MaplePrimes Questions

What technique would you try to solve the following non-linear differential equation for real g(x) given h(x) is real (finite positive) polynomial?      h(x) = g(x) + g'(x)/g(x),      where g'(x) = dg(x)/dx

I want to convert the maple result 

arctan(y/x) to arctan(y,x), 

so if x is equal zero i obtain pi/2 and not diveided by zero 

 

 

i have two euations including integration which has two unkwnon x1 and x2.
how can i get these equations solved, thanks for the help.

restart:

with(DirectSearch)

[BoundedObjective, CompromiseProgramming, DataFit, ExponentialWeightedSum, GlobalOptima, GlobalSearch, Minimax, ModifiedTchebycheff, Search, SolveEquations, WeightedProduct, WeightedSum]

(1)

with(LinearAlgebra):

with(Student:-Calculus1):

with(Student:-NumericalAnalysis):

A:=convert(taylor(exp(Q),Q,6),polynom);

1+Q+(1/2)*Q^2+(1/6)*Q^3+(1/24)*Q^4+(1/120)*Q^5

(2)

Q:=a[11]*(E[r])^2+a[22]*(E[theta])^2+2*a[12]*E[r]*E[theta];

E[r]^2*a[11]+2*E[r]*E[theta]*a[12]+E[theta]^2*a[22]

(3)

psi:=0.5*c*(exp(Q)-1);

.5*c*(exp(E[r]^2*a[11]+2*E[r]*E[theta]*a[12]+E[theta]^2*a[22])-1)

(4)

F:=Matrix(3,3,[[lambda[r],0,0],[0,lambda[theta],0],[0,0,lambda[z]]]);

Matrix(3, 3, {(1, 1) = lambda[r], (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = lambda[theta], (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = lambda[z]})

(5)

sigma[r]:=-p+diff(psi,E[r])*F[1,1]^2;

-p+.5*c*(2*E[r]*a[11]+2*E[theta]*a[12])*exp(E[r]^2*a[11]+2*E[r]*E[theta]*a[12]+E[theta]^2*a[22])*lambda[r]^2

(6)

sigma[theta]:=-p+diff(psi,E[theta])*F[2,2]^2;

-p+.5*c*(2*E[r]*a[12]+2*E[theta]*a[22])*exp(E[r]^2*a[11]+2*E[r]*E[theta]*a[12]+E[theta]^2*a[22])*lambda[theta]^2

(7)

sigma[z]:=-p+diff(psi,E[z])*F[3,3]^2;

-p

(8)

p1:=diff(psi,E[r])*F[1,1]^2;#Pressure is constituted form 3 parts, one part is p1, other part is p2 and a constant p0

.5*c*(2*E[r]*a[11]+2*E[theta]*a[12])*exp(E[r]^2*a[11]+2*E[r]*E[theta]*a[12]+E[theta]^2*a[22])*lambda[r]^2

(9)

E[r]:=0.5*(lambda[r]^2-1);

.5*lambda[r]^2-.5

(10)

E[theta]:=0.5*(lambda[theta]^2-1);

.5*lambda[theta]^2-.5

(11)

E[z]:=0.5*(lambda[z]^2-1);

.5*lambda[z]^2-.5

(12)

lambda[r]:=x2*sqrt((r^2-x1)/x2)/r;

x2*((r^2-x1)/x2)^(1/2)/r

(13)

lambda[theta]:=r/sqrt((r^2-x1)/x2);

r/((r^2-x1)/x2)^(1/2)

(14)

lambda[z]:=1/x2;

1/x2

(15)

sigma[r];

-p+.5*c*(2*(.5*x2*(r^2-x1)/r^2-.5)*a[11]+2*(.5*r^2*x2/(r^2-x1)-.5)*a[12])*exp((.5*x2*(r^2-x1)/r^2-.5)^2*a[11]+2*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)*a[12]+(.5*r^2*x2/(r^2-x1)-.5)^2*a[22])*x2*(r^2-x1)/r^2

(16)

sigma[theta]:

sigma[z]:

#p2:=int((sigma[r]-sigma[theta])/r,r):%Pressure is constituted form 2 parts, one part is p1, other part is p2 and and a constant p0

Digits:=10:

c:=790000:

a[11]:=0.539:

a[22]:=0.368:

a[12]:=0.653:

p_in:=10000:

p_out:=0:

r_in:=5.4e-3:

r_out:=6.1e-3:

