Maple 18 Questions and Posts

These are Posts and Questions associated with the product, Maple 18

sys := {x^3*a[1]*b[0]+x*a[1]^3*b[0]-x^3*a[0]-x*a[0]*a[1]^2+omega*a[1]*b[0]-omega*a[0] = 0, -x^3*a[-1]*b[-1]^2*b[0]+x^3*a[0]*b[-1]^3-omega*a[-1]*b[-1]^2*b[0]+omega*a[0]*b[-1]^3-x*a[-1]^3*b[0]+x*a[-1]^2*a[0]*b[-1] = 0, -4*x^3*a[1]*b[0]^2+4*x^3*a[0]*b[0]+8*x^3*a[1]*b[-1]+2*x*a[0]*a[1]^2*b[0]+2*x*a[1]^3*b[-1]+2*omega*a[1]*b[0]^2-8*x^3*a[-1]-2*x*a[-1]*a[1]^2-2*x*a[0]^2*a[1]-2*omega*a[0]*b[0]+2*omega*a[1]*b[-1]-2*omega*a[-1] = 0, 4*x^3*a[1]*b[-1]*b[0]^2-4*x^3*a[-1]*b[0]^2-32*x^3*a[1]*b[-1]^2+4*omega*a[1]*b[-1]*b[0]^2-32*x^3*a[-1]*b[-1]+4*x*a[-1]*a[1]^2*b[-1]+4*x*a[0]^2*a[1]*b[-1]-4*omega*a[-1]*b[0]^2+4*omega*a[1]*b[-1]^2-4*x*a[-1]^2*a[1]-4*x*a[-1]*a[0]^2-4*omega*a[-1]*b[-1] = 0, 4*x^3*a[-1]*b[-1]*b[0]^2-4*x^3*a[0]*b[-1]^2*b[0]+8*x^3*a[1]*b[-1]^3-8*x^3*a[-1]*b[-1]^2-2*omega*a[-1]*b[-1]*b[0]^2+2*omega*a[0]*b[-1]^2*b[0]+2*omega*a[1]*b[-1]^3-2*x*a[-1]^2*a[0]*b[0]+2*x*a[-1]^2*a[1]*b[-1]+2*x*a[-1]*a[0]^2*b[-1]-2*omega*a[-1]*b[-1]^2-2*x*a[-1]^3 = 0, x^3*a[1]*b[0]^3-x^3*a[0]*b[0]^2-18*x^3*a[1]*b[-1]*b[0]+omega*a[1]*b[0]^3-5*x^3*a[-1]*b[0]+23*x^3*a[0]*b[-1]+x*a[-1]*a[1]^2*b[0]+x*a[0]^2*a[1]*b[0]+5*x*a[0]*a[1]^2*b[-1]-omega*a[0]*b[0]^2+6*omega*a[1]*b[-1]*b[0]-6*x*a[-1]*a[0]*a[1]-x*a[0]^3+5*omega*a[-1]*b[0]-omega*a[0]*b[-1] = 0, -x^3*a[-1]*b[0]^3+x^3*a[0]*b[-1]*b[0]^2+5*x^3*a[1]*b[-1]^2*b[0]+18*x^3*a[-1]*b[-1]*b[0]-23*x^3*a[0]*b[-1]^2-omega*a[-1]*b[0]^3+omega*a[0]*b[-1]*b[0]^2+5*omega*a[1]*b[-1]^2*b[0]-x*a[-1]^2*a[1]*b[0]-x*a[-1]*a[0]^2*b[0]+6*x*a[-1]*a[0]*a[1]*b[-1]+x*a[0]^3*b[-1]-6*omega*a[-1]*b[-1]*b[0]+omega*a[0]*b[-1]^2-5*x*a[-1]^2*a[0] = 0}

Hi all,

I have a system of nonlinear equations with for equations, 4 variables I want to solve for, and 2 parameters. All of the variables and parameters must be non-negative.

The code I used to try to do this is:

Where eqi (i = 1, ... , 4) are expression (not equations in themselves). For example, eq1 is:


When I try to run this code I get the following error:

"Error, (in SolveTools:-Inequality:-Piecewise) piecewise takes at least 2 parameters"


Can anyone help me how I can make Maple do what I want here? :)


Thank you in advance,



p_{x} = a*p + u*q;    (1)

q_{x} = -conjugate(u)*p - a*q;  (2)     # where a is complex parameter, p_{x} mean derivative of p w.r.t x

Define, u=p^2+conjugate(q)^2,  (3)  

Now take the derivative of (3) w.r.t x and by using (1) and (2), we get

 u_{x}=2(a*p^2 - conjugate(a)*conjugate(q)^2) + 2*(p^2 + conjugate(q)^2)*(p*q - conjugate(p)*conjugate(q)).  (4)

How to calculate the results (4) on maple?

I want to calculate (partialH/partialq). But I encountered an error when I evaluate it.

alias(q = q(x, t), p = p(x, t));
                              q, p
H := lambda*p*q+conjugate(lambda)*conjugate(p)*conjugate(q)+(1/2*(p^2+conjugate(q)^2))*(conjugate(p)^2+q^2);
diff(H, q(x, t));

maple shows error:

Error, invalid input: diff received q(x, t), which is not valid for its 2nd argument
How to fix this issue?


