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i am trying to solve 6 ODE with boundary condition


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with*plots

with*plots

(1)

Eq1 := (1-theta(eta)/theta[r])*(diff(f(eta), eta, eta, eta))+(diff(f(eta), eta, eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(1-theta(eta)/theta[r])*(diff(diff(diff(f(eta), eta), eta), eta))+(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(2)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(3)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(4)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(5)

Eq5 := (1+s*theta(eta))*(diff(theta(eta), eta, eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(1+s*theta(eta))*(diff(diff(theta(eta), eta), eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(6)

Eq6 := 2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

(7)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0;

f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

(8)

fixedparameter := [M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1];

[M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1]

(9)

Eq7 := eval(Eq1, fixedparameter);

(1+(1/10)*theta(eta))*(diff(diff(diff(f(eta), eta), eta), eta))-(1/10)*(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))+(1+(1/10)*theta(eta))^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-.5*(diff(f(eta), eta))+.5*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(10)

Eq8 := eval(Eq2, fixedparameter);

G(eta)*(diff(F(eta), eta))+F(eta)^2+.5*F(eta)-.5*(diff(f(eta), eta)) = 0

(11)

Eq9 := eval(Eq3, fixedparameter);

G(eta)*(diff(G(eta), eta))+.5*f(eta)+.5*G(eta) = 0

(12)

Eq10 := eval(Eq5, fixedparameter);

(1+.1*theta(eta))*(diff(diff(theta(eta), eta), eta))+.1*(diff(theta(eta), eta))^2+f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)+.3333333333*H(eta)*(theta[p](eta)-theta(eta)) = 0

(13)

Eq11 := eval(Eq6, fixedparameter);

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+.5*theta[p](eta)-.5*theta(eta) = 0

(14)

bcs2 := F(10) = 0;

F(10) = 0

(15)

bcs3 := G(10) = -f(10);

G(10) = -f(10)

(16)

bcs4 := H(10) = n;

H(10) = n

(17)

bcs5 := theta(10) = 0;

theta(10) = 0

(18)

bcs6 := theta[p](10) = 0;

theta[p](10) = 0

(19)

L := [.2];

[.2]

(20)

for k to 1 do R := dsolve(eval({Eq10, Eq11, Eq4, Eq7, Eq8, Eq9, bcs1, bcs2, bcs3, bcs4, bcs5, bcs6}, n = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta[p](eta)], numeric, output = listprocedure); Y || k := rhs(R[5]); YP || k := rhs(R[6]); YJ || k := rhs(R[7]); YS || k := rhs(R[2]) end do

``


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with*plots

with*plots

(1)

Eq1 := (1-theta(eta)/theta[r])*(diff(f(eta), eta, eta, eta))+(diff(f(eta), eta, eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(1-theta(eta)/theta[r])*(diff(diff(diff(f(eta), eta), eta), eta))+(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(2)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(3)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(4)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(5)

Eq5 := (1+s*theta(eta))*(diff(theta(eta), eta, eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(1+s*theta(eta))*(diff(diff(theta(eta), eta), eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(6)

Eq6 := 2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

(7)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0;

f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

(8)

fixedparameter := [M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1];

[M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1]

(9)

Eq7 := eval(Eq1, fixedparameter);

(1+(1/10)*theta(eta))*(diff(diff(diff(f(eta), eta), eta), eta))-(1/10)*(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))+(1+(1/10)*theta(eta))^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-.5*(diff(f(eta), eta))+.5*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(10)

Eq8 := eval(Eq2, fixedparameter);

G(eta)*(diff(F(eta), eta))+F(eta)^2+.5*F(eta)-.5*(diff(f(eta), eta)) = 0

(11)

Eq9 := eval(Eq3, fixedparameter);

G(eta)*(diff(G(eta), eta))+.5*f(eta)+.5*G(eta) = 0

(12)

Eq10 := eval(Eq5, fixedparameter);

