©©§§§



1gY XI

z = K1 ln y + K3

For any point (x, y, z), there is a unique characteristic curve links(x0, y0, 0) and (x, y, z).

And we solve

x0 = e−

K2K1

y0 = e−

K3K1

And this point (x0, y0, 0) satisfies the initial conditionu(x, y, 0) = xy, i.e.

u = e−

K2+K3K1

©©© AAA^ ©©©§§§



1gY XII

Recall that u = K1 and K2 = z − u lnx, K3 = z − u ln y. We have

u = xye−2z

u

Another method:

z = K1 lnx + K2

y = K3x

x0 = e−

K2K1

y0 = K3e−

K2K1

©©© AAA^ ©©©§§§



1gY XIII

u = K3e−

2K2K1

Recall that u = K1 and K2 = z − u lnx, K3 = yx . We have

u = xye−2z

u

The usual mistake is like this

u = ye−

K2K1

Recall that u = K1 and K2 = z − u lnx . We have

u = xye−z

u

©©© AAA^ ©©©§§§

apde4 3 - PKUdsec.pku.edu.cn/~tlu/lecture notes/apde4_3.pdf · We derived the conservation laws of...

Documents

Transcript of apde4 3 - PKUdsec.pku.edu.cn/~tlu/lecture notes/apde4_3.pdf · We derived the conservation laws of...