lecture 14, 19 oct 2010

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Part I

First and Second Order Partial

Differential Equations

1

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DRAFTCHAPTER 1Partial Differential Equations

(PDEs): Generalities

Consider a space and time, defined on a given domain and denoted by coor-

dinates x , y , z , . . . , t . For example, the domain can be a surface, a cube or acylinder as shown in Figure 1.1.

One of the coordinates, usually denoted by t is time.

Figure 1.1: domain

A vector function u = u(t , x , y , z , . . .) ; = 1, 2, ...d is defined on ad-dimensional space and, in general, describes some physical phenomena. For

example, u(x , y, z, t) can be the pressure gradient at a point, or the velocity ofa fluid, or the electric potential and so on. The function u(x , y, z, t) is usuallyspecified only implicitly by determining the rate of change of the quantity as one

moves around in space and time. For example, conservation of mass and mo-

mentum determines how the fluid velocity varies over space and time. Physical

considerations lead to a general equation, relating the rate of change with time

to rates of change in space. Define u = (u1, . . . , ud). Let

u

t= ut ,

u

x= ux ,

2u

tx= utx , etc. (1.1)

3

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4 1: Partial Differential Equations (PDEs): Generalities

A general PDE has the form

F[x, y, . . . , t, u, ut, ux, uxt, uxy, . . .] = 0 (1.2)

x , y , . . . , t : independent variables and u, ut, ux, . . . : dependent variablesA PDE depends on two or more variables, unlike an ordinary differential that

depends only on one variable. The highest partial derivative is the order of the

PDE.

The domain D of the PDE is defined by the range of the independent vari-

ables x , y , z , . . . , t for which u(x,y,..,t) is to be determined. See Figure 1.2.

u(x,y)

Figure 1.2: u(x, y)

All dependent functions u(x , y, z, t) are assumed to have finite partial deriva-tives, called a strong solution. Weak solutions are also allowed for which the

partial derivatives may become singular at certain regions, allowing for discon-

tinuities in the behavior of the dependent functions u(x , y, z, t).The PDE is linear if the function F[t, x, y, . . . , u, ut, ux, uxt, uxy, . . .] is a lin-

ear function ofu, ut, ux, uxt, . . .. A PDE is quasi-linear if it is linear in the high-est order derivative and the other coefficient functions of the terms u, ut, ux, uxt, . . .are nonlinear functions.

The PDE give in Eq. 1.2 yields a unique solution only if some conditions

are imposed on the solution, termed as boundary values. The solution of an

ordinary differential equation is unique if some constraints are specified, such

as the initial and final value of the sought for solution. In contrast, for a PDE,

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1.1: Diffusion Equation 5

an entire boundary value function has to be specified on a subdomain ofD

for the sought solution. For example, ifu(x, t) depends on time and space, theboundary (initial) value ofu(x, t) is specified at initial time t = 0 for all x, suchthat u(x, 0) = f(x). See Figure. 1.3.

u(t,x)

f(x)

Figure 1.3: u(t, x) and boundary

In this course, we focus only first order and second order partial differential

equations; to simplify the discussions to its bare essential we focus on study-ing scalar functions in two dimensions, with derivatives in higher dimensions

addressed only for special cases.

1.1 Diffusion Equation

To illustrate the content of a typical PDE, a derivation is given of the diffusion

equation, which is an important exemplar of parabolic PDEs. Consider the con-

centration of particles n(z, t) in a medium under the influence of gravity. Theparticles will tend to move downwards. The configurations depends only on

height z. The medium has a time distribution of the particles that we now deter-mine.

