An hour or two ago, I answered a question in which it was a question of plotting a complex-valued function of 2 real variables. But the question itself and also my answer to it disappeared somewhere. Therefore, I send my answer here below.

There are two options for plotting:
1. Graphs of real and imaginary parts (as 2 surfaces in 3D).
2. Graph of the absolute value of this function (one surface in 3d) .

restart;
f:=(1+cosh(2*x))*exp(-4*I*t):
plot3d([Re,Im](f), x=0..1, t=0..1, color=[red,blue]);
plot3d(sqrt(add([Re,Im](f)^~2)), x=0..1, t=0..1, color=green);

I want to made a comparison via plots of RK-4, NSFD and LWM.

I have noticed a few times now with Maple 2019. It looses kernel connection when it is sitting there idly. This time I observed it. Had saved a document after an intensive calculation. The memory used was about 30Gig. shortly after saving the cpu fan was running hard. I checked task manager and cpu was cycling to 100%, it was mserever. Then the memory usage droped to about 6gig and message as shown. During this time Maple screen down in the LH corner displayed "Ready", so it didn't think it was doing anything.
 

Dear Community,

I run a MapleSim model from Maple. The simulation runs fine giving me the correct graphical plots, but I wonder how could I obtain also the numerical values of the probes vs. time? Sorry I could not figure it out. A matrix format would be perfect: 1st column time, the other columns probe 1 .. n values vs. time. (Files attached)

Tx for the kind help in advance,

best regards

Andras

RunMapleSim.mw , RCNetwork.msim

Hello,

How I can take variation from left-hand side of  5, and reach to right-hand side of  5. After by using integral by part obtained  7?

Thank you

Maple pdsolve supports periodic boundary conditions. So I was hoping it will be able to solve the heat PDE inside disk with periodic boundary conditions. But I am not able to make it work. 

Is there a trick to make Maple solve this, is there something I need to add or adjust something else? or it is just the functionality is not currently implemented?

This is what I tried

restart;

pde := diff(u(r,theta,t),t)=diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2);
bc1 := u(a,theta,t)=0;
bc2 := eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi);
bc3 := u(r,-Pi,t)=u(r,Pi,t);
ic  := u(r,theta,0)=f(r,theta);
sol := pdsolve([pde, bc1,bc2,bc3, ic], u(r, theta, t), HINT = boundedseries(r = 0)) assuming a>0,r>0

I solved this analytically by hand using standard separation of variables method. The issue of telling Maple the solution is bounded at center of disk, I assume is being handled automatically by the HINT=boundedseries(r = 0).

If I remove the hint, it also does not solve it. 

Maple 2019, Physics package 338

Hellow, help required to remove the errors

 How to  obtain the pressure drop i am unable to get the output. I am uploading the file  and the equations

help_dp.mw

For this problem

I'd like to see if Maple can give, or simplify the solution it now gives to look like this solution 

The one it currently gives is

restart;

pde:=diff(w(x,t),t)+c*diff(w(x,t),x)=0; 
ic:=w(x,0)=f(x);
bc:=w(0,t)=h(t);
sol:=pdsolve([pde,ic,bc],w(x,t))  assuming t>0,x>0,c>0

And I did not know how to simplify it or obtain the simpler one. I tried strip and TWS hints.  I also do not understand why Maple gives an integral with 0 as upper limit there (the second integral).

Using Physics package cloud version 338 and Maple 2019. On windows 10.

Thank you

Hi,

I have to build a simple model in MapleSim that allows to simulate a mass that goes down along an inclined plane with a certain friction.

Is there a component that simulates the inclined plane? I tried to use a prismatic joint but it has some predifined translational directions so I can't impose the movement along the plane.

Is there a way to take the laplacian of 1/r and get the "physics" answer of -4*pi*delta(\vec{r})?

with(plots):R := 5; alpha := (1/9)*Pi;
C1 := plot([R*cos(t), R*sin(t), t = 0 .. 2*Pi], color = blue);
A := [R*cos(alpha), R*sin(alpha)]; B := [R*cos(alpha+Pi), R*sin(alpha+Pi)]; AB := plot([A, B], scaling = constrained);
display({AB, C1}, scaling = constrained);# bad drawing

https://aws.amazon.com/getting-started/projects/deploy-elastic-hpc-cluster/

is it possible to use maple on high performance clusters?

i can only think to use c program to call cmaple with MPI in linux to use high performance clusters.

is there any other official method to do this?

if i upload my maple 2015 version to amazon for this computing, will it used up all license in this first chance of installation leading to that i can not install maple 2015 linux version to other machine?

https://docs.aws.amazon.com/AWSEC2/latest/WindowsGuide/ConfigWindowsHPC.html#ComputeNode

which virtual machine should i install the maple 12? on one virtual machine or all compute nodes?

how many compute nodes are need to compute dsolve 100,000 systems which may or may not have solution in maple 12?

Hi all

We denote the collecction of sets determined by the first k coin tosses

Suppose the imitial stock price is ,with up and down facter being and .

Up : S1(H)=u S0 and S1(T)=d S0

S_{N+1}= alpha S_N

where alpha =u or d

Let the probability of each and be and and   the sigma-lgebra generated by the coin tosses up to (and inchudling) time t:

After three coin tosses.

