Hi,

I solve laplace equation in a square. All the lines of my code is okay.

Please just look to the last part of my code titled procedure:

When I run my code without (last funciton f #f := (x,y) -> 0;) see last lines to find "f". It's runing, there is no problem. But when I put add f, there is an error. Many think  for any help.

Procedure
Using the previous suty in section stencil we can write the procedure to solve the Laplace equation in [0,1]*[0,1] with the boundary condition Neumann conditions on the vertical boundary and Dirichlet boundary condition on the horizontal baoundary. In our study we will use the same stepsize h in x and y direction.

PoissonSolve:=proc(N,_f)
local Z,i,h,y,x,sys,w,f,sol,j,u,Data;
# define basic grid parameters
Z := i -> (1/(N+1))*i;
x[0] = Z(0),x[N+1] = Z(N+1),y[0] = Z(0),y[N+1] = Z(N+1);
 h := evalf(Z(1)-Z(0));
# Fix the boundary data and the source matrix
for i from 0 to N+1 do:
    # Neumann boundary condition
     u[N+1,i] :=  u[N,i] ;    
     u[0,i] := u[1,i];
     # Dirichlet boundary condition
     u[i,0] := 0;
     u[i,N+1] := 0;
   od:
   f := Array(0..N+1,0..N+1,[seq([seq(evalf(_f(Z(i),Z(j))),i=0..N+1)],j=0..N+1)],datatype=float);
# Write down the system of equations to solve and solve them
     sys := [seq(seq(Stencil(h,i,j,u,f),i=1..N),j=1..N)];
  w := [seq(seq(u[i,j],i=1..N),j=1..N)];
  sol := LinearSolve(GenerateMatrix(sys,w));
   # parse the solution vector sol back into "matrix" form
   for i from 1 to N do:
     for j from 1 to N do:
        u[i,j] := sol[(j-1)*N+i]:
     od:
   od:
# generate a 3D plot of the solution using the surfdata command
   Data := [seq([seq([Z(i),Z(j),u[i,j]],i=0..N+1)],j=0..N+1)]:
surfdata(Data,axes=boxed,labels=[`x`,`y`,`u(x,y)`],shading=zhue,style=patchcontour);
end proc:



Here is an example of the output when the source function is set to zero
                                 f(x, y) = 0
; i.e., when  reduces down to Laplace's equation:
#f := (x,y) -> 0;
#PoissonSolve(10,f);

Question8.mw

Hi

I need you to help me in writting procedure with input "r" ( order of derivative) and some coefficients seq(alpha[i],i=1..N).  My code work very well, need only put all the element in procedure with output The Taylor series obtained in last line of my code and the order of error.  I want the procedure return the coefficients beta used in the series and the order of Error  and the coefficients beta[i]

May thinks.

Second_Question.mw

I will not modify my previous question: only I add this remark. I tried to write these lines of procedure.
Add_display.mw

I need only to add( if the procedure is true) these lines in my procedure.  These lines gives the degree of error computed in the code, but when I put these lines in the codes, there is an error. Thanks for your help.

Degree := degree(Error,stepsize);
    if (showorder) then:
       print(cat(`This stencil is of order `,Degree));
    fi:
    if (showerror) then:
       print(cat(`This leading order term in the error is `,Error));
    fi:
    convert(D[r$n](f)(vars) = stencil,diff);

Dear all,

Please, I need help to remplace in the code attached,  addadd(beta[i]*u(x+[i]*h),i=0..N)+Error; 
by add(beta[i]*u(x+alpha[i]*h),i=0..N)+Error;  I see, that we must add a function, alpha, but how.

I think I must modify this rows: but how:

stencil := add(a[ii[i]]*subsop(r=op(r,phi)+ii[i]*stepsize,phi),i=1..N);

Question2.mw

Many thinks

R dJ(t)/dt+J(t)/C=f(t)

where f(t) is a driving electromotive force. Use the fourier transform to analyze this equation as follows.

Find the transfer function G(alpha)  then find g(t) .

 Thanks ....

