MaplePrimes Questions

I am preparing a document where in I want to show half circle inide the drawing. I can plot a half circle, but can not bring it inside the drawing canvass. Is there a way to show smooth curves inside the drawing. Pencil tool helps to show my curve, but not a smooth one! In any picture tool, bring forward, sent backward options will be there. Are there such tools in maple?
Thanks for answers.

Cheers.

Ramakrishnan V
 

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Download drawing_semicircle.mw

Hello everyone

I have the solution of diffusion equation from Help of maple website. I put the code here

*****************************

restart: with(plots):
 

unprotect(D);
 

alias(c[0]=c0, c[1]=c1, c[2]=c2);
PDE:=diff(C(x,t),t)=D*diff(C(x,t),x,x);
IBC:={C(x,0)=cx0, C(0,t)=ct0, D[1](C)(10,t)=0};
ct0:=1;
cx0:=0;
D:=1;
pds:=pdsolve(PDE,IBC,numeric);
L1:=[0.01, 0.1, 1, 5, 10];
L2:=[red, green, yellow, blue, magenta, black];
for i from 1 to 5 do
 pn[i] := pds:-plot(t=L1[i], color=L2[i]):
end do:
display({seq(pn[i], i=1..5)}, title=`Numerical solution at t=0.01, 0.1, 1, 5, 10`);

****************************

 

the code is working perfectly. But, My question is how can I found the diffusion constant (D) if I have the solution ( C(x,t) ).  Probably it should be an algorithm which use least square method to find (D) based on the data C(x,t).

I am looking for a fast and efficient algorithm if there is any.

thank you so much for your kind attentions in advance

Sincerely yours,

Amir

This ODE turns out to be a simple separable ODE. With one solution, if the ODE is rewritten.

But in its original form, Maple dsolve gives 6 complicated looking solutions with complex numbers in some of them. Even though all 6 solutions are valid.

Any one knows why Maple did that and not give the one simple solution instead? 

I used isolate to solve for y' from the original ODE. Verfiied that only one solution exist.  The ODE then became y'(x)= 3*y(x)/(2*x). Which by uniqueness theorem, should have one unique solution in some region in the RHS or in some region in the LHS that does not inculde x=0 ?

Just wondering what is going on, and why Maple did not give same simpler solution, so I can learn something new. That is all.

restart;

Typesetting:-Settings(typesetprime=true):

ode:= 1/2*(2*x^(5/2)-3*y(x)^(5/3))/x^(5/2)/y(x)^(2/3)+1/3*(-2*x^(5/2)+3*y(x)^(5/3))*diff(y(x),x)/x^(3/2)/y(x)^(5/3) = 0;

(1/2)*(2*x^(5/2)-3*y(x)^(5/3))/(x^(5/2)*y(x)^(2/3))+(1/3)*(-2*x^(5/2)+3*y(x)^(5/3))*(diff(y(x), x))/(x^(3/2)*y(x)^(5/3)) = 0

DEtools:-odeadvisor(ode);

[[_1st_order, _with_linear_symmetries], _exact, _rational]

maple_sol:=dsolve(ode);

y(x) = (1/3)*2^(3/5)*3^(2/5)*(x^(5/2))^(3/5), y(x) = (1/3)*(-(1/4)*5^(1/2)-1/4-((1/4)*I)*2^(1/2)*(5-5^(1/2))^(1/2))^3*2^(3/5)*3^(2/5)*(x^(5/2))^(3/5), y(x) = (1/3)*(-(1/4)*5^(1/2)-1/4+((1/4)*I)*2^(1/2)*(5-5^(1/2))^(1/2))^3*2^(3/5)*3^(2/5)*(x^(5/2))^(3/5), y(x) = (1/3)*((1/4)*5^(1/2)-1/4-((1/4)*I)*2^(1/2)*(5+5^(1/2))^(1/2))^3*2^(3/5)*3^(2/5)*(x^(5/2))^(3/5), y(x) = (1/3)*((1/4)*5^(1/2)-1/4+((1/4)*I)*2^(1/2)*(5+5^(1/2))^(1/2))^3*2^(3/5)*3^(2/5)*(x^(5/2))^(3/5), x/y(x)^(2/3)+y(x)/x^(3/2)+_C1 = 0

map(x->odetest(x,ode),[maple_sol])