(1+.1*theta(eta))*(diff(diff(theta(eta), eta), eta))+.1*(diff(theta(eta), eta))^2+f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)+.3333333333*H(eta)*(theta[p](eta)-theta(eta)) = 0

(13)

Eq11 := eval(Eq6, fixedparameter);

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+.5*theta[p](eta)-.5*theta(eta) = 0

(14)

bcs2 := F(10) = 0;

F(10) = 0

(15)

bcs3 := G(10) = -f(10);

G(10) = -f(10)

(16)

bcs4 := H(10) = n;

H(10) = n

(17)

bcs5 := theta(10) = 0;

theta(10) = 0

(18)

bcs6 := theta[p](10) = 0;

theta[p](10) = 0

(19)

L := [.2];

[.2]

(20)

for k to 1 do R := dsolve(eval({Eq10, Eq11, Eq4, Eq7, Eq8, Eq9, bcs1, bcs2, bcs3, bcs4, bcs5, bcs6}, n = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta[p](eta)], numeric, output = listprocedure); Y || k := rhs(R[5]); YP || k := rhs(R[6]); YJ || k := rhs(R[7]); YS || k := rhs(R[2]) end do

``


then i get this error

Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

i dont know where i need to change after view it one by one..

Download hydro.mw

Dear Mapleprimes,

I have been struggling with a problem in the last couple of days. I wish to export a Maple plot to LaTeX while ensuring font consistency. While searching for solutions online, I found the psfrag package in LaTeX. So far, however, I have been unsuccesful in making this work. As as test, I attempted to export plot(x^2) to LaTeX. I used the following code to convert to .eps which worked fine:

plotsetup(ps, plotoutput = `plot1.eps`, plotoptions = `portrait, noborder,height=5in,width=5in`);plot(x^2);

Then in LaTeX, I have:

\documentclass{article}

\usepackage{graphicx}

\usepackage{psfrag}

\begin{document}

\begin{figure}[!h]
\centering
\psfrag{x}{$ \alpha $}
\includegraphics[scale=0.5]{plot1.eps}
\end{figure}
\end{document}

However, no replacements are made. After intense Google searching I found the following post http://www.mapleprimes.com/posts/43255-Trouble-Replacing-Maple-Axes-Labels which to sum up argues that this was only possible with earlier versions of Maple.

Does anyone know if the problem has been resolved?

Does anyone know any other ways to ensure font consistency for plots imported from Maple to LaTeX?

Thank you very much in advance!

C

Hi all,

 

Say I have some list like this,

tmp:=[[0, 0, 1], [0, 1, 0], [0, 1, 1], [0, 1, 2], [1, 0, 0], [1, 0, 1], [1, 0, 2], [1, 1, 0], [1, 1, 1], [1, 1, 2], [1, 2, 0]];

 

And I have worked out some probabilities for each of them, a,b,c,d, ect.

I want to print them like this

Pr( 001 ) = 1

Pr( 010 ) = 1-phi[2]+phi[2]*(1-p[3])*(1-phi[3])

Pr( 011 ) = phi[2]*p[3]*(1-phi[3])

and so on.

I there a way to do that?

The probabilities can be extracted from a Vector. I have no problem to print them.

I dont know how to convert the 0,1,2 into the desired format as shown above.

 

This is the best I can do.

 

Also, is it possible to convert all the subscripte [] to _ when printing the output?

and get ride of all * as well.

Thanks,

 

casperyc

 

Dear all;

Please Have some one an idea to transform or convert 2nd order ODE to system of First ODE ( of course using maple).

