One WayWave EquationSee Logan's Book

This is a first order linspace and finePDE Wefocus on its solute

Let c constant

PDE U.ttChat u U O t o

P x en possiblyI c UH D fix xer

Thesimplest versionCnsider Xe IR and c a given constant

We can arfirm thatuh 4 flex etis a solutanto CP since

Ut of'fx et

Ux f x et

ex fix BoosKx A is instant

n x t A as kex ct

note Thet ke e w

we call w the frequencyh the wavenumberc the phasespeed

Note also that is 2

X is thewavelength

Describes a wave ofconstantshapethat propagates with constant

speed c

If c 0 wave propagates to therightCeo wave propagates tothe left

A useful fact

dugyttigulxctht.DE 3

by PDE c s

dua

Ux e t Ut s 0

by PDE

Consider

Ut t CH t Ux so xc.IR t oCP

uh 07 46dugUx Ct Ut O

dxq.sccx.tlThis is the characteristic equator

which generates a characteristic curve

In space ku Along C

ff suxdxq.tutsuxdx.tk4 0dt

ex utt2tux oxc.IR tooCP Wyo se XEIR

Here chat Et

using du h 2ttUt O

7 2t too

yieldsthe characteristic

Integrating r

Xs Et k ka astat

t a

ey.catgivenby x Et3O X

3,0

since uis instant along curve the

ulxit expl57 expfcx.ttfor tso and XEIR

wavespeedsup atrate 2T butretainsits

shape

Nonlinear Waves

Utt Clu x t Ux 0 x.cl2 tooCP

Wyo GCD xeR

Here c depends once itself nonlinear

Thwhet follows assume that car x t O

dug

0 again but

dxq.sc u x t f for E

i e dug3 dux thx t t3 O

u is constant eloy characteristics and

these are since dxldt c.hr tethatd2x

diff sda.clucxittx.tl f so

so along characteristics the acceleration

depends on the gradient of teespeed c

w r t u itself

To find E through Lx t we w te ther

dxg

C u 13,0 X t

Ix clueDixit c 4G x tdt

Applying A at G o after

integrating we obtain

x c lots tts twhichdefies 3 3Exit implicitly

So uh t 4Gwhere 3 is givenby thesehelmto f

InviscidBurger's EgnatiaLx Ut the Ux O XER t oCP

loan uh D fIx YeEunX I

z th

GHx

I

since c cu u th characteristics are

straighthies emanating frm 13,0 with

speed dollsD G

Since c 0 the wave travelstothe

right but the 33 co travel atspeed2

and 3 I travel atrate 1

t

Z

9 goi

f I l r X3 Z I O l 2

As weshall see the characteristicswill

not cross in this example for 0 s t s l

We will saythat to I isthe shocktire in our example Seebelow

Cupt

wave

itsteepens

O X

At tst wave becomesmulti rakedThis is called a shock

The solution of the CPiswlugeratradihualfunction aftertheshock we call it a weak

solution

To findthe solution

for tsl hat 2 for xe2tUH 4 1 for X ttl

for2tsxcttlx.sc

4csDtt3or 5 12 3 ft's

we can solvefer 3 X Ztl t

uh t ofG yields

hunt 2taxettl til

8 the general solution to CP

Nyt 2 for Xc 2TUH 4 1 for X ttl

what 3 ztexettl tal

Howdo we find the shock tire

Forth case with Ctu o

Ut t Clu Ux O t o

l ult ol Cx xc.IR

The shockforblowup hhe to isfound as follows

if c'Cu 0 with a 30 04 20for sufficiently longtinct thesolutionbearesmultivalued i e Ux is unbounded

To find Ux differentiate

x doll tis wrt x

I CHAD t t 3

Solving for3 1

It 0447443 t

teen for us 463

Ux 0431

It c'lol t

so whenthe denominator 0 we

get

to MY f HGDwhichwillbemeaningful so long asto Of the initial tire

In the example club u 4151 2 3I U

443 c'fol s l D l

to I tle shockfire

Thus for xc.IR the CP What happenswhen we have a finitedomain or semi infinite

donor a

i if c O

Ulo t gli i

xHt 1 x0 UHd 4G L

ea f if es 0

uh H htt

i n0 UHd 4G L

Forced one way wave equation

alx.tn uxtblx tulUt ftx t u

S'pse a b f e C D

D is a space the domain

WLOG ansider

acx.t.uluxtut fhat.nl xt fCP Wx Dsg

As before

dig ah t n

If a text s fait a

Rule before 1 0

Set x 3 at t.co

le les D g13

and is a systemof differentialequationswith a solutionthatdepends on 2arbitraryconstants

Along characteristics

Xs Flt c CaUs Gct G ca

and the instants can beevaluated

by construing solenhu to obey I C

C E C 3 9 413

then x Fl t c Gl Gtsk

w Glt c 131,4137

and 3 3Cx t via Soluhu of

Then issubstituted into A

obtaining free solution

ex Let's try thismethod

on the following example

This is a nonlinear wave with a

reaction term proportionedto u

Utt telex the O xc.IR t O

UH o s I XER

Doyle dxFl a su

u or bus dt

SolvingUs Ge

t Since u D Hz46,07 c and 413,07

32 So att O

413,07 c 32

Solving xst s 3ze ti.cz

u

Now we find 02

X 6 3 Itcz 3 9 332

X et33g 3 e t

Next solve for 3 3 x

3 243 e tSince us 3ge t

uh t ej j Blowupto

occurs when3eto I O or to bog Ks

exercise trytheabovemethod to solve

2xaux 2tUUt s U x tuh olsgG I