Nonlinear boundary problems

Nonlinear boundary problems

Some denitions and examples

A boundary problem is called nonlinear if either the dierential equation

or the boundary value problem is nonlinear The concept of a nonlinear

dierential equation is clear see sections and For the solutions of these

equations the superposition principle does not hold From the standpoint of

the physicist or engineer nonlinear partial dierential equations appear if the

underlying phenomenon itself is nonlinear or if parameters or coecients in

the dierential equation depend on the solution function In electrodynamics

very often the material parameter depends on the solution of the equations

of electromagnetism

A boundary condition is called nonlinear if the condition itself depends on

the solution function of the actual dierential equation

Let us view some typical nonlinear dierential equations

the diusion equation with variable diusion coecient or

the Carrier equation for a string with varying tension

the Euler equation of motion

Other examples are the partial dierential equations describing the cooling

of a nuclear reactor where the neutron density nx t and the temperature

T x t are coupled through n vT

x

T

t

n

t

n

xx

T n where v is

the ow velocity of the coolant and the coolant density Another example is

given by the equation of motion of glaciers Finsterwalder equation

n ku

n

a

u

x

a

u

t

Here ux t is the vertical thickness of the glacier which glides downhill on

an inclined plane of angle where k tan and n

Nonlinear boundary conditions are found when considering the super uidity

of He

The pertinent partial dierential equation u

t

u

xx

and the initial

condition ux are linear but the boundary condition is nonlinear

by vx y z t r where The partial dierential equation is

linear but the boundary condition that there exists atmospheric pressure p

on the water surface is not linear The Bernoulli equation reads

v

t

r

v

p

gz

r

t

r

p

gz

Here g is the gravitational acceleration p pressure density of water and

z is the coordinate orthogonal on the water surface Formal integration with

respect to the three space coordinates x y z results in

t

r

p

gz const

On the ground z h of the lake the normal component v

n

vanishes so

that z

zh

On the surface of the lake z one has

t

r

p

const

Integration of together with the two boundary conditions results in elliptic

functions snoidal and cnoidal waves A linearized approximate solution

of the problem will be discussed in problem Moving boundaries are mainly

nonlinear although very often it is dicult to recognize the nonlinearity

Consider an in nite halfspace orthogonal to the x axis which is lled with

water On the boundary plane x the temperature is cooled down to

T

C Thus the water will freeze there As time passes on a freezing

front will penetrate into the water and freeze it If we designate by T

x t

the temperature within the ice and by T

x t within the water and let T

s

be

the temperature within the front then continuity demands

T

s

T

x Xt t T

x Xt t or dT

dT

HereXt designates the actual location of the freezing front Now we consider

the two heat conduction equations Subscript refers to ice and subscript

to water respectively

T

x t

t

c

T

x t

x

T

x t

t

c

T

x t

x

where kgdm

and c kJkgK designate density and speci c heat K is

the thermodynamic temperature WmK is the thermal conductivity heat

conductivity Boundary conditions exist mainly within the freezing front

Energy conservation demands that the relative rate of heat ow through the

front is equal to the melting heat L kJkg transported away

T

x Xt t

t

T

x Xt t

t

L

dX

dt

where

Stefan boundary condition Furthermore the con

ditions at x and x have to be taken into account

T

x T

for x

Having a short look at the foregoing equations it is not immediately clear

that they represent a nonlinear problem But using

dT

i

T

i

t

dt

T

i

x

dx for i

and

dx

dt

dX

dt

T

t

T

t

T

X

T

X

and insertion into yield the nonlinear boundary condition

T

X

T

X

T

X

T

X

L

T

t

T

t

The nonlinearity is hidden in the fact that the moving front boundary Xt

depends on the temperature T x t We will present a solution of the Stefan

problem in the next section

Another example of a nonlinear boundary problem appears in connection

with the disposal of radioactive waste One must know if the heat produced

by the radioactivity of the waste may be removed by the surrounding rocks

For the timedependent thermal diusivity at ttct the heat

conduction equation reads

ux t

t

at

ux t

x

x t