[0, 0, 0, 0, 0, 0]

solve(ode,diff(y(x),x),AllSolutions)

(3/2)*y(x)/x

ode2:=isolate(ode,diff(y(x),x)); #solve for y' first

diff(y(x), x) = -(3/2)*(2*x^(5/2)-3*y(x)^(5/3))*y(x)/(x*(-2*x^(5/2)+3*y(x)^(5/3)))

ode2:=simplify(ode2)

diff(y(x), x) = (3/2)*y(x)/x

DEtools:-odeadvisor(ode2);

[_separable]

sol:=dsolve(ode2)

y(x) = _C1*x^(3/2)

odetest(sol,ode2)

0

 

Download strange_ode_answer.mw

Maple 2019.1

Physics 395

Hi, I am trying to plot the phase potrait for this as follow:

s0 := 3*10^5;
d := 10^(-3);
delta := 10^4;
b := 5*10^6;
lamda := 4.16;

DEplot([diff(I(t), t) = s0 + I(t)*(-d - delta*Q(t)/(b + Q(t))), diff(Q(t), t) = -lamda*Q(t)], [I(t), Q(t)], t = 0 .. 10, I = 0 .. 100, 0, Q = 0 .. 100, 0, dirfield = 400, arrows = smalltwo, number = 2, [[0, 4, 0.1], [0, 0.2, 4.1], [0, 7, 0.2], [0, 0.2, 7]], color = red, linecolor = blue, numsteps = 100)

 

But, there is an error saying "Error, (in DEtools/DEplot) vars must be declared as a list, e.g. [x(t),y(t),...]". However, I did the same for other problem and worked well tho. I have no idea what the problem is for above.

Dear Maple users

 

I am just curious about how far Maplesoft is updating the Mac version of Maple to 64 bit (Catalina). This version of the Mac OS will hit the shelves in late September this year. As I have been told, no program built on 32 bit will be able to run on this new version of Mac OS. I am pretty sure there will be a lot of software troubles for students upgrading to this version. We can recommend the students not to upgrade immediately, but it would be interesting to hear how far Maplesoft is creating a 64 bit Maple-installer for Mac?

 

Regards,

Erik V.

Please download 1.txt.

Integrand := parse(FileTools[Text][ReadFile]("1.txt")):

int(Integrand, [z = -R .. R, y = 0 .. R], numeric);

plots[implicitplot3d](Phi = phi, z = -R .. R, y = 0 .. R, Phi = 0 .. 0.1e-1, color = ColorTools[Gradient]("Red" .. "Blue", best)[4], grid = [50, 50, 20]);

Why the integrand has positive real amounts in the domain [z = -R .. R, y = 0 .. R] for R=0.5, but the integral value is negative?

Hi guys

 

I've created a joint display which plots a spacecurve and an arrow which represents the tangent to the curve:

I've written this in the animation format so I can move the tangent arrow along the curve, however when I replace

arrow(subs(s=0,p),subs(s=0,ptan),width=0.3,length=4)

with

arrow(subs(s=A,p),subs(s=A,ptan),width=0.3,length=4)

I get the error:

Error, (in Plot:-AnalyzeData:-StandardizeData) points cannot be converted to floating-point values

I have tried animating the arrow alone(see below) and this works fine, so the problem is coming from try to pass the animation parameter A into the display array of multiple plots.

DG.mw

Thanks

Hi

I have trouble with solving this ODE system using dsolve command:

and 

 

This system have following solutions:

where

and

C's and A are constants of integration.

 

They're equations from this paper https://arxiv.org/abs/1710.01910 (45 and 47). 
               

However, my solution differs from correct one - in output there are hypergeometric functions everywhere.

Is there any way to fix/convert this solution? Or to get rid of these functions (my f1 solution looks very close to original one but with coupled hypergeometric function). 
 

