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nm 11458
September 06 2024
4 4

Update

Here is an updated version wich now gets exact result in book. I used applyrule few places to force specific transformation.  Book uses C and D for constants. I used c__2 and c__3.

 

The idea is to replace G'/G^2  by u(t) and solve the resulting ode then replace u(t) back by G'/G^2

 

> 

restart;
> 

eq_47:= diff( diff(G(zeta),zeta)/G(zeta)^2,zeta)=mu+lambda*(diff(G(zeta),zeta)/G(zeta)^2)^2;
the_sub_1:=diff(G(zeta),zeta)/G(zeta)^2 = u(zeta);
the_sub_2:=dsolve(the_sub_1);
PDEtools:-dchange({the_sub_2},eq_47,{u},params={lambda,mu});
new_ode:=simplify(%);
u_sol:=dsolve(new_ode);
u_sol:=lhs(u_sol)=applyrule(tan(x::anything)=sin(x)/cos(x),rhs(u_sol));
 

(diff(diff(G(zeta), zeta), zeta))/G(zeta)^2-2*(diff(G(zeta), zeta))^2/G(zeta)^3 = mu+lambda*(diff(G(zeta), zeta))^2/G(zeta)^4

(diff(G(zeta), zeta))/G(zeta)^2 = u(zeta)

G(zeta) = 1/(Int(-u(zeta), zeta)+c__1)

(2*u(zeta)^2/(Int(-u(zeta), zeta)+c__1)^3+(diff(u(zeta), zeta))/(Int(-u(zeta), zeta)+c__1)^2)*(Int(-u(zeta), zeta)+c__1)^2-2*u(zeta)^2/(Int(-u(zeta), zeta)+c__1) = mu+lambda*u(zeta)^2

diff(u(zeta), zeta) = mu+lambda*u(zeta)^2

u(zeta) = tan((mu*lambda)^(1/2)*(c__1+zeta))*(mu*lambda)^(1/2)/lambda

u(zeta) = sin((mu*lambda)^(1/2)*(c__1+zeta))*(mu*lambda)^(1/2)/(cos((mu*lambda)^(1/2)*(c__1+zeta))*lambda)

> 

rules:=[ 'sin(x::anything)=sin('expand'(x)), cos(x::anything)=cos('expand'(x))'];
u_sol:=lhs(u_sol)=applyrule(rules,rhs(u_sol));

[sin(x::anything) = sin(('expand')(x)), cos(x::anything) = cos(('expand')(x))]

u(zeta) = sin(c__1*(mu*lambda)^(1/2)+zeta*(mu*lambda)^(1/2))*(mu*lambda)^(1/2)/(cos(c__1*(mu*lambda)^(1/2)+zeta*(mu*lambda)^(1/2))*lambda)

> 

rules:=[ sin(a::anything+b::anything)=sin(a)*cos(b)+sin(b)*cos(a), cos(a::anything+b::anything)=cos(a)*cos(b)-sin(a)*sin(b)]:
u_sol:=lhs(u_sol)=simplify(applyrule(rules,rhs(u_sol)));

u(zeta) = (sin(c__1*(mu*lambda)^(1/2))*cos(zeta*(mu*lambda)^(1/2))+sin(zeta*(mu*lambda)^(1/2))*cos(c__1*(mu*lambda)^(1/2)))*(mu*lambda)^(1/2)/((cos(c__1*(mu*lambda)^(1/2))*cos(zeta*(mu*lambda)^(1/2))-sin(c__1*(mu*lambda)^(1/2))*sin(zeta*(mu*lambda)^(1/2)))*lambda)

> 

u_sol:=eval(u_sol,[sin(c__1*sqrt(mu*lambda))=c__2,cos(c__1*sqrt(mu*lambda))=c__3]);
u_sol:=eval(u_sol,rhs(the_sub_1)=lhs(the_sub_1));
u_sol:=lhs(u_sol)=applyrule(sqrt(mu*lambda)/lambda=sqrt(mu/lambda),rhs(u_sol));

u(zeta) = (c__2*cos(zeta*(mu*lambda)^(1/2))+sin(zeta*(mu*lambda)^(1/2))*c__3)*(mu*lambda)^(1/2)/((c__3*cos(zeta*(mu*lambda)^(1/2))-c__2*sin(zeta*(mu*lambda)^(1/2)))*lambda)