At each instant of time, the total number of particles is constant. The number

of particles in the shaded portion of Figure 1.4(b) is equal to n(t, z)Adz. Sincethere is conservation of mass, the rate of change of n(z, t)Adz must be equalto the difference between the fluxes of particles entering from surface at z andlearning through surface at z + dz, as shown in Figure 1.4. Hence

t[n(t, z)Adz] = AJ(z) AJ(z + dz) (1.3)

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n(t,z)Adz

AJ(z)

AJ(z + dz)

Figure 1.4: (a) Area A; (b) Particles moving in the upward direction.

where J(z) = rate per unit area and per unit time that particles cross surface atposition z. Simplifying yields,

n(z, t)

t= J

z(1.4)

J=-D n z

Figure 1.5: J(z)

Ifn =constant, we have J = 0; hence a good approximation is to set

J = D n(z, t)z

(1.5)

where D is the diffusion constant and is a characteristic of the medium. Hence

n(z, t)

t=

z Dn(z, t)

z (1.6)For constant diffusion constant D we have

n(z, t)

t= D

2n(z, t)

z2: Diffusion Equation (1.7)

1.2 Special cases of PDEs

Before starting the systematic study of PDEs, we mention a few special cases

whose solutions can be found by inspection.

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1.2: Special cases of PDEs 7

1. u(x, y)/y = 0 ; u(x, y) = (x)

2. 2u(x, y)/xy = 0 ; u(x, y) = (x) + v(y)

3.

uxy = f(x, y) (1.8)

u(x, y) = (x) + v(y) +

x0

d

y0

df(, ) (1.9)

4.

ux

uy

= 0 (1.10)

Clearly, this looks like case 1, i.e. u/z = 0, where x, y are related to z bya change of variables. Let

x =1

2( + ); y =

1

2( )

= x + y, = x yu(x, y)

x

=

x

u

+

x

u

=u

+u

u(x, y)

x=

y

u

+

y

u

=

u

u

ux uy = u = 0 u = f()u(x, y) = f(x + y) : wave moving to the left

and shown in Figure 1.6.

V(x)y = 0

V(x)y = 1

x= 1

Figure 1.6: (a) y = 0 (b) y = 1

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In general, for , constant, consider the PDE

ux + uy = 0 (1.11)

A change of variables

= x + y (1.12)

= x y (1.13)yields the solution

u = 0 ; u(x, y) = f(x y) (1.14)

5. Consider the wave equation

2u

x2

2u

y2= 0

or (

x+

y)(

x

y)u = 0

= x + y, = x y0 = 2xu 2yu = 4u u(x, y) = V1(x + y) + V2(x y)

which yields left-running and right-running waves.

Similarly

c2uxx uyy = 0 (1.15) u(x, y) = V1(x + cy) + V2(x cy) (1.16)

6. Consider the eikonal equation

u2x + u2y = 1 (1.17)

Assume, in analogy with the linear case, that

u(x, y) = (x) + (y) (1.18)and which yields (prime is differentiation)

((x))2 + ((y))2 = 1

((x))2 = 1 [(y)]2 = 2 : constant of separation (x) = ; (y) =

1 2

u(x, y) = x + y

1 2 + where is a constant of separation and is a constant of integration.

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DRAFTCHAPTER 2First order PDEs andCharacteristics

Consider a general first order partial differential equation given by

a(x, t)V

x+ b(x, t)

V

t= c(t, x)V + d(x, t) (2.1)

We want a solution subject to the solution equaling a given initial value along

some contour in the (t, x) plane. Let have coordinates x(), t(), as shownin Figure 2.1, and be defined parametrically, namely

: x(); t() : [a, b] (2.2)V() = f() (2.3)

Figure 2.1: : a curve in domain D.

9

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10 2: First order PDEs and Characteristics

We want to find the unique value of u(t, x) in the domain D so that the

solution agrees with the pre-fixed values on specified by f().Using the method of characteristics, we will study the following three special

types of first order PDEs.

(i) ut + cux = 0 : rigid translation with no damping

(ii) ut + cux = u : rigid translation with damping (no distortion)(iii) ut + uux = 0 : translation with distortion

2.1 Characteristic Curves

A wide class of linear and quasi-linear first order PDEs can be solved by finding

special curves in the xt-plane, called characteristic curves, or characteristics inshort. On these curves, the PDE simplifies to an ordinary differential equation.

The task of solving the PDE decomposes into the task of finding the characteris-

tics and then integrating the ODE along the characteristics from a parametrized

initial value.

Define the characteristic curve, parametrized by s, and defined as follow

dx(s)ds

= a(x, t) ; dt(s)ds

= b(x, t) (2.4)

Figure 2.2 shows a typical characteristic.

