Perturbation in PDE

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# Perturbation in PDE

Hey,

I really need help. Can anyone explain me the way to solve this:

I think this is ment to be solved using Lindstedt method.

Consider the PDE u_tt-u_xx+k^2*u+eps*u^3=0.

a) We assume k>0, and look for a solution u(t,x)=U(w(eps)t,x), where U=U0+eps*U1+eps^2*U2+O(eps^2), and w(eps)=1+eps*w1+eps^2*w2+O(eps^3).

We assume u(0,x)=A*cos(x)+eps*B*cos(3x)+O(eps^2), and u_t(0,x)=0. A~=0, and U_j is bounded for all j.

We look for a positive value k0 of k, for which not all resonant terms can be eliminated from the equation for U2. Show that for k~=k0 all resonant terms have been eliminated from the equations for U1 and U2.

b) Again assume that k>0. Let u(t,x,eps) be a solution of the PDE that is periodic in x with period 2*pi, periodic in t with minimal period P(eps) and is even in x and in t. Assume that as eps -> 0 the solution u(t,x,eps) converges to a nonzero function u0(t,x) and its period P(eps) converges to a nonzero P0. Show that there is a countable set E of positive real numbers such that if k is not in E then there exist a nonzero A and an integer n such that u0(t,x)=A*cos(sqrt(n^2+k^2)*t)*cos(n*x).

c) Is k0 from (a) is in E from (b)?

This would mean the world if anyone would help me!

You can also contact me by mail:

w.jhon1984@gmail.com

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## I’m think k0 come from the

I’m think k0 come from the cutoff frequency designing , the media . Hence, k0 from (b) process !.

## Happy

Happy to see Pahio is again active on this site!

## Happy

Happy to see Pahio is again active on this site!

## Happy

Happy to see Pahio is again active on this site!

## Happy

Happy to see Pahio is again active on this site!

## An open invitation

Join maths corner on facebook. Procedure: join fb and I can add you as member - your contributions are welcome.

## Fermat's theorem

Fermat’s theorem works even if the base is a Gausssian integer subject to a) the prime under consideration is of shape 4m+1 and b) the exponent and base are co-prime.

((2+3*I)^16-1)/17

## Fermat's theorem

Fermat’s theorem works even if the base is a Gausssian integer subject to a) the prime under consideration is of shape 4m+1 and b) the exponent and base are co-prime.

((2+3*I)^16-1)/17

## Fermat's theorem

Fermat’s theorem works even if the base is a Gausssian integer subject to a) the prime under consideration is of shape 4m+1 and b) the exponent and base are co-prime.

((2+3*I)^16-1)/17 = -47977440 - 803040*I

## Euler's generalisation of Fermat's theorem .......

This works even when the base is a Gaussian integer:

Reading GPRC: gprc.txt …Done.

GP/PARI CALCULATOR Version 2.6.1 (alpha) i686 running mingw (ix86/GMP-5.0.1 kernel) 32-bit version compiled: Sep 20 2013, gcc version 4.6.3 (GCC) (readline v6.2 enabled, extended help enabled)

Copyright (C) 2000-2013 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER.

Type ? for help, \qto quit. Type ?12 for how to get moral (and possibly technical) support.

parisize = 4000000, primelimit = 500000 (17:50) gp ¿ ((14+15*I)^104-1)/105 (17:51) gp ¿

## Euler's generalisation of Fermat's theorem .......

This works even when the base is a Gaussian integer:

Reading GPRC: gprc.txt …Done.

GP/PARI CALCULATOR Version 2.6.1 (alpha) i686 running mingw (ix86/GMP-5.0.1 kernel) 32-bit version compiled: Sep 20 2013, gcc version 4.6.3 (GCC) (readline v6.2 enabled, extended help enabled)

Copyright (C) 2000-2013 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER.