(diff(G(zeta), zeta))/G(zeta)^2 = (c__2*cos(zeta*(mu*lambda)^(1/2))+sin(zeta*(mu*lambda)^(1/2))*c__3)*(mu*lambda)^(1/2)/((c__3*cos(zeta*(mu*lambda)^(1/2))-c__2*sin(zeta*(mu*lambda)^(1/2)))*lambda)

(diff(G(zeta), zeta))/G(zeta)^2 = (mu/lambda)^(1/2)*(c__2*cos(zeta*(mu*lambda)^(1/2))+sin(zeta*(mu*lambda)^(1/2))*c__3)/(c__3*cos(zeta*(mu*lambda)^(1/2))-c__2*sin(zeta*(mu*lambda)^(1/2)))

> 

 


 

Download simplification.mw

 

This is for Eq 49. Same thing but using hyperbolic trig relations

> 

restart;
> 

eq_47:= diff( diff(G(zeta),zeta)/G(zeta)^2,zeta)=mu+lambda*(diff(G(zeta),zeta)/G(zeta)^2)^2;
the_sub_1:=diff(G(zeta),zeta)/G(zeta)^2 = u(zeta);
the_sub_2:=dsolve(the_sub_1);
PDEtools:-dchange({the_sub_2},eq_47,{u},params={lambda,mu});
new_ode:=simplify(%);
u_sol:=dsolve(new_ode);
u_sol:=simplify(u_sol) assuming lambda*mu<0;

(diff(diff(G(zeta), zeta), zeta))/G(zeta)^2-2*(diff(G(zeta), zeta))^2/G(zeta)^3 = mu+lambda*(diff(G(zeta), zeta))^2/G(zeta)^4

(diff(G(zeta), zeta))/G(zeta)^2 = u(zeta)

G(zeta) = 1/(Int(-u(zeta), zeta)+c__1)

(2*u(zeta)^2/(Int(-u(zeta), zeta)+c__1)^3+(diff(u(zeta), zeta))/(Int(-u(zeta), zeta)+c__1)^2)*(Int(-u(zeta), zeta)+c__1)^2-2*u(zeta)^2/(Int(-u(zeta), zeta)+c__1) = mu+lambda*u(zeta)^2

diff(u(zeta), zeta) = mu+lambda*u(zeta)^2

u(zeta) = tan((mu*lambda)^(1/2)*(c__1+zeta))*(mu*lambda)^(1/2)/lambda

u(zeta) = -tanh((-mu*lambda)^(1/2)*(c__1+zeta))*(-mu*lambda)^(1/2)/lambda

> 

u_sol:=lhs(u_sol)=applyrule(tanh(x::anything)=sinh(x)/cosh(x),rhs(u_sol));

 

u(zeta) = -sinh((-mu*lambda)^(1/2)*(c__1+zeta))*(-mu*lambda)^(1/2)/(cosh((-mu*lambda)^(1/2)*(c__1+zeta))*lambda)

> 

rules:=[ 'sinh(x::anything)=sinh('expand'(x)), cosh(x::anything)=cosh('expand'(x))'];
u_sol:=lhs(u_sol)=applyrule(rules,rhs(u_sol));

[sinh(x::anything) = sinh(('expand')(x)), cosh(x::anything) = cosh(('expand')(x))]

u(zeta) = -sinh((-mu*lambda)^(1/2)*c__1+(-mu*lambda)^(1/2)*zeta)*(-mu*lambda)^(1/2)/(cosh((-mu*lambda)^(1/2)*c__1+(-mu*lambda)^(1/2)*zeta)*lambda)