(x(s),t(s))

Figure 2.2: x(s) and t(s)

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2.1: Characteristic Curves 11

Then, from Eqs. 2.1 and 2.4

dV

ds=

dx(s)

ds

V

x+

dy(s)

ds

V

y

= aV

x+ b

V

y(2.5)

Hence, we have reduced solving the first order PDE to solving the following

system of ordinary differential equations

x(s)

ds= a ,

dt(s)

ds= b (2.6)

dVds

= cV + d (2.7)

The existence and uniqueness of ODEs, for smooth a,b,c, and d, guaran-tees that exactly one solution curve x(s), t(s), V(s) passes through a given pointx0, t0, V0. To obtain a unique solution, one needs to solve the initial value prob-lem. Recall the curve is characterized by parameter so that

x() ; t() ; V = f() (2.8)

(x(s), t(s))

Figure 2.3: (a) Labeling characteristic curves by intersection with . (b) Char-acteristic (x(s), t(s)) intersection at .

Typically, all the characteristics intersect with the curve , as shown in Figure2.1. For each value of, there is a unique characteristic curve intersecting , asshown in Figure. 2.3(a). 1 Let us label the different characteristic curves by the

different point at which they intersect , as shown for a given characteristic inFigure 2.3(b).

1The case when this is not true is discussed in the last Section of this Chapter.

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The coordinates s, provide a coordinate system, equivalent to x, t, that

specify all points on the plane, as shown in Figure ??. For a non-singular coor-dinate transformation from x, t to s, , we need to have a non-zero Jacobian Jdefined by

dxdt = Jdsd

J =x

s

t

x

t

s(2.9)

(t(),x

())

= 1

= 2

= 3

= 4

s = 0

s =1

s = 2

s = 3

(x(, s),t(, s))

u(t,x)

(t, x

)

Figure 2.4: (a) The parameter labels the different characteristic curves by spec-ifying their intersection with . (b) Specifying u(x, t) = u(x(s, ), t(s, )).

2.2 Solution on Characteristics

As discussed in previous Section, every point of the xt-plane can be labeled by(s, ), as shown in Figure 2.4(a), which in turn yields the solution u(s, ), asshown in Figure 2.4(b). The solution of the PDE is uniquely specified on the

xt-plane by specifying V(s, ), as shown in Figure 2.5; the characteristic curvesare labeled by x(s, ), t(s, ) and the solution is V(s, ), such that

x(0, ) = x() (2.10)

t(0, ) = t() (2.11)

V(0, ) = f() (2.12)

On the characteristic curve we have

dV(s)

ds= d(x(s)t(s)) + c(t(s), x(s))V(s) (2.13)

d(s) + c(s)V(s) (2.14)

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2.2: Solution on Characteristics 13

V

t

0

V(0,0 ) = f(0 )

V(s,0)

Figure 2.5: The solution V(s, ) specified on the characteristics.

Using the integrating factor exp( s0

dscs(s)) yields

d

ds(e

s

0dsc(s)V(s)) = e

s

0dsc(s)d(s) (2.15)

Therefore, integrating from 0 to s yields

V(s, ) = V(0) + es0 ds

c(s

)s0

dses

0 ds

a(s

)d(s) (2.16)

Note the solution satisfies the boundary condition since

V(0) = V(x(, 0), t(, 0)) (2.17)

= f(t(), x()) (2.18)

= f() (2.19)

We invert x(, s), t(, s) to obtain = (t, x), s = s(t, x), and obtain the sought

after solution

V(x, t) = V(s(x, t), (x, t)) (2.20)

Consider the special case of c and d beings constants; we have

V(t, x) = f(t, x) +d

c(1 ecs) (2.21)

where s = s(t, x)

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2.3 Rigid Translation

We discuss two cases of rigid translation of a wave form.

With damping.

Without damping.

