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I am trying to use solve to determine 5 unknowns from 12 equations each with seperate data points); however, maple requires you to have equal number equations and variables. Is there any way around this?

Hi there,

I've got the following differential equation system:,

dU/dt = delta·dotD -lambda·U - kappa·U^2
dL/dt = (1-phi)·lambda·U + 1/4 ·kappa·U^2


being phi, delta, kappa, lambda, kappa some fixed parameters of the system, and where dotD (the derivative wrt time of a function D), which is defined a piecewise funtion:

dotD(t)=1/(3·T1)·DT for t in [0,T1]

dotD(t)=2/(3·(T2-T1-T))·DT for t in [T1+T,T2]

where T and DT are also known, and T1 approaches 0, and T2 approaches T1+T.

Setting the equation system in Maple and trying to solve it, gives a NULL result. However, trying to solve each piece separately seems to work fine.

Why is this?

 

Furthermore, taking limits for the [T1+T,T2] part (having solved each piece separately) yields an invalid limits point error. Ain't the possibility to take limits for both parameters at the same time?

Any ideas?

 

This is the Maple worksheet: MaplePrimes_LQ_model_solve.mw

Thank you.

jon

Hi

y''+(4/x)*y'-a*y^n=0

that a=constant


Hi there,
I have a set of differential equations whose solution, Jacobian matrix and its eigenvalues, direction field, phase portrait and nullclines, need to be computed.

Each of the equations has a varying parameter.

I know how to get the above for a single parameter value, but when I set a range of values for the parameters, Maple is not able to handle all cases as I would expect: solving the differential equation system:

eq1 := x*(1.6*(1-(1/100)*x)-phi*y)
eq2 := (x/(15+x)-0.3e-1*x-.4)*y+.6+theta
desys := [eq1, eq2];
vars := [x, y];
steadyStates := map2(eval, vars, [solve(desys)])

already yields an error:
Error, (in unknown) invalid input: Utilities:-SetEquations expects its 2nd argument, equations, to be of type set({boolean, algebraic, relation}), but received {-600*y+(Array(1..2, {(1) = 8400, (2) = 15900})), Array(1..5, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0})}


The equations are the following:
de1 := diff(x(t), t) = x(t)*(1.6*(1-(1/100)*x(t))-phi*y(t));
de2 := diff(y(t), t) = (x(t)/(15+x(t))-0.3e-1*x(t)-.4)*y(t)+.6+theta

the parameters being:
phi:=[0 0.5 1 1.5 2]
theta:=[5. 10.]

How can I handle the situation so that Maple computes each of the above for each combination of the parameters?

I would like to avoid using two for loops and having to store all results in increasingly bigger and complicated arrays.

The worksheet at issue is this: MaplePrimes_Tumour_model_phi_theta_variation.mw


Thanks,
jon

Hi all,

I am trying to solve the following differential equation numerically using dsolve,

 

y * abs (y''') = -1

y(0) =1, y'(0) = 0, y''(0)=0

 

it works fine when tthe absolute function is not there, i.e. yy''' = -1. 

Do you have any suggestion?

Hello friends;

I uploaded the file. Firstly i solved the 4. order ode. I converted it to trigonometric functions from exponantial.

I have 4 equation and 4 unkowns. I thought , it is sufficient to solve this. As you see , i can not solve the system .

May you help me please where i am doing wrong. 

Thanks.

odev4.mw

(PS:i downloaded the file again correctly)

Hi all!

 

I do a small calculation and get a system of 6
nonlinear equations.
And "n" is the degree of the equation is float.

Here are the calculations that lead to the system.

