Mariusz Iwaniuk

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8 years, 223 days

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These are replies submitted by Mariusz Iwaniuk

@Dkunb 

I doubt  be  can solvelable in Maple even near future,unless the people from Maple they will get to hard work.

Even Mathematica can't solve.

@Joseph Poveromo 

In my opinion it should be: int(y'(x)*x^2/(x^2-1),x) = int(sqrt(y(x)),x)^(-2/3)

 int[y'(x)* (x^2)/[(x^2)-1],x)  =  (int[sqrt(y(x)])^(-2/3)

Maple give me errors ?

@Prakash J 

Use:

dsn := dsolve([eq1, eq2, eq3, eq4, ics], numeric, delaymax = tau, method = ck45);
dsn := dsolve([eq1, eq2, eq3, eq4, ics], numeric, delaymax = tau, method = rkf45);

From Help:

"rkf45 finds a numerical solution using a Fehlberg fourth-fifth order Runge-Kutta method with degree four interpolant. This is the default method of the type=numeric solution for initial value problems when the stiff argument is not used.  The other non-stiff method is a Runge-Kutta method with the Cash-Karp coefficients, ck45.
"

@yaseentareq 

restart;
InverseElzakiTransform := proc(F, v) inttrans:-invlaplace(eval(F/v, v = 1/v), v, t); end proc;
F := v^(n + 2);
(InverseElzakiTransform(F, v) assuming (0 < n));

#t^n/GAMMA(1 + n)

If inverse laplace transform can't solve we have to use another way or another mathematical mainpulation.

 

InverseElzakiTransformBYInverseFourier := proc(F, v) (limit(exp(-Z*t)*inttrans:-invfourier(eval(eval(F/v, v = 1/v), v = Z + w*I), w, t), Z = 0, right) assuming (0 < t)); end proc;
F := v^3/(v^2 + 1);
InverseElzakiTransformBYInverseFourier(F, v);

#sin(t)

InverseElzakiTransformBYInverseMellin := proc(F, v) (simplify(eval(inttrans:-invmellin(eval(F/v, v = 1/v), v, t, 1 .. infinity), t = exp(-t))) assuming (0 < t)); end proc;
F := v^3/(v^2 + 1);
(InverseElzakiTransformBYInverseMellin(F, v) assuming (0 < n));

#sin(t)

I can only write in Maple Integral Transform are very Very weak.

Please share the code in copyable form, so that other users can play with it. No one wants to retype all this code (and double check for correct transition). This will raise your chances for getting quick and competent help. 

R0 := 1/(a - sqrt(b + c*cos(x)));
R1 := (int(R0, x = 0 .. 2*Pi) assuming (0 < a, 0 < b, 0 < c)):
a := 0.9;
b := 4.5;
c := 0.1;
R1;

#-5.145818656 - 0.*I

 

See........................... here.

@QM 

Ok. You want Maple to compare with Mathematica,for indefine integrals you may see on this web site.

 

Regards M.I.

Why are you asking if you know how to calculate it ?

@somestudent 

Try:

INV := invztrans((z - 1)^2/(a*z^2 + b*z + c), z, n);
INVnew := simplify(allvalues(INV));
simplify(evalf(subs(a = 2, b = 1, c = 4, eval(subs(n = 16, INVnew)))), zero);

Regards M.I.

@Axel Vogt 

That's where in my codes I should use the fnormal to make Maple solved my PDE ?

I just wanted to compare the results between Mathematica and Maple and nothing else.

I got the  PDE equation from this weblink.  and maybe it will clear everything up.

Thanks.

 I separate real and imaginary parts in PDEs, but plot3d is doesn't it look the same? 

What have I done wrong?


 

restart

psi := u(r, t)+I*v(r, t); evalc(I*(diff(psi, t)) = -exp(-psi)*(diff(psi, r, r)-(diff(psi, r))^2+2*(diff(psi, r))/r)); EQ := expand(simplify(`~`[evalc]({`~`[Re, Im](%)})))

{-(diff(v(r, t), t)) = -(diff(diff(u(r, t), r), r))*cos(v(r, t))/exp(u(r, t))-(diff(diff(v(r, t), r), r))*sin(v(r, t))/exp(u(r, t))+(diff(u(r, t), r))^2*cos(v(r, t))/exp(u(r, t))+2*(diff(u(r, t), r))*(diff(v(r, t), r))*sin(v(r, t))/exp(u(r, t))-2*(diff(u(r, t), r))*cos(v(r, t))/(exp(u(r, t))*r)-(diff(v(r, t), r))^2*cos(v(r, t))/exp(u(r, t))-2*(diff(v(r, t), r))*sin(v(r, t))/(exp(u(r, t))*r), diff(u(r, t), t) = sin(v(r, t))*(diff(diff(u(r, t), r), r))/exp(u(r, t))-cos(v(r, t))*(diff(diff(v(r, t), r), r))/exp(u(r, t))-(diff(u(r, t), r))^2*sin(v(r, t))/exp(u(r, t))+2*(diff(u(r, t), r))*cos(v(r, t))*(diff(v(r, t), r))/exp(u(r, t))+2*(diff(u(r, t), r))*sin(v(r, t))/(exp(u(r, t))*r)+(diff(v(r, t), r))^2*sin(v(r, t))/exp(u(r, t))-2*cos(v(r, t))*(diff(v(r, t), r))/(exp(u(r, t))*r)}

(1)

NULL

ICB := {u(inf, t) = 0, u(r, 0) = (1/4)*exp(-(1/2)*r^2), v(inf, t) = 0, v(r, 0) = 0, (D[1](u))(eps, t) = 0, (D[1](v))(eps, t) = 0}

{u(inf, t) = 0, u(r, 0) = (1/4)*exp(-(1/2)*r^2), v(inf, t) = 0, v(r, 0) = 0, (D[1](u))(eps, t) = 0, (D[1](v))(eps, t) = 0}

(2)

eps := 10^(-10); inf := 1000

sol := pdsolve(EQ, ICB, numeric, time = t, range = eps .. inf, abstol = 0.1e-2)

valsU := sol:-value(u(r, t), output = listprocedure); valsV := sol:-value(v(r, t), output = listprocedure)

U := eval(u(r, t), valsU); V := eval(v(r, t), valsV)

abs(U(1, 1)+I*V(1, 1))

.2338656056

(3)

plot3d(4*abs(U(r, t)+I*V(r, t)), t = 0 .. 2, r = eps .. 3)

 

NULL

NULL


 

Download PDE_with_Imaginary_Unit_ver2.mw

@Alex0099 

Try Wolfram Alpha ?

@Alex0099 

It is clear and accurately conveys the information that Wolfram Alpha is based on the Mathematica engine, that it has step-by-step solutions available for a fee, and that the author is uncertain if step-by-step solutions work for double summations, as the author is not registered on the platform.

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