Hello, my problem is that I've changed in tools maple default mode to "Maple input" from "2-d" I've aplied it globbaly and it doesnt'work. Still when type Enter it switch automatically to 2-d Math, and I have to change it by Ctrl-M to 1-d. Any ideas how to solve it? I want 2-d just to output.

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 

i solved my equation as follow , i want to know a compelet describtion that which method of pde/numeric methods is using .

thanks 

In Maple15, in spherical coordinates, what statements plot covariant vector [1,r^2*sin(theta),sin^2(theta)] and its contravariant equivalent [1,sin(theta),1/r^2] within their respective bases vectors. Assume r = 2, theta = pi/6 and phi = pi/4. I presume each plot will display the same space vector.

M := Matrix([[0,1,1,1,1],[1,0,1,1,1],[1,1,0,1,1],[1,1,1,0,1],[1,1,1,1,0]]);

how to multiply M[1] which is [0,1,1,1,1] with Matrix([[x0], [x1], [x2], [x3], [x4]])
to become [0,x1,x2,x3,x4]

may not use Matrix([[x0], [x1], [x2], [x3], [x4]])

just want to use first element as x0 when it is not 0

second element as x1, etc

Hello friends,

I try to write a loop in Maple.

The equation is f (x) = x + 3

x is a variable and should be from 50 to 70 with dx = 5 can be used in the function, so that f (x) is to be determined.

How can I write such a loop?

Hi

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

that a=constant

LM:=proc(n)
local L;
uses combinat;
L:=permute([1$(n*(n-1)/2), 0$(n*(n-1)/2)], n*(n-1)/2);
[seq(Matrix(n,{seq(seq((i+1,j)=L[k][(i-1)*i/2+j], j=1..i), i=1..n-1)}, shape=symmetric), k=1..nops(L))];
end proc:
M := LM(5);
N := nops(M);
append("E:\\mm.txt");
for i from 1 to N do
ExportMatrix( "E://mm.txt", M[i]);
od:

hope to export to a text file

and show

for example

matrix([[1,1,1],[1,1,0],[0,0,0]]);

matrix([[1,1,1],[1,1,0],[0,0,0]]);

matrix([[1,1,1],[1,1,0],[0,0,0]]);

...

etc

is it possible to generalize a function to a combinatorial level for approximate axioms

for example, first 100 or 1000 data points satisfy axioms

or 100% satisfy a axioms which means satisfy to infinity

because i find data always not exactly satisfy the axioms,
i guess it only satisfy to some limit, this may explain why data has decimal number

or conversely is it possible to generalize some axioms which approximate the original exact axioms
then data can exactly satisfy the approximate axioms

can generalize a nested forloop to achieve this goal?

how can it be done in algebra point of view?

For example:

x*y = for loop -> for loop -> i*j

it can change for loop expression into algebra

for i from 1 to 10 do
for j from 1 to 10 do
print i*j
od:
od:

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