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Can somebody help me to find the solution?

I think there is something wrong with the definition of bvw1. If I use dsolve (in soln) with only bvw as Initial Condition,

I get a solution but if I also insert bvw1 as an Initial condition soln won't appear.

Here's what's written in the image:

'Imagine the course of a planet around a star with L=0.5 and e=0.7'

Solve Keppler's differential equation with Initial Conditions:'

HI, I am trying to solve two PDEs but in boundry conditions there is arising an error plz help.
Nazi.mw

In Maple 15 it seems that plottools:-transform only accepts this form of conditional statement:  

`if`(conditional expression, true expression, false expression).

Is there any way to have plottools:-transform process more than one condition? Do later versions of Maple permit this?

In the following problem though b and c are same (except the way denominator 2 is hanfled), command ' a-b ' readily answers zero, but a-c not so. Why? Only on condition of assumption real it gives zero!

a := (1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega:

b := (1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega:

a-b;

0

(1)

c := (1/2)*(kappa*omega^2+omega^3)*(Y+(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*(kappa*omega^2+omega^3)))^2/omega:

a-c;

(1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega-(1/2)*(kappa*omega^2+omega^3)*(Y+(-N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*kappa*omega^2+2*omega^3))^2/omega

(2)

"(->)"

0

(3)

``

Why the answer is not given as zero?

``

``

 

Download what_is_the_difference_between_b_and_c.mw

What difference therms b and c make for Maple? Are they not same?

Ramakrishnan V

rukmini_ramki@hotmail.com

Hi

 

f := x->x^2:

AB(f, 0, 5);

But my proc is not true for below example:

f := x -> x^3-2*x;
AB(f, -1, 4);
I think c has two value and it is not true.

How can I add a condition to my proc that c be between a and b (a<c<b)?

 

 

restart;

with(DETools, diff_table);

kB := 0.138064852e-22;

R := 287.058;

T := 293;

p := 101325;

rho := 0.1e-2*p/(R*T);

vr := diff_table(v_r(r, z));

vz := diff_table(v_z(r, z));

eq_r := 0 = 0;

eq_p := (vr[z]-vz[r])*vr[] = (vr[]*(vr[r, z]-vz[r, r])+vz[]*(vr[z, z]-vz[z, r]))*r;

eq_z := 0 = 0;

eq_m := r*vr[r]+r*vz[z]+vr[] = 0;

pde := {eq_m, eq_p};

IBC := {v_r(1, z) = 0, v_r(r, 0) = 0, v_z(1, z) = 0, v_z(r, 0) = r^2-1};

sol := pdsolve(pde, IBC, numeric, time = z, range = 0 .. 1);

 

what am I doing wrong?

it's telling me: Error, (in pdsolve/numeric/par_hyp) Incorrect number of boundary conditions, expected 3, got 2
but i did just as in the example :-/

Hi everyone,

I am trying to solve the equation of heat tranfer, time dependent, with particular Initial and boundary conditions but I am stuck by technical problems both in getting an analytical solution and a numerical one.

The equation

the equation.

I defined a and b numerically. domain is : and I defined surf_power numerically.

The initial condition is : , T0 defined numerically

The boundary condition is : , because it has a shperical symetry.

To me, it looks like a well posed problem. Does it look fine ?

Problem in analytical solution :

It doesn't accept the boundary condition so I only input the initial condition and it actually gives me back an expression that can be evaluated but it never does : I can't reduce it more than an expression of fourier which I can't eval. The solution :
The solution calculated in (0,0). I was hoping T0...

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ?

The numerical solution problems :

Sometimes it tells me that my boundary condition is equivalent  to 0 = 0, and I don't see why. Some other times it tells me I only gave 1 boundary/initial condition even if I wrote both. Here is what I wrote for example :

(because it kept asking me to add these two options : 'time' and 'range')

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ? I must at least have syntax problems because even if I keep reading the Help, it's been a long time since I used Maple.

Thank very much for any indication you could give me !

Simon

Hi everyone.

I have been experiencing a problem trying to solve a coupled system of 3 differencial equations

My problem is that a got a message back as I try to solve the system:

"Error, (in pdsolve/numeric/process_IBCs) improper op or subscript selector"

by apply this point that ''all dependent variables must be functions of the same independent variables''

i again accost with another error ''

Error, (in pdsolve/numeric/process_IBCs) initial/boundary conditions can only contain derivatives which are normal to the boundary, got (D[1, 1](w))(x, -3/400000000)

''

please help me.....very very thanks

I have the following characteristic equation by use of maple. How do I find a condition on x, that will return real eigenvalues and complex eigenvalues?

