# Items tagged with helphelp Tagged Items Feed

### Error : System is undetermined ...

April 17 2014
1 2

Hi everybody.

There are several weeks I'm on a school project and for it we need a Maplesim Simulation. It contains a sine voltage (source), a diode bridge (simplify), a condenser and a resistor. The problem is : when I want to simulate it, an error message appears : System is undetermined. I dopn't understand why I've this and from where's the problem...

If you could help me asap, it will be very very nice.

Have a nice day.

### How to plot a graph...

April 16 2014
1 4

i solve 4 ODE with boundary condition.. i try to plot a graph F(eta) with different value of M.. but it doesnt comeout.. anyone can help me please??

 (1)

 (2)

 (3)

 (4)

 (5)

 (6)

 (7)

 (8)

 (9)

 (10)

 (11)

April 09 2014
0 8

Hello ,

2014_PDE_project.mw

I was using maple as a documentation software and was preparing some notes for my exam.

The file now opens but most of the content is gone. Even the one which was properly saved.

I see Joe Riel 5676 has solved an earlier case like this. could somebody look at this and correct the file and upload it again please.

### Need to solve system of PDEs...

April 07 2014
2 9

Hi, everyone!

I need help.

There are a system of 2 pde's:

diff(Y(x, t), x\$2) = exp(-2*x*b)*(A(x, t)-Y(x, t)), diff(A(x, t), t) = exp(-2*x*b)*(Y(x, t)-A(x, t))

and initial and boundary conditions:

A(x, 0) = 0, Y(0, t) = 0.1, (D[1](Y))(0, t) = 0.

Goal:
For each b = 0, 0.05, 0.1.
1)to plot 3-d  Y(x,t): 0<=x<=20,0<=t<=7.
2)to plot  Y(x,4).

Are there any methods with no finite-difference mesh?

I realized the  methods such as  pds1 := pdsolve(sys, ibc, numeric, time = t, range = 0 .. 7)  can't help me:

Error, (in pdsolve/numeric/match_PDEs_BCs) cannot handle systems with multiple PDE describing the time dependence of the same dependent variable, or having no time dependence

I found something, that can solve my system analytically:
pds := pdsolve(sys), where sys - my system without initial and boundary conditions. At the end of the output: huge monster, consisted of symbols and numbers :) And I couldn't affiliate init-bound conditions to it.

I use Maple 13.

### How do i assing for loop results to table or matr...

April 06 2014
1 6

Hy i need your help experts,

I have à small program for some variables calculations and im intrested to plot loop result,

Here is the program:

I just want to assign the results of my loop to a matrix to be easily ploted.

Thak you for help.

### Fix a syntax error in my simple function (please h...

March 31 2014
0 1

firstly apologies in advance for stuff in this question such as "triangle symbol",  my computer is pretty old.

ok so i was confused a bit here, what i'm trying to do is write a maple procedure that computes Af for a given f contained in V . except we only need to correct the bug in the script below. This script demonstrates such a procedure in the case that omega is a square. The domain is given here as the negative set of a function F contained in V .  I have left in notes where/what i think we need to do but i dunno how to...

N:=10 ; # Global Var
F:=(x,y)->sgn(abs(x-N/2)+abs(y-N/2)-N/4);
Average := proc(F, f0) local f, i, j;
f := f0; # !!!!!!!!!!!!!! something is bad here...
for i to N do for j to N do
if F(i, j) < 0 then
f[i, j] := (f0[i - 1, j] + f0[i + 1, j] + f0[i, j + 1] + f0[i, j - 1])/4 ;
end if;
end do;end do;
return f;
end proc;
f0:=Matrix(N,F); # just to have something to test the procedure
Average(F,f0); # does not return the expected average, modifies f0

the necessary information we were given to produce this so far was..