F_a:=0.381846:

p21:=(c/2)*(((r^2-x1)/r^3)-(x2*r/(r^2-x1)))*exp(Q);

395000*((r^2-x1)/r^3-x2*r/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)

(17)

p22:=(c/2)*Int(exp(Q)*(3*x2*r^6-r^6+5*x1*r^4-x1*x2*r^4-7*x1^2*r^2+3*x1^3)/(r^4*(r^2-x1)^2),r=r_in..r_out);#This is the part that should be maintained as an integral until the final solution

395000*(Int(exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*(3*r^6*x2-r^6-r^4*x1*x2+5*r^4*x1-7*r^2*x1^2+3*x1^3)/(r^4*(r^2-x1)^2), r = 0.54e-2 .. 0.61e-2))

(18)

p2:=p21-p22;#p2 is computed using the integration by part method

395000*((r^2-x1)/r^3-x2*r/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)-395000*(Int(exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*(3*r^6*x2-r^6-r^4*x1*x2+5*r^4*x1-7*r^2*x1^2+3*x1^3)/(r^4*(r^2-x1)^2), r = 0.54e-2 .. 0.61e-2))

(19)

 

p0:=(p_out+eval(p2,r=r_out));#p0 is the constant which is defined form the boundary conditions either p_out=subs(r=r_out,sigma[r]) or p_in=subs(r=r_in,sigma[r])

395000*(-4405655.099*x1+163.9344262-0.61e-2*x2/(-x1+0.3721e-4))*exp(.539*(13437.24805*x2*(-x1+0.3721e-4)-.5)^2+1.306*(13437.24805*x2*(-x1+0.3721e-4)-.5)*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)+.368*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)^2)-395000*(Int(exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*(3*r^6*x2-r^6-r^4*x1*x2+5*r^4*x1-7*r^2*x1^2+3*x1^3)/(r^4*(r^2-x1)^2), r = 0.54e-2 .. 0.61e-2))

(20)

 

p:=p1+p2+p0;#p is the total pressure

395000.0*(.5390*x2*(r^2-x1)/r^2-1.1920+.6530*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*x2*(r^2-x1)/r^2+395000*((r^2-x1)/r^3-x2*r/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)-790000*(Int(exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*(3*r^6*x2-r^6-r^4*x1*x2+5*r^4*x1-7*r^2*x1^2+3*x1^3)/(r^4*(r^2-x1)^2), r = 0.54e-2 .. 0.61e-2))+395000*(-4405655.099*x1+163.9344262-0.61e-2*x2/(-x1+0.3721e-4))*exp(.539*(13437.24805*x2*(-x1+0.3721e-4)-.5)^2+1.306*(13437.24805*x2*(-x1+0.3721e-4)-.5)*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)+.368*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)^2)

(21)

#p:=H+H00;

eq1:=Int((sigma[r]-sigma[theta])/r,r=r_in..r_out);

Int((395000.0*(.5390*x2*(r^2-x1)/r^2-1.1920+.6530*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*x2*(r^2-x1)/r^2-395000.0*(.6530*x2*(r^2-x1)/r^2-1.0210+.3680*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*r^2*x2/(r^2-x1))/r, r = 0.54e-2 .. 0.61e-2)

(22)

eq2:=Int(2*Pi*sigma[z]*r,r=r_in..r_out);

Int(2*Pi*(-395000.0*(.5390*x2*(r^2-x1)/r^2-1.1920+.6530*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*x2*(r^2-x1)/r^2-395000*((r^2-x1)/r^3-x2*r/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)+790000*(Int(exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*(3*r^6*x2-r^6-r^4*x1*x2+5*r^4*x1-7*r^2*x1^2+3*x1^3)/(r^4*(r^2-x1)^2), r = 0.54e-2 .. 0.61e-2))-395000*(-4405655.099*x1+163.9344262-0.61e-2*x2/(-x1+0.3721e-4))*exp(.539*(13437.24805*x2*(-x1+0.3721e-4)-.5)^2+1.306*(13437.24805*x2*(-x1+0.3721e-4)-.5)*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)+.368*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)^2))*r, r = 0.54e-2 .. 0.61e-2)

(23)

#eq1:=Quadrature((sigma[r]-sigma[theta])/r,r=r_in..r_out,method=gaussian[5],output=value):

#eq2:=Quadrature(2*Pi*sigma[z]*r,r=r_in..r_out,method=gaussian[5],output=value):

eq1=evalf(p_out-p_in)

Int((395000.0*(.5390*x2*(r^2-x1)/r^2-1.1920+.6530*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*x2*(r^2-x1)/r^2-395000.0*(.6530*x2*(r^2-x1)/r^2-1.0210+.3680*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*r^2*x2/(r^2-x1))/r, r = 0.54e-2 .. 0.61e-2) = -10000.