Dear all

I have a first-order complex ODE, I would like to use the polar coordinates to find the associated system of the ordinary differential equation then I determine the equilibrium point of the system and deduce that there is at least one limit cycle.

many thanks for your help


I found that i can use  simplify under assumption to gain the result :

simplify((-p^3)^(1/3), assume = negative);

simplify((p^3)^(1/3), assume = positive);

But confusing ! I expected the simplify command just with the option=symbolic works at the same manner ?


My purpose :

simplify((-p^7)^(1/7), assume = negative);



simplify((p^7)^(1/7), assume = positive);



Without negative sign the simple symbolic result appears :

simplify((p^3)^(1/3), symbolic);



In power 3 the Imaginary part is included too !

simplify(((-p)^3)^(1/3), symbolic);




simplify((p^7)^(1/7), symbolic);



simplify((-p^7)^(1/7), symbolic);



simplify((-p^7)^(1/7), symbolic, radical);








n := 0
u[0] := x^3+(1/2)*A*x^2
for k from 0 to n do
A[k] := sum((Diff(u[i], x))*(Diff(u[k-i], x)), i = 0 .. k);
A[k] := sum((diff(u[i], x))*(diff(u[k-i], x)), i = 0 .. k)
end do;

it gives A[0]:=0 which is incorrect. why? and how it will give correct answer?

 Mathematical model for the temperature profiles of steel pipes quenched by water cooling rings


 Can anyone solve the given equations along with the boundary conditions analytically with Maple and draw the graphs ???????????

Dear maple users,

How to plot a figure for different values of rk?

like rk=5,10,15,20:

Thanking you,

Waiting for replay.


how can i use curve fitting tools for get m and n in this equation  for excel data


i want to get m and n?



Dear all

If we travel in straight lines in R^3. We begin at the point A=(1, 2, 3)
in the direction of the vector (1, 2, 2) and we end at the final point (10, 11, 12).

We made a single 90-degree turn.

Can we sketch a figure and we show the position where we take the turn?



Dear all

I hope to solve a linear system AX=bk where A is a nxn matrix and X is a nx1 vector and b is a vector from the canonical basis of R^n ( for example in R^3 : b1=[ 1 0 0], b2 =[ 0 1 0] and b3=[ 0 0 1]


Dear all

I have a function g that I compute its limit at x0 

but when I compute series(g(x),x=x0) then I compute the limit of the series I get different limit


Many thanks for your help

Dear all

Can we use maple to compute integral using the residual theorem


We can consider the contour a circle that contains one pole (Ipi/2)

Many thanks


Dear maple users,
I am solving an ode problem with an analytical solution.
programming running properly, but my plot not exact with the already existing article plot. 
how to get the exact plot.

Thanking you.


N := 3;





dsolve(diff(f(x), `$`(x, 3)));

f(x) = (1/2)*_C1*x^2+_C2*x+_C3


Rf := 2*(diff(f[m-1](x), x, x, x))-(2*mh*mh)*(diff(f[m-1](x), x))+sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)-bet*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(f[t](x), x, x))+f[m-1-n](x)*f[n-t](x)*(diff(f[t](x), x, x, x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(f[t](x), x, x)), t = 0 .. n), n = 0 .. m-1));

2*(diff(diff(diff(f[m-1](x), x), x), x))-2*(diff(f[m-1](x), x))+sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)-.2*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x))+f[m-1-n](x)*f[n-t](x)*(diff(diff(diff(f[t](x), x), x), x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x)), t = 0 .. n), n = 0 .. m-1))


dsolve(diff(f[m](x), x, x, x)-CHI[m]*(diff(f[m-1](x), x, x, x)) = h*H*Rf, f[m](x));

f[m](x) = Int(Int(Int(CHI[m]*(diff(diff(diff(f[m-1](x), x), x), x))+2*h*(diff(diff(diff(f[m-1](x), x), x), x))-2*h*(diff(f[m-1](x), x))+h*(sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1))-(1/5)*h*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x))+f[m-1-n](x)*f[n-t](x)*(diff(diff(diff(f[t](x), x), x), x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x)), t = 0 .. n), n = 0 .. m-1)), x), x)+_C1*x, x)+_C2*x+_C3


f[0](x) := 1-exp(x);



for m to N do CHI[m] := `if`(m > 1, 1, 0); f[m](x) := int(int(int(2*CHI[m]*(diff(f[m-1](x), x, x, x))-(2*h*H*mh*mh)*(diff(f[m-1](x), x))+h*H*(sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)), x)-h*H*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(f[t](x), x, x))+f[m-1-n](x)*f[n-t](x)*(diff(f[t](x), x, x, x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(f[t](x), x, x)), t = 0 .. n), n = 0 .. m-1))*bet, x)+_C1*x, x)+_C2*x+_C3; s1 := evalf(subs(x = 0, f[m](x))) = 0; s2 := evalf(subs(x = 0, diff(f[m](x), x))) = 0; s3 := evalf(subs(x = 1, f[m](x))) = 0; s := {s1, s2, s3}; f[m](x) := simplify(subs(solve(s, {_C1, _C2, _C3}), f[m](x))) end do:

f(x) := sum(f[l](x), l = 0 .. N);








Analytical solution approach:





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