Let N be a positive integer and [N] = {i contained in N | 1<= i <=N }  Let "Omega" C {(i,j) contained in [N] x [N] | 2<=i,j<=N-1} be a subset. Let V = R^([N]x[N]) be the vector space of real valued functions [N]x[N] -> R
and A, "triangle symbol":V->V (average) and "triangle symbole" (Laplacian) be the linear maps such that
[Af](i; j) = f(i; j)      if (i; j) not contained in "Omega"   OR

[f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 if (i,j) is contained in "Omega"

["traingle symbol"f](i,j) =  0 if (i,j) isnt contained in "Omega"   OR

( f(i,j) - [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 )    if (i,j) is contained in "Omega"

### problem with convergence ...

March 30 2014
0 13

Hi all

I dont know why some Z1 appears on the screen and the code does not converge.
thanks alooooot

restart;
n:=3;
nn:=3;
m:=1;
BB:=1;
BINF:=5:
pr:=7;
digits:=10;
>
eq1:=diff(f(tau),tau\$3)+((3/5)*f(tau)*diff(f(tau),tau\$2))-(1/5)*(diff(f(tau),tau\$1))^2+((2/5)*tau*diff(h(tau),tau\$1))-((2/5)*h(tau))-BB*diff(f(tau),tau\$1)=0;
eq11:=(1/pr)*diff(h(tau),tau\$3)+(3/5)*f(tau)*diff(h(tau),tau\$2)=0;

h(tau):=sum(p^i*h[i](tau),i=0..nn);
f(tau):=sum(p^i*f[i](tau),i=0..n);

H1:= p*(diff(f(tau),tau\$3)+((3/5)*f(tau)*diff(f(tau),tau\$2))-(1/5)*(diff(f(tau),tau\$1))^2+((2/5)*tau*diff(h(tau),tau\$1))-((2/5)*h(tau))-BB*diff(f(tau),tau\$1))+(1-p)*(diff(f(tau),tau\$3)):
H11:= p*((1/pr)*diff(h(tau),tau\$3)+(3/5)*f(tau)*diff(h(tau),tau\$2))+(1-p)*(diff(h(tau),tau\$3)):
>
eq2:=simplify(H1):
eq22:=simplify(H11):
eq3:=collect(expand(eq2),p):
eq33:=collect(expand(eq22),p):
eq4:=
convert(series(collect(expand(eq2), p), p, n+1), 'polynom');
eq44:=
convert(series(collect(expand(eq22), p), p, n+1), 'polynom');
for i to n do
s[i] := coeff(eq4, p^i) ;
print (i);
end do:
for i to nn do
ss[i] := coeff(eq44, p^i) ;
print (i);
end do:
s[0]:=diff(f[0](tau), tau\$3);
ss[0]:=diff(h[0](tau), tau\$3);
ics[0]:=f[0](0)=0, D(f[0])(0)=0, D(f[0])(BINF)=0;
icss[0]:=h[0](BINF)=0, D(h[0])(0)=1, D(h[0])(BINF)=0;

dsolve({s[0], ics[0]});
f[0](tau):= rhs(%);
#dsolve({ss[0], icss[0]});
h[0](tau):= -exp(-tau); #;rhs(%);

>
>
for i to n do
f[ii-1](tau):=convert(series(f[ii-1](tau), tau, nn+1), 'polynom');
h[ii-1](tau):=convert(series(h[ii-1](tau), tau, nn+1), 'polynom');
s[i]:=simplify((s[i]));
ics[i]:=f[i](0)=0, D(f[i])(0)=0, D(f[i])(BINF)=0;
dsolve({s[i], ics[i]});
f[i](tau):=rhs(%);
ss[i]:=(ss[i]);
icss[i]:=h[i](BINF)=0, D(h[i])(0)=0, D(h[i])(BINF)=0;
dsolve({ss[i], icss[i]});
h[i](tau):=rhs(%);

end do;

f(tau):=sum((f[j])(tau),j=0..n);
with(numapprox):

plot(diff(f(tau),tau),tau=0..5,color=blue,style=point,symbol=circle,symbolsize=7,labels=["tau","velocity"]);

### Differential Equations - Quasi Period...

March 20 2014
1 3

Hi, can I get some help with this?

The question is:

Consider the following IVP for a mass of m = 2 kg attached to a spring with a spring constant k = 9 N/m. The spring mass system is in a medium with damping constant b.

2y" + by' + 9y = 0

y(0) = 0, D(y)(0) = -3

It then asks find three values b1, b2, b3 where b1 is underdamped, b2 is critical, b3 is over.