(24)

eq2=F_a

Int(2*Pi*(-395000.0*(.5390*x2*(r^2-x1)/r^2-1.1920+.6530*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*x2*(r^2-x1)/r^2-395000*((r^2-x1)/r^3-x2*r/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)+790000*(Int(exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*(3*r^6*x2-r^6-r^4*x1*x2+5*r^4*x1-7*r^2*x1^2+3*x1^3)/(r^4*(r^2-x1)^2), r = 0.54e-2 .. 0.61e-2))-395000*(-4405655.099*x1+163.9344262-0.61e-2*x2/(-x1+0.3721e-4))*exp(.539*(13437.24805*x2*(-x1+0.3721e-4)-.5)^2+1.306*(13437.24805*x2*(-x1+0.3721e-4)-.5)*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)+.368*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)^2))*r, r = 0.54e-2 .. 0.61e-2) = .381846

(25)

fsolve({eq1=evalf(p_out-p_in),eq2=F_a},{x1,x2});

fsolve({Int((395000.0*(.5390*x2*(r^2-x1)/r^2-1.1920+.6530*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*x2*(r^2-x1)/r^2-395000.0*(.6530*x2*(r^2-x1)/r^2-1.0210+.3680*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*r^2*x2/(r^2-x1))/r, r = 0.54e-2 .. 0.61e-2) = -10000., Int(2*Pi*(-395000.0*(.5390*x2*(r^2-x1)/r^2-1.1920+.6530*r^2*x2/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*x2*(r^2-x1)/r^2-395000*((r^2-x1)/r^3-x2*r/(r^2-x1))*exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)+790000*(Int(exp(.539*(.5*x2*(r^2-x1)/r^2-.5)^2+1.306*(.5*x2*(r^2-x1)/r^2-.5)*(.5*r^2*x2/(r^2-x1)-.5)+.368*(.5*r^2*x2/(r^2-x1)-.5)^2)*(3*r^6*x2-r^6-r^4*x1*x2+5*r^4*x1-7*r^2*x1^2+3*x1^3)/(r^4*(r^2-x1)^2), r = 0.54e-2 .. 0.61e-2))-395000*(-4405655.099*x1+163.9344262-0.61e-2*x2/(-x1+0.3721e-4))*exp(.539*(13437.24805*x2*(-x1+0.3721e-4)-.5)^2+1.306*(13437.24805*x2*(-x1+0.3721e-4)-.5)*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)+.368*(0.18605e-4*x2/(-x1+0.3721e-4)-.5)^2))*r, r = 0.54e-2 .. 0.61e-2) = .381846}, {x1, x2})

(26)

SolveEquations([eq1=evalf(p_out-p_in),eq2=F_a]);

Warning, objective function returns unfeasible value HFloat(undefined) for initial point [x1 = .9, x2 = .9]; trying to find a feasible initial point

 

Error, (in DirectSearch:-Search) cannot find feasible initial point; specify a new one

 

 

 

 

fsolve_problem.mw

For this integro-differential equation,

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

Maple is able to obtain an exact intrinsic solution

from which an exact solution can be extracted, namely,

ExtrinsicSolution:= y(x) = sqrt(3)*(-8*_C1*x^(8/3) + 12*x^2 - 3)^(3/4)

My question concerns how was this solution obtained.

Even more, specifically, 'odeadvisor' suggests converting the

equation in question to the form

ode:= y = G(x,diff(y(x),x));

However, I cannot reconcile how this can be applied to an equation which

contains two integrals. (Regretably, I am not able to directly, attach my

Maple worksheet directly on to this sheet). The situation is that after

applying 'dsolve' to the above 'Equation', Maple comes back with an

intrinsic solution which can was used to obtain the 'ExtrinsicSolution' in 

the above.  So it is the missing steps between applyingthe dsolve command

to Equation and the intrinsic solution which MS provides which, in turn, leads

to the 'Extrinsic Solution' above. I would greatly, appreciate if anyone can 

fill in the missing steps.

