Partial Differential of complex number
Ayo
2iwD1A + 3A^2(A*)=0
Where D1 represent partial differential, for instance of A, A* is the complex conjugate of A
and A = (1/2)a exp(iG) with real a and G.
Seperation into real and imaginary parts is required to obtain
D1a = 0 and wD1G + 3/8a^2 = 0
Musatov
Look at the results below very carefully:
http://www.meami.org/?cx=000961116824240632825%3A5n3yth9xwbo&cof=FORID%3A9%3B+NB%3A1&ie=UTF-8&q=2iwD1A+%2B+3A^2%28A*%29%3D0+proof&sa=Search#305
#
Math Forum Discussions
Jul 6, 2009 ... I would be glad to receive help on the proof of the
solution of this equation; 2iwD1A + 3A^2(A*)=0. Where D1 represent
partial differential, ...
http://mathforum.org/kb/thread.jspa?threadID=1963361&tstart=0 - 12
hours ago
#
Date: Jul 6, 2009 3:45 PM Author: Ayo Subject: Partial ...
Jul 6, 2009 ... be glad to receive help on the proof of the solution
of this equation; 2iwD1A + 3A^2(A*)=0 ... and A = (1/2)a exp(iG) with
real a and G. ...
http://mathforum.org/kb/plaintext.jspa?messageID=6775877 - 12 hours
ago
Simultaneity?
Musatov
> On Jul 6, 12:45 pm, Ayo <ayode...@yahoo.com> wrote:
>
> > I would be glad to receive help on the proof of the solution of this equation;
>
> > 2iwD1A + 3A^2(A*)=0
>
> > Where D1 represent partial differential, for instance of A, A* is the complex conjugate of A
> > and A = (1/2)a exp(iG) with real a and G.
> > Seperation into real and imaginary parts is required to obtain
>
> > D1a = 0 and wD1G + 3/8a^2 = 0
>
> Look at the results below very carefully:
>
>
> #
> Math Forum Discussions
> Jul 6, 2009 ... I would be glad to receive help on the proof of the
> solution of this equation; 2iwD1A + 3A^2(A*)=0. Where D1 represent
> hours ago
> #
> Date: Jul 6, 2009 3:45 PM Author: Ayo Subject: Partial ...
> Jul 6, 2009 ... be glad to receive help on the proof of the solution
> of this equation; 2iwD1A + 3A^2(A*)=0 ... and A = (1/2)a exp(iG) with
> real a and G. ...http://mathforum.org/kb/plaintext.jspa?messageID=6775877- 12 hours
> ago
>
> Simultaneity?
Let me show you how it works:
2iwD1A + 3A^2(A*)=0. Where D1 represent partial differential, for
instance of A, A* is the complex conjugate of A.
2iwD1A + 3A^2(A*)=0 Where D1 represent partial differential, ...
#
Parametrically Excited Systems
5.1.3. A DOUBLE PENDULUM. The two examples described above are systems
having a single ..... 3a a A3 a A3. 4w0. 0oao woao. We seek a solution
for. (5.1.33) ...
doi.wiley.com/10.1002/9783527617586.ch5
#
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Do you see what can be done with this system?
Ken Pledger
<5228972.81236.12469095...@nitrogen.mathforum.org>,
Ayo <ayod...@yahoo.com> wrote:
I can see that you've made an effort to state the problem
carefully, but a few things are still not clear. (A small point is that
your subject should be "Partial _derivative_ of a complex _function_".)
You haven't mentioned the independent variable(s). I'll assume
that a, G and possibly w are functions of several _real_ variables,
and your D1 represents partial differentiation with respect to the
first of those variables. Also your 3/8a^2 seems to mean (3/8)a^2
rather than 3/(8a^2).
Since A = (1/2)a.exp(iG), the product and chain rules give
D1(A) = (1/2)D1(a).exp(iG) + (1/2)ai.exp(iG).D1(G).
Also (A^2)(A*) = (1/4)(a^2).exp(2iG).(1/2)a.exp(-iG)
= (1/8)(a^3).exp(iG).
Hence your d.e. becomes
iw.D1(a).exp(iG) + a(i^2)w.exp(iG).D1(G) + (3/8)(a^3).exp(iG) = 0.
Cancelling the non-zero factor exp(iG) then gives
iw.D1(a) - aw.D1(G) + (3/8)(a^3) = 0.
Now you should be able to separate the real and imaginary parts, and
(assuming neither w nor a is zero) get almost the solution you
mentioned. But you'll find there's a sign error somewhere.
Ken Pledger.
MeAmI.org
> In article
> <5228972.81236.1246909546522.JavaMail.jaka...@nitrogen.mathforum.org>,
What is less real or imaginay in the below set?