 

 

 

Dear All,

I am solving 6 ODE equations with boundary conditions using Runge kutta Felbergh 45 (Maple 12). then, i got this problem.. any suggestion??

Thank you :)

ISPC3.mw

``

restart; with(plots); M := 3; k = .2; blt := 6; r := 2; l := .1; Pr := 6.8; Ec := 2; N := .5; rho := .5; Tv := .5; Tt := .5; c := 1; cm := .1; cp := .1

Eq1 := diff(f(eta), eta, eta, eta)+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;

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-3*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta))) = 0

(1)

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

(2)

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

(3)

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

(4)

Eq5 := diff(theta(eta), eta, eta)+Pr*(f(eta)*(diff(theta(eta), eta))-2*(diff(f(eta), eta))*theta(eta))+N*Pr*(theta1(eta)-theta(eta))/(rho*c*Tt)+N*Pr*Ec*(F(eta)-(diff(f(eta), eta)))^2/(rho*Tv) = 0;

diff(diff(theta(eta), eta), eta)+6.8*f(eta)*(diff(theta(eta), eta))-13.6*(diff(f(eta), eta))*theta(eta)+13.60000000*theta1(eta)-13.60000000*theta(eta)+27.20000000*(F(eta)-(diff(f(eta), eta)))^2 = 0

(5)

Eq6 := 2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+cp*(theta1(eta)-theta(eta))/(c*cm*Tt) = 0;

2*F(eta)*theta1(eta)+G(eta)*(diff(theta1(eta), eta))+2.000000000*theta1(eta)-2.000000000*theta(eta) = 0

(6)

bcs1 := f(0) = r, (D(f))(0) = -1, (D(f))(blt) = 0, F(blt) = 0, G(blt) = -f(blt), H(blt) = k, theta(0) = 1, theta(blt) = 0, theta1(blt) = 0;

f(0) = 2, (D(f))(0) = -1, (D(f))(6) = 0, F(6) = 0, G(6) = -f(6), H(6) = k, theta(0) = 1, theta(6) = 0, theta1(6) = 0

(7)

L := [0.1e-2];

[0.1e-2]

(8)

for k to 1 do R := dsolve(eval({Eq1, Eq2, Eq3, Eq4, Eq5, Eq6, bcs1}, B = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta1(eta)], numeric, output = listprocedure); Y || k := rhs(R[2]); YP || k := rhs(R[3]); YR || k := rhs(R[4]); YQ || k := rhs(R[5]) end do

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

 

R

R

(9)

print([(YP || (1 .. 1))(0)]);

[YP1(0)]

(10)

``

P1 := plot([YP || (1 .. 1)], 0 .. 14, labels = [eta, (D(f))(eta)]):

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

 

plots:-display([P1]);

 

``

``


Download ISPC3.mw

I would like to look at a 3d plot including an condition about the two variables of the plot

For this simple case, I only want to see the plot with the condition that x>0. Is that possible?

plot3d(x*y,[x,0,1],[y,0,1]);

I am trying to simplify the eigenvalues of a 2x2 matrix [[a,b],[c,d]] subject to the condition a,b,c, and d are integers such that a+b = c+d. Why do the following commands not achieve this?

with(LinearAlgebra):

A:=Matrix[[a,b],[c,d]]):

Eigenvalues(A) assuming a::integer,b::integer,c::integer,d::integer,a+b=d+c

How might I achieve what I need?

Hello,

Since I was working in Matlab with Galerkin method which implies periodic boundary conditions I was wondering how to implement this in maple.

I tried this:

restart;

pde2 := diff(u(x, t), t)+3*(diff(u(x, t)^2, x))+diff(u(x, t),x$3) = 0

IBC := {u(0, t) = u(2, t), u(x, 0) = sech(50*(x-1/2))^2+2*sech(30*(x-1))^2, (D[1](u))(0, t) = (D[1](u))(2, t), (D[2](u))(0, t) = (D[2](u))(2, t)}

pds := pdsolve(pde2, IBC, numeric, time = t, range = 0 .. 2)

But it's telling me: 

Error, (in pdsolve/numeric/process_IBCs) initial/boundary conditions can only contain derivatives which are normal to the boundary, got (D[2](u))(0, t)

So what's wrong?

Dear all:

hello everybody;

I need your help to solve the system f(x,y)=0, and g(x,y)=0, such that there some parameter in the system, also all the parameter are positive and also our unkowns  x and y are also positive.

I try to write this code. I feel that under some condition we can have four solution or three or two. I need your help. Many thinks.

 

Systemsolve.mw

i am solving 3 ODE with boundary condition.. with boundary condition

 

b.mw

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/b.mw .

Download b.mw

 

then i got this error

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

i dont know where i need to change.. could you help me..

 

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