Hi

 Mt := 1-(sum(4*l^2*exp(-beta[n]^2*tau)/(beta[n]^2*(l^2+beta[n]^2)), n = 1 .. infinity))

where the beta[n]s   are  roots of :

beta[n]*BesselJ(1, beta[n])-l*BesselJ(0, beta[n]) = 0

for  l=1,10,20,40,50,100

I want to plot Mt vs. tou for these l 's  in one diagram

Hi,

I have a problem solving two equations.  They are as follows:

s := 1/(273.16+50); s1 := 1/(273.16+145); s3 := 1/(273.16+250); s2 := 1/(273.16+197.5); gamma0 := 0.1e-3; gamma1 := .5; gamma2 := 0.15e-2; beta := -3800:

c := 300; n := 200; tau1 := 99; tau2 := 120;

Delta := solve(1-exp(-(gam0*tau1+(1/2)*gam1*tau1^2)*exp(beta*s1)) = 1-exp(-(gam0*a+(1/2)*gam1*a^2)*exp(beta*s2)), a);
a := Delta[1];

Theta := solve(1-exp(-(gam0*(a+tau2-tau1)+(1/2)*gam1*(a+tau2-tau1)^2)*exp(beta*s2)) = 1-exp(-(gam0*b+(1/2)*gam1*b^2)*exp(beta*s3)), b);
b := Theta[1];

n1 := int((gam1*t+gam0)*exp(beta*s1)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s1)), t = 0 .. tau1);
n22 := (n-n1)*(int((gam1*t+gam0)*exp(beta*s2)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s2)), t = a1 .. a1+tau2-tau1));
n2 := eval(n22, a1 = a);
n33 := (n-n1-n2)*(Int((gam1*t+gam0)*exp(beta*s3)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s3)), t = b1 .. c));
n3 := eval(n33, a1 = a);
n4 := n-n1-n2-n3;

g1 := -n1*(Int((1/(gam1*t+gam0)-t*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-n2*(Int((1/(gam0+gam1*(a+t-tau1))-(a+t-tau1)*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2))-n3*(Int((1/(gam0+gam1*(b+t-tau2))-(b+t-tau2)*exp(s3))*(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3)), t = tau2 .. c))+(n-n1-n2-n3)*(1/(gam0+gam1*(b+c-tau2))-(b+c-tau2)*exp(s3))*(gamma0+gamma1*(b+c-tau2)+gamma2*(b+c-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+c-tau2)+(1/2)*gamma1*(b+c-tau2)^2+(1/3)*gamma2*(b+c-tau2)^3)*exp(beta*s3));

g2 := -n1*(Int((t/(gam1*t+gam0)-(1/2)*t^2*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-n2*(Int(((a+t-tau1)/(gam0+gam1*(a+t-tau1))-(1/2)*(a+t-tau1)^2*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2))-n3*(Int(((b+t-tau2)/(gam0+gam1*(b+t-tau2))-(1/2)*(b+t-tau2)^2*exp(s3))*(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3)), t = tau2 .. c))+(n-n1-n2-n3)*((b+c-tau2)/(gam0+gam1*(b+c-tau2))-(1/2)*(b+c-tau2)^2*exp(s3))*(gamma0+gamma1*(b+c-tau2)+gamma2*(b+c-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+c-tau2)+(1/2)*gamma1*(b+c-tau2)^2+(1/3)*gamma2*(b+c-tau2)^3)*exp(beta*s3));

solve({g1 = 0, g2 = 0}, {gam0, gam1});

Warning, solutions may have been lost.

What do I do wrong?

Thanks for advice in advance.

Hi there,

I am trying to maximize a function given a set of values to a parameter in the function. The function is an differential equation belonging to a system of two differential equations.

I have a for loop to state different values to the parameter.

Maple yields the error:

Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

When trying to maximize the function.

Supposed that I was doing something wrong in the loop, if I reproduce the contents of the loop outside, and set a value for the parameter. If I plot the solution of the ordinary differential equation, I can see where the maximum lies.

