can someone give help or a hint on how to overcome the faulty result in the pdsolve problem with radiation bc described in the following worksheet? I tried also with different method, but no success. the result from pdsolve makes physical no sense, because under pure radiation bc (h=0) the body should cool down and not heat up!

maple_heat_rad_pdsol.mw

thanks :-)

btw, the problem was reported befor, see:

How to compute the highest total degree of a polynomial in which a variable, say x, appears?

For example, f := x^2*y+3*x*y^2+x^3*y^3-x^5+y^4*z^5;
then the highest total degree in x is 6.

The degree command does not seem to do what I would like to do, since degree(f,x) will give 5 instead.

Thank you in advance.

Dyson_Mclean_V3.mw

Dyson_Mclean_V5.mw

Hi All,

I have a question that I see has been seldom discussed except certain posts by Mr. Lopez. This concerns the estimation of optimum parameters of the solution of a system of differential equations to fit experimental data. I referred to the Maple document provided by Mr. Lopez on the Maplesoft website.

Good morning.

I want to do a program in a worksheet that it can recall other workeet for plotting with display.

I don´t know how can recall other worksheets.

Please help me.

P.D.: I send a worksheet 01-06-2011-robot-man.mw

I'm learning to handle the output of pdsolve. It is a module and I'm quite new to that. The objective is to plot the solution for different values of, say, time t, in a fairly systematic way. I'll consign here my progress for reference. Feel free to comment if you have suggestions.

 sol := pdsolve
  ( diff(u(x, t), t) = (1/10)*(diff(u(x, t), x, x))
  , {u(0, t) = 0, u(x, 0) = 1, (D[1](u))(1, t) = 0}
  , numeric
  );

Hi,

I am tring to solve two coupled pde's with piecewise/Heaviside conditions. I am puzzled with the maple error...Rhop, Rhos are constants n given some values..

> pde_sys:=[diff(c(m,t),t)=(2/m)*diff(c(m,t),m)+diff(c(m,t),m,m)-(convert(piecewise(d(m,t)<Rhop,(Rhop-d(m,t))*d(m,t)*c(m,t)),Heaviside)+(Rhos-d(m,t))*c(m,t)*convert(piecewise(d(m,t)>Rhop,Rhop,d(m,t)),Heaviside)), diff(d(m,t),t)=convert(piecewise(d(m,t)<Rhop,(Rhop-d(m,t))*d(m,t)*c(m,t)),Heaviside...

Like when I type

1/2^2;

it gives

1/4

but if I type

0.5^2;

it gives

0.25

I want it to consider 0.5 as 1/2, because sometimes I make computations on large equations and I want to have access to the symbolic output, not the numeric output.

Hello all,

I am trying to solve a pde system with piecewise conditions numerically but unsucessful. Here's what i am doing:

> restart;

> sys:=[diff(c(x,t),t)=v(x,t)*diff(c(x,t),x$2),diff(v(x,t),t)=piecewise(c(x,t)>v(x,t),c(x,t)*diff(v(x,t),x$2),0)];

> bc:={c(0,t)=2,c(1,t)=1,c(x,0)=0,v(0,t)=1,v(1,t)=2,v(x,0)=0};

> sol:=pdsolve(sys,bc,type=numeric,time=t);

I am getting the following error.......

Error, (in pdsolve/numeric/process_PDEs...

Hello,

I am trying to solve a transient heat transfer problem.

The problem is that I have an insulated pipe that is immersed in cold water.  At time 0, the fluid filling the pipe is at a constant temperature.  The insulation at one end of the pipe is different to the rest of the insulation.  I need to find out the time that it would take any fluid to reach a certain temperature, called the cool down time.

A graphic representation of the...

Is there an easy way to find the limit of the following equations as Pec_i approches infinity.

 Equation 1

(Q*(R*S+1-S)/(1-R*R3*k*(1-R1)/C/Pec_i*S)*(H_vap+1/(1-exp(Pec_i)))*exp(C*Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*S/k)-H_vap)*(1-exp(Pec_i*(1-S)/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S))) = 1;

 Equation 2

 Theta0/C = (exp(Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*(1-S))-1)*(H_liq+1/(1-exp(-C/k*Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*S)));

Does the series sum((-1)^floor(ln(n)^2)/n, n = 1 .. infinity) converge?

Please help me to make sense of the ways to use the simplify function. In this particular case Maple does some computation and gives me some huge output which I paste below. When I try to simplify the huge output Maple just hangs. But if I use varied commands of simplify detailed below such as simplify(huge_output,symbolic)  or  simplify(huge_output,size)  Maple gives me an output but none of the output are equal to each other and I also noticed that in one instance...

 Okay so I can see how to get the poles. But I have some further questions.

1) For beta AND Q both GREATER THAN 0 WHICH poles have positive imaginary part and is there a PROGRAMMATIC way of selecting these ones?

2) How do I calculate the residue at these poles? The approach given by Alec above seems useful and is much...

Dear fellow users,

I am trying to call Maple functions from Matlam .m file. How can I do it?

I am using Matlab 7.12 (R2011a) and Maple 12.

Thanks in advance..

I'm trying to write a for loop that will generate several procedures, which doesn't quite seem to work as expected.

Here is the sample code:

for j from 1 to 2 do
f[j]:=proc(x,y)
x+y^(j)
end proc;
od;

Then if you run the following, you get:

f[1](x,y) = x+y^3

f[2](x,y) = x+y^3

Whereas, I'd like to have:

f[1](x,y) = x + y^1

f[2](x,y) = x + y^2

Any suggestions on why the j value...