Thanks

 

how to convert decimal to fraction without simplify

for example

convert(0.25, fraction)

expect 25/100, but not 1/4

Hi,

I have a problem with dsolve in the following code

restart;
>
n:=20;
m:=1;
cc:=-200;
zzeta:=0.1;
sefr1:=0.3;
sefr:=0.2;
MM:=0;
lambda:=0.1;
Br:=1;
nn:=3;
>
>
#u(tau):=tau;
u(tau):=421.7129935*tau-2217.587728*tau^2+8897.376593*tau^3-27612.59182*tau^4+64248.00336*tau^5-1.083977605*10^5*tau^6-10.57029600-1.080951714*10^6*tau^13+7.999517316*10^5*tau^14-4.788741005*10^5*tau^15+2.309563748*10^5*tau^16+26511.11102*tau^18-5959.001794*tau^19+1.148523882*10^5*tau^7-95.23809524*tau^21+4.545454545*tau^22-9435.563781*tau^8-2.587683745*10^5*tau^9+6.473880128*10^5*tau^10+948.0272727*tau^20-88660.41892*tau^17-1.008692404*10^6*tau^11+1.175504242*10^6*tau^12;
>
>
B := 1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2*(1-zzeta)));
eq4 := 4*B*u(tau)-(1+zzeta)*(diff(tau*(diff(theta(tau), tau)), tau))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2;


theta(tau):=sum(p^ii*theta[ii](tau),ii=0..nn);
HH:= p*((4*(1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2-(1/2)*zzeta))))*u(tau)-(1+zzeta)*(diff(theta(tau), tau)+tau*(diff(theta(tau), tau, tau)))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2)+(1-p)*(diff(theta(tau),tau$2)):
eq5:=simplify(HH):
eq6:=collect(expand(eq5),p);

eq7:=
convert(series(collect(expand(eq5), p), p, nn+1), 'polynom');


for ii to nn do
ss[ii] := (coeff(eq7, p^ii)) ;
print (ii);
end do;

ss[0]:=diff(theta[0](tau), tau, tau);

icss[0]:=theta[0](zzeta/(2*(1-zzeta)))=0, D(theta[0])(1/(2*(1-zzeta)))=1;

dsolve({ss[0], icss[0]});
theta[0](tau):= rhs(%);


for ii to nn do
ss[ii]:=evalf[5](ss[ii]);
icss[ii]:=theta[ii](zzeta/(2*(1-zzeta)))=0, D(theta[ii])(1/(2*(1-zzeta)))=0;
dsolve({ss[ii], icss[ii]});
theta[ii](tau):=rhs(%);
end do;

I would be most grateful if you help me to find this problem.

Thanks for your attention in advance

 

Dear all

I would like to convert Matlab code to Maple, is there anu idea, this is the code.

 

% Usage: [y t] = abm4(f,a,b,ya,n) or y = abm4(f,a,b,ya,n)
% Adams-Bashforth-Moulton 4-th order predictor-corrector method for initial value problems
% It uses
% Adams-Bashforth 4-step method as a precdictor,
% Adams-Moulton 3-step method as a corrector, and
% Runge-Kutta method of order 4 as a starter
%
% Input:
% f - Matlab inline function f(t,y)
% a,b - interval
% ya - initial condition
% n - number of subintervals (panels)
%
% Output:
% y - computed solution
% t - time steps
%
% Examples:
% [y t]=abm4(@myfunc,0,1,1,10);          here 'myfunc' is a user-defined function in M-file
% y=abm4(inline('sin(y*t)','t','y'),0,1,1,10);
% f=inline('sin(y(1))-cos(y(2))','t','y');
% y=abm4(f,0,1,1,10);

function [y t] = abm4(f,a,b,ya,n)
h = (b - a) / n;
h24 = h / 24;

y(1,:) = ya;
t(1) = a;

m = min(3,n);

for i = 1 : m % start-up phase, using Runge-Kutta of order 4
    t(i+1) = t(i) + h;
    s(i,:) = f(t(i), y(i,:));
    s2 = f(t(i) + h / 2, y(i,:) + s(i,:) * h /2);
    s3 = f(t(i) + h / 2, y(i,:) + s2 * h /2);
    s4 = f(t(i+1), y(i,:) + s3 * h);
    y(i+1,:) = y(i,:) + (s(i,:) + s2+s2 + s3+s3 + s4) * h / 6;
end;