Hi,

I try to solve this equation using pdsolve but there is no results.
restart:
with(PDEtools):
with(plots):

Eq:=diff(u(t,x), t$2) =diff(u(t,x),x$2)+sin(u(t,x));

pdsolve(Eq);

Thank you for your help.

I have system of equation contain of two equations

eq1:=a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1:
eq2:=a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2: 

I want to find the number k so that the equation

a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1 + k*(a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2) = 0

can be factor, where k satisfy

a:=a1+k*a2:
b:=b1+k*b2:
c:=c1+k*c2:
d:=d1+k*d2:
e:=e1+k*e2:
f:=f1+k*f2:

I tried

A:=a*x^2+b*y^2+c*x*y+d*x+e*y+f:
collect(A,x);
B:=collect(discrim(A, x), y);
C:= discrim(B,y);

and got the expression

-16*(4*a*b*f-a*e^2-b*d^2-c^2*f+c*d*e)*a=0.

For example 

> restart:

a1:=14:

b1:=-21:

c1:=0:

d1:=-6:

e1:=45:

f1:=-14:

a2:=35:

b2:=28:

c2:=0:

d2:=41:

e2:=-122:

f2:=56:

a:=a1+k*a2:

b:=b1+k*b2:

c:=c1+k*c2:

d:=d1+k*d2:

e:=e1+k*e2:

f:=f1+k*f2:

P:=c*d*e+4*a*b*f-a*e^2-b*d^2

-f*c^2:

eq1:=a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1:

eq2:=a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2:

with(RealDomain):

Q:=solve(P=0,k);

factor(eq1+Q*(eq2));

solve([

eq1=0,eq2],[x,y]);

Where, Q is k which I want to find. 

My question is, if I have the system of equations

eq1:=a1*x^3+b1*y^3+c1*x^2*y+ d1*x*y^2 + e1*x+f1*y+g1:
eq2:=a2*x^3+b2*y^3+c2*x^2*y+ d2*x*y^2 + e2*x+f2*y+g2: 

How can I get a similar to the 

-16*(4*a*b*f-a*e^2-b*d^2-c^2*f+c*d*e)*a=0?

the sequence is non-decreasing up to some point after which it is non-increasing. Note that i can be 1 or n. A constant sequence is considered to be unimodal.

Examples of unimodal lists:

[1, 1, 1, 1, 1],
[1, 2, 2, 3, 4, 5, 5, 5],
[5, 5, 4, 4, 3, 3, 1],
[1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1],
[1, 2, 2, 3, 3, 3, 4, 4, 2, 2, 1, 1, 1]

Examples of lists that are not unimodal:

[1, 0, 1, 0],
[1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1],
[1, 1, 2, 2, 3, 4, 5, 2, 2, 6, 4, 2, 2, 1, 0]

i don't have a clue

Hi MaplePrime-ers,

I'm using the following piece of code to (i) solve the system of symbolically, so I can (ii) evaluate equations quickly at many points of time.  This works quite well for 4 defined values, but I'm having problems adding a 5th defined value.  Specifically, solve leaves the "solution may be lost" message after taking forever.  As the symbolic solution will be run mulitple times by a optimziation algorithm, I'd ideally like to get the solve time under 2 minutes.  I've attached both executed worksheets.  Is there anything I can do to have solve work as I intend?

This first code snippet achieves what I would like to do Series_noGear.mw:

 

#Interconnection Equations
eq2[1] := FD_T + EM2_T = 0;
eq2[2] := ICE_T + GEN_T = 0;
eq2[3] := EM2_A + GEN_A + BAT_A = 0;
eq2[4] := -FD_W + EM2_W = 0;
eq2[5] := -ICE_W + GEN_W = 0;
eq2[6] := -EM2_V + GEN_V = 0;
eq2[7] := -EM2_V + BAT_V = 0;

#ICE
eq_c[1] := ICE_mdot_g= ICE_T * ICE_W;

#BAT
eq_c[2] := BAT_V = 271;