At rst sight this looks linear But the initial condition ux and the

boundary condition

u t ft at

u t

x

gt

where ft and gt are known contain nonlinearities It has been shown that

this problem is actually nonlinear

Such nonlinear boundary problems occur for ordinary dierential equations

too Collatz gives some examples

y

y

y y

y

fy y y

yl y

y

y y

cosx y y y

y

Problems

Solve the problem of water waves on a lake by linearization This yields

t

g

z

for z

Use the setup

x y z t A expik

x

x k

y

y tUz

and the boundary conditions

U

z

for z h

g

t

for z

Solution Uz cosh kz h k

k

x

k

y

U g

dU

dz

gk sinh kh

cosh kh for z

so that the wave dispersion k is given by

c

Ph

k

r

g

k

tanhhk k

Solve Solution y x

Hint use the method how to

solve the equation of motion m%x Kx Try DSolve

Solve by the same method Equation of motion

Try to solve

Moving and free boundaries

It seems to be clear that moving boundaries may appear only for partial dif

ferential equations containing the independent variable time These partial

dierential equations are mostly of parabolic type like Free bound

aries however occur mainly with elliptic partial dierential equations and

very often describe equilibrium situations

We rst would like to solve the Stefan problem The heat conduction

equations are exactly of the form in section There we

found that the similarity transformation

p

d

T

d

dT

d

Using u T

we obtain a separable equation like

du

d

u

which yields after integration

Z

du

u

Z

d lnu

lnC

u const exp

and

T const

Z

exp

d

constAerf T

erf

x

p

at

Thus we have for ice

T

x t T

Aerf

x

p

a

t

and for the water

T

x t T

Berfc

x

p

a

t

Here the function erf is the error function and erfc erf is the

complement error function As is well known one has

erf erfc erf erfc

The solutions and satisfy the boundary conditions For

a solution of Mathematica yields

T C

p

C

erf

Returning to we nd that the initial condition is also satis ed

due to Now the two unknown integration constants A and B will be

used to satisfy the boundary condition Let us make a setup for the

path of the freezing front

Xt t

dX

p

T

A erf

p

a

T

s

T

B

erf

p

a

If is assumed to become known later we can nd an equation for A and B

A

T

s

T

erf

p

a

B

T

s

T

erfc

p

a

Now using the setup we can satisfy We calculate

T

x t

x

xXt

and

T

x t

x

xXt

and

d erf

d

p

e

Insertion of these expressions into results in

T

s

T

exp

a

p

a

erf

p

a

T

s

T

exp

a

p

a

erf

p

a

p

L

For given parameters

a

a

L for ice and water and assumptions for

T

s

C T

C equation is a transcendental equation determin

ing giving something like see problem Then the solutions

read

T

x y T

T

s

T

erf

p

a

erf

x

p

a

t

x Xt

T

x t T

T

s

T

erfc

p

a

erfc

x

p

a

t

x Xt

Moving boundaries may also occur for elliptic partial dierential equations

The incompressible steadystate D viscous ow between two plates

in a distance a the socalled HeleShaw problem is an example Finite

elements have been used to solve it The plates are arranged in such a

way that gravity acts parallel to the y axis and main ow is in the x direction

Then the two velocity components u v are driven by pressure p and gravity g

ux y

a

px y

x

vx y

a

px y

y

g

and the velocity potential reads

x y

a

px y gy

by uid What then is the boundary condition in a steady state on the

interface between the two uids Apparently the mean velocity u

and u

respectively will be given by

u

a

gradp

gy grad

u

a

gradp

gy grad

One can assume that the components of u

v

normal to the interface between

the two uids are continuous Special applications of this problem are of

industrial interest injection of a uid into a cell electrochemical machining

seeping of water through a dam lubrication of a bearing cavitation etc may

be formulated as a HeleShaw problem

The taste of cheese depends on the diusion of chemical substances pro

duced by typical bacteria or mold fungi Thus the cheese producers are

interested in diusion problems within solids The speed of the moving front

determines the storage time and consequently price and quality of the product

For the steady state of a diusion process one has

D

c

x

m

cx is the local concentration of the chemical substance andm is a production

or absorption rate moles or gramssec Let x describe the cheese

surface and designate by x

t the actual depth of penetration of the diusing

substance The steady state will be described by

cx t

c

x

for x x

and during the maturity process one has on the cheese surface

c t c