`` ``

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``

 

``

sysode := 2*q*(3*q-1)*f1(tau)/tau^2+2*q*(diff(f1(tau), tau))/tau+diff(f1(tau), tau, tau)+(kappa^2+f2(tau))*(1+omega)*(tau/t0)^(-(3*(3+omega))*q) = 0, (54*q^3-30*q^2+4*q)*f1(tau)/tau^3+(24*q^2-4*q)*(diff(f1(tau), tau))/tau^2+11*q*(diff(f1(tau), tau, tau))/tau+diff(f1(tau), tau, tau, tau)-3*omega*(1+omega)*(kappa^2+f2(tau))*q*(tau/t0)^(-(3*(1+omega))*q)/tau = 0;

2*q*(3*q-1)*f1(tau)/tau^2+2*q*(diff(f1(tau), tau))/tau+diff(diff(f1(tau), tau), tau)+(kappa^2+f2(tau))*(1+omega)*(tau/t0)^(-3*(3+omega)*q) = 0, (54*q^3-30*q^2+4*q)*f1(tau)/tau^3+(24*q^2-4*q)*(diff(f1(tau), tau))/tau^2+11*q*(diff(diff(f1(tau), tau), tau))/tau+diff(diff(diff(f1(tau), tau), tau), tau)-3*omega*(1+omega)*(kappa^2+f2(tau))*q*(tau/t0)^(-3*(1+omega)*q)/tau = 0

(1)

``

``

simplify(dsolve([sysode], build));

{f1(tau) = _C1*tau^(-q+1/2-(1/2)*(-20*q^2+4*q+1)^(1/2))+_C2*tau^(-q+1/2+(1/2)*(-20*q^2+4*q+1)^(1/2))+_C3*tau^(-9*q+2)*hypergeom([-(1/12)*(16*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(-16*q+(-20*q^2+4*q+1)^(1/2)+3)/q], [-(1/12)*(4*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(-4*q+(-20*q^2+4*q+1)^(1/2)+3)/q], -(1/2)*(tau/t0)^(6*q)*omega), f2(tau) = (-695520*(q^2+(11/21)*q+2/21)*(tau/t0)^(3*q*(omega+5))*_C3*(q-3/10)*omega*q*(q^2-(25/69)*q+2/69)*tau^(-9*q)*hypergeom([-(1/12)*(4*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(-4*q+(-20*q^2+4*q+1)^(1/2)+3)/q], [-(1/12)*(-8*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(8*q+(-20*q^2+4*q+1)^(1/2)+3)/q], -(1/2)*(tau/t0)^(6*q)*omega)-89424*(q^2*(tau/t0)^(3*q*(omega+7))*omega^2*tau^(-9*q)*_C3*(q^2-(25/69)*q+2/69)*hypergeom([-(1/12)*(-8*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(8*q+(-20*q^2+4*q+1)^(1/2)+3)/q], [-(1/12)*(-20*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(20*q+(-20*q^2+4*q+1)^(1/2)+3)/q], -(1/2)*(tau/t0)^(6*q)*omega)+(7/3)*(q^2+(11/21)*q+2/21)*(hypergeom([-(1/12)*(16*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(-16*q+(-20*q^2+4*q+1)^(1/2)+3)/q], [-(1/12)*(4*q+(-20*q^2+4*q+1)^(1/2)-3)/q, (1/12)*(-4*q+(-20*q^2+4*q+1)^(1/2)+3)/q], -(1/2)*(tau/t0)^(6*q)*omega)*_C3*(tau/t0)^(3*(3+omega)*q)*(q^2-(25/69)*q+2/69)*tau^(-9*q)+(1/69)*kappa^2*(1+omega)))*(q^2-(7/9)*q+2/9))/((1+omega)*(4*q-(-20*q^2+4*q+1)^(1/2)-3)*(8*q-(-20*q^2+4*q+1)^(1/2)+3)*(8*q+(-20*q^2+4*q+1)^(1/2)+3)*(4*q+(-20*q^2+4*q+1)^(1/2)-3))}

(2)

NULL

NULL

``

NULLNULL

NULL

NULL

NULL

NULL

``


 

Download question.mw

When I apply the uses function with the Physics package in a procedure, the commands in this package are not restricted to the inside of the procedure, but are applied globally. See the example below:

gds := proc(LL, qi, t)

 local ta,i;  

uses Physics;

ta := sec(diff(diff(LL, diff(qi[i](t), t)), t), i = 1 .. nops(qi));

