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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