> 

rules:=[ sinh(a::anything+b::anything)=sinh(a)*cosh(b)+cosh(a)*sinh(b), cosh(a::anything+b::anything)=cosh(a)*cosh(b)+sinh(a)*sinh(b)]:
u_sol:=lhs(u_sol)=simplify(applyrule(rules,rhs(u_sol)));

u(zeta) = -(sinh((-mu*lambda)^(1/2)*c__1)*cosh((-mu*lambda)^(1/2)*zeta)+cosh((-mu*lambda)^(1/2)*c__1)*sinh((-mu*lambda)^(1/2)*zeta))*(-mu*lambda)^(1/2)/((cosh((-mu*lambda)^(1/2)*c__1)*cosh((-mu*lambda)^(1/2)*zeta)+sinh((-mu*lambda)^(1/2)*c__1)*sinh((-mu*lambda)^(1/2)*zeta))*lambda)

> 

u_sol:=eval(u_sol,[sinh(sqrt(-mu*lambda)*c__1)=c__2,cosh(c__1*sqrt(-mu*lambda))=c__3]);
u_sol:=eval(u_sol,rhs(the_sub_1)=lhs(the_sub_1));
 

u(zeta) = -(c__2*cosh((-mu*lambda)^(1/2)*zeta)+c__3*sinh((-mu*lambda)^(1/2)*zeta))*(-mu*lambda)^(1/2)/((c__3*cosh((-mu*lambda)^(1/2)*zeta)+c__2*sinh((-mu*lambda)^(1/2)*zeta))*lambda)

(diff(G(zeta), zeta))/G(zeta)^2 = -(c__2*cosh((-mu*lambda)^(1/2)*zeta)+c__3*sinh((-mu*lambda)^(1/2)*zeta))*(-mu*lambda)^(1/2)/((c__3*cosh((-mu*lambda)^(1/2)*zeta)+c__2*sinh((-mu*lambda)^(1/2)*zeta))*lambda)

> 

 


 

Download simplification_rest.mw

 

For eq (50), I get zero in RHS. I do not know how book got what they show there.

 

reply...

nm 11458
September 02 2024
1 5

I do not wish to find \n in the answer

    \n is special. It does not require escaping. So just change all your \\n  in your input to \n and you will not see \n anymore but will get new line instead, which is I assume what you wanted.

by adding extra \  you basically have scaped the next \. You do not want to do this with \n

update

Looks like OP just want to see "X","Y","buff"  each on its own row.

So why not just a Matrix? Forget about using printf for making display.


 

> 

restart;
> 

intersections := proc(P, Q, T)
local R, W, w, t, a, b, sol, buff, v;
local my_collection:=Array(1..0):
my_collection ,= "X","Y","buff";

sol := NULL;

if T = Y then
   W := X;
else
   W := Y;
fi;

R := resultant(P, Q, T);

#print(`Résultant :`); print(R);

w := fsolve(R, W);
t := NULL;

for v in [w] do
    t := t, fsolve(subs(W = v, P), T);
od;

for a in {w} do
    for b in {t} do
        if T = Y then
           buff := abs(subs(X = a, Y = b, P)) + abs(subs(X = a, Y = b, Q));
           my_collection ,= a,b,buff;

           #printf(`X=%a,   Y=%a   --->  %a\\n`, a, b, buff);
           if buff < 1/100000000 then
              sol := sol, [a, b];
           fi;
        else
            buff := abs(subs(X = b, Y = a, P)) + abs(subs(X = b, Y = a, Q));
            my_collection ,= a,b,buff;
            #printf(`X=%a,   Y=%a   --->  %a\\n `, a, b, buff);
            if buff < 1/100000000 then
               sol := sol, [b, a];
            fi;
        fi;
    od;
od;

#printf(`Nombre de solutions :  %a\\n`, nops({sol}));
#print({sol});

my_collection:=convert(my_collection,list);
my_collection:=Matrix( (nops(my_collection)/3,3),my_collection);
RETURN(my_collection);

end proc:
> 

intersections(X^2 + Y^2 - 1, X - Y, X);

Matrix(5, 3, {(1, 1) = "X", (1, 2) = "Y", (1, 3) = "buff", (2, 1) = -.7071067812, (2, 2) = -.7071067812, (2, 3) = 0., (3, 1) = -.7071067812, (3, 2) = .7071067812, (3, 3) = 1.414213562, (4, 1) = .7071067812, (4, 2) = -.7071067812, (4, 3) = 1.414213562, (5, 1) = .7071067812, (5, 2) = .7071067812, (5, 3) = 0.})

> 

 


 

Download display_stuff.mw

 

reply...

nm 11458
August 28 2024
2 7

I do not see how the book solution your wrote can be the same as Maple's solution. 