Having plot it, the Optimizamtion:-Maximize works as expected.

However, omitting the plot has a weird effect: I only get the same result depending on the bounds I set for the Maximization:

de1 := diff(A(t), t) = r*m*(1-g)*A(t)-piecewise(t < 8, r*A(t), t >= 8, (r+k)*A(t));
de2 := diff(G(t), t) = r*m*g*A(t)-l*G(t);

ics := A(0) = 25.0, G(0) = 0.;
num := dsolve({de1, de2, ics}, {A(t), G(t)}, type = numeric, output = listprocedure, parameters = [g]);

num(parameters = [g = .15]);
val := eval(G(t), num);

# odeplot(val, [t, G(t)], t = 0 .. 100);

Maximize(val);
Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

val2 := Maximize(val);

Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

val3 := Maximize(val(t), t = 0 .. 60);

  [10267.824035766165, [t = 8.25727747134303]]

val4 := Maximize(val(t), t = 0 .. 100);

[6.863211343195069e-9, [t = 59.84184367042171]]

The right answer is [10267.824035766165, [t = 8.25727747134303]]: Why do I get two different answers even if in that range there is only one relative maximum?

I ignore whether the way I am specifying the arguments for the Maximize function is correct. val is a procedure.

What am I missing?

Attached is the worksheet: MaplePrimes_malaria_param_variation_2.mw

Thanks,

jon

Hi all!

F is a delta function:

F:=delta(x-x[0])*delta(y-y[0])

I want it be expaned through trigonometric series:

F:=sum(sum(Q[k*l]*sin(l*Pi*x/a)*sin(k*Pi*y/b), k = 1 .. infinity), l = 1 .. infinity)

So I want to get every Q:

Q[k, l] := `assuming`([4*(int(int(f[z1]*sin(l*Pi*x/a)*sin(k*Pi*y/b), x = 0 .. b), y = 0 .. a))/(a*b)], [k::posint, l::posint, a > 0, b > 0])

But it result in (when x[0]:=a/2, y[0]:=b/2):

4*(int(int(F[0]*exp(I*omega*t)*delta(x-x[0])*delta(y-y[0])*sin(l*Pi*x/a)*sin(k*Pi*y/b), x = 0 .. b), y = 0 .. a))/(a*b)

I wonder HOW CAN I GET THE EXACT RESULT:Q[k, l] := 4*sin(l*Pi/a)*sin(k*Pi/b)/(a*b)

THANKS!

http://en.wikipedia.org/wiki/Hypersurface

http://people.cs.uchicago.edu/~niyogi/papersps/surfacesampling.pdf

hypersurface is a homogenous polynomial

f(x,y) = 0

i do not understand how sampling hypersurface can generate this kind of polynomial

Dear All,

i am solving a system of pde with boundar conditons then i got this error...

Error, (in pdsolve/numeric/plot) unable to compute solution for tau>HFloat(0.0):

Thank.

jeffrey_fluid.mw

Download jeffrey_fluid.mw

Good day everyone, could you please help use Gauss Elimination method for these system of equations. See the worksheet here F1.mw

Thanks.

restart; with(linalg); with(stats); with(plots); with(Statistics); with(LinearAlgebra); 

s := 1/(273.16+50); s1 := 1/(273.16+145); s3 := 1/(273.16+250); s2 := 1/(273.16+197.5); gamma0 := 0.1e-3; gamma1 := .5; gamma2 := 0.15e-2; beta := -3800;
c := 300; n := 200; tau1 := 99; tau2 := 120;

Delta := solve(1-exp(-(gam0*tau1+(1/2)*gam1*tau1^2)*exp(beta*s1)) = 1-exp(-(gam0*a+(1/2)*gam1*a^2)*exp(beta*s2)), a);
a := Delta[1];

