## 135 Reputation

13 years, 252 days

## for 3d and its contour...

Dear Carl

I try to plot

f:=epsilon=(lambda1+sqrt(-lambda2^2+Y*(lambda3^2+(lambda4-lambda5*Y)^2)))^2:

as 3d

implicitplot3d(f, Y = 0 .. 0.4,epsilon=0..20,Delta=-25*Pi..25*Pi,labels=[Y,E1,Delta],tickmarks=[3,3,3]);
can u plz check if it is correct

and the contour

with(Student[Calculus1]):contourplot(f,Delta =-25*Pi .. 25*Pi,Y=0..1,contours=[10],axes=boxed,thickness=2,color=black,font=[1,1,18],tickmarks=[5, 5],linestyle=1,view=[0..1,0..0.6]);

## plot Y agianst (epsilon^2/(4Pi^2))...

Thanks Carl

Deos it make a difference if I plot Y agianst (epsilon^2/(4Pi^2))

## @Christian Wolinski  many thankx f...

why you are calculating the roots..

## thanks...

Im really grateful...

I excpect optical bistability shape...

but many thanks

## thanks but it takes time to run...

thanks for ur code

------------------------- Defining the nature of the variables used ----------------------

epsilon:=epsilon1/(2*Pi):gamma1:=8*Pi:gamma2:=0.002*Pi:x1:=100*Pi:omega2:=200*Pi:delta1:=0.2*Pi:Delta:=25*Pi:G:=20*Pi:Omega:=0:

lambda1:=G*Omega*gamma1/(2*(0.25*gamma1^2+Delta^2)):lambda2:=G*Omega*Delta/(0.25*gamma1^2+Delta^2):lambda3:=gamma1+lambda1:lambda4:=delta1-G^2*Delta/(0.25*gamma1^2+Delta^2):lambda5:=2*x1^2*omega2/(omega2^2+gamma2^2):c:=a^2+b^2:
f:=epsilon-(lambda1-I*lambda2)=(a+I*b)(lambda3+I*(lambda4+lambda5*(c))):
f:=convert(epsilon-(lambda1-I*lambda2)=(a+I*b)*(lambda3+I*(lambda4+lambda5*(a^2+b^2))),rational):
f1:=(lhs-rhs)(evalc(f)):

B:=coeff(f1,I):

A:=expand(f1-B*I):

F:=t->fsolve({eval(A,epsilon=t),B}):

plot('eval(a^2+b^2,F(t))', t=0..5, labels=[epsilon,a^2+b^2], view=[0..5,0..0.5], labelfont=[times,16],color=black,axes=boxed,font=[1,1,18],tickmarks=[4,2],thickness=2);
Warning,  computation interrupted

## @acer  thanks.. but how can i know...

thanks.. but how can i know the  approximation function lambda2=ao+a1lambda1+a2lambda1^2

## @acer  thanks for ur answer,, wha...

what i mean is for example we have y=x^2 the approximation function is : y=ao+a1x+a2x^2+a3x^3

all we need 4 pair (x,y) which are (0,0),(1,1),(2,4),(3,9)

and then subsitue the points in the approximation function to find ao,a1,a2,a3

in the plot below all I need 4 point from the figure and substitue in y=ao+a1x+a2x^2+a3x^3

contourplot(4*lambda2*result^2/(Pi*(lambda2+1)^2)-lambda1,lambda1=0..1,lambda2=0..1,contours=[0],axes=boxed,title=tit,titlefont=[SYMBOL,16],thickness=1,color=black,font=[1,1,18],tickmarks=[2, 4],linestyle=1,view=[0.002..1,0.002..1]);

how did u got it

mamy thanks

## result that used in the plot...

k1:=sqrt(1-lambda1^2):
result:=int(sqrt(1-k1^2*sin(x)*sin(x)),x=0..Pi/2):

## @Rouben Rostamian   Many thanks th...

Many thanks that is useful

## phase portraits...

In the Deplot3D it must have initial conditions, but the example I have is plot of the phase portraits for the system where the inc are unkown

## still the results not symbolic...

/          (1/2)\         (1/2)
|/    2   \     | /  1   \      / 2    \
2 EllipticE||- ------|     | |------|      \a  - 1/
||   2    |     | | 2    |
\\  a  - 1/     / \a  - 1/
/          (1/2)\         (1/2)
|/    2   \     | / 2    \
2 EllipticE||- ------|     | \a  - 1/
||   2    |     |
\\  a  - 1/     /
/                 (1/2)\         (1/2)
| (1/2) /    1   \     | / 2    \
2 EllipticE|2      |- ------|     | \a  - 1/
|       |   2    |     |
\       \  a  - 1/     /

## thankx...

that is easy many thanks

I was thinking of (wich is work)

eq1:=u=(3*x-1)*(1-x):
eq2:=v=2*x*(1-2*x):
U:=[solve(eval(eq1),x)]:
vua:=eval(solve(eq2,v));
V:=eval~(vua,x=~U):
P3:=plot(V,u=-1..1,view=[-1..1,-2..2],color=red,labels=[u,v],axes=boxed,numpoints=9000,linestyle=1,font=[1,1,18],tickmarks=[3,4],thickness=2);

## roots will be different....

thanks .. true the roots (for these fixed parameters) are between 0..1 & 8..10.

and in general (changing the parameters) the roots will be different.

many thanks indeed.

## @tomleslie  1- the plots of the +ve...

1- the plots of the +ve roots aganst arbitrary so(4..5) is done by carl.

2 - want to subsitute these positive roots in the expressions xs,zs also for arbitrary so for example:

if so=4.3 then we sebstitue it in Q --> we get two +ve roots and  then subsitute them in xs hence, zs ..etc then plot xs against so, plot zs against so.

many thanks for ur patients.

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