alejandrito20

the problem says:
"show that the zero solution is nonlinear stable. For this, find the change of variable that transforms this system in a linear system"

[math] \frac{dx}{dt}=-x + \beta (x^2+ y^2) [/math]

[math] \frac{dy}{dt}=-2y + \gamma x y [/math]

i tried with the method of eigenvalues of Jacobian matrix, and both eigenvalues are negatives, but my teacher says that this method is incorrect. Help please

The integral of the derivative of a periodic function over a period will always be zero. But as Timo observes, you have imposed additional conditions that make the function non-differentiable, in contrtadiction to your problem statement.

 

You have something messed up in the problem statement.

in the text (http://arxiv.org/PS_cache/hep-th/pdf/0011/0011225v2.pdf eq2.13) says:

"in a compact internal space without boundary, the integral vanishes"

the inicial space is [math] z \in -\infty, \infty[/math], but [math] z = r \phi [/math] , whit r radio konstant

I think you meant that A' is discontinuous in -pi and pi, not A. Why not simply integrate from 1 to (1+2pi), and not have your problem?

 

yes , is discontinuous in A'

wy not simply integrate from 1 to (1+2pi), and not have your problem?

 

your says [math]\int_1^{1+2\pi} dy (A' exp(A))' [/math]?????

 

EDIT: Oh, I overlooked something: If A' is discontinuous: How can (A' exp(A))' exist?

(A' exp(A))' there is Not exist in -pi,0,pi

i need to evaluate the follows integral:

[math]\oint dy (A' e^{A})' [/math]

where ' is derivation respect [math]y[/math] and [math]A(y)=A(y+2\pi)[/math] and too [math]A(y)'=A(y+2\pi)'[/math]

where [math]y \in (-\pi,\pi)[/math] ia a angular coordinate , and [math]A' [/math] is discontinuous in [math]-\pi,0,\pi[/math]

the result is zero, but i think that is NOT CORRECT to say:

[math]\oint dy (A' e^{A})' = A' (\pi)e^{A(\pi)}- A' (-\pi)e^{A(-\pi)} =0 [/math], since [math]A' [/math] is discontinuous in -pi and pi

In a space time [math]5D[/math], the action for the brane [math]4D[/math] is:

[math] \int dx^4 \sqrt{-h}[/math]

In the Randall Sundrum the action for the hidden brane is:

[math] V_0\int dx^4 \sqrt{-h}[/math], where [math]V_0[/math] is the tension on the brane hidden.

follow the stress energy tensor

[math] T_{MN}= V_0 h_{uv} \delta^u_M \delta^v_N \delta(\phi)[/math], where [math]\phi[/math] is the extra dimention.

In other paper, where [math]T_{MN}[/math], for example in the friedman equation in http://arxiv.org/abs/hep-th/0303095v1 (page 6)...

[math] T_{00}= -\rho \delta(\phi)[/math]

[math] T_{ii}= p \delta(\phi)[/math]

the other component are zero.

I understand thar [math]\rho , p[/math] are energy density and presion

If , i use other embedding my energy stress tensor is

[math] T_{00}= - \delta(\phi)[/math]

[math] T_{ii}= \delta(\phi)[/math]

[math] T_{0 \phi}= \delta(\phi)[/math]

[math] T_{\phi \phi}= \delta(\phi)[/math]

¿can i to multiply the each component of the stress tensor by differents constants???...for example:

[math] T_{00}= - k_1 \delta(\phi)[/math]

[math] T_{ii}= k_2 \delta(\phi)[/math]

[math] T_{0 \phi}= k_3 \delta(\phi)[/math]

[math] T_{\phi \phi}= k_4 \delta(\phi)[/math]

In the serie [math] \sum_0^{\infty} a_n (x - c)^n [/math], the radius of convergency is:

[math]R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|[/math].

My problem is : Find the radius of convergency when:

[math] \sum_0^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot x^{2n+1} [/math]

i don't understand mainly who is [math]a_n[/math].

