Math-dashti

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These are questions asked by Math-dashti

i want to plot the function appear in P but there is a variable t which appear from variable which my system is depending on u,v and must v appear in my function instead of exponential so exponential contain the t variable  there is any way for fixing this?

restart;

with(plots):

 

f := (u,v) -> -u+u^3;
g := (u,v) -> -2*v;

proc (u, v) options operator, arrow; -u+u^3 end proc

 

proc (u, v) options operator, arrow; -2*v end proc

(1)

 

equilibria := solve({f(u,v)=0, g(u,v)=0}, {u,v});

{u = 0, v = 0}, {u = 1, v = 0}, {u = -1, v = 0}

(2)

 

de1 := diff(u(t),t) = f(u(t),v(t));
de2 := diff(v(t),t) = g(u(t),v(t));

diff(u(t), t) = -u(t)+u(t)^3

 

diff(v(t), t) = -2*v(t)

(3)
   

PDEtools:-ConservedCurrents({de1, de2}, [u(t), v(t)]); P1 := -(1/2)*op(1, rhs(op(%)))

[_J[t](t, u(t), v(t)) = f__1((-u(t)^2+1)*exp(-2*t)/u(t)^2, v(t)*exp(2*t))]

 

-(1/2)*(-u(t)^2+1)*exp(-2*t)/u(t)^2

(4)

P := -(-u^2+1)*exp(-2*t)/(2*u^2)

-(1/2)*(-u^2+1)*exp(-2*t)/u^2

(5)

 

equilibria;

{u = 0, v = 0}, {u = 1, v = 0}, {u = -1, v = 0}

(6)

 

p1 := contourplot(P, u=-1.5..2, v=-1.5..1.5, scaling=constrained,
    colorscheme="DivergeRainbow", contours=[seq](x, x=-0.4..0.4,0.1)):

Error, (in plot/iplot2d:-Levels) could not evaluate expression

 

 

p2 := pointplot([[0,0],[1,0],[-1,0]], symbol=solidcircle, symbolsize=15, color=black):

 

p3 := fieldplot([f(u,v), g(u,v)], u=-1.5..2, v=-1.5..1.5,
        arrows=medium, fieldstrength=fixed(0.4), grid=[10,10], labels=["",""]):

 

display(p1,p2,p3, axes=normal);

 

 

Download p1-1.mw

it is apear by sign of taking derivative from equation and substitute in it the sign less zero is stable otherwise is unstable but i want to shown by graph like shown in graph

plot-stablity-1equation.mw

what is problem in here

restart;

with(plots):

with(LinearAlgebra):

with(DEtools):

diff(u(x), x) = f(u(x), v(x)), diff(v(x), x) = g(u(x), v(x)) for the two differential equations.

f := (u,v) -> u+cos(v);
g := (u,v) -> u*v-v+sin(v);

proc (u, v) options operator, arrow; u+cos(v) end proc

 

proc (u, v) options operator, arrow; v*u-v+sin(v) end proc

(1)

The equilibria:

equilibria := solve({f(u,v)=0, g(u,v)=0}, {u,v},explicit);

{u = 1, v = Pi}, {u = -cos(RootOf(sin(_Z)*_Z^2+sin(_Z)-2*_Z)), v = RootOf(sin(_Z)*_Z^2+sin(_Z)-2*_Z)}

(2)

allvalues(RootOf(sin(_Z)*_Z^2+sin(_Z)-2*_Z))

RootOf(sin(_Z)*_Z^2+sin(_Z)-2*_Z, 1.306542374), RootOf(sin(_Z)*_Z^2+sin(_Z)-2*_Z, -1.306542374), RootOf(sin(_Z)*_Z^2+sin(_Z)-2*_Z, -2.331122370), RootOf(sin(_Z)*_Z^2+sin(_Z)-2*_Z, 2.331122370), 0

(3)
 

NULL

Download remove.mw

when we do the linearization we do transfer the critical point to the orgin, i want see the plot of nonlinear system before the linearization and after linearization which is critical point is tranfered to orgin, also i try to find `conserved quantity` and make a good plot with the critical point shown in graph how i can do the linearization by maple, also  we  by hand when we solved we show that in new system when we give him variable we have to write for each new system  u=x-x[0] when is u is new system for linearization we need that or not? becuase result is not change when we take jacobian result is same !

 

Linearization.mw

i want plot a system of differential equation and do phaseportrait i did but when i want make it a little bit more clear and colorfull like rainbow when i find the C.Q i don't know how set the option for ploting?

e1.mw

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