const

Apparently

c

m

D

x x

x

Dc

m

is a solution of to As soon as the ripening time of the

cheese is over its surface x will be sealed At this moment the location

of the diusion front can be designated by x

Then the mathematical

problem to be solved reads

c

t

D

c

x

m x x

c

for x t

cx

t c

x

x x

t

cx t c

m

D

x x

x x

t

Various numerical methods have been used to solve this implicit prob

lem It is called an implicit problem because x

t is absent A transformation

c ut may create an explicit problem or the boundary condition

can be replaced by cx ft for x giving

cx t

X

p

exp

p

t

p

cos px

x

Ablation is the stepbystep removal of matter ice fuel for thermonuclear

fusion metals mainly by direct transition of the solid into the gaseous state

This is of interest for laserinduced fusion In this process the ablation front

penetrates into the material and one again has a moving boundary inter

face between two thermodynamic phases The penetration process can be

described by x

t st Let l be the thickness of the layer in which the

thermal ablation energy qt is supplied then the local temperature

T x t obeys

T

t

T

x

x st

T x

t gt

T

x

dx

dt

qt x

st

and

T x for x l t

T ft for x t s l

First x and ft are unknown Solving for T x t for a given path st is

called the inverse Stefan problem A transformation on a comoving coordi

nate system xst is useful see the remarks on progressing waves in the

next section

Free boundaries occur on surfaces of a free jet ow where the atmospheric

pressure determines the boundary A very simple example of this type can be

described by conformal mapping The function

z e

x u e

u

cos v y v e

u

sin v

describes the pouring out of a liquid of an in nite half space tank Borda

outlet For the steadystate building of the jet the nonlinear boundary con

dition is needed

u

u

const C

streamlines It is not possible to directly solve for ux y vx y We

make an hodograph transformation as in section We replace the dependent

variables ux y vx y by the independent variables xu v yu v becoming

then the new dependent variables Thus

du u

x

dx u

y

dy dv v

x

dx v

y

dy

dx x

u

du x

v

dv dy y

u

du y

v

dv

so that

dx v

y

du u

y

dv D dy v

x

du u

x

dv D

where

D

u

x

u

y

v

x

v

y

Comparing the expressions for dx and dy in as well as in

results in

x

u

v

y

D x

v

u

y

D

and

x

u

y

u

u

x

v

y

C

These equations now express the free nonlinear boundary condition on the

free surface of the jet But how do we nd a mathematical expression for the

free surface We make a setup for the surface v const using an unknown

function F u

x

u

F u y

u

p

F

u

After some trial and error and inserting into one obtains

F u e

u

y

u

p

e

u

Integration yields

xu u e

u

yu const

p

e

u

e

u

arcsin

p

e

u

An investigation of the solution demonstrates that the ow velocity makes a

discontinuous jump on the free surface from vsurface to v in air Thus

the potential ux y is discontinuous too LeviCivita potential Problems

of this kind occur in plasma physics hose instability If the pressure in

uid jets drops below the saturation vapour pressure of the uid cavitation

sets in the uid evaporates and gas bubbles are formed If they implode

screws

Seepage ow through a retaining dam if an earth ll dam is an important

engineering problem connected with hydropower generation and ood disaster

protection The surface of seepage water ow presents free boundary that is

contained within the dam Seepage of a uid through a porous material in

the x direction is described by the Darcy law

v rp gy r

Here is the porosity of the dam material and gy the weight of the water

Let x be the waterside of the dam and d its thickness then the boundary

conditions on the solids read

x y H for y H

x d y h for y h

x d y y for y h

Here H is the height of the water level and h is the height of the dam h H

Furthermore one can assume that there is no seepage at x d y h

y

for x d y

x

for x d y h

Within the dam the free seepage water surface y

x is described by

y n

or by the Bernoulli equation One is now interested in the shape of the free

boundary

y

fx f H fd h

d

dx

fx

x

d

dx

fx

xd

The problem described by the last equations to can be solved

using a Baiocchi transformation

wx y

H

Z

y

x d

leading to a partial dierential equation for w This high mathematics is

modied

x

k

x y

x

k

x y

y

y

k

x y

x

k

x y

y

For an isotropic homogeneous dam k

k

k

k

constants

would be valid for an isotropic inhomogeneous dam one would have functions

k

x y k

x y k

x y k