Rigid translation with no damping

Consider the case of rigid translation, namely

cVx + Vt = 0 (2.22)

with initial vaue V(0, x) = f(x). The characteristic curve is given by

dx

ds= c ,

dt

ds= 1 (2.23)

and

dV

ds= 0 : constant on characteristics (2.24)

Let be the boundary value curve parameterized by such that

x = , t = 0 , V = f() (2.25)

Then, from Eqs. 2.23 and 2.25

x(s, ) = cs + ; t(s, ) = s (2.26)

and

V(s, ) = f() (2.27)

Solving for s, yields

s = t ; = x ct (2.28)

Hence the solution of the first order PDE is given by

V(s, ) = f((x, t)) = f(x ct) (2.29)

The initial value f() propagates unchanged on the characteristic curve.

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2.3: Rigid Translation 15

Rigid translation with damping

cux + uy = u ; u(x, 0) = f(x) (2.30)

Boundary curve and value is given by

= {(x, y) : y = 0, x = , } (2.31)u(x, 0) = u(, 0) = f() (2.32)

ey

Figure 2.6: Wave dissipating while propagating along the x-direction

The characteristic curves are given by

dx

ds= c ;

dy

ds= 1 (2.33)

x = cs + ; y = s (2.34)

s = y ; = x cy (2.35)

On the characteristic curve

duds

= u u(s, ) = u(0, )es = f()es

Converting from s, to x, t yields the solution

u(x, y) = f(x cy)ey

As shown in Figure 2.6, u(x, y) is a traveling wave that is exponentially damped.

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2.4 Linear PDEs

Consider the linear PDE

ux + 3x2uy = 2xu (2.36)

The boundary curve is given by parametrized by , namely

= {x(), y()| [a, b]} (2.37)

The boundary value is given by

u(x, y) = g() (2.38)

The characteristic curve is fixed by

dx

ds= 1 ;

dy

ds= 3x2 (2.39)

The characteristic curves are given by

x(, s) = s + x() ; y(, s) = s3 + y() (2.40)

The characteristic curves intersect at s = 0 and hence

x(, 0) = x() (2.41)

y(, 0) = y() (2.42)

a

b

x

y

Figure 2.7: Boundary curve : x = 0; a < y < b.

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2.4: Linear PDEs 17

Define a specific by

= {(x, y) : x = 0, a < y < b} x = 0 , y() = ; a < < b

and hence the characteristics are given by

x(, s) = s (2.43)

y(, s) = s3 + ; a < < b (2.44)

Inverting Eq. 2.40 yields

s(x, y) = x (2.45)

(x, y) = y x3 (2.46)

On the characteristic curve

du

ds= 2su du

u= 2s (2.47)

Note the characteristic curves given in Eq. 2.40 are labeled by , which specifiesthe boundary curve . On characteristic curves

u = k()es2 = k()ex2 (2.48)

Boundary Condition on yields

u() = u(0, y) = g() = k (2.49)

or k = g(y x3) (2.50)

Hence,

u(x, y) = g(y x3)ex2 (2.51)

Verification

uy =u

y= ex

2

g(y x3) (2.52)

ux =u

x= [3g(y x3)x2 + 2xg(y x3)]ex2 (2.53)

ux + 3x2uy = 2xe

x2g(y x3) = 2xu (2.54)

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2.5 Fan-like Characteristics

Consider the PDE

xux + yuy = 1 + y2 (2.55)

with boundary curve = {(x, y); y = 1, x = , } and boundaryvalue given by

u(x, 1) = f() = + 1 (2.56)

The characteristic curve equation is given by

dxds

= x , dyds

= y (2.57)

x(, s) = h1()es ; y(, s) = h2()e

s (2.58)

The boundary conditions are

: x(, 0) = = h1() (2.59)

y(, 0) = 1 = h2() (2.60)

x(, s) = es ; y(, s) = es (2.61)

and yields the characteristics

x(, s) = es ; y(, s) = es (2.62)

x = yThe characteristics are shown in Figure 2.8. Inverting Eq. 2.62 yields

s = ln y ; =x

y(2.63)

From Eq. 2.55

du

ds= 1 + y2 = 1 + e2s

u = c + s +1

2e2s

u(, 0) = c +1

2= + 1

c = + 12

=x

y+

1

2

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2.6: Expansion Wave 19

y = 1

x= y

Figure 2.8: Fan-like characteristic curves x = y.