#EM2
EM2_ReqPow_eq := (-148.3) + (4.267)*abs(EM2_W) + (12.77)*abs(EM2_T) + (-0.0364)*abs(EM2_W)^2 + ( 1.16)*abs(EM2_W)*abs(EM2_T) + (-0.258)*abs(EM2_T)^2 + ( 0.0001181)*abs(EM2_W)^3 + (-0.0005994)*abs(EM2_W)^2*abs(EM2_T) + ( 0.0001171)*abs(EM2_W)*abs(EM2_T)^2 + (0.001739 )*abs(EM2_T)^3 + (-1.245e-07 )*abs(EM2_W)^4 + ( 1.2e-06)*abs(EM2_W)^3*abs(EM2_T) + ( -1.584e-06)*abs(EM2_W)^2*abs(EM2_T)^2 + ( 4.383e-07)*abs(EM2_W)*abs(EM2_T)^3 + (-2.947e-06)*abs(EM2_T)^4;
eq_c[3] := EM2_P = piecewise( EM2_T * EM2_W = 0, 0, EM2_W*EM2_T < 0,-1 * EM2_ReqPow_eq, EM2_ReqPow_eq);
eq_c[4] := EM2_A = EM2_P/EM2_V;

#GEN
GEN_ReqPow_eq:= (-5.28e-12) + ( 3.849e-14)*abs(GEN_W) + (-71.9)*abs(GEN_T) + (-1.168e-16)*abs(GEN_W)^2 +(1.296)*abs(GEN_W)*abs(GEN_T) + (2.489)*abs(GEN_T)^2 + (1.451e-19)*abs(GEN_W)^3 + (0.0001326)*abs(GEN_W)^2*abs(GEN_T) + (-0.008141)*abs(GEN_W)*abs(GEN_T)^2 + (-0.004539)*abs(GEN_T)^3 +(-6.325e-23)*abs(GEN_W)^4 + (-2.091e-07)*abs(GEN_W)^3*abs(GEN_T) + ( 3.455e-06)*abs(GEN_W)^2*abs(GEN_T)^2 + ( 2.499e-05)*abs(GEN_W)*abs(GEN_T)^3 + (-5.321e-05)*abs(GEN_T)^4;

eq_c[5] := GEN_P = piecewise(GEN_T * GEN_W = 0, 0, GEN_W*GEN_T < 0,-1 * GEN_ReqPow_eq, GEN_ReqPow_eq);
eq_c[6] := GEN_A = GEN_P/GEN_V;

#assumptions
assume(BAT_V::nonnegative);
assume(FD_W::nonnegative);

termeqs := {eq_c[1],eq_c[2],eq_c[3],eq_c[4],eq_c[5],eq_c[6]};

sys_eqs2 := termeqs union convert(eq2,set);

drivers2:= {ICE_T,ICE_W,FD_T,FD_W};
symvarnames2:=select(type,indets(convert(sys_eqs2,list)),name);
notdrivers2:=symvarnames2 minus drivers2;

sol2:=solve(sys_eqs2,notdrivers2) assuming real;

symb_sol2:=unapply(sol2,[drivers2[]]);

symb_sol2(1,2,3,5);

#Enumerate (there will generally be about 40, not 3)

count := 0;
for i1 from 1 to 3 do
     for i2 from 1 to 3 do
          for i3 from 1 to 3 do
               for i4 from 1 to 3 do
                    count := count + 1;
                    solsol2(count) := symb_sol2(i1,i2,i3,i4);
               od; 
          od;
     od;
od;
count;

This second code snippet includes the changes in bold, which make solve take forever Series_addGear.mw:

#Interconnection Equations
eq2[1] := FD_T + EM2_T = 0;
eq2[2] := ICE_T + GBb_T = 0;
eq2[3] := EM2_A + GEN_A + BAT_A = 0;
eq2[4] := -FD_W + EM2_W = 0;
eq2[5] := -ICE_W + GBb_W = 0;
eq2[6] := -EM2_V + GEN_V = 0;
eq2[7] := -EM2_V + BAT_V = 0;
eq2[8] := GBa_T + GEN_T = 0;
eq2[9] := -GBa_W + GEN_W = 0;

#ICE
eq_c[1] := ICE_mdot_g= ICE_T * ICE_W;