 

restart;
 with(DirectSearch):
 B:=1: 
 q:=1: 
 l:=1: 
 n:=4.7:
 V:=0.05:
 N:=1200:
 
 
 kappa:=Vector(N+1,[]):
 theta:=Vector(N+1,[]):
 u:=Vector(N,[]):
 M:=Vector(N,[]):
 Z:=Vector(N,[]):
 
 M_F:=q*(6*l*(z-l)-z^2/2):
 M_1:=piecewise((z<l), l-z, 0):
 M_2:=piecewise((z<2*l), 2*l-z, 0):
 M_3:=piecewise((z<3*l), 3*l-z, 0):
 M_4:=piecewise((z<4*l), 4*l-z, 0):
 M_5:=piecewise((z<5*l), 5*l-z, 0):
 M_6:=6*l-z:
 M_finish:=(X_1,X_2,X_3,X_4,X_5,X_6,z)->M_1*X_1+M_2*X_2+M_3*X_3+M_4*X_4+M_5*X_5+M_6*X_6+M_F:
 
 
 kappa_old:=0:
 theta_old:=0:
 u_old:=0:
 M_old:=0:
 
 
 step:=6*l/N:
 u[1]:=0:
 kappa[1]:=0:
 theta[1]:=0:
 
 
 
 
 for i from 2 to N do
 
 z:=i*step:
 kappa_new:=kappa_old+B/V*(M_finish(X_1,X_2,X_3,X_4,X_5,X_6,z))^n*step:
 
 theta_new:=theta_old+1/2*(kappa_old+kappa_new)*step:
 
 u_new:=u_old+1/2*(theta_old+theta_new)*step:
 
 Z[i]:=z:
 kappa[i]:=kappa_new:
 theta[i]:=theta_new:
 u[i]:=u_new:
 kappa_old:=kappa_new:
 theta_old:=theta_new:
 u_old:=u_new:
 
 end do:
 
 So,my system:


 u[N/6]=0;
 u[N/3]=0;
 u[N/2]=0;
 u[2*N/3]=0;
 u[5*N/6]=0;
 u[N]=0;

 

I want to ask advice on how to solve the system.
I wanted to use Newton's method, but I don't know the initial values X_1..X_6.

Tried to set the values X_1..X_6 and to minimize the functional
Fl:=(X_1,X_2,X_3,X_4,X_5,X_6)->(u[N/6])^2+(u[N/3])^2+(u[N/2])^2+(u[2*N/3])^2+(u[5*N/6])^2+(u[N])^2:

with the help with(DirectSearch):
GlobalOptima(Fl);
But I don't know what to do next

Please, advise me how to solve the system! I would be grateful for examples!

 

Hello people in mapleprimes,

 

I have a question of labels of equations.

For example, suppose that there is a following description in a worksheet.

>2*x+y=5;

  2*x+y=5   (1)

>3*x+y=6;

  3*x+y=6   (2)

>solve({(1),(2)},{x,y});

 {x=1,y=3}          (3)

subs((3),100+x+100*y)

    400   

 

And, when I moved (2) to the next to (1) with dragging, and then, clicked !!! to execute it, then the worksheet 

changes as follows, which is wrong as accompanying ??

>2*x+y=5;3*x+y=6;

  2*x+y=5

 3*x+y=6   (1)

>solve({(1),??},{x,y})

{x = x, y = -3*x+6}          (2)

>subs({(1),??},100*x+100*y)

-200*x+600

 

Do you know some good way for ?? not to appear in the above sitiation.

I will appreciate when you tell me some measures about this.

 

Best wishes.

 

taro

 

 

 

 

 

 

I want to solve maximize of equation,but the maximize failed to solve it,who can help me.thanks.

c[1] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)}:

c[2] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+[1/(z^2+1)]^(3/2)+1}+(1/8)*{y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1)}:

t[1] := diff(c[1], x);

(1/8)*w*{-(3/2)*(1/((x+y+z)^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/((x+y)^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/((x+z)^2+1))^(1/2)*(2*x+2*z)/((x+z)^2+1)^2-3*(1/(x^2+1))^(1/2)*x/(x^2+1)^2}+(1/8)*{1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2}

(1)

t[2] := diff(c[2], y);

(1/8)*w*{-(3/2)*(1/((x+y+z)^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/((x+y)^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/((y+z)^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}+(1/8)*{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}