for i = m + 1 : n % main phase
    s(i,:) = f(t(i), y(i,:));
    y(i+1,:) = y(i,:) + (55 * s(i,:) - 59 * s(i-1,:) + 37 * s(i-2,:) - 9 * s(i-3,:)) * h24; % predictor
    t(i+1) = t(i) + h;
    y(i+1,:) = y(i,:) + (9 * f(t(i+1), y(i+1,:)) + 19 * s(i,:) - 5 * s(i-1,:) + s(i-2,:)) * h24; % corrector
end;

number10:=`466d06ece998b7a2fb1d464fed2ced7641ddaa3cc31c9941cf110abbf409ed39598005b3399ccfafb61d0315fca0a314be138a9f32503bedac8067f03adbf3575c3b8edc9ba7f537530541ab0f9f3cd04ff50d66f1d559ba520e89a2cb2a83`:

number8:=`315c4eeaa8b5f8bffd11155ea506b56041c6a00c8a08854dd21a4bbde54ce56801d943ba708b8a3574f40c00fff9e00fa1439fd0654327a3bfc860b92f89ee04132ecb9298f5fd2d5e4b45e40ecc3b9d59e9417df7c

I first define

f:=x->convert(x, decimal, hex):

with(Bits):
str1:=convert( `Xor(f(number8), f(number10))`, bytes);

now how can I get back the alphabets, since again use of convert with bytes return the inital argument.

Moreover, I would really appreciate if someone could explain the difference between 

convert(`expr`, bytes)

convert( [expr], bytes)

 

Many regards!!

 

As the title, how to convert "sin(x)+cos(x)" and this kinds into "sqrt(2)*sin(x+pi/4)"?

Sum of two sine functions with the same cycle should can be converted into one sine function, with some amplitude gain and phase offset. 

 

> assume(a < 0);
> convert(cosh(sqrt(a)), sincos);
print(`output redirected...`); # input placeholder
/ (1/2)\
cos\(-a) /

This is what I expected.

Now

> assume(L > 0);
> assume(K > 0);
> assume(mu > 0);
> assume(mu^2 < 4*L*k);
> assume(t > 0);
> convert(cosh((1/2)*t*sqrt(mu^2-4*L*k)/L), sincos);
print(`output redirected...`); # input placeholder
/ (1/2)\
| / 2 \ |
|t \mu - 4 L k/ |
cosh|--------------------|
\ 2 L /

I wanted to obtain again the cos function. Could someone help me?
(What is the reason that convert does not work "well" in later case?)

 Thanks,  Sandor

 

Is it possible that this expression has an elementary one (specifically the dilog's):