Hence

u =x

y+

1

2+ ln y +

1

2y2 (2.64)

To verify the solution note

ux =1

y; uy =

1

y+ y x

y2(2.65)

Therefore

xux + yuy = 1 + y2 (2.66)

2.6 Expansion Wave

Consider the PDE

ut + uux = 0 (2.67)

with

u(x, 0) = h1, x > 0h2, x < 0.

where h2 < h1. This equation emerges in the study of a bursting dam (or mem-brane), with the water from the burst dam forming what is called an expansion

wave. The boundary curve is

=

(x, t) : t = 0, x = ,

(2.68)

Define x(s, ), t(s, ) by

dt

ds= 1,

dx

ds= u (2.69)

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h2

h1

x

u(x,0)

Figure 2.9: Initial value u(x, 0).

and hence

du(s, )

ds= 0 u(s, ) = u(0, ) (2.70)

From Equation 2.69 we have

t = s, x(s, ) = su(s, ) + (2.71)

x

t x=h2t

x=h1t

u=h1

u=h2

Figure 2.10: Solution of the PDE with no characteristics in the wedge.

The initial condition u(x, 0) = f(x) yields

u(0, ) = f() =

h1; > 0h2; < 0.

Hence, the characteristic curves are

x(s, ) = su(0, ) + =

h1s + ; > 0h2s + ; < 0.

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2.6: Expansion Wave 21

The characteristic are hence straight lines and shown in Figure 2.10. The solution

is given by

u(x, t) =

h1, x > h1t ; ( > 0)h2, x < h2t ; ( < 0).

Note that domain bounded by x = h2t and x = h1t has no characteristics.The reason is the discontinuity at x = 0 in the initial condition. In particular, weneed characteristics for h2t < x < h1t.

x

t

x=h2t

x=h1t

t0

xx=h2t0 x=h1t0

u=h1

u=h2

u=x/t0

u

Figure 2.11: (a) Characteristic curves inside the fan. (b) The solution u(x, t0),including the portion inside the fan.

One assumes that all the values in interval [h1, h2] are present at for u(x, 0)at point x = 0 and thus yields the characteristics that fill up the blank domain.2

Namely

dx

ds= u x = us ; h1 < u < h2 (2.72)

Hence, along the characteristic

u =x

s=

x

t, h

2t < x < h

1t (2.73)

This yields a fan-like structure for characteristics given in Figure 2.11(a); the

solution is an expansion wave given by

u(x, t) =

h1 ; x > h1tx/t ; h2t < x < h1t

h2 ; x < h2t.2By considering the case of a continuous initial condition, it can be shown that our assump-

tion emerges by taking the limit of a discontinous initial condition at x = 0.

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The values of u(x, t) for a value of t > 0 is shown in Figure 2.11. As can be

seen from Figure 2.11(a) many characteristics emerge from the origin x = 0 =t; Figure 2.11(b) shows how, for a typical time t0 > 0, the solution u(x, t0)smoothly increases from u = h2 for x < 0 to u = h1 for x > 0.

2.7 Traffic Flow

To exemplify the interpretation of characteristics, consider the simple model of

traffic flow, with density and velocity of of cars, denoted by , u respectively arerelated due to conservation of number of cars by the following first order

partial differential equation

t + (u)x = 0 ; t > 0 ; < x <

After a long red light at x = 0, the roadway is empty for x > 0, and is full ofstationary bumper-to-bumper cars for x < 0. At time t = 0 the light turns green.The initial data, as shown in Figure 2.12(a) is given by3

(x, 0) =

m x < 00 x > 0.

corresponds to a stop light at x = 0, which turns green at t = 0.

0 x

(x,0)

m

-1/n 1/n x

n(x,0)

m

Figure 2.12: (a) Initial traffic distribution. (b) Initial traffic distribution as a limit

of a continuous distribution.

As the signals effect propagates up road, the cars accelerate in turn. One canask: at what time will the car at x = a be able to start moving? and where andwhen will this car attain speed V /2? We analyze the characteristics to addressthese and similar questions.

The traffic flow equation is given by

t+

x(u) = 0, t > 0, < x < . (2.74)

3The initial discontinuous traffic distribution given in Figure 2.12(a) can be obtained as the

limit, ofn , of the initial distribution shown in Figure 2.12(b).