In Maple solution, there is "n" in the denominator inside the sum. In your book solution you have removed this "n".

Now the "n" in Maple solution does cancel with one of the terms in the sum in the numerator, the one with cos(nt)*n but the second term in the numerator has no n in it to cancel. It only has sin(n*t) so you can't just cancel the n .

So your book solution is either wrong or you copied it wrong.

You can see below Maple solution and the book solution give different answer.  You can change infinity to as many terms your want and pick arbitrary x and t values and arbitray functions f(x) and g(x).

Btw, why do you not also insert content when you post your worksheet, instead of just adding link? It is small. 
 

> 

 

> 

 

> 

restart;
> 

 

> 

# Definieer de parameters
L := Pi:
c := 1: # Je kunt c aanpassen aan je specifieke probleem

# Stel de golfvergelijking op
pde := diff(u(x,t),t,t) = c^2 * diff(u(x,t),x,x):

# Definieer de randvoorwaarden
bc := u(0,t) = 0, u(L,t) = 0:

# Definieer de beginvoorwaarden
ic := u(x,0) = f(x), D[2](u)(x,0) = g(x):

# Los de PDE op met begin- en randvoorwaarden
sol := pdsolve([pde, bc, ic], u(x,t)):

# Toon de oplossing
simplify(sol);

u(x, t) = 2*(Sum(sin(n*x)*((Int(sin(n*x)*f(x), x = 0 .. Pi))*cos(n*t)*n+(Int(sin(n*x)*g(x), x = 0 .. Pi))*sin(n*t))/n, n = 1 .. infinity))/Pi

> 

eval(rhs(sol),infinity=5);
simplify(value(%)):
value(eval(%,[t=2,f(x)=x^3,g(x)=x^2])):
evalf(eval(%,x=1));

Sum(2*sin(n*x)*((Int(sin(n*x)*f(x), x = 0 .. Pi))*cos(n*t)*n+(Int(sin(n*x)*g(x), x = 0 .. Pi))*sin(n*t))/(Pi*n), n = 1 .. 5)

10.00905987

> 

book_sol:=2*Sum(sin(n*x)*(Int(sin(n*x)*f(x), x = 0 .. Pi)*cos(n*t) + Int(sin(n*x)*g(x), x = 0 .. Pi)*sin(n*t)), n = 1 .. infinity)/Pi:
eval(book_sol,infinity=5);
simplify(value(%)):
value(eval(%,[t=2,f(x)=x^3,g(x)=x^2])):
evalf(eval(%,x=1));

2*(Sum(sin(n*x)*((Int(sin(n*x)*f(x), x = 0 .. Pi))*cos(n*t)+(Int(sin(n*x)*g(x), x = 0 .. Pi))*sin(n*t)), n = 1 .. 5))/Pi

12.43548172

> 

 


 

Download not_same.mw

reply...

nm 11458
August 27 2024
2 2

Use StringTools and make sure y is "y"

fx:=3*x^2-9*y;
str:=String(fx);
hasy:=StringTools:-Has(str,"y")

   true

reply...

nm 11458
August 25 2024
3 1

The `->` confuses the translator. It is a `Rule` in Mathematica.  For correct translation use what Mathematica uses in its internal FullForm of the expression instead of the displayed Form. i.e replace `->` by `Rule` which is what it is.

So your code now works and becomes

> 

restart
> 

MmaTranslator:-FromMma("Subscript[x, 0] -> 2 (y - z)^2"); # should return: x[0] = 2*(y - z)^2

x[0] = (2*y-2*z)^2

> 

MmaTranslator:-FromMma("Rule[ Subscript[x, 0],2 (y - z)^2]"); # should return: x[0] = 2*(y - z)^2

x[0] = 2*(y-z)^2

> 

MmaTranslator:-FromMma("a -> 2*(y - z)^2");

a = (2*y-2*z)^2

> 

MmaTranslator:-FromMma("Rule[a,2*(y - z)^2]");

a = 2*(y-z)^2

> 

 


 

Download use_rule.mw

On a side note, I would avoid translating code with Subscript to Maple. Maple translates Subscript[x, 0] to indexed variable. Which on the screen looks like x__0 in Maple. But it is not the same. This can cause errors in your translated code if you mix it with your own Maple code which uses x__0 and not x[0].

i.e. Subscript[x, 0]  is translated to x[0] in Maple and not to x__0, and these are different types in Maple.