Y0:=(1/16)*(s*t*(exp(2*t)*s+exp(4*t)+1)*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^16*(1+(-s^2+1)^(1/2))^16/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^16*(1-(-s^2+1)^(1/2))^16))+s^3*t*(exp(4*t)+1)*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^8*(1+(-s^2+1)^(1/2))^8/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^8*(1-(-s^2+1)^(1/2))^8))+exp(2*t)*t*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^32*(1+(-s^2+1)^(1/2))^32/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^32*(1-(-s^2+1)^(1/2))^32))+4*((exp(4*t)+1)*s+2*exp(2*t))*(s^2+2)*dilog((-exp(2*t)*s+(-s^2+1)^(1/2)-1)/(-1+(-s^2+1)^(1/2)))-4*((exp(4*t)+1)*s+2*exp(2*t))*(s^2+2)*dilog((exp(2*t)*s+(-s^2+1)^(1/2)+1)/(1+(-s^2+1)^(1/2)))+((32*s^2*t+64*t)*exp(2*t)+16*(((t+1/8)*s^2+2*t+2)*exp(4*t)-(5/4)*s*exp(-2*t)-(1/8)*exp(-4*t)*s^2+(5/4)*s*exp(6*t)+(1/8)*s^2*exp(8*t)+(t-1/8)*s^2-2+2*t)*s)*arctanh((exp(2*t)-1)*(-1+s)/((-s^2+1)^(1/2)*(exp(2*t)+1)))+8*(-s^2+1)^(1/2)*((1/8)*s*(exp(4*t)+1)*ln((exp(4*t)*s+2*exp(2*t)+s)^12/s^12)+(1/8)*exp(2*t)*ln((exp(4*t)*s+2*exp(2*t)+s)^24/s^24)+(s^2-6*t-3)*exp(2*t)+((-(1/8)*s^2-3*t)*exp(4*t)+s*exp(-2*t)+(1/8)*exp(-4*t)*s^2+s*exp(6*t)+(1/8)*s^2*exp(8*t)-(1/8)*s^2-3*t)*s))/((s*exp(-2*t)+exp(2*t)*s+2)*(exp(4*t)*s+2*exp(2*t)+s)*((-s^2+1)^(1/2)+2*arctanh((-1+s)/(-s^2+1)^(1/2))))

Also I'm wondering since Y0 should solve the ode

-(diff(diff(y(t), t), t))+(4-12/(1+s*cosh(2*t))+8*(-s^2+1)/(1+s*cosh(2*t))^2)*y(t) = C/(1+s*cosh(2*t))

with some constant C but I only get rubbish.

I ask this because I found that in another context this seems to be correct:

f1:=-(1/12)*Pi^2*((-s^2+1)^(1/2)-arccosh(1/s))/(-s^2+1)^(3/2)+(1/12)*arccosh(1/s)^3/(-s^2+1)^(3/2)-(1/4)*arccosh(1/s)^2/(-s^2+1)

f2:=(1/2)*((-s^2+1)^(1/2)*(polylog(2, s/(-1+(-s^2+1)^(1/2)))+polylog(2, -s/(1+(-s^2+1)^(1/2))))-polylog(3, s/(-1+(-s^2+1)^(1/2)))+polylog(3, -s/(1+(-s^2+1)^(1/2))))/(-s^2+1)^(3/2)

and f1=f2

but maple doesnt convert it.

Also maple has trouble to convert

2*arctanh(sqrt((1-s)/(1+s)))=arccosh(1/s)

everywhere: 0<s<1

hi,

     there is a common  differential equation in my maple note,the solution of the eq. can be expressed by

associated Legendre function(s),but i get a result by hypergeometric representation.how i can translate the later into a  single Legendre fun?

 Thank you in advance  

ode := 'sin(theta)*(diff(sin(theta)*(diff(Theta(theta), theta)), theta))'/Theta(theta)+l*(l+1)*sin(theta)^2 = m^2

sin(theta)*(diff(sin(theta)*(diff(Theta(theta), theta)), theta))/Theta(theta)+l*(l+1)*sin(theta)^2 = m^2

(1)

dsolve(ode)

Theta(theta) = _C1*((1/2)*cos(2*theta)-1/2)^((1/2)*m)*sin(2*theta)*hypergeom([(1/2)*m+(1/2)*l+1, (1/2)*m-(1/2)*l+1/2], [3/2], (1/2)*cos(2*theta)+1/2)/(1-cos(2*theta))^(1/2)+_C2*hypergeom([(1/2)*m-(1/2)*l, (1/2)*m+(1/2)*l+1/2], [1/2], (1/2)*cos(2*theta)+1/2)*(-2*cos(2*theta)+2)^(1/2)*((1/2)*cos(2*theta)-1/2)^((1/2)*m)/(1-cos(2*theta))^(1/2)

(2)

`assuming`([simplify(dsolve(ode))], [l::posint, m::integer, l >= m])