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2.7: Traffic Flow 23

We assume the linear velocity-density relation(empirical):

u = V(1 m

) (2.75)

Thus when = 0 (open road) the car travels at velocity V; when = m(traffic jam) the car does not move.

Substituting in and differentiating we have the standard form for the PDE

given by

t + c()x = 0 (2.76)

where

c() = V(1 2m

). (2.77)

The initial distribution ofc() is shown in Figure 2.13.

0 x

c/((x,0))=V[1-(2/m)(x,0)]

V

-V

Figure 2.13: The initial distribution ofc() = c((x, 0).

This quasi-linear first-order PDE has two simplifying features. The coeffi-

cient of unity in front of t allows us to forego parametrization, with t = s.The homogeneity of the equation means that along a characteristic (x, t) is aconstant.

Hence the equation of the characteristic in the (x, t)-plane yields

dt

ds= 1 ;

dx

dt= c();

d

dt= 0 ; (x, t) = (x, 0) (2.78)

dxdt

= c((x, 0)) = constant (2.79)

Hence, all the characteristics are straight lines, with slopes depending upon

the (constant) value of(x, 0).Graphs of(x, 0) and c

(x, 0)

versus x appear below, as well as a sketch of

the characteristics in the (x, t)-plane. From the sketch it is evident that (x, t) =

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0 for x > V t and (x, t) = m for x x0

(2.125)

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The boundary curve is given by

= {(x, 0)|t() = 0; x() = , +}and hence

f(x) = f() =

0, 0 < < x01, > x0

The characteristics, shown in Figure 2.19(b), are given by

dx(s, )

ds= et;

dt(s, )

ds= 1

x(s, ) = 1 + es

; t = s t = ln(x c) ; c = 1

On the characteristics, the PDE yields

du(s, )

ds= 0 ; u(s, ) = u(0, ) = f()

Expressing s, in terms ofx, t yields the solution

u(x, t) = 0, x < et + x0 11, x > et + x0

1

As shown in Figure 2.19(b), the discontinuity in f(x) at x0 creates a disconti-nuity in u(x, t) that propagates along the characteristic curve passing throughx0.

2.11 Intersection of Characteristic Curves

The intersection of characteristic curves creates discontinuities since on each

characteristic curve, the PDE propagates the value picked up from the boundary

value specified by , with a typical case shown in Figure 2.20.

Hence along the intersection, the value of u(x, y) has to become multi-valued. What this means is that the value ofu(x, y) has discontinuities acrossthe line of intersection.5

Consider the PDE

u

x+ u

u

y= 0 (2.126)

5Note for the case when the solution u(x, y) has discontinuities (shock waves) the equiva-lence of Eqs. 2.99 and 2.98, no longer holds. The intersection of characteristic curves give rise

to discontinuous solutions.

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2.11: Intersection of Characteristic Curves 33

Figure 2.20: Intersection of characteristic curves

with initial condition, shown in Figure 2.21(a), as follows (a > 1)

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

1 x2, < x < a1 a2, x > a (2.127)

f(x)

a

x=1

x= axs

= 1

x= 1

x< a x> a

= 0 = 1 = 2

ts

Figure 2.21: (a) Initial value f(x). (b) Intersecting characteristic curves.

From Eq. 2.105

u(x, y) = xy 1

1 + 4y(y x)2y2 (2.128)Recall, from Eq. 2.91, that the characteristic curves are

x(s, ) = su(s, ) + (2.129)

y(s, ) = s (2.130)

Since

u(s, ) = u(0, ) = f() (2.131)

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we have

x(, s) = sf() + ; y(, s) = s (2.132)

Hence, for < a, the characteristic curves are given by

x(s, ) = s(1 2) + ; y(s, ) = s (2.133) x y(1 2) = (2.134)

and for > a the characteristic curves are given by

x y(1 a2) = , a > 1 (2.135)The intersecting characteristic curves are shown in Figure 2.21(b), with the first

two characteristic curves to intersect being the one for = a coming from theleft and the right.

2.12 Shock Waves

Consider the initial value problem

(x, t)

t+ 2

(x, t)

t= 0 (2.136)

with initial value

(x, 0) = f(x) =

4, x < 03, x > 0

(2.137)

3

4

Figure 2.22: Initial value of(x, 0).

