adel-00

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11 years, 63 days

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These are replies submitted by adel-00

@Carl Love I am not sure if I can exapnd it either with Bessel function or somthing simmilar

if i integrate sin(erf(t)) e^(a+ib)t

@acer thanks I got it

@Preben Alsholm 

and with that

restart:Digits:=70:
------------------------- Defining the nature of the variables used ----------------------
N:=0:M:=0:N1:=1+N:w:=10:


ini1:= x(0)=0.5,y(0)=0.5,z(0)=0;
            ini1 := x(0) = 0.5, y(0) = 0.5, z(0) = 0
var:={x(t),y(t),z(t)}: 
dsys:={diff(z(t),t)=-(N1+M*cos(2*w*t))*z(t)-1+f*(x(t)+y(t)), diff(x(t),t)=-(N1-I*w-2*M*exp(-2*I*w*t))*x(t)-f*(N1+(z(t)))-2*f*M*exp(2*I*w*t),diff(y(t),t)=-(N1+I*w-2*M*exp(2*I*w*t))*y(t)-f*(N1+(z(t)))-2*f*M*exp(-2*I*w*t)}:
zd:=subs(dsys,diff(z(t),t));
               zd := -z(t) - 1 + f (x(t) + y(t))
res:=dsolve(dsys union {x(0)=0.5,y(0)=0.5,z(0)=0},numeric,output=listprocedure):
Warning, The use of global variables in numerical ODE problems is deprecated, and will be removed in a future release. Use the 'parameters' argument instead (see ?dsolve,numeric,parameters)
#tit:=sprintf("F=%g,N=%g",f,N):
plot3d(z(t,f), f=-1..1, t=0..3, grid=[29,49]);

@acer 

Many thanks for ur response 

If I want to plot z(t) 

CodeTools:-Usage( plot3d(zfun(t,f), f=-1..1, t=0..3, grid=[29,49]) );

@Preben Alsholm 

N:=0:M:=0:N1:=1+N:w:=10:


ini1:= x(0)=0.5,y(0)=0.5,z(0)=0;
            ini1 := x(0) = 0.5, y(0) = 0.5, z(0) = 0
var:={x(t),y(t),z(t)}: 
dsys:={diff(z(t),t)=-(N1+M*cos(2*w*t))*z(t)-1+f*(x(t)+y(t)), diff(x(t),t)=-(N1-I*w-2*M*exp(-2*I*w*t))*x(t)-f*(N1+(z(t)))-2*f*M*exp(2*I*w*t),diff(y(t),t)=-(N1+I*w-2*M*exp(2*I*w*t))*y(t)-f*(N1+(z(t)))-2*f*M*exp(-2*I*w*t)}:
zd:=subs(dsys,diff(z(t),t));
                 zd := -z(t) - 1 + x(t) + y(t)
res:=dsolve(dsys union {x(0)=0.5,y(0)=0.5,z(0)=0},numeric,output=listprocedure):

 

I mean plot 3d of z(t),t,f 

many thanks

and for plotting 3d g(z(t),t,f)

 

@Carl Love 

Dear Carl

If want to plot

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

epsilon against the sqrt(Y)

is it correct like this

r1:=plot([rhs(f), sqrt(Y), sqrt(Y)= 0..1],axes=boxed,thickness=2,color=black,font=[1,1,20],tickmarks=[3, 3],linestyle=1,labels= [typeset(epsiloni), sqrt(Y)]);

@Carl Love 

restart:with(plots):
Digits:=35:
------------------------- Defining the nature of the variables used ----------------------
assume(Y,real);
Omega:=5*Pi:
gamma1:=8*Pi:
gamma2:=0.002*Pi:
x1:=100*Pi:
omega2:=200*Pi:
gamma0:=0.2*Pi:
#Delta:=25*Pi:
G:=20*Pi:

#tit:=sprintf("G=%g,delta1=%g,gamma1=%g",G,delta1,gamma1):
lambda1:=(1/(2*Pi))^2*(G*Omega*gamma0/(2*(0.25*gamma0^2+Delta^2))):
lambda2:=(1/(2*Pi))^2*(-G*Omega*Delta/((0.25*gamma0^2+Delta^2))):
lambda3:=(1/(2*Pi))^2*gamma1+lambda1:
lambda4:=(1/(2*Pi))^2*(0.2*omega2-G^2*Delta/(0.25*gamma0^2+Delta^2)):
lambda5:=(1/(2*Pi))^2*(2*x1^2*omega2/((omega2^2+gamma2^2))):
epsilon-(lambda1+sqrt(-lambda2^2+Y*(lambda3^2+(lambda4-lambda5*Y)^2)))^2:
#implicitplot(f,epsilon=0..200,Y=0..1,numpoints=1000,axes=boxed,thickness=2,color=black,font=[1,1,20],tickmarks=[3, 3],linestyle=1):
implicitplot3d(f, epsilon=0..20,Y = 0 .. 0.4,Delta=-25*Pi..25*Pi,labels=[E1,Y,Delta],tickmarks=[3,3,3]);
with(Student[Calculus1]):contourplot(f,Delta =-25*Pi .. 25*Pi,Y=0..1,contours=[0],axes=boxed,thickness=2,color=black,font=[1,1,18],tickmarks=[5, 5],linestyle=1,view=[0..1,0..0.6]);

@acer 

sorry

I thought it is not clear here

i have done the implicit plot for 3d, but i dont know how to do the contour for the implicit expression 

 

@Carl Love 

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]);

 

@Carl Love 

Thanks Carl

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

@Christian Wolinski 

many thankx for ur reply...

why you are calculating the roots..

 

Im really grateful...

I excpect optical bistability shape...

but many thanks

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 the  approximation function lambda2=ao+a1lambda1+a2lambda1^2

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