Theta(theta) = ((1/2)*cos(2*theta)-1/2)^((1/2)*m)*(sin(2*theta)*hypergeom([(1/2)*m+(1/2)*l+1, (1/2)*m-(1/2)*l+1/2], [3/2], (1/2)*cos(2*theta)+1/2)*_C1+2^(1/2)*(1-cos(2*theta))^(1/2)*hypergeom([(1/2)*m-(1/2)*l, (1/2)*m+(1/2)*l+1/2], [1/2], (1/2)*cos(2*theta)+1/2)*_C2)/(1-cos(2*theta))^(1/2)

(3)

convert(Theta(theta) = _C1*((1/2)*cos(2*theta)-1/2)^((1/2)*m)*sin(2*theta)*hypergeom([(1/2)*m+(1/2)*l+1, (1/2)*m-(1/2)*l+1/2], [3/2], (1/2)*cos(2*theta)+1/2)/(1-cos(2*theta))^(1/2)+_C2*hypergeom([(1/2)*m-(1/2)*l, (1/2)*m+(1/2)*l+1/2], [1/2], (1/2)*cos(2*theta)+1/2)*(-2*cos(2*theta)+2)^(1/2)*((1/2)*cos(2*theta)-1/2)^((1/2)*m)/(1-cos(2*theta))^(1/2), `2F1`)

Theta(theta) = (1/2)*_C1*((1/2)*cos(2*theta)-1/2)^((1/2)*m)*sin(2*theta)*Pi^(1/2)*GAMMA(-(1/2)*m-(1/2)*l)*JacobiP(-(1/2)*m-(1/2)*l-1, 1/2, m, -cos(2*theta))/((1-cos(2*theta))^(1/2)*GAMMA(1/2-(1/2)*m-(1/2)*l))+_C2*Pi^(1/2)*GAMMA(1-(1/2)*m+(1/2)*l)*JacobiP(-(1/2)*m+(1/2)*l, -1/2, m, -cos(2*theta))*(-2*cos(2*theta)+2)^(1/2)*((1/2)*cos(2*theta)-1/2)^((1/2)*m)/((1-cos(2*theta))^(1/2)*GAMMA(-(1/2)*m+(1/2)*l+1/2))

(4)

``

 

Download question_12.19.mw

 

It would be nice if Maple had a procedure which could turn a procedurelist into a listprocedure.
I use these words in the sense they are used in dsolve/numeric.
Thus by a procedurelist I mean a procedure returning lists (of numbers).
By a listprocedure I mean a list of procedures all having the same formal parameters and each returning one number.
Thus a `convert/listprocedure`should accept a procedurelist p as input and give as output the corresponding listprocedure [p1,p2, ... pN] where N is the number of elements in the output from p.

I tried making a `convert/listprocedure` myself and in doing so found that it was not totally trivial.
I had lots of problems but did end up with something that seems to work.

convert-listprocedur.mw

But my main point is that Maple ought to have some such facility either as described available to the user or by changing fsolve, complexplot or what have you, so that procedurelists are accepted.

Hello.

I am novice here and I have an question about export animations frames into sequence of postscript files. I don't googled nothing about this theme.
Here is my question:

Is there an option to convert the animated picture into  sequence of separate images and then (eg using the program cycle) save this images as separate PostScript images? I know the possibility of exporting to GIF file and then converting them into PS file. But I do not have a PS file with a sequence of bitmap images. I would sufficed for me to be able to see a separate frame of animation (through any index).

Thanx Jaroslav Hajtmar

Hi All,

 

I tried to convert the following hypergeometric function into BesselJ function. But I failed to do so. Could any one let me know the  reference or procedure to convert the Hypergeom function into bessel function.

 

Following is the integral I am intended to do.

Result:

I need to convert the result into equivalent bessel function.

 

If at all there is a way to co-relate the generalized corelation between bessel function <-> hypergeom function.

 

Direct me to any books you come across.

 

Thanks

 

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