The answer is [math]R= \infty[/math]

thanks :=)

groups, singer, or instrumen, in particular, of music listen to you, when you doing calculus of theoretical physics?

sorry, but I dont speack english very good.

Hello

The problem is

find the value of [math]\lambda[/math] for [math]\lim_{n \to \infty} \frac{a_{n+1}}{a_n} \nonumber\ [/math] < 1, where

[math] a_n= \frac{(\lambda^nn!)^2}{(2n+1)!} \nonumber\ [/math] con [math]\lambda >0 [/math]

I tried to do:

[math]\frac{a_{n+1}}{a_n}=\frac{(\lambda^{n+1}(n+1)!)^2}{(2n+3)!}\cdot \frac{(2n+1)!}{(\lambda^nn!)^2}[/math]

[math]=( \frac{\lambda^{n+1}}{\lambda^n}\frac{(n+1)!}{n!})^2\frac{(2n+1)!}{(2n+3)!} [/math]

[math]\frac{(2n+1)!}{(2n+3)!}= \frac{3 \cdot 5 \cdot 7........(2n+1)}{5 \cdot 7 ........(2n+1) \cdot (2n+3)}[/math]

[math]= ( \lambda (n+1))^2\frac{3}{(2n+3)} [/math]

but the Answer is [math]\lambda \in {0,2}[/math]

in a pipe , open and closed at the ends, the frecuency is

[math]f=(2n-1)\frac{v}{4L}[/math],

if [math]L[/math] decreases, then [math]f[/math] increasses.

I don't understand why in a text of physics of music (in spanish) says:

"in a pipe, closed in the vocal fold, and open in lips....in the lips there is antinodes of velocity...

if the lips widen, then frecuency increasses"

someone tell me the mathematics asociated to this afirmation?

i need examplees of spacetimes where the ricci scalar is constant but nonzero . Particulary i search examples of line element.

there is correct the expresion [math]\int^{-\pi+\epsilon}_{\pi-\epsilon} d\theta[/math]....where [math]\theta[/math] is a angular coordinate between [math](-\pi,\pi)[/math]....¿what means this?...

i believe that this mean that the angular coordinate theta runs from [math]\pi-\epsilon[/math] to

[math]-\pi+\epsilon[/math] in the sense anti clock (figure)

If the brane tension in RS is [math]T=24M^3_5 \sqrt{\frac{-\Lambda}{24M^3_5}}[/math] eq1

¿whats units have the tension of brane???

and , ¿whats units have the cosmological constant and planck scale [math]M_5[/math]??

I have tried with units of [math]\Lambda =\frac{1}{L^2}[/math] (L lenght), but if E have units of (E=energy), then en eq1 units of T is [math]\sqrt {\frac{E^3}{L^2}}[/math]

In the einstein equation

[math]R_{uv}-0,5 R g_{uv}+ \Lambda g_{uv} = \frac{8\pi G}{c^4}T_{uv}[/tex][/math]

i understand that units of [math]g_{uv}=L^2[/math] and then [math]R=\Lambda=\frac{1}{L^2}[/math]

¿[math]R_{uv}[/math] is dimensional less??

[math]G=\frac{L^3}{T^2 M}[/math] and [math]\frac{G}{c^4}=\frac{T^2}{M}[/math] then

¿¿¿[math]T_{uv}=\frac{M}{T^2}[/math]?????

Assuming something like [math]t =x^{0} [/math] the metric you present looks conformally equivalent to the Minkowski metric. (Not sure if the [math]f(t)= 0[/math] or the [math]f(t)= \infty[/math] course trouble, so maybe mod that and the statement that the function is positive).

yes [math]t =x^{0} [/math] , [math]f(t=0)=\infty[/math],[math]f(t=\infty)=0[/math], f(t) is positive.

the metric in t= 0 is [math]\infty \eta_{uv}dx^u dx^v[/math], in [math]t=\infty[/math] is [math]0\eta_{uv}dx^u dx^v[/math].....

physically....¿what would mean this???

hello

i understand that in a flat space the metric is [math]\eta_{uv}dx^udx^v[/math]...i know that this means that the light follows straight geodesic in this space time...

but ¿what would means that metric is [math]f(t)\eta_{uv}dx^udx^v[/math] where f(t)=infinite in t=0 and f(t)=0 in t=infinite.....obvious i understand the matematics, but physically ¿what means?.....for example..¿what means that in bing bang in t=0 f(t)= infinite????