Type ? for help, \qto quit. Type ?12 for how to get moral (and possibly technical) support.

parisize = 4000000, primelimit = 500000 (17:50) gp ¿ ((14+15*I)^104-1)/105 = -249662525598174865517621222098021785366399633335910441957688800663877876192221716937263714468906280908614454012799368615180549371243472 - 118511838209654103558982122027130965758920275429164915998560474682902951765213030198935065103035392002339412087987613469408163154998032*I (17:51) gp ¿

## Fermat's theorem works when the base is a Gaussian integer

What puzzles me is that the theorem works when the base is a prime in the ring of Gaussian integers and the exponent is a prime of shape 4m + 1 but does not work when the exponent is a prime of shape 4m+3.Can any one throw some light on this?

## Fermat theorem works

Hi Deva, perhaps the entry ’theorem on sums of two squares by Fermat’ may explain it or help this problem,

Jussi

## Fermat theorem works

Hi Jussi. Thanks will try.

## Fermat theorem works

Hi Jussi. Thanks will try.

## conjecture pertaining to Gaussian integers

Let the base be a Gaussian integer = a + ib. Let p be a prime number of shape 4m + 3. Then ((a + ib)^(p^2-1) - 1) == 0 (mod p). This is subject to the base, a + ib and p being co-prime,

## conjecture pertaining to Gaussian integers

Let the base be a Gaussian integer = a + ib. Let p be a prime number of shape 4m + 3. Then ((a + ib)^(p^2-1) - 1) == 0 (mod p). This is subject to the base, a + ib and p being co-prime,

## conjecture pertaining to Gaussian integers

Let the base be a Gaussian integer = a + ib. Let p be a prime number of shape 4m + 3. Then ((a + ib)^(p^2-1) - 1) == 0 (mod p). This is subject to the base, a + ib and p being co-prime,

## conjecture pertaining to Gaussian integers

## conjecture pertaining to Gaussian integers

A couple of examples given below:Reading GPRC: gprc.txt …Done.

GP/PARI CALCULATOR Version 2.6.1 (alpha) i686 running mingw (ix86/GMP-5.0.1 kernel) 32-bit version compiled: Sep 20 2013, gcc version 4.6.3 (GCC) (readline v6.2 enabled, extended help enabled)

Copyright (C) 2000-2013 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER.

Type ? for help, \qto quit. Type ?12 for how to get moral (and possibly technical) support.

parisize = 4000000, primelimit = 500000 (10:53) gp ¿ ((2+I)^8-1)/3 (10:54) gp ¿ ((2+I)^48-1)/7 (10:55) gp ¿ ((2+I)^120-1)/11 (10:55) gp ¿

## conjecture pertaining to Gaussian integers

Happy to report that ”Nick”, on mersenneforum.org, has stated that my conjecture can be taken as proved.

## conjecture pertaining to Gaussian integers

Happy to report that ”Nick”, on mersenneforum.org, has stated that my conjecture can be taken as proved.

## Modified Fermat's theorem

Modified Fermat’s theorem: Let a belong to the ring of Gaussian integers Then a^(p^2-1)= = 1 (mod p). Here p is a prime number with shape 4m+1 or 4m+3.

## modidied Fermat's theorem

Nice theorem! How do you prove it?

## Modified Fermat's theorem

Before replying to Pahio’s call for proof would like to add that I forgot to add the condition: a and p should be co-prime.

## Modified Fermat's theorem

Pahio, happy to say ”Nick” of Mersenneforum.org has given a simple proof.

## Modified Fermat's theorem

Pahio, happy to say ”Nick” of Mersenneforum.org has given a simple proof.

## Modified Fermat's theorem

Pahio, happy to say ”Nick” of Mersenneforum.org has given a simple proof.

## Modified Fermat's theorem

Pahio, happy to say ”Nick” of Mersenneforum.org has given a simple proof.

## Modified Fermat's theorem

Pahio, happy to say ”Nick” of Mersenneforum.org has given a simple proof.

## Modified Fermat's theorem

Pahio, happy to say ”Nick” of Mersenneforum.org has given a simple proof.

## modified Fermat theorem

I don’t see such a proof. Can you please write it in PlanetMath?