x y Engineering values of and

the k

ij

may vary in nature

In recent years there has been great interest in moving elastic boundaries

as in the blood streaming in an artery

Problems

Using the command FindRoot nd from You can assume

ice

kgdm

water

kgdm

c

ice

kJkg K c

water

kJkg K

ice

Wm K

water

Wm K

a

ice

cm

sec a

water

cm

sec

T

s

C T

C

Remember Joule J m

kg s

Watt W Js

m

kg s

International Union of Pure and Applied Physics

Vary T

s

T

Plot and taking numerical values from problem

Use T

C

C

C and x Do you observe a salient

point on the x axis Does it depend on L Yes

Plot and separately For the methods see section

Waves of large amplitudes Solitons

Phenomena with large amplitudes are mainly described by nonlinear equations

and cannot be described by approximately linearized dierential equations

nuity equation for the onedimensional time dependent compressible ow of a

gas without sources According to one has

t

u

x

u

x

The Euler equation assumes the form

u

t

uu

x

p

x

The assumption of adiabatic behavior and the denition of the velocity

of sound allows us to rewrite in the form or

u

t

uu

x

a

x

Thus the calculations to yield the onedimensional potential

ow equation in the form

a

x

xx

x

xt

tt

Again ux t x tx and the velocity of sound a is given by

in the form

a

a

x

t

This is a consequence of and or can be read o from

If

x

t

and

xt

are small one obtains a constant sonic speed a a

and

becomes the linear wave equation

xx

tt

a

To solve the nonlinear partial equation we follow the lines of thinking

in section and apply a Legendre transformation In full analogy to

we make the setup where now x and t are dependent variables

x

u

t

q

xx

u

x

tt

q

t

d

x

d

x

t

dt udx qdy

u q ux qt x t

d

u

du

q

dq xdu tdq

u

x

q

t

uu

x

u

uq

t

q

xx

qq

D

xt

qv

D

tt

uu

D

where

D

uu

uq

uq

qq

We then obtain the exactly linearized potential equation in the form

obtains

w

dx

dt

u au

which cannot be used since the wave velocity w depends on the unknown

solution ux t But the characteristics for

dq

du

u a

are independent from q and can be used to solve as we shall see in the

next section

Since the wave speed w depends on u a wave propagates faster in relation

to the speed of ow velocity u This means that a pressure wave amplitude

will steepen up during propagation This continual steepening of the wave

front can no longer be described by single valued functions of location One

has to introduce a discontinuity into the ow to get over this diculty This

discontinuity is called a shock wave across which the ow variables change

discontinuously This jump is described by the Hugoniot equation

The discontinuity vanishes if the basic uid equations are modied

to contain viscosity and heat conduction Including these terms Bechert

derived a nonlinear partial dierential equation of fth order which could

be solved by a similarity transformation This solution is no longer

discontinuous but describes a continuous but very steep transition between

the values of the ow parameters before and after the wave front

It seems that Preiswerk in his thesis was rst to draw attention to

the fact that the same gasdynamic equation that we just discussed describes a

water ow with hydraulic jumps water surges in rivers when a water lock is

suddenly opened upstream Pressure jumps occurring in hydroelectric power

stations obey similar equations Preiswerk has shown that gasdynamic

equations can be applied on water surface ow if c

p

c

is put in these

equations Formally the temperature T of the gas corresponds to the water

depth

We now have the mathematical tools to undertake the problem of gravity

waves on the surface of a lake described by and equations

An approximate linearized solution has been discussed in problem of

section The nonlinearized solution starts with the Bernoulli equation

which we write in the form

t

x

z

gh h

Here h

is the constant depth of the unperturbed free water surface and

hx z t describes the disturbed lake surface Now z designates the

ground of the lake This designation is in contradiction to section Since

z

for z

On the free surface the boundary condition reads

z

h

t

x

h

x

for z hx z t

The nonlinear boundary condition is given by Some authors add

a term describing wind pressure or capillary tension to describe the initial

excitation of the waves We now have two unknown functions x z t and

hx z t to be determined from and the three boundary conditions

Several methods exist to solve this problem Using the

transformations

p

x c

t

t

p