Theta := solve(1-exp(-(gam0*(a+tau2-tau1)+(1/2)*gam1*(a+tau2-tau1)^2)*exp(beta*s2)) = 1-exp(-(gam0*b+(1/2)*gam1*b^2)*exp(beta*s3)), b);
b := Theta[1];

n1 := n*(int((gam1*t+gam0)*exp(beta*s1)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s1)), t = 0 .. tau1));
200. - 200. exp(-0.01119474511 gam0 - 0.5541398828 gam1)
n2 := (n-n1)*(int((gam1*t+gam0)*exp(beta*s2)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s2)), t = a .. a+tau2-tau1));

g1 := -n1(gam0, gam1)*(int((1/(gam1*t+gam0)-t*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-evalf(n2(gam0, gam1)*(int((1/(gam0+gam1*(a+t-tau1))-(a+t-tau1)*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2)))

g2 := -n1*(int((t/(gam1*t+gam0)-(1/2)*t^2*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-evalf(n2*(int(((a+t-tau1)/(gam0+gam1*(a+t-tau1))-(1/2)*(a+t-tau1)^2*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2)))

solve({g1 = 0, g2 = 0}, {gam0, gam1})

I want to find the answer of gam0 and gam1. It takes me 20 hours until now...and still evaluating...

Please Help ..

I can't directly copy Maple code into messages but have to insert into Word first, apply some changes that for better readability and copy it from there via Word-paste. Is there a direct way?

I am trying separation of variables in Maple. I get an equation that has the form   

And now I want to tell Maple to assign the terms with 1/R in them to one variable, say eq21, and the term with 1/Z to second variable, say eq22 The idea is that I can later more easily work with each separate ode. 

I do not know to separate those apart. I could offcourse copy and paste by hand, but I want to automate this.

I tried match() and patmatch() but I am not seeing the way. Here is the code:

restart;
T:=  (r,z)-> Z(z)*R(r);
eq1:= diff(T(r,z),r$2)+1/r*diff(T(r,z),r)+diff(T(r,z),z$2);
eq2:=expand(eq1/(Z(z)*R(r)));

#need now way to break the above into 2 different variables.

ps. I know I can do this:

restart;
T:=  (r,z)-> Z(z)*R(r);
eq1:= diff(T(r,z),r$2)+1/r*diff(T(r,z),r)+diff(T(r,z),z$2);
eq2:=expand(eq1/(Z(z)*R(r)));
eq3:=collect(eq2,1/R(r));
eq21:=op(1,eq3);
eq22:=op(2,eq3);

But this for me is not the right way to do it. I think there should be more algebraic way.

there is a solution of equation,so the equation can be divided by the solution,but because the equation is complex,it can't be simplify by the soution,can anyone give me some help?thanks a lot.

Buenas

Alguien tiene el planteamento de la ecuacion 1D HEAT EQUATION en maple?

Me pueden enviar?

Gracias

Saludos

En anexo esta el enunciado

a curve has residual p if it is linked, in a complete intersection, to a curve with residual p-1

0 residual if is a complete intersection of two surfaces

do complete intersection means two surfaces totally overlapped?

why they are not the same one if complete intersection?

test.mw

In this file, I tried my best to solve the pde. But the answer is still rather non-informative. I need some help to simplify it.

I did notice that my Maple might need reinstallation, due to a "bug" in the 18.02 update.

My ultimate aim is try to use some similar techniques to solve this, test2.mw, which has a similar type pde.

The standard pdsolve(pde) would just not work.

UPDATE:

I used the same file in Maple 17 on a differnt Machine, which can be solved by pdsolve. So I guess it's just that  the 18.02 update package is broken itself. I have tried to uninstall and reinstall twice.

Thanks,

casper

I want to solve a system of 11^3 equations  (each equation is quite small) in 11^2 unknows. The equations are all viewed mod 2. I only want to know the possible solutions for 11 of those variables.

That is given variables a[i], 1 from 1 to 11 and b[n] for the remaining 11^2-11  variables.

I want to know all the  distinct consistent solutions for the a[i]'s. I dont care what the b's are as long as the solution to the a[i]s is consistent.

I am VERY sure that the equations will quickly show that there are no solutions but I am having difficulty comming up with how to code this.