#BAT
eq_c[2] := BAT_V = 271;

#EM2
EM2_ReqPow_eq := (-148.3) + (4.267)*abs(EM2_W) + (12.77)*abs(EM2_T) + (-0.0364)*abs(EM2_W)^2 + ( 1.16)*abs(EM2_W)*abs(EM2_T) + (-0.258)*abs(EM2_T)^2 + ( 0.0001181)*abs(EM2_W)^3 + (-0.0005994)*abs(EM2_W)^2*abs(EM2_T) + ( 0.0001171)*abs(EM2_W)*abs(EM2_T)^2 + (0.001739 )*abs(EM2_T)^3 + (-1.245e-07 )*abs(EM2_W)^4 + ( 1.2e-06)*abs(EM2_W)^3*abs(EM2_T) + ( -1.584e-06)*abs(EM2_W)^2*abs(EM2_T)^2 + ( 4.383e-07)*abs(EM2_W)*abs(EM2_T)^3 + (-2.947e-06)*abs(EM2_T)^4;
eq_c[3] := EM2_P = piecewise( EM2_T * EM2_W = 0, 0, EM2_W*EM2_T < 0,-1 * EM2_ReqPow_eq, EM2_ReqPow_eq);
eq_c[4] := EM2_A = EM2_P/EM2_V;

#GEN
GEN_ReqPow_eq:= (-5.28e-12) + ( 3.849e-14)*abs(GEN_W) + (-71.9)*abs(GEN_T) + (-1.168e-16)*abs(GEN_W)^2 +(1.296)*abs(GEN_W)*abs(GEN_T) + (2.489)*abs(GEN_T)^2 + (1.451e-19)*abs(GEN_W)^3 + (0.0001326)*abs(GEN_W)^2*abs(GEN_T) + (-0.008141)*abs(GEN_W)*abs(GEN_T)^2 + (-0.004539)*abs(GEN_T)^3 +(-6.325e-23)*abs(GEN_W)^4 + (-2.091e-07)*abs(GEN_W)^3*abs(GEN_T) + ( 3.455e-06)*abs(GEN_W)^2*abs(GEN_T)^2 + ( 2.499e-05)*abs(GEN_W)*abs(GEN_T)^3 + (-5.321e-05)*abs(GEN_T)^4;

eq_c[5] := GEN_P = piecewise(GEN_T * GEN_W = 0, 0, GEN_W*GEN_T < 0,-1 * GEN_ReqPow_eq, GEN_ReqPow_eq);
eq_c[6] := GEN_A = GEN_P/GEN_V;

#GB
eq_c[7] := GBb_T = -1/GB_R * GBa_T;
eq_c[8] := GBb_W = GB_R * GBa_W;

assume(BAT_V::nonnegative);
assume(FD_W::nonnegative);
assume(GB_R::nonnegative);

termeqs := {eq_c[1],eq_c[2],eq_c[3],eq_c[4],eq_c[5],eq_c[6],eq_c[7],eq_c[8]};

sys_eqs2 := termeqs union convert(eq2,set);

drivers2:= {GB_R,ICE_T,ICE_W,FD_T,FD_W};
symvarnames2:=select(type,indets(convert(sys_eqs2,list)),name);
notdrivers2:=symvarnames2 minus drivers2;

sol2:=solve(sys_eqs2,notdrivers2) assuming real;

symb_sol2:=unapply(sol2,[drivers2[]]);

 

Does assume make solve work faster, or just complicate things?  Any help is greatly appreciated!

Series_addGear.mw

Series_noGear.mw

Hello there

I have to plot wavelet frame functions (framelet) using Maple. Can I send you the figure and help me to learn how to plot it and send me the code. I will pay for you if you can do it. 

Thank you

hi,i can not upload anything to mapleprimes.com every time i want to upload a file, i face error generating page . what should i do ?! 

Hi,

Maybe someone can give me a nice answer without Maple.

I am given a fourier series:
ln|cosx|=Co - sum( (-1)^k/k * cos2kx,k=1..infinity)
and am asked what this tells me about the chevychev series for ln(u).