and the abbreviation

c

p

gh

one obtains

zz

z

for z

z

h

c

h

for z h

c

z

gh h

Using a perturbation setup

h h

h

h

one gets

zz

nzz

n

z

z

z

c

h

for z h

z

h

zz

h

c

h

h

for z h

c

gh

c

gh

Elimination results in the Kortewegde Vries equation

h

c

h

h

h

c

h

h

Writing it in simplied form

vx t v x c

t

one obtains

v

v c

v

After two integrations one gets

v

c

v v

v

c

v

v

so that

v

Z

v

max

dv

p

c

v

v

p

This results in a soliton solution

vx t

c

sech

p

c

x c

t

A soliton or solitary wave is a nonlinear wave propagating without change of

shape and velocity In these waves an exact balance occurs between the non

linearity eects steepening the wave front due to the increasing wave speed

and the eect of dispersion tending to spread the wave front Soliton solutions

have been discussed in plasma physics solid state physics in supracon

ductors in the FermiUlam problem in nuclear power stations

in ionospheric problems in nonlinear mechanics pendulum in general

relativity in lasers as well as in the elementary particle theory

Modulation of the amplitude of a wave presents an important physical and

technical problem with many applications The eect of modulation is a vari

ation of the amplitude and the phase of a wave Let us rst investigate these

eects on a generalized wave equation We shall investigate nonlinear

ity dispersion and dissipation To do this we consider the weakly nonlinear

wave equation

c

tt

xx

b g

t

V

N

t

G

where V

dVd N

n

and G are nonlinear functions and c b g

are constants which may depend on the frequency is a small parameter

We now dene a phase surface

x t const

which has the property that all points x t on it have the same value of the

wave function x t We thus have

dx

dt

t

x

see a constant phase Dening wave number k and frequency by

x

k

t

or

t

we nd that is the phase speed In the threedimensional case we

have r

k wave vector

curl

k

which indicates that wave crests are neither vanishing nor splitting o The

last two equations result in the conservation of wave crests

k

t

r

A point moving with the group velocity

dx

dt

g

d

dk

sees unchanged The equation can be classied as follows

N G V

the equation is linear k and are independent

of x und t

b g no dispersion k no dissipation

A expikx it ck

b g frequency dispersion k no dissipation

A expikx it k c

p

k

b

b g dissipation complex

A expikx it

ic

g c

p

k

c

g

b g dispersion and dissipation

A expikxit D k

c

k

big

N G V

the wave equation is nonlinear V

describes

strong nonlinearity N weak nonlinearity and G weak dissipation For

weak nonlinearity dened by

h

xx

i

Z

xx

d h

tt

i

x

k const

t

Two conclusions can be drawn

For a nonlinear nondissipative wave equation the amplitude A

is constant and x t kx tx x tt yields a dispersion relation

For any nonlinear dissipative wave equation the frequency is

not modied by the dissipative terms in rst order of

Now we investigate a modulated wave We assume a sinusoidal wave

a

cos

k

x

t with the amplitude ax t and phase x t varying

slowly in x and t

x t k

x

t x t

k

a

According to we redene

x t

t

t

kx t

x

k

x

For weak modulation one may write

a

a

a

k

k k

k

k k

If one makes the replacement

by i

t

k k

by i

x

one obtains the socalled nonlinear Schroedinger equation

i

a

t

k

a

x

k

a

x

a

jaj

a

In a frame of reference moving with the group velocity equation

becomes

i

a

a

jaj

a

a

k

In plasma physics the nonlinear Schroedinger equation describes electron

waves It also appears in nonlinear optics If one inserts

a U c expik i jaj

U

into one gets as the real part

c

Z

dU

p

k

U

U

C

C

which represents an elliptic integral and then a Jacobi function cnoidal

wave For C

one obtains an envelope soliton

U c const sech

k

c

and the real part of the solution of reads

a const sech

k

c

cosk

since

Z

dx

x

p

x

arsechx

The steepening of a wave front occurs not only for plane waves as discussed

earlier but also for spherical waves Such waves may be inward running

compression waves like in a kidney stone destroyer or in an antitank rocket

launcher or in outward running rarecation waves explosion waves

In spherical geometry the equations and read

t

u

r

u

r

u

r

u

t

uu

r

p

r

But these waves are no longer adiabatic so is no longer valid and has

to be replaced by the polytropic equation

d

dt

p

n

t

p

n

u

r

p

n

given by

n

c c

p

c c

V

c c

p

c

V

are the specic heats for polytropic isobaric and isochoric thermody

namic changes of state respectively A progressing wave simple wave setup