Here is example

> 

restart;
> 

e:=MmaTranslator:-FromMma(`Subscript[x, 0]`);

x[0]

> 

whattype(e);

indexed

> 

lprint(e);

x[0]

> 

e:=x__0;

x__0

> 

whattype(e);

symbol

> 

lprint(e);

x__0


btw, I never ever use Subscripted variables in Mathematica. This has nothing to do with translation to Maple, but these cause their own share of problems in Mathematica itself. stackexchange Mathematica forum has dozens of questions related to using Subscripted variables in Mathematica and almost eveyone says to avoid them in Mathematica.

 

Download use_rule_V2.mw

 

reply...

nm 11458
August 18 2024
2 7 
plots:-polarplot(4, coordinateview = [0 .. 6,0 .. 2*Pi], thickness = 3,tickmarks=[default, 2*Pi/(Pi/12)]);

 

reply...

nm 11458
August 02 2024
2 2

 

> 

pde:=diff(u(x,t),x$2)-(v-2)*u(x,t)-(v-v1/mu)*(-u(x,t)^3+u(x,t)^5)=0

diff(diff(u(x, t), x), x)-(v-2)*u(x, t)-(v-v1/mu)*(-u(x, t)^3+u(x, t)^5) = 0

> 

xx:=solve(zeta=sqrt(k)*(x-c*t),x);

(k^(1/2)*c*t+zeta)/k^(1/2)

> 

tr:={x=xx,u(x,t)=U(zeta)};
> 

 

{x = (k^(1/2)*c*t+zeta)/k^(1/2), u(x, t) = U(zeta)}

> 

pde2:=PDEtools:-dchange(tr,pde,{U,zeta},'known'={u(x,t)},'unknown'={U(zeta)})

(diff(diff(U(zeta), zeta), zeta))*k-(v-2)*U(zeta)-(v-v1/mu)*(-U(zeta)^3+U(zeta)^5) = 0

> 

pde3:=PDEtools:-dchange({U(zeta)=sqrt(W(zeta))},pde2,{W},'known'={U(zeta)},'unknown'={W(zeta)})

(-(1/4)*(diff(W(zeta), zeta))^2/W(zeta)^(3/2)+(1/2)*(diff(diff(W(zeta), zeta), zeta))/W(zeta)^(1/2))*k-(v-2)*W(zeta)^(1/2)-(v-v1/mu)*(-W(zeta)^(3/2)+W(zeta)^(5/2)) = 0

> 

expand(pde3*sqrt(W(zeta)))

-(1/4)*k*(diff(W(zeta), zeta))^2/W(zeta)+(1/2)*(diff(diff(W(zeta), zeta), zeta))*k-W(zeta)*v+2*W(zeta)+W(zeta)^2*v-W(zeta)^3*v-W(zeta)^2*v1/mu+W(zeta)^3*v1/mu = 0

> 

collect(%,W(zeta));

(-v+v1/mu)*W(zeta)^3+(v-v1/mu)*W(zeta)^2+(-v+2)*W(zeta)+(1/2)*(diff(diff(W(zeta), zeta), zeta))*k-(1/4)*k*(diff(W(zeta), zeta))^2/W(zeta) = 0

> 

 


 

Download transformation.mw

reply...

nm 11458
July 21 2024
1 1

convert the result to set, this will automatically remove duplicates.