Can we propose a code computing the element of the filtration F1 and F3 and sigma(S3) (the sigma algebra generated by S3).

For example by hand we have F1={ emptyset, Omega, AH, AT}

Where AH={ w: w1=H}

AT={w: w1=T}

Can we compute

restart;
with(Finance);
S := [7.9, 7.5, 7.1, 6.5, 5., 3.7, 3.3, 2.95, 2.8];
         [7.9, 7.5, 7.1, 6.5, 5., 3.7, 3.3, 2.95, 2.8]
T := BinomialTree(3, S, .3);
TreePlot(T, thickness = 2, axes = BOXED, gridlines = true);

many thanks

Hello all,

I'm trying to do kinetic modeling of sequential dissociations with DE. I'm hitting a snag when modeling the third dissociation. The population should start at zero at t=0, but some of my model functions are non-zero at t=0. Is there anyway to fix this to force the funtions to go through zero?

Scheme:
PPPP -> intermediates -> PPP -> intermediates -> PP -> intermediates -> P  
(where P is a subunit and intermediates are confirmational changes before dissociation of a subunit)

a'..d' is the first dissociation
e' is the second dissociation
f'..l' is the third dissociation
Fits are evaluated by the residual sum of squares.

sol := dsolve([a' = -k1*a(x), b' = k1*a(x)-k1*b(x), c' = k1*b(x)-k1*c(x), d' = k1*c(x)-k1*d(x),
e' = k1*d(x)-k2*e(x), 
f' = k2*e(x)-k3*f(x), g' = k3*f(x)-k3*g(x), h' = k3*g(x)-k3*h(x), i' = k3*h(x)-k3*i(x), j' = k3*i(x)-k3*j(x), k' = k3*j(x)-k3*k(x), l' = k3*k(x)-k3*l(x), 
a(0) = 1, b(0) = 0, c(0) = 0, d(0) = 0, e(0) = 0, f(0) = 0, g(0) = 0, h(0) = 0, i(0) = 0, j(0) = 0, k(0) = 0, l(0) = 0],
{a(x), b(x), c(x), d(x), e(x), f(x), g(x), h(x), i(x), j(x), k(x), l(x)}, method = laplace);

f1 := sol[6];
f1 := rhs(f1);
g1 := sol[7];
g1 := rhs(g1);
h1 := sol[8];
h1 := rhs(h1);
i1 := sol[9];
i1 := rhs(i1);
j1 := sol[10];
j1 := rhs(j1);
kk := sol[11];
kk := rhs(kk);
l1 := sol[12];
l1 := rhs(l1);

xdata := Vector([0,10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160,170,180,200,210,220,230,240,250,260,270,280,290,300,310,320,330,340,350,360,370,380,390,400], datatype = float);
ydata := Vector([0.0034,0.00392,0.00184,0.00782,0.01873,0.03683,0.11016,0.09838,0.18402,0.24727,0.20901,0.2972,0.37635,0.49235,0.57845,0.4457,0.50285,0.5672,0.62783,0.57264,0.54918,0.44792,0.49795,0.55218,0.47512,0.46473,0.37989,0.32236,0.3323,0.20894,0.28473,0.21273,0.19855,0.13548,0.12725,0.13277,0.0784,0.07969,0.06162,0.03855], datatype = float);

k1 := 0.391491454107626e-1; 
k2 := 0.222503562261129e-1; 

z1:=f1;
z2:=f1+g1;
z3:=f1+g1+h1;
z4:=f1+g1+h1+i1;
z5:=f1+g1+h1+i1+j1;
z6:=f1+g1+h1+i1+j1+kk;
z7:=f1+g1+h1+i1+j1+kk+l1;

Statistics[NonlinearFit](z1,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z1,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z2,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z2,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z3,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z3,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z4,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z4,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z5,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z5,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z6,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z6,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z7,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z7,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

3rd_diss.mw

Example of Duffing equation with boundary conditions.
y'' + 0.2y' + y^3 - 0.3cos(s) = 0;
y(0) = y (2Pi);
y'(0) = y'(2Pi);
For convenience, we replace the original equation with a system of two first order equations:
--------------------------------------------------------------------------
x1'(t) = 2*Pi*x2(t);
x2'(t) = - 0.4*Pi*x2(t) - 2*Pi*x1(t)^3 +0.6*Pi*cos(2*Pi*t);
x1(0) = x1(1);
x2(0) = x2(1);
--------------------------------------------------------------------------
I have long wanted to apply an optimization package to solve a boundary value problem for ODE. The decision helped procedure for solving ODE, written by forum member vv.
It seems to me that two solutions have been found and that the solutions are weakly sensitive to the initial approximations. These are two closed trajectories. For example, these are points that belong to these solutions:
(0.5966963,  1.0482816) , ( - 0.3132584, 0.0664941).
I am wondering: are the solutions right, and how justified is the use of optimization methods for such tasks?
At the end of the program, the solution is checked on the original Duffing equation using standard Maple functions.   Duffing_equation_BC.mw

(In the figures, the trajectory bypass occurs three times.)