based on similarity transformations can be written down to solve

equations

ur t rt

U pr t

r

t

P

r

r

t

Ordinary di erential equations will be obtained and spherical shock waves

standable since the magnetic permeability H may depend on the solution

for the magnetic eld H Furthermore in ferroelectric material or in plas

mas the electric permittivity may depend on the electric eld E It might

be therefore of interest to investigate if the electromagnetic pulse EMP due

to nuclear explosions could be explained by an electromagnetic shock wave

in a plasma A transverse electromagnetic wave E

x

Ez t in a plasma

satises a nonlinear wave equation

Ez t

z

Ez t

t

where

dDdH

The solution for the Ewave propagating with a phase speed

may be

Ez t

z

t

This wave will steepen if Ez or

We thus investigate

E

z

t

e

m

H

P

e

H

m

If we designate

z t

then one obtains for the critical time t

c

t

c

As soon as this time is over steepening will occur EMP measurements gave

so that

becomes the critical condition For a homogeneous

isotropic nondissipative cold plasma this condition is satised since

P

e

H

m

Here

P

is the plasma frequency and m the electron mass One might thus

speculate that the EMP is an electromagnetic shock wave

The inverse scattering method is an ingenious method to solve nonlin

ear partial di erential equations We will discuss the solution of the Korte

wegde Vries equation which we write in the form

v

t

vv

x

v

xxx

The inverse scattering method does not directly solve this nonlinear partial

di erential equation but instead it solves two linear equations that have the

same solution as For this purpose we consider the onedimensional

timeindependent Schroedinger equation

n

k

n

t as well as continuous values E k

E To solve two

setups may be made

n

c

n

t expk

n

x expikx Rk t expikx

The bound states E

n

are normalized by

R

n

dx and the free states

E correspond to waves inciding from x on the potential v The

part R of the particles presented by will be reected from the potential

and the part T k t expikx penetrates into the potential Due to particle

conservation one has jRj

jT j

In the next step one solves for

v giving

v

xx

E

Insertion into results in

E

t

x

Q

x

x

Q

x

where

Q

t

x

v E

x

is a formal abbreviation We rst consider only the

n

Since

n

expk

n

x

forx and

R

dx one obtains

E

n

t

k

n

t k

n

This indicates that the E

n

do not depend on t and is simplied Two

integrations yield

n

t

n

x

v E

n

x

F

n

t

n

D

n

t

n

x

Z

n

dx

The integration constants D

n

t must vanish since

n

diverges for x

Then is a linear partial di erential equation containing the still

unknown solution v of Multiplication of by

n

integration

from to and use of result in

d

dt

Z

n

dx F

n

t

Z

n

dx

Then becomes

n

n

v E

n

n

n

n

d

dt

c

n

t k

n

c

n

t thus c

n

c

n

expk

n

t

If one repeats the calculation for T k t expikx one receives a dif

ferential equation for T k t from which Rk t Rk expik

t follows

Quantum theory teaches that the knowledge of c

n

t k

n

t and Rk t is suf

cient to construct the scattering potential v which is actually a solution of

the Kortewegde Vries equation The construction of the potential v may

be made using the GelfandLevitan integral equation Let

Bx t

Z

Rk t expikxdk

X

n

c

n

t expk

n

tx

then the integral equation

Kx y t Bx y t

Z

x

By y

tKx y

tdy

determines K and the solution of the Kortevegde Vries equation is given

by

vx t

dKx y t

dx

for y x

The inverse scattering method can be used for the solution of many nonlinear

partial di erential equations

Problems

The steepening up of the wave front of a largeamplitude wave in a

viscous gas can be described by

ax

u

u

u

lnu u

u

u

u

lnu

u

This solution describes a continuous variation between two asympyotic

states ux u

and ux u

Plot the function ux

for u

u u

x

Hint use the command ContourPlot Use a u

u

Hint plot Sech

x for x Then x x c

t in the argument

Plot the envelope soliton

Hint

Solve for k

C

C

by integration

Derive the ordinary di erential equations for U P resulting

from

Derive the Hugoniot equation also called shock adiabatic curve In a

shock wave front the conservation of mass momentum and energy writ

ten for the normal components of the ow velocity connect the domains

before and after the front

v

v

p

v

p

v

i

v

i

v

where i c

p

T is enthalpy see next section For given p

and V

the Hugoniot curve p

V

is given by

i

i

V

V

p

p

Plot this curve and the adiabatic curve pV

const c

p