Thanks

Hi, this question is for Samir Kahn or other maplesoft staff.

Is there a way to use modelica biochem or any other chemical path library into maplesim? Do you plan to integrate this library in future versions of maplesim? Wolfram systemModeler has this library already integrated as does symbiology in matlab. Mr Kahn, I hope that your background in chemical engineering help you push for this new development in maplesim and help this product be even better than the competion. thank you and let me know as soon as possible.

juan

Hello,

Is it possible to use Maple 15 Student Edition with the same license information and purchase code both on Windows and Ubuntu?

I did a few Windows reinstalls (guess about 3) hence I used the same purchase code to activate the license that many times. Today I tried installing and using Maple 15 on Ubuntu 12.04.3 LTS 32-bit and to my dismay the license manager said, "Not enough activations left on this account." Furthermore, I have exhausted my download links and would like to know how many times Maple 15 can be activated and if the download links can be refreshed?

Thanks

I have found the value of t1 and T, but I am unsure of the value. t1 and T in this case is the initialization time at the which the inventory level to zero in Reaches
warehouse. whether possible negative value? is there a lack of proper writing of parameters in this case? so that the output of t1 and T is negative? thanks a lot

> TC := (p*A*((lambda-sigma)*(exp(lambda*t2)-1)+lambda*(exp((lambda-sigma)*T)-exp((lambda-sigma)*t2)))/(lambda*(lambda-sigma))-Ca-A*exp(lambda*t1)*((beta*a2+b2)*(exp(beta*t1)-1)/beta^2-(lambda*a2-b2)*(1-exp(-lambda*t1))/lambda^2)/(beta+lambda)+A*exp(lambda*t1)*b2*t1/(lambda*beta)-W*((alpha*a1+b1)*(1-exp(-alpha*t1))-alpha*b1*t1*exp(-alpha*t1))/alpha^2+A*(exp(lambda*t2)*((a1+b1*t2)*lambda*alpha+b1*(lambda-alpha))-exp(lambda*t1)*(b1*(alpha^2-lambda^2*exp((lambda+alpha)*(t2-t1)))-(a1+b1*t1)*lambda*alpha*(alpha+lambda*exp((lambda+alpha)*(t2-t1))))/(lambda+alpha))/(lambda^2*alpha^2)+Cs*A*exp((lambda-sigma)*t2)*((T-t2)*(lambda-sigma)-exp((T-t2)*(lambda-sigma))+1)/(lambda-sigma)^2-L*A*((exp(lambda*T)-exp(lambda*t2))/lambda-(exp((lambda-sigma)*T)-exp((lambda-sigma)*t2))/(lambda-sigma))-C*(A*(exp((beta+lambda)*t1)-1)/(beta+lambda)+W+A*(exp((lambda-sigma)*T)-exp((lambda-sigma)*t2))/(lambda-sigma)))/T;
>
> TC2 := subs(t2 = (ln(W*(lambda+alpha)/A)+(lambda+alpha)*t1)/(lambda+alpha), TC);
>
> x1 := diff(TC2, t1) = 0;
>
> x2 := diff(TC2, T) = 0;
>
> x3 := evalf(simplify(lhs(eval(subs(A = 200, lambda = 0.3, a2 = 2.4, b2 = 0.01, a1 = 0.9, b1 = 0.01, alpha = 0.01, beta = 0.04, W = 100, Ca = 90, L = 7, C = 10, p = 15, Cs = 3, sigma = 0.25, x1)))));
>
> x4 := evalf(simplify(lhs(eval(subs(A = 200, lambda = 0.3, a2 = 2.4, b2 = 0.01, a1 = 0.9, b1 = 0.01, alpha = 0.01, beta = 0.04, W = 100, Ca = 90, L = 7, C = 10, p = 15, Cs = 3, sigma = 0.25, x2)))));

> x5 := fsolve({x3, x4}, {T, t1});
          x5 := {T = -15.72808468, t1 = -12.55374671}

You have clickable calc. Today I saw in your email that you have clickable physic but in french. Do you have it in english too>?

Thank you

manolo