In which case "gradient field" is the term you want to look up. The vanishing of the integral over any closed path is a property of a gradient field.

but, i read that :

If A = grad F,The line integral over a gradient field is equal to the difference between the values of the potential function at the end-points ...the integral round a closed curve has the value zero...

i think that [math]\oint \nabla W \cdot dr=0[/math]

but [math]\oint \nabla \cdot (W^\zeta \nabla W) dr[/math], with z a number, ¿is zero too???????

Do you know the equation of the path?

No

If so, are you sure that the bounded integral is zero

YES

Merged post follows:

Knowing what path would certainly be helpful as well.

in the article ( arxiv.org/abs/0907.1321v1) says: "Note also that the left hand side of the equation (10) vanishes upon a line integration over a closed path along the compact internal space."

i need to evaluate [math](A'e^{nA})'[/math],where [math]A=kte |\phi|[/math],with [math]\phi[/math] a angular coordinate between [math](-\pi,\pi)[/math] upon a line integration over a closed path along the coordinate angular [math]\phi[/math].......¿why the result is zero?

....' is the derivate on [math]\phi[/math]

It will get to the result, alejandrito. Where is the difficulty? Is it the right hand side?

to multiplicate by [math]W^{-2} g^{uv}[/math]

[math]R-W^{-2}\tilde{R}=-\frac{\nabla^2 (W^{p+1})}{W^{p+1}}[/math]

[math]\nabla^2(W^{p+1})=P(P+1)W^{P-1}\nabla W \nabla W+(P+1) W^P \nabla ^2 W[/math]

Ok......Sorry

I tried very time.......gap in my mind..........i'm crazy.........

ansatz is [math]g_{mn}dy^mdy^n+W^2(y)g_{uv}(x)dx^udx^v[/math]

if [math]R_{uv}=\tilde{R}_{uv}-\frac{g_{uv}}{(p+1)W^{p-1}}\nabla^2W^{p+1}[/math] eq1

why eq 1 is???

[math]\frac{1}{p+1}(\tilde{R}W^{-2}-R)=pW^{-2}\nabla W \nabla W+W^{-1}\nabla^2W[/math]

i tried to multiplicate eq1 by [math]W^{-2}g^{uv}[/math] but i don't go to the result

in a D space time of coordinates [math]x^u,y^m[/math], with u=...p+1, m=...D-p-1, and A=.....D

if [math]R_{MN}=8 \pi G_D (T_{MN}-\frac{G_{MN}T^P_P}{D-2})[/math] (eq1)

I don't understand why

[math]R^u_u=\frac{8\pi G}{D-2}((D-p-3)T^u_u-(p+1)T^m_m)[/math] eq 2

[math]R^m_m=\frac{8\pi G}{D-2}((p-1)T^m_m-(D-p-1)T^u_u[/math] eq 3

i tried to multiplicate eq1 by [math]g^{MN}[/math] and then

[math]R=8\pi G_D(T-\frac{T}{D-2})[/math]

but, i don't go to the eqs 2 and 3

For a MSc degree here in the UK, you do not need to find anything new. Is that different in Chile?

My teacher says that my thesis have to be new.....

in chile

undergraduate physics : 4 years...the last courses are : MQ1, electrodinamics, Clasical Mechanics1 (MArion), Termodinamic......

MSc : MQ2 and Statistical Mechanics....and two electives (General Relativity and Clasical MEchanics2 (Goldstein) + thesis 2 semester.

PhD : 4 electives + thesis 6 semester

¿in england the MSc have a thesis or MSc is only asignatures????????

Is this for a PhD?

not , is for a thesis of master of physics in chile, sudamerica. The level of PhD is most hight that the level of master.

and...¿there is a suggest or idea for my thesis?