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These are questions asked by Al86

I want to use maple2015 to perform a Monte-Carlo simulation of a one dimensional Harmonic Oscillator.

Consider a particle having displacements, x=na where n is an integer, i.e 0 plus minus one, etc.

The potential energy of the particle is given by U(x)= 1/2 kx^2 = 1/2 ka^2 n^2.

For this simulation will choose ka^2 = 0.01 k_B T where k_B is Boltzmann's constant, and T is the temprature in Kelvin.

So we should perform a Monte-Carlo simulation starting from x(0)=20a, and perfroming individual steps by selecting an attmept to move x=+-a ,with equal probability, and choose whether to make the actual step according to energy change expected in that step.

I need to draw particle's position and x(t), and particle's energy E(t).

After that I need to calculate and draw the autocorrelation functions \phi_x(t) and \phi_E(t).

I found some code that might help me, but I am nor sure how to implement the above two lines on maple; your help is appreciated.


I have the PDE u_{xx}+u_{yy} = 1 with BC: u|_{x^2+y^2=1} =0 ;


how to write down the command of the BC in solving this PDE?, btw can I make maple show me how to solve this PDE analytically?


Thanks in advance.


Here are the lines that I wrote so far:

pde := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 1;

ans := pdsolve(pde)


how to add the BC correctly to pdsolve? I am not sure how to write the condition x^2+y^2=1 and that u will get a value on this boundary.


I read in the net that the method used in pdsolve numeric is the theta method, my question: is it the most efficient with regard to rate of convergence of the numerical solution of the PDE?

If not then why is it used as the default method?




I have the following code:

PDE := diff(u(x, t), t) = diff(u(x, t), x, x)-sin(x+t)+cos(x+t);
IBC:= D[1](u)(0,t)=-sin(t),
D[1](u)(1,t)=-(u(1, t))^4+(cos(1+t))^4-sin(1+t),

pds := pdsolve( PDE, [IBC], numeric, time = t, range = 0 .. 1,

spacestep = 0.1e-1, timestep = 0.1e-1,


And I want to plot the difference |pds(t,x) - cos(x+t)| in maple for x=1 and t=0..5


I thought to use the following piece of commands but I get an error:

P:=unapply(pds, t,x);
Q:-plot(x=1, t=0..1);

I get an error that Q isn't a module.

I thought that unapply is used for this case, can you help me with this simple task?


Thanks in advance.


I have the following two PDEs:

PDE := diff(u(x, t), t) = diff(u(x, t), x, x)+sin(x+t)-cos(x+t);

IBC:= D[1](u)(0,t)=-sin(t),

pds := pdsolve( PDE, [IBC], numeric, time = t, range = 0 .. 1,
spacestep = 1/32, timestep = 1/32,


PDE2 := diff(v(x, t), t) = diff(v(x, t), x, x);
IBC2:= D[1](v)(0,t)=0,
D[1](v)(1,t)=-0.000065*v(1, t)^4,

pds1 := pdsolve( PDE2, [IBC2], numeric, time = t, range = 0 .. 1,
spacestep = 1/32, timestep = 1/32,


Now, what I want to do with these two PDEs is the following:


For each h=timestep=spacestep  = 1/16 , 1/32 , 1/64 , 1/128 , 1/256

Calculate the error norm ||E||_h = sqrt(sum_{j=0}^{1/h} h* |u(j*h,tval)-v(j*h,tval)|^2)

where tval is some chosen point between 0 and 1 (this value is fixed for each spacestep chosen).


And then plot the graph of log ||E||_h vs. log h above.


What I don't know is how to extract each time the spacestep and its PDE's two solutions, does someone have a suggested script to use here?



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