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complicated wave equation ...

Asked by: mostibra 20
pde wave
May 12 2015
0  1

now the equation is

d2u/dt2-(2*d2u/x2)+d2u/dxdt=0    

initial condition: u(x,0)=1-(xsign(x)), abslute x<1,0 otherwise. Assume sign(x)=-1 for x<0, 1for x>0 

 Ut(x,0)=cos(pix), bslute x<1, 0 otherwise , he didnt give any B.Cs

so I would like to know the analytical and numerical sols, and plots for the wave at t=2,4

for Numerical:   delta x=0.1, delta t=0.025, range 0..4

How can error convergence be corrected?...

Asked by: emmantop 40
numeric bvp
May 12 2015
0  9

Good day,

How can this be corrected ''Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging'' see the worksheet here VT.mw

from a procedure to sub_procedure?????? how can I ...

Asked by: ben maths 20
procedure programming
May 10 2015
0  11

Hi all,
can you help in that please?
How can I use this small procedure (root_of_cheb) as a sub-procedure in  the next procedure (EvalInt) ?
is it possible?

restart:
root_of_cheb:=proc(n)
   local xk,b,k:
   xk:=(k,n)->cos((2*k-1)*Pi/(2*n));
   sort([seq(evalf(xk(k,n)),k=1..n)]):
end:
EvalInt:=proc(f,n)
   local xk:
   xk:=(k,n)->cos((2*k-1)*Pi/(2*n));
   evalf((Pi/n)*add(f(xk(i,n)),i=1..n)):
end:

Thank you

Hfloat error in solving partial differential equat...

Asked by: mobiusinfi 40
pdsolve
May 10 2015
1  7

with(PDEtools);
Es := 0.117108e12;
Ef := 0.78125e11;
l := 0.150e-6;
s := 0.500000e-3;
f := 0.5898334197e-6;
o := 0.9e-5;
d := 0.10e-17;
cb := 0.1e7/(19.9);
R := 8.3144621;
T := 298;





PDE := diff(u(x, t), t)-(diff(u(x, t)+o^2*Es*cb*u(x, t)^2/(9*R*T), x, x)) = 0;
IBC := {u(1, t) = 1, u(x, 0) = 0, (D[1](u))(1, t) = l*f/(d*cb)};
S := pdsolve(PDE, IBC, numeric, time = t, range = 0 .. 1, timestep = 0.1e-4, spacestep = 0.1e-6);
p1 := S:-plot(t = .1, numpoints = 100);
Error, (in pdsolve/numeric/plot) unable to compute solution for t>HFloat(0.0):
matrix is singular
p2 := S:-plot(t = .2, numpoints = 50, color = green);
Error, (in pdsolve/numeric/plot) unable to compute solution for t>HFloat(0.0):
matrix is singular
p3 := S:-plot(t = .3, numpoints = 50, color = blue);
Error, (in pdsolve/numeric/plot) unable to compute solution for t>HFloat(0.0):
matrix is singular
plots[display]({p1, p2, p3});
Error, (in plots:-display) expecting plot structures but received: {p1, p2, p3}


Is there a way to increase Maple's hard limits, or...

Asked by: spark1631 65
stacklimit
May 07 2015
0  3

I want to increase the stack limit. but i can not raise it above the hard limit....

 

so i wonder whether there is a way to increase the hard limit, or at least tell me how much is it?

How to use Verlet algorithm to derive the orbit of...

Asked by: physics0150 0
homework dynamics stationary
May 06 2015
0  0

Derive the orbit of the Moon around the Earth by doing a Verlet algorith of Molecular Dynamics simulation. Use one hour for your step τ. Place the stationary Earth at the origin of the Cartesian system. For initial conditions, use the position and the speed of the Moon when it is at its apogee (furthest from Earth). Plot the orbit.

How to solve Cauchy sequences...

Asked by: laporte bernard 45
May 05 2015
0  2

(a) Show that if {an} ∞ n=1 is Cauchy then {a 2 n} ∞ n=1 is also Cauchy. (b) Give an example of a Cauchy sequence {a 2 n} ∞ n=1 such that {an} ∞ n=1 is not Cauchy

Open maple worksheet...

Asked by: Annonymouse 160
worksheet incomplete-question
May 04 2015
0  0

I am running Maple in a windows virtual machine, on a mac computer.

I have a number of worksheets on its disk

Windows advised me to run its error checking utility (chkdsk)

when I try and open them it gives me a number of options:

maple text

plain text 

and maple input

 

None of these are the same as the original files. What has happened? and how can i fix it?



Length of output exceed limit 1000000...

Asked by: Sofey 10
May 03 2015
0  3

can anybody help me? i want to check the consistency of my scheme. My equation is too long if i check manually, so i used maple 13 to simplify my equation. But it cannot simplify it because of length of output exceed limit 1000000

restart

eqn1 := u+(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2)+(u-(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2))-u-(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)-u+(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3))*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3));

(1-exp(-m))^2*u[tt]-(1-exp(-m))^2*u[xx]+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-16*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+4*(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3))

 (1)

a := simplify(eqn1);

`[Length of output exceeds limit of 1000000]`

 (2)

``


Download consistency_expmle_4.mw

.

how to prove with maple...

Asked by: laporte bernard 45
student
May 03 2015
0  1


> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

How can PDE be solved in terms of Bessel function?...

Asked by: emmantop 40
plot
May 02 2015
0  4

Good day everyone,

please how can one solve this pde in terms of Bessel function or any other analytic solution with the plot.

See the file ID.mw

Thanks.

how to copy and paste from maple 12 a quite big qu...

Asked by: laporte bernard 45
April 30 2015
0  4

Hi, as I can't manage to copy and paste on mapleprimes, I would be glad to get a hint ...

Maple V please help SOS urgently need help...

Asked by: c4meleon 30
April 30 2015
0  0

Hello,
I am looking here for a tutor, that can help me doing some Maple V (5) programming.
I am a mathematic student and we use Maple programming.
I will be happy to pay a small amount of money for each exercice you help me doing,

if anyone is interested, please contact me here.

(The procedures that we usually have to write are for example:
 Newton-Raphson Method, Chebyshev Polynomial,...  I don't think it is hard for you.
Thank you very time for your time and your help.

convert 2-D expression to 1-D...

Asked by: habex 15
2dmath worksheet editing
April 26 2015
2  1

Hello, I would like to customize the context menu in Maple 18 and looking for a way, to convert 2-D expressions to classical maple input using command(s) (same function as "2-D Math > Convert To > 1-D Math Input"). My main problem is, that using the standard procedures in the context menu I get the result of an expression, not the expression itself.

Can you give me an idea, how to do this? Thanks, Csaba

error appear in maple ...

Asked by: franky12 10
ode numeric
April 26 2015
0  1

@Carl Love Dear Sir i am trying to solve the system of nonlinear ODE equation with semiinfinite domain by using shooting method . but i am receiving following error 

"Error, (in isolate) cannot isolate for a function when it appears with different arguments"

how can i remove this error. i am unable to find the mistake. kindly help me 

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