 

> 

restart;
> 

interface(rtablesize=70);


result:=Array(1..0):
phi := (p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi) -> (p__1*cosh(xi) - p__2*sinh(xi))/(p__3*cosh(xi) + p__4*sinh(xi));
xi := x:
for p__1 in [1, -1, I, -I] do
    for p__2 in [1, -1, I, -I] do
        for p__3 in [1, -1, -I, I] do
            for p__4 in [0] do
                for q__1 in [1, -1, I, -I] do
                    for q__2 in [1, -1, I, -I] do
                        for q__3 in [1, -1, I, -I] do
                            for q__4 in [1, -1, I, -I] do
                                result ,=evalf(phi(p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi)):
                                #print('phi'(p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi) = result1);
                            od:
                       od:
                    od:
                od:
            od:
         od:
     od:
od:

[10, 10]

proc (p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi) options operator, arrow; (p__1*cosh(xi)-p__2*sinh(xi))/(p__3*cosh(xi)+p__4*sinh(xi)) end proc

> 

print("Before, the size was ",numelems(result));
result_no_duplidates:=convert(convert(result,set),list):
print("After remvoing duplicates, the size is ",nops(result_no_duplidates));
map(X->[''phi(p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi)'' = X],result_no_duplidates):
Matrix(%);

"Before, the size was ", 16384

"After remvoing duplicates, the size is ", 64

Matrix(%id = 36893489808398873292)

> 

 


 

Download A.mw

reply...

nm 11458
July 19 2024
3 2

I do not think there is conversion builtin in Maple for this, at least I could not find one.  But see if this does what you want for now. I only tested it on your input !

e0:=int(x^(1/2),x);
evalindets(e0,sqrt,F->sqrt(op(1,F)^numer(op(2,F))))

Update

For more general cases, (for your other comment below also) you can try this function called mySimplify


 

> 

restart;
> 

mySimplify :=e->evalindets(e,`^`,F->`if`( op(2,F)::fraction, surd( op(1,F)^numer(op(2,F)), denom(op(2,F))),F)):
test_cases:=[  int( (x^2)^(1/3),x), int(x^(1/2),x), x^(2/3) ]:
Matrix(map(X->[X,mySimplify(X)], test_cases))

Matrix(3, 2, {(1, 1) = (3/5)*x*(x^2)^(1/3), (1, 2) = (3/5)*x*(x^2)^(1/3), (2, 1) = (2/3)*x^(3/2), (2, 2) = (2/3)*sqrt(x^3), (3, 1) = x^(2/3), (3, 2) = (x^2)^(1/3)})

> 

 


 

Download mysimplify.mw

reply...

nm 11458
June 29 2024
0 5

while running my code i am not able to edit it

If you mean you are running something in the worksheet, and while it is sitll running you can't edit or do anything? If so, then this is by design. Maple interface can't handle making any changes to worksheet while it is running something. You have to wait until computation is done.

Make sure you set each worksheet to separate engine. This is a nice feature in Maple.

This way while one sheet is busy, you can always open new worksheet and use that because each will be using new engine. See options on how to change this setting.

 

although my code is working for one pde but while changing pde it is giving me error why?

Good chance you have made some error typing the new pde or the new command. The error message you get should make it clear why.

 some times it takes alot of time to evaluate is there any error in the code or should i change my laptop?

You should use the command timelimit look it up in help. You set a timeout how long some computation should take. The general format for doing this can be something like

try
   result := timelimit(30, some_maple_command); #30 seconds timeout
catch:
  print("Warning, The command has timed out");
  result:= FAIL;
end try;

You do not need to get new laptop. Many CAS computations take long time or hang. So you should always use timelimit to guard against this. Ofcourse having faster PC with more memory is not going to hurt if you can afford it. 

 

reply...

nm 11458
June 21 2024
3 1

PDEtools:-Infinitesimals

compute the infinitesimals of symmetry groups admitted by a given PDE system

25944

> 

pde:= diff(u(t,x,y),t$2)+2*a(t)*diff(u(t,x,y),x)^2+2*b(t)*u(t,x,y)*diff(u(t,x,y),x$2)+diff(u(t,x,y),y$2)=0;
PDEtools:-Infinitesimals(pde,u(t,x,y))

diff(diff(u(t, x, y), t), t)+2*a(t)*(diff(u(t, x, y), x))^2+2*b(t)*u(t, x, y)*(diff(diff(u(t, x, y), x), x))+diff(diff(u(t, x, y), y), y) = 0

[_xi[t](t, x, y, u) = 0, _xi[x](t, x, y, u) = 1, _xi[y](t, x, y, u) = 0, _eta[u](t, x, y, u) = 0], [_xi[t](t, x, y, u) = 0, _xi[x](t, x, y, u) = 0, _xi[y](t, x, y, u) = 1, _eta[u](t, x, y, u) = 0], [_xi[t](t, x, y, u) = 0, _xi[x](t, x, y, u) = x, _xi[y](t, x, y, u) = 0, _eta[u](t, x, y, u) = 2*u]

 

 

Download lie_pde.mw

reply...

nm 11458
June 15 2024
1 3

Do not know if this will work for what you are doing or not.
 