c

V

The rupture of an embankmenttype water dam

A typical nonlinear engineering problem is the calculation of the possible

rupture of an embankment dam At the occasion of a rupture a water surge

shall propagate into the channel downstream of the dam Such surge may

generate heavy destructions along the channel or river Let us assume that

the channel has a width B and extends in the direction of the x axis Let

the water depth H in the storage lake be H m and the water depth

h in the channel is assumed to be h

m The water level after a dam

rupture will be designated by hx t Thus the local water mass is given by

Bhx qx

where

is the constant water density Let the storage

have the initial conditions for h and the stream velocity ux t at t

hx H m ux x m

hx h

m ux x

This indicates that at t and x m a vertical water wall of a height

H h

m exists At the other end of the lake x no ow is present

The relevant equations describing the evolution in time of these nonlinear

onedimensional phenomena are

the continuity equation

t

qx t

x

ux tqx t q

t

u

x

q q

x

u

and the equation of motion

u

t

uu

x

p

x

The local hydrostatic pressure px t per unit length is given by

px t

gqx tB

Then we can now write for

u

t

uu

x

g

B

q

x

We thus have two nonlinear partial di erential equations and

for the two unknown functions ux t and qx t We use the method of char

acteristics developed in section for such a system of two partial di erential

equations of rst order We compare our system of two partial equations with

and read o

a

a

a

a

b

u b

gB b

q b

u

Then and yield the propagation speed of small waves

dx

dt

u

r

gq

B

u

p

gh downstream

dx

dt

u

r

gq

B

u

p

gh upstream

and equations and result in

r

gq

du

g

dq

the characteristics The problem is now that we cannot use or integrate

the characteristics equation because they contain the still unknown solutions

ux t and qx t We rst make a transformation to a new variable ms

We dene

d

r

g

B

dq

p

q

q

q

Z

dq

p

q

r

g

B

h

Z

r

g

h

dh

p

gh

Then we use the Riemann invariants dened by We use

r u s u u r s r s

du d u const

n

r

s

The r s or u q plane is called state plane by some authors We now consider a

mapping between the linear state plane r s and the nonlinear physical plane

described by x t Let us discuss the correspondence between the two planes

We allocate the point P r

s

of the state plane to the dam location point

P of the physical plane This expresses the fact that in the point

P x t a local water wall of absolute height h H h

or relative height m above the normal water level in the channel exists

with streaming velocity u u

According to the height

h

H m corresponds to r

u

s

p

gh

At the other end of the lake x one has u and h H

m Thus the point

Q corresponds to Qr

Q

s

Q

where u

Q

Q

p

gH r

Q

Q

s

Q

Q

r

Q

s

Q

Inserting numbers for hH and g

ms

we receive for

P

p

u

r

s

all measured in ms

On the other end of the lake we have for

Q

Q

u

Q

r

Q

s

Q

At the time t of the rupture of the dam the same physical states exist at

x and x But at this time the dam breaks down and elementary

waves composing later on a steepening surge downstream and a rarefaction

wave upstream start at x Replacing in the dx x dt t

we can write for the wave speeds

x

t

u

p

gh u

Thus the rst elementary wave running to the left to x and upstream

reduces the water levelH in the lake It has a wave speed xt

ms

and s

const r

The wave running to

and s

r

const Waves running to the left from P

to Q transfer their svalue to Q since s const is valid for waves running

to the left s

Q

s

Waves running to the right downstream from P to

x transfer their rvalue so that the whole domain righthand of the

dam x always has the same rvalue At the time t t the

next two elementary waves start Both waves now run into domains where

the states had been modied by the rst two waves The upstream wave

enters an area where the water level had been reduced from H to H h

and might be reected at the lake end x It will no longer return with

ms

because water level and driving pressure had been lowered The

second elementary downstream wave starting at t t will be faster than the

rst one because in the channel the water level had been increased by the rst

downstream wave and the water had started to stream to x In order

to be able to calculate the u etc we need to have some knowledge

about the nal state t x For an innitely long channel we dene

the point Rr

N

s

N

in the state plane Apparently the nal conditions will

read for R qx t h

m

N

p

and