(2)

eliminate({t[1], t[2]}, w);

[{w = -{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}/{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}}, {{1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}-{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(x^2+2*x*z+z^2+1))^(1/2)*(2*x+2*z)/((x+z)^2+1)^2-3*(1/(x^2+1))^(1/2)*x/(x^2+1)^2}}]

(3)

w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*sqrt(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*sqrt(1/(x^2+2*x*y+y^2+1))*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*sqrt(1/(y^2+2*y*z+z^2+1))*(2*y+2*z)/((y+z)^2+1)^2-3*sqrt(1/(y^2+1))*y/(y^2+1)^2);

w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2)

(4)

sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[1]);

sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)})

(5)

subs(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[2]);

-(1/8)*(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+[1/(z^2+1)]^(3/2)+1}/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2)+(1/8)*{y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1)}

(6)

"#"Iwant to maximize the equation (5)and (6),under the conditon of x,y,z are negative or positive at the same time.

 

NULL

 

Download maximize.mw

Could anyone assist in rectifying this error ''Error, (in fsolve) {f[1], f[2], f[3], f[4], f[5], f[6], f[7], f[8], f[9], f[10], f[11], theta[11]} are in the equation, and are not solved for''. Here is the worksheet FDM_Revisit_1.mw

I have a characteristic equation. some times It has polar roots . sometimes It has real roots and sometimes both of them.

I want to extract real roots and extract polar roots if they are.

for instance:

q:=m3*r^3+m2*r^2+m1*r+m0:

rot:=solve(q=0,r);

I want to know how can I use if in this part ?

Dear all;

I need your help to get all the solution of this nonlinear  system  with four parameters: r1, r2, q1, q2  assumed to be positives. Let the system:

the first equation is: r1*x*(1-x/q1)=x*y/(1+x)

the second equation: r2*y*(exp(-y)-q2)=-x*y/(1+x)

How can get the positive solution ( x, y) of the previous system.

Thank you for your help.

Hello people in mapleprimes,

I want to solve the next system of equation for B/A and C/A.

eq1:=A+B=F+G;
eq2:=k*(A-B)=kappa*(F-G);
eq3:=F*exp(I*kappa*a)+G*exp(-I*kappa*a)=C*exp(I*k*a);
eq4:=kappa*F*exp(I*kappa*a)-kappa*G*exp(-I*kappa*a)=k*C*exp(I*k*a);


But, though it is well-known, solve({eq1,eq2,eq3,eq4},{B/A,C/A})
does not work well, as the values I want to solve it for are
expressions: B/A and C/A not variables.

Then, you might thing the next works well.
eq:=subs({B=A/t,C=A/u},{eq1,eq2,eq3,eq4}):
solve(eq,{t,u});

But, this doesn't work well, with the answer was
only the ratio of t and u expressed as the following:

t = t, u = exp(I*k*a)*(exp(-I*kappa*a)*k^2-exp(I*kappa*a)*k^2-exp(-I*kappa*a)*kappa^2+exp(I*kappa*a)*kappa^2)*t/(4*kappa*k*exp(I*kappa*a)*exp(-I*kappa*a))

Isn't there nice way to solve the above system of equation, except that
sol1:=solve({eq3,eq4},{F,G});assign(sol1);
sol2:=solve({eq1,eq2},{A,B});assign(sol2);

Best wishes
taro

hello, i went solve these equation ,with a & b take any value

b*x*ln(x)-x*ln(a)+a=0

thank you

Hello all,

I have the following equation:

N*exp(-(1/2)*eta*epsilon*(N*alpha*epsilon*w+2*N*w*C[max]-alpha*epsilon*z-2*Q1*alpha)/(w*N))*S1*upsilon*w-N*S1*upsilon*w+K1^2*alpha*eta*z*epsilon+K1*alpha*eta*z*epsilon*S1 = 0

in which I need to find solution for epsilon (analytical solution) when epsilon>0.  

Thanks,

Dmitry

 

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