26000

> 

restart;

26000

> 

interface(typesetting=extended);

extended

> 

make_nice:=proc(k::anything)
     evalindets(k,`&*`(anything,'specfunc(ln)'),F->InertForm:-MakeInert(F));
end proc;

proc (k::anything) evalindets(k, `&*`(anything, 'specfunc(ln)'), proc (F) options operator, arrow; InertForm:-MakeInert(F) end proc) end proc

> 

k:=3/8*ln(55/52);

(3/8)*ln(55/52)

> 

make_nice(k)

`%*`(`%/`(3, 8), %ln(`%/`(55, 52)))

> 

k:=3/8*ln(55/52)+sin(x)+3/4*exp(x);

(3/8)*ln(55/52)+sin(x)+(3/4)*exp(x)

> 

make_nice(k)

`%*`(`%/`(3, 8), %ln(`%/`(55, 52)))+sin(x)+(3/4)*exp(x)

> 

 


 

Download make_nice.mw

reply...

nm 11458
June 14 2024
4 1

From help it says subs does not work on contravariant indecies. Only on covariant. So need to use SubstituteTensorIndices from Physics package, See help page on SubstituteTensorIndices for more info.  It says to use subs on contra need to use ~=~ format then it should work.

I am using V 2024. If this does not work for you, may be it is version issue then,

20936

> 

restart;

20936

> 

with(Physics):
Setup(mathematicalnotation = true);
Define(A, B):
e:=g_[alpha, mu] * A[~mu] * g_[~alpha, ~nu] * B[nu, sigma, ~rho];

[mathematicalnotation = true]

`Defined objects with tensor properties`

Physics:-g_[alpha, mu]*A[`~mu`]*Physics:-g_[`~alpha`, `~nu`]*B[nu, sigma, `~rho`]

> 

subs(alpha = beta,e); #does not change contra

Physics:-g_[beta, mu]*A[`~mu`]*Physics:-g_[`~alpha`, `~nu`]*B[nu, sigma, `~rho`]

> 

SubstituteTensorIndices(alpha = beta,e); #changes contra and covariant

Physics:-g_[beta, mu]*A[`~mu`]*Physics:-g_[`~beta`, `~nu`]*B[nu, sigma, `~rho`]

> 

#help says this should work.
subs(~alpha = ~beta,e); #changes contra only OK

Physics:-g_[alpha, mu]*A[`~mu`]*Physics:-g_[`~beta`, `~nu`]*B[nu, sigma, `~rho`]

 

 

Download tensor_june_14_2024.mw

reply...

nm 11458
June 01 2024
2 1

I never heard of the method of isodine. May be you meant isoclines  ?

To make slope plot do

ode:= x+y(x)*diff(y(x),x)=0;
DEtools:-DEplot(ode,y(x),x=-2..2,y=-3..3)

To get the solution, the command is 

dsolve(ode)

To get step by step solution the command is

Student:-ODEs:-ODESteps(ode);

reply...

nm 11458
May 27 2024
1 2 
evalb(expand(sin(x + y))= sin(x)*cos(y) + cos(x)*sin(y)); 

    true

Maple does not expand by default. (which is a good thing)

Can you recommend a good way to find out if two terms are equivalent?

There are few ways. One easy way is to do simplify(A-B) and see if you get zero

simplify(sin(x + y) - (sin(x)*cos(y) + cos(x)*sin(y)))
 
      0

But ofcourse this depends how good the simplify command is. Also you can help the simplify with assumptions if needed.

There is also the command is and coulditbe to try

is(sin(x + y) = sin(x)*cos(y) + cos(x)*sin(y))

              true

There might be other ways.  I had few false positives from coulditbe so be careful with it.

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