for

R u

N

r

N

s

N

The whole phenomenon

of the breakdown of the dam occupies a square in the s r plane The four

corners are given by Rs

N

r

N

s

N

r

s

r

N

and P Qs

r

Now

it the accuracy and our will that have to decide how many steps i N

we will calculate For this decision we consider the pressure di erence from

p

gqx

gBhx

Bx

B

down to p

N

Br

N

This concerns the variation

N

r

r

N

etc If we choose N

pressure steps then each elementary wave carries r

jsj This corresponds to an accuracy of Table

describes the situation in detail

Table Pressure steps for a downstream wave

in front of the wave behind the wave

Nr s u xt s u

the x t plane for every point in the r s plane the values ux t qx t in the

x t plane are dened by the equations to Interpolation within

the grid delivers any wanted ux t qx t and thus the solution of equations

Gas ow with combustion

Combustion of petrol or gunpowder within a gas ow has many practical ap

plications turbogas exhauster jet engines ram jets rocket propulsion and

nally guns Depending on the type of propellant or gunpowder the combus

tion or explosion process has quite di erent characteristic features Usually

combustion is dened as the burning of a fuel associated with the genera

tion of heat The spreading out of a combustion may excite a combustion

wave A detonation or explosion is a very rapid chemical reaction of an ox

idizer and a fuel with large release of heat and pressure waves A de agra

tion is the burning of explosives or fuel at a rate slower than a detonation

Chemical reactions and thermodynamics enter into the description of these

processes Combustion of gases and in gases is always connected with a gas

dynamic compressible ow In some combustion and detonation processes

it may be necessary to include new source terms like in and

into the basic equations A source term gx t describing the increase of

the gas mass by combustion may be gx t D where D describes the gas

production g cm

s

due to combustion Let ux t be the ow veloc

ity of the generated gas then the source term fx t in may read

fx t Dux t ux t describing a jolting acceleration of the new gas

masses The energy theorem may connect the area in front and behind the

combustion front

u

i

u

i D

where i c

p

T designates the enthalpy dened by U p U is the thermody

namic internal energy c

V

T Since we do not intend to start an exposition

of thermodynamics we stop the presentation at this point We just want to

show that the characteristics method discussed in section can be applied

on combustion phenomena too

In the frame of a research contract we had the opportunity to investigate

the intake stroke and the compression stroke of a Diesel engine of type JW

The comparison between the values pt calculated by the charac

teristics method as described in section and the measured values of the

Problems

Calculate the e!ciency of a ram jet Lorin engine Assume that the

heat generated by the fuel combustion is given by Q c

p

T

T Here

G fw is the mass ow ratio f is the cross section of the tube w

ow velocity per unit mass T

T is the temperature increase Energy

conservation results in

c

p

T

w

c

p

T

w

and c

p

T

w

c

p

T

e

w

e

Here the subscripts and e designate the values at innity and at the

exhaust of the tube Assume that compression due to the stagnation

pressure of the ying ram jet and the consecutive expansion are free of

losses so that entropy is conserved Then TT

T

T

c

and

QG c

p

T

e

T

w

e

w

c

p

T T

e

T

T

w

e

w

T

T

The useful power is dened by the thrust power Gw

w

e

w

Fw

where F is the propulsion force Thus the solution for the

e!ciency is

Fw

Q

T

T

w

w

e

w

Whereas a ram jet produces thrust only at ight a jet engine produces

thrust even at rest This is due to the supercharger of the engine

The turbine generates the thrust If however one assumes that the

compressor power equals the turbine power then are

again valid But now the temperature T before the combustion does

depend on the compressor power and the engine works even for w

Which engine has the higher e!ciency"

Calculate the maximum exhaust speed v

max

of a rocket Assume that

the whole enthalpy of the exhaust gases is transformed into kinetic

energy

Hints assume adiabatic behavior pp

TT

for

the change of state during the exhaust c

p

c

V

is the ratio of the

specic heats Use the Bernoulli equation in the form

v

v

p

Z

p

dp

v

p

p

p

v

max

p

p

This is the Saint VenantWantzel formula Calculate v

max

for air

T

K Result ms

A rocket motor will reach

this exhaust speed in space p