RETURN(ta) end:

sxy := diff(x(t), t)^2 + diff(y(t), t)^2:

gds(sxy, [x, y], t);

Error, (in Physics:-diff) name expected for external function
 

On the other hand, when I apply the uses function with the LinearAlgebra package in a procedure, the commands in this package are restricted to the inside of the procedure only.
dst:=proc(MM) 

local DA; 

uses LinearAlgebra;

DA:=Determinant(MM); 

RETURN(DA) end:

dst(<<1 | 2>, <3 | 4>>);

                  -2

Determinant(<<1 | 2>, <3 | 4>>);

                         Determinant(Matrix(2, 2, [[1, 2], [3, 4]]))

This could be a bug in Maple 2019?

Hello, I am wondering if Maple is capable of generating a subgraph for a directed, weighed graph with the GraphTheory package. The online resources I can find only include undirected, unweighed graphs. 

can you please include an example with commands that is able to perform the said task?

hi everyone:

how can I solve this below equation?

w:=(x)->sin(lambda*x)+b*cos(lambda*x)-sinh(lambda*x)-b*cosh(lambda*x);

b := -(sin(lambda*L)+sinh*lambda*L)/(cos(lambda*L)+cosh*lambda*L);

L:=10;

equation:=int(w(x)^2,x=0..L)=1;

lambda=????

tnx..

Hi everybody?

how can I solve this PDE with Runge-Kutta method and 2D plot in terms of w(x,t) , t and 3D plot in terms of t, x, w(x,t)?

code1.mw

how this integral can be calculated in the simplest form ? the second question is what exactly is done when using assuming? for example when using assuming real, all the functions or parameteres are affected? thanks in advanced

 

restart:with(IntegrationTools):

(int(int(exp(-(y-beta[0]-beta[1]*x-b0-b1*x)^2/(2*sigma^2))*exp(-b0)*exp(-b1),b0=0..infinity),b1=0..infinity,continuous=true) assuming real)

signum(limit(-(1/2)*Pi^(1/2)*exp((1/2)*(2*b1+2*beta[1])*x+(1/2)*sigma^2-y-b1+beta[0])*2^(1/2)*sigma*(erf((1/2)*(sigma^2+(b1+beta[1])*x-y+beta[0])*2^(1/2)/sigma)-signum(sigma)), b1 = infinity))*infinity

(1)

 


 

Download error_function.mw

I know you can call python from Maple, I am thinking if there is the other way around. That is use Maple (and its toolbox) as backend engine to do calculations (e.g. Global Optimization), and say manipulate the data in Python as the front-end.

This is an ode from text book.  The little tricky part on this is the right hand has abs on the dependent variable, otherwise it is trivial.

restart;
ode:=diff(y(x),x) = abs(y(x))+1;
sol:=dsolve(ode,y(x))

Gives

I am not able to get odetest to give zero on either of the two solutions.  

odetest(sol[1],ode);
odetest(sol[2],ode);

None give zero. I tried assumptions on x>0, x<0 and tried simplify(...,symbolic), nothing gives zero.

Now the book gives the solutions without constant of integration, which is strange. This is what the book gives as solution

                 y(x) = exp(x)-1   x>=0
                 y(x) = 1-exp(-x) x<0

Which is the same as Maple's solution, but without the constant of integration! So when I removed _C1 from both solutions and added the assumptions on x, then I got zero

odetest(subs(_C1=1,sol[1]),ode) assuming x<0;
                      0

odetest(subs(_C1=1,sol[2]),ode) assuming x>=0
                      0

I solved this by hand, and got same solution as Maple actually (may be I made the same mistake as Maple? :) 

But the above shows these solution are not correct? I do not now know what happend to the constant of integration in the book solution since it only shows final solution. It looks like book just used C=1 for the constant of integration. But Maple also thinks the book solution is correct.

fyi, the implicit solution by Maple does verify with no problem:

ode:=diff(y(x),x)=abs(y(x))+1;
sol:=dsolve(ode,y(x),'implicit');
odetest(sol,ode)

             0

Any one can shed some light what is going on? Is Maple solution correct? If so, why it does not verify? Should Maple have given the book solution?

this is problem 9, page 15, "Elementary differential equations" by William F. Trench, 2001

Maple 2019.1

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