## 35 Reputation

12 years, 72 days

## I know why, I decalred dx as the sqrt o...

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.

## I know why, I decalred dx as the sqrt o...

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.

## Yes!...

Thanks, it works!

## Yes!...

Thanks, it works!

## Variable type problem...

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?

## Variable type problem...

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?

## Good to know...

So after these nswers it is better to discretize everything.

## Good to know...

So after these nswers it is better to discretize everything.

## Explanation...

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.

## op operator...

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

thanks

## op operator...

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

thanks

## Excellent...

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.

## Excellent...

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...

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...

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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