I set b1 as 1, b2 as sqt 72, b3 as 9.

Then it asks to find the quasi period.

I can't get my quasi period right. My answer is 2pi/ sqrt (4.5).

Any help?

### plot curves fucntion help...

March 20 2014
0 3

Hi. Thanks to help me in ths question.

t:=[1,2,3];

f(t[1])=[3,4];  f(t[2])=[11,12];   f(t[3])=[41,1];

How can plot  f  versus t

### problem with system of ode's or maple ?...

March 19 2014
0 2

hi all.
i have a system of ODE's including 9 set of coupled OED's .

i have  converted second deravaties to dd2 , in other words : diff(a[i](t),t,t)=dd2[i](t) . i =1..9 :

and i have set these 9 equations in form of vibrational equations such :  (M.V22)[i]+(K(t).V(t))[i]+P(t)[i] = eq[i] , where M is coefficient Matrix of second  derivatives , V22 is Vector of second derivaties , for example V22[1] = diff(a[1](t),t,t) , and  P(t) is the numeric part of equations ( they are pure number and do not contain any symbolic function ) and K(t).V(t) is the remaining part of equations such that : (K(t).V(t))[i] = eq[i] - (M.V22)[i] - P(t)[i]  , and V(t) are vector of a[i](t)'s which V(t)[1] = a[1](t) ,

i have used step by step time integration method (of an ebook which i have attachted that part of ebook here), when i set time step of solving process to h=0.01 , i can solve this system up to time one second or more, but when i choose h=0.001 or smaller, the answer diverges after 350 steps . i do not know whether the problem is in my ODS system, or maple can not handle this ?the answer about the time t=0.3 are the same in both steps, but after that, the one with stpe time h=0.001 diverges. my friend has solved this in mathematica without any problem, could any body help me ?! it is urgent for me to solve this problem,thnx everybody.

ebook.pdf  step_=_0.001.mw  step_=_0.01.mw

### Matrix maple help...

March 17 2014
0 2

Dear all

Please I need your help to simplify by the coefficient a in this Matrix

I have The matrix A defined by  A:=Matrix(2,2, [[a,a],[3*a,4*a]]);

I want with maple transform A to  A:=a*Matrix(2,2, [[1,1],[3,4]]);

### mapleprimes.com does not send email notifications...

March 15 2014
2 7

why mapleprimes.com does not send me email notifications ?! could anyone help ? i do not receive email notification of related posts.

### Plot Error of ODE. ...

March 14 2014
0 5

Dear all,

h: stepsize;

x in [0,x0];

I give all the step of my code, but I think there is a mistake. I wait for your Help.

I would like to compute the error between  Method Huen with step size h and step size 2h using the definition of epsilon given below:

## The error written epsilon(x0,h)= sqrt(1/(N+1) * sum_i=0^N  (y_i^{2h}-y_(2i)^h)^2 ), where y_i^(2h) is the approximation of y(i*2*h).

## We want : loglog epsilon versus h.

f:=(x,y)=1/(1+cos(y));

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

ic:=y(0)=1;  h:=x0/(2*N);

## Method Heun with step size 2h

> Heun1 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (2*i+2)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[i], y[i]], i = 0 .. N)];

end proc;

### Now Heun with step size h  ( the same h)

> Heun2 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (i+1)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[2*i], y[2*i]], i = 0 .. N)];

end proc;

### IF then statement with nested loop....

February 19 2014
0 1

Rewrite the code that counts the number of primes less than using an if-then statement.  Implement    your code where j goes from 2 to 15.

j:=1;
1
for i from 2 to 10
while ithprime(j)<2^i do
j:=j+1
end:
print(2^i,primes=j-1):
2048, primes = 9

I need to edit this code to satisfy a IF then Statement. can any one help me out?

\

regards "Geordi"

### Error in writing own function...

February 10 2014
0 3

M:= Matrix ([[a,b], [c,d]])

If a.d-b.c=0 then print (Matrix(2)); else print (Matrix ([[(d/a.d-b.c), (-b/a.d-b.c)], [-c/a.d-b.c, a/a.d-b.c]]); end if;

I don't understand where I have gone wrong. If anybody could help I would be greatly appreciative

Thank you

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