The characteristics are given by

dt(, s)

ds= 1 ;

dx(, s)

ds= 2 (2.138)

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2.12: Shock Waves 35

Hence

d

ds= 0 (, s) = (, 0) = f() (2.139)

t(, s) = s + h1() ; x(, s) = 2f()s + h2() (2.140)

The initial curve is given by

= {(x, t) : x = , t = 0; } (2.141)Hence

t(, s) = s ; x(, s) = 2f()s + (2.142)

or, x(, s) = 2f()t(, s) + : characteristics (2.143)

The characteristics, shown in Figure 2.23(a), can be seen to intersect.

tS0

xS0

1

2

Figure 2.23: (a) Intersecting characteristic curves define the shock front. (b) The

time and position xs0 , ts0 at which the shock starts.

The explicit expression for the characteristics are given by

x = 2f()s + = 8t + , < 06t + , > 0

(2.144)

The shock starts at xs0, ts0 , as shown in Figure 2.23(b), where ts0 is the firstinstant when the characteristics intersect. To find ts0 note that, we have, in gen-eral

xs0 = g(1)ts0 + 1

xs0 = g(2)ts0 + 2

ts0 = 2 1

g(2) g(1)

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The first characteristic curve, after 1, to intersect with 1 is clearly 1+. Hence,

ts0 = 1

g(1)

xs0 = 1 g(1)

g(1)= 1 + g(1)ts0

Note since ts0>0, for a shock wave to form we must have

g(1) < 0 (2.145)

Conversely, ifg()>0 for all , there is no shock front formation, and the waveis said to be an expansion wave.

For our case 1 < 0 and 2 > 0; hence, from Eq. 2.144

g(1) = 2f(1) = 8 ; 1 = 0 (2.146)g(2) = 2f(2) = 6 ; 1 = 0+ (2.147)

and hence

ts0 = 2 16 8 =

1

2(2 1) 0 (2.148)

and xs0 = 1 + 0 0 (2.149)In summary the shock starts at the origin, namely (xs = 0, ts = 0) and has aconstant velocity.

Shock front; shock velocity

We rewrite the PDE as follows

0 =

t+ c()

x=

t+

q

x(2.150)

where c =q

The PDE breaks down at the intersection of the characteristics giving mul-

tiple solutions at the intersection. However, it is assumed that the conserved

quantities continue to be conserved at the intersection. In effect, what this meansis that we assume the integrated version of the PDE continues to be valid at the

intersection.

Consider the PDE in the range of a < x < b. We expect that the integratedversion of Eq. 2.150 holds across the shock wave (discontinuity), namely

0 =

ba

dx

t+

q

x

q(a, t) q(b, t) =ba

dx

t=

d

dt

ba

dx(x, t)

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2.12: Shock Waves 37

xs

xs+

xs

Figure 2.24: Shock located at xs; value to the left xs and right xs+ ofxs.

xs(t) = 7t

Figure 2.25: Shock wave located at xs(t) = 7t.

Since there is a discontinuity in the density , we break up the integral andcompute the following

q(a, t) q(b, t) = ddt

xs(t)a

dx(x, t) +

bxs(t)

dx(x, t)

=dxs(t)

dt(xs, t) +

xs(t)a

dx

t dxs(t)

dt(xs+, t) +

bxs(t)

dx

t

=dxsdt

[(xs , t) (xs+, t)] +

q(a, t) q(xs, t) q(xs+, t) q(b, t)

where xs and xs+ are defined in Figure 2.24.Hence

dxsdt

= q(xs, t) q(xs+, t)(xs, t) (xs+, t)

(2.151)

Consider the following PDE

t+ 2

t= 0

We have q = 2 and hence

dxsdt

=2() 2(+)

() (+) =42 324 3 = 7

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The shock front trajectory, shown in Figure 2.25, is a straight line given by

xs(t) =dxs(t)

dt(t ts0) + xs0

Since xs0 = 0 = ts0 and dxs/dt = 7, we obtain

xs(t) = 7t

lecture 14, 19 oct 2010

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