Full text of "FISHERIES REVIEW VOL.35 NO.3 1990"B Appl , 43 rue
Cuvier, 75231 Paris Cedex 05, France] FR 35{3) 8 1990 ...... A, D and
S. P. Musatov MODIFICATION OF SOCIAL BEHAVIOR IN PROCHiLODUS
NICRtCANS. ...... Gdym, Poland] 90-004210 Kubasch, A and A. W. Rourke
ARTERIOSCLEROSIS IN ...... Anthias taematus D1 strlbuti on/Geographi
cal 90-004045 Meristfc Studies ...
www.archive.org/stream/.../fisheriesreviewv019681mbp_djvu.txt
The Interaction of the Solar Wind With VenusDepletion of the plasma by
the loss of energetic (1.5-3.0 keV) ions is ...... Gringauz, K. I.,
Bezrukikh, V. V., Musatov, L. S., and Breus, T. K. 1968 ...... Icarus
3 8:192-211. Schubert, G., Covey, C., Del Genio, A., Elson, .....
Strickland, D. 1. 1973. The 01 1304 and 1356-A emissions from the
atmosphere of Venus ...
dawn.ucla.edu/personnel/russell/papers/interact_solwind/
Recent Observations of the Moon by SpacecraftThe effective dielectric
constant was found to be 3.0 ± 0.2 at depths from 25 cm to many
meters. ...... Sasser, J. H. and Patteson, A. W.: 1967, Potential
Lunar Landing Areas .... Cherkasov, I. I., Gromov, V. V., Zobachev, N.
M., Musatov, A. A., ...... D1-82-0475, Boeing Scientific Research
Labs., Seattle, Wash. ...
adsabs.harvard.edu/full/1969SSRv....9..491J
by LD Jaffe - 1969 - Cited by 2 - Related articles
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voltage V (Figure D.1). ...
www.scribd.com/doc/13100665/Appliedmaterialsscienceebook-EEn
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P., and Musatov, L. S.: 1966, Kosm. ...
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by JV Kovalevsky - 1971 - Cited by 3 - Related articles
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(I.W. H, It I trtn nllii iihilKftxIiiiiti klkNNlknv niMr)i! ...... 145
p, MUSATOV, A. L [Sloop patlis; a tale] Krulye tropy; povest
1 . ...... Klein, D. 1. - Results of and prospects for work an the
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a< W 3 26 W'+6 XW WI'>S' (5.19) (5.20) 5 4 5 rf) ...
link.aps.org/doi/10.1103/RevModPhys.54.437
Ayo
Ayo
Given:
2iwD1A + 3A^2(A*) = 0 (1)
Where A is a function of T1, T2 and D1 is a first partial derivative of A and A* being a complex conjugate of A.
Also given;
U = A(T1,T2)exp(iwTo) (2)
Find the second order partial derivative of U i.e D1^2(U)?
Here are my steps:
From (1);
D1A = (-3A^2A*)/2iw (3)
Such that;
D1U= D1Aexp(iwTo)
Substituting for the value of D1A eq (3)
D1U= (-3A^2A*)/(2iw) exp(iwTo) (4)
Now D1^2 becomes;
D1^2(U)=(-3.2.A.(D1A).A*/2iw) exp(iwTo)
=(-6AA*.(D1A)/2iw) exp(iwTo)
=(-6AA*/2iw).(-3A^2A*/2iw). exp(iwTo)
=(-18A^3A*^2/4w^2) exp(iwTo) (final ans)
Could you look again into this and comfirm if I am wrong expecially with the execution of D1^2(U)
Ken Pledger
I've been busy starting off a lecture course, but now have a bit
of time to look at this.
In article
<2673054.10025.12471360...@nitrogen.mathforum.org>,
Ayo <ayod...@yahoo.com> wrote:
> ....kindly have a further look at
> this also, this time I showed the steps that i took, I would welcome your
> comment where any mistake is observed.
>
> Given:
> 2iwD1A + 3(A^2)(A*) = 0 (1)
> Where A is a function of T1, T2 and D1 is a first partial derivative of A and
> A* being a complex conjugate of A.
>
> Also given;
> U = A(T1,T2)exp(iwTo) (2)
> Find the second order partial derivative of U i.e D1^2(U)?
> Here are my steps:
> From (1);
> D1A = (-3(A^2)A*)/2iw (3)
assuming w != 0
> ....
> D1U= D1Aexp(iwTo)
This assumes that both w and To are independent of T1. Are
they? If not, you need to use the product and chain rules.
> Substituting for the value of D1A eq (3)
> D1U= (-3(A^2)A*)/(2iw) exp(iwTo) (4)
> Now D1^2 becomes;
> D1^2(U)=(-3.2.A.(D1A).A*/2iw) exp(iwTo)
This ignores the dependence of A* on T1, so you're in trouble
form here on.
Ken Pledger.
Primes?
patlis; a tale] Krulye tropy; povest 1 . ...... Klein, D . 1. -
Results of and prospects for work an the nature