60 Reputation

5 Badges

4 years, 107 days

MaplePrimes Activity

These are replies submitted by Al86

I am attaching below my code that I changed a little bit from the code in the link in my OP, I added autocorrelation functions and the graphs of theirs and of the displacement and energy.

I get the following error:

Error, cannot determine if this expression is true or false: 0.2071500000e-4*x2^2 <= 0. or 149.5449674 < exp(-0.5000000000e-2*x2^2)

It seems that something is wrong in the if conditional, how to fix it?


Also, the values of <E^2>,<E>,<x>,<x^2>,C_V are all zero, which doesn't seem to me to be probable.


Can anyone help me with my code?


@vv where can I find this Poisson formula for 2d Poisson equation?


In Evans' book on PDE second edition I have the following solution for n=2:


u(x) = -1/2pi \int_{\mathbbb{R}^2} log(|x-y|)f(y)dy



@tomleslie , hi tom; I changed a little bit your code (the natural logs I changed to log10, tval changed to 5, and I am now looking at the error norm between cos(x+t) and the solution of the PDE:

diff(v(x, t), t) = diff(v(x, t), x, x)-sin(x+t)+cos(x+t)

D[1](v)(1,t)=-v(1, t)^4+(cos(1+t))^4-sin(1+t),

the analytical solution of this PDE is cos(x+t))

Now in order to calculate the slope of the log10 log10 graph I use:

LeastSquares([[log10(1/16), log10(E[1])], [log10(1/32),log10(E[2])],[log10(1/64), log10(E[3])],[log10(1/128),log10(E[4])],[log10(1/256),log10(E[5])]], v);

I get a slope that is smaller than 2 while if I change the boundary conditions to:


(its analytical solution is also cos(x+t))

I get a slope which is approximately 2.

My colleague told me that theoretically the slope should be 2, why is that if I change to the first boundary conditions that the slope of the log log graph decreases so drastically?

The analytical solution of the two problems is identical, isn't it?

Thanks in advance, I appreciate your help.



Forgot to add that now we are comparing the numerical solutions of the PDEs with the analytical solution which in this case is cos(x+t).



Hi, if I would like to use leastsquares command and plot the function log ||E||_h as a function of log h, how would I change your code?


I mean we have:

pA[2]:= plot( [ seq( [ log(hvals[j]), log(E[j]) ],
labels=["space\\timestep", "Norm"],
labelfont=[times, bold, 16],
labeldirections=[horizontal, vertical],
title="Norm vs step (log10-log10)",
titlefont=[times, bold, 20]



So I would like to generate a least squares approximation with the xdata being log h=log 1/16 ,..., log 1/256

and the ydata would be the corresponding value of log ||E||_h at the points of h=1/16 ,...,1/256

and then to plot the least square approximation.


So the least squares part would be:


LeastSquares([[log(1/16), log(E[1])], [log(1/32),log(E[2])],[log(1/64), log(E[3])],[log(1/128,log(E[4])],[log(1/256),log(E[5])]], v);

Now, how to plot this least squares function?

as log ||E||_h vs log h.


@tomleslie thanks Tom.

I appreciate your help, if I were to use the same analysis just instead of the error between the two PDEs, only to calculate the norm of the solution of the first PDE, all I need to change in your code is this line:


E[p]:= sqrt
( h*add
( (sol[1](j*h))^2,


Or do I need to change also other lines of your code?


@tomleslie the correct form syntactically is:

||E||_h = sqrt( h*add(abs(u(j*h,tval)-v(j*h,tval))^2, j=0..1/h));


I appreciate your help.


@Markiyan Hirnyk yes I asked on PDEs, I forgot that I should be using 2^(-k) here.


@tomleslie thanks, so much syntax to learn to use here.


@tomleslie I am sorry that I am asking another question.


But how can I plug a value into mf after the substituion of the numerical paramters, I mean I want to plug in the IBC after the PDEs values of u(x,0) = 1+mf(at t=0) and u(x,0) = 1+myPoly(at t=0) respectively, how to implement this?


 I found how to do this for mf, just subs[eval](t=0,mf)=:mf1 I am sure it will work also for the polynomial.


@tomleslie , I have another request, how do I use the coefficients of the two approaches (of a*exp(-b*t)-c*t and of the sixth polynomial) in any other further commands; i.e I want to use mf as an explicit function with the numerical coefficients obtained before and the same with the polynomial in further operations.


Here's an example of what I want to do:

PDE2 := diff(u(x, t), t) = diff(u(x, t), x, x)+sin(x+t)-cos(x+t)-(diff(mf, t));

IBC2 := (D[1](v))(0, t) = 0, (D[1](v))(1, t) = -0.65e-4*(v(1, t)+mf)^4, v(x, 0) = 1+a

pds2 := pdsolve(PDE2, [IBC2], numeric, time = t, range = 0 .. 1, spacestep = 0.1e-1, timestep = 0.1e-1, errorest = true)*pds2:-plot(x = 1, t = 0 .. 1)


How to make the polynomial and a*exp(-b*t)-c*t with the numerical coefficients available to me?

Thanks in advance.



@tomleslie thanks.

The polynomial seems like a close to accurate approximation.


@tomleslie are there any other methods? how about polynomial interpolation?


I don't understand how do I extract the coefficients a,b,c from the fitting a*exp(-b*t)-c*t?

@Preben Alsholm Hi, I am reading the book: Finite elements using maple.

They give a code for collocation approximation to a simple ODE with weights, can we use this approach here with my nonlinear integral equation?

with 100 collocation points x[i]?


Here's a preview of the book:

on pages 79-80 there's the code.

A description of collocation method is on page 68.


@tomleslie my programming skills are weak, I know, I am trying to ammend it.


@tomleslie It looks good.


BTW for the case 

PDE := diff(u(x, t), t) = diff(u(x, t), x, x)+sin(x+t)-cos(x+t);
IBC:= D[1](u)(0,t)=0,
D[1](u)(1,t)=-0.000065*(u(1, t))^4,


How would I get the slope at each interval of the log ||e||_h vs log h graph, I mean I get a linear piecewise graph, and I want to extract from the graph the slopes at each interval where the graph seems still as a line in that interval (before it breaks into another line).


Thanks in advance.

I looked at maple's help, but I don't seem to find an appropriate command to extract the slopes.


1 2 3 4 Page 2 of 4