I was trying this ode with Maple

Do you agree this solution is not correct by Maple?

restart;

ode:=diff(y(t),t)+y(t)=Dirac(t);
ic:=y(0)=1;
sol:=dsolve([ode,ic],y(t),method='laplace');

It gives  y(t) = 2*exp(-t)

But from the discussion in the above link we see this is wrong solution. Maple also does not verify it:

odetest(sol,[ode,y(0)=1])

[-Dirac(t), -1]

Would this be considered a bug I should report or not? Note this result is only when using Laplace method. The default method gives better solution.

ode:=diff(y(t),t)+y(t)=Dirac(t);
ic:=y(0)=1;
sol:=dsolve([ode,ic],y(t));
odetest(sol,[ode,y(0)=1])

 

Maple 2023.2.1

How to convert barycentric coordinates to cartesian ? Thank you

I want to approximate a positive function that is decreasing in Gamma, say f(Gamma), that is very complicated yet very smooth. I need this in order to obtain a tractable and compact version of its derivative, which enters in the partial derivative of another (very simple) function.

Along the way, three related questions emerge: Derivatives_and_Approximations.mw

Thanks a lot!

Hi,

I am attempting to illustrate various solids in Maple. How can I do this with ? (Figure 3 on my worksheet). Thank you

S5SolidQ.mw

Is there a way to manipulate an equation so that it is in the form of (Expression of Primary Variables)*(Expression of Secondary Variables)
In the example below from Video 1: Fast Analytical Techniques for Electrical and Electronic Circuits (youtube.com), the primary variables are R1 and R2

PS. When I type ctrl-v to insert an image, I always get 2 copies.

I got the proportional symbol to work once, typing "proportional" + CRTL + Space.  Went for wlak came back and could not get it to work at all.

Does it actually work or am I imagining things?

What's the correct command to return the number of elements in this command? It returns an expression sequence, but I didn't manage to get anything useful in return.

StringTools:-SearchAll("aba", "abababababababababab")

Greetings All,

This is an application for control theory, specifially using Maple to solve control problems in the area of Interconnection and Damping Assignment Passivity Based Control (IDA-PBC).

- Assuming two variables (iL and Vo), there is a potential function that I am trying to solve for called "Ha".  I have two equations here, and I want to solve for Ha using the pdsolve() command:  

eq1 := diff(Ha(iL, Vo), iL) = rhs(result[1]);
eq2 := diff(Ha(iL, Vo), Vo) = rhs(result[2]);
pdsolve( {eq1, eq2  } );

Once I do this, Maple gives me an expression for Ha that has arbitrary functions in it (I understand where these are coming from).  So far, so good.

--> In order to get help solving for these arbitrary functions, I also want to tell Maple some constraints.  For example:

"the Hessian matrix of Ha must be positive definite"

Is there a way to do this?

with(geometry);
with(LinearAlgebra);
xA := 1;
yA := 0;
xB := 0;
yB := 0;
xC := 0;
yC := 1;
Mat := Matrix(3, 3, [xA, xB, xC, yA, yB, yC, 1, 1, 1]);
Miv := MatrixInverse(Mat);
phi := (x, y) -> Transpose(Multiphy(Miv), <x, y, 1>);
for i to 6 do
    B || i := phi(xA || i, yA || i);
end do;
Error, (in LinearAlgebra:-Transpose) invalid input: too many and/or wrong type of arguments passed to LinearAlgebra:-Transpose; first unused argument is Vector(3, {(1) = xA1, (2) = yA1, (3) = 1})
How to correct this error ? Thank you.

First, I should mention that I am new to Maple. When I try to solve the below system of ODEs

sys_ode := diff(x(t), t) = -x(t)^2/(4*Pi*y(t)*(x(t)^2 + y(t)^2)), diff(y(t), t) = y(t)^2/(4*Pi*x(t)*(x(t)^2 + y(t)^2))

with initial conditions of 

ics := x(0) = 1, y(0) = 1

using the command 

sol_analytic := dsolve([sys_ode, ics])

I receive the below error of 

Error, (in dsolve) numeric exception: division by zero

Any help or guidance to resolve this is greatly appreciated.

Certainly a standard question.

I have an integer n*n matrix A (the entries are explicitly integers; there is no variable -type x- in the matrix). I want the Smith normal form of A, that is A=UDV where U,V are integer matrices with determinant +-1 and D is a diagonal matrix with -eventually- some zero and positive integers d_i s.t. d_i divides d_{i+1}.

"SmithForm()" doesn't work directly (I get rational -non integer- matrices). Maybe it is necessary to declare the matrix A as 'Matrix(integer)' ...
Thank you in advance for your help.

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