mah00

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12 years, 129 days

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These are replies submitted by mah00

I know why,

I decalred dx as the sqrt of that quantity then I take the derivative of the square (dx^2 is what you called f). When I do this, the answer is in expanded form.

When I declare dx without the sqrt, it comes in factorized form, that's why you aren't able to find the same result.

 

Problem solved.

 

Thanks for your time.

I know why,

I decalred dx as the sqrt of that quantity then I take the derivative of the square (dx^2 is what you called f). When I do this, the answer is in expanded form.

When I declare dx without the sqrt, it comes in factorized form, that's why you aren't able to find the same result.

 

Problem solved.

 

Thanks for your time.

Thanks, it works!

Thanks, it works!

Thank you for your answer.

However I have this error after defining the matrix C: "Error, invalid input: LB expects its 1st argument, m, to be of type Vector, but received Matrix(12, 1, [[...],[...],[...],[...],[...],[...],[...],[...],[...],[...],[...],[...]], datatype = anything)"

That's why I did all those conversions in my first code.

You don't have this error?

Thank you for your answer.

However I have this error after defining the matrix C: "Error, invalid input: LB expects its 1st argument, m, to be of type Vector, but received Matrix(12, 1, [[...],[...],[...],[...],[...],[...],[...],[...],[...],[...],[...],[...]], datatype = anything)"

That's why I did all those conversions in my first code.

You don't have this error?

So after these nswers it is better to discretize everything.

Thank you all for your answers.

So after these nswers it is better to discretize everything.

Thank you all for your answers.

Well this is my starting point so I don't know how to start. What I know is this:

Ksi1:=WienerProcess(IdentityMatrix(2)):
> ksi:=SamplePath(Ksi1(t),t=0..tf,timesteps = 2*tf):
 

ksi will be an array containing the generated opints from the wiener process. What I want is to create a sample path of a wiener process which is continuous. In this case, ksi(1.5) does not exist for example.

 

Hope this is more clear.

@Preben Alsholm, Can you give some explanation on why you used the "op" in dsolve?

thanks

@Preben Alsholm, Can you give some explanation on why you used the "op" in dsolve?

thanks

Thank you very much. It is exactly what I wanted.

I will try to understand what you did and come back if something is not clear.

Thank you very much. It is exactly what I wanted.

I will try to understand what you did and come back if something is not clear.

Ok, I understand better now. thanks for your time.

It is true that the explicit dependance on t makes the system more complicated.

 

 

For future consultation of this topic, x and y represents here a particle in a system of three vortices of equal strength.

You can use the following code to plot the Hamiltonian and see the vortices:

f2:=evalc(Re((I/(2*Pi))*((1/(subs(j=1,z)-(x+I*y)))+(1/(subs(j=2,z)-(x+I*y)))+(1/(subs(j=3,z)-(x+I*y)))))):

g2:=evalc(-1*Im((I/(2*Pi))*((1/(subs(j=1,z)-(x+I*y)))+(1/(subs(j=2,z)-(x+I*y)))+(1/(subs(j=3,z)-(x+I*y)))))):

H1 := evalc(int(f2, y)-(int(g2, x))-(int(diff(int(f2, y), x), x)));

contourplot(subs(t=0,H1),x=-2..2,y=-2..2,contours=20);

 

Ok, I understand better now. thanks for your time.

It is true that the explicit dependance on t makes the system more complicated.

 

 

For future consultation of this topic, x and y represents here a particle in a system of three vortices of equal strength.

You can use the following code to plot the Hamiltonian and see the vortices:

f2:=evalc(Re((I/(2*Pi))*((1/(subs(j=1,z)-(x+I*y)))+(1/(subs(j=2,z)-(x+I*y)))+(1/(subs(j=3,z)-(x+I*y)))))):

g2:=evalc(-1*Im((I/(2*Pi))*((1/(subs(j=1,z)-(x+I*y)))+(1/(subs(j=2,z)-(x+I*y)))+(1/(subs(j=3,z)-(x+I*y)))))):

H1 := evalc(int(f2, y)-(int(g2, x))-(int(diff(int(f2, y), x), x)));

contourplot(subs(t=0,H1),x=-2..2,y=-2..2,contours=20);

 

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