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Here's a short tensor manipulation which goes totally bananas. Basically I have a metric, and I define a vector k, and right at the end I calcualte the covariant derivative of it. In the metric and elsewhere I have a constant epsilon, a function of time a(t), and 2 functions of all coords, Phi and psi. Then at the end it gives the covariant derivative of k, but now with epsilon and a as functions of all coordinates. 

Any idea what's going on??

 

code:

 

restart;
with(Physics);with(PDEtools):
Setup(coordinatesystems=spherical);

ds2:=expand(a(t)^2*(-(1+2*epsilon*Phi(r,theta,phi,t))*dt^2+(1-2*epsilon*psi(r,theta,phi,t))*(dr^2+r^2*dtheta^2+r^2*sin(theta)^2*dphi^2)));
declare(%,(H)(t),(k0,k1,k2,k3)(X));
Setup(metric = ds2);

Define(k[~mu]=[-1/a(t)^2+epsilon*k1(X),epsilon*k2(X),epsilon*k3(X),1/a(t)^2+epsilon*k0(X)]);

Define(NullGDE[nu]=(k[~mu](X)*D_[mu](k[nu](X))),OpticalDeformationMatrix[mu,nu]=(D_[mu](k[nu](X))));

(OpticalDeformationMatrix[1,1,nonzero]);

 

the final output is Matrix(1, 1, [[(-(2*((diff(epsilon(X), r))*(psi(X))(X))+2*epsilon(X)*(diff((psi(X))(X), r)))*(epsilon(X)*(k1(X))(X)*((a(t))(X)^2)-1)*a(t)-(2*epsilon(X)*(psi(X))(X)-1)*((diff(epsilon(X), r))*(k1(X))(X)*((a(t))(X)^2)+epsilon(X)*((diff((k1(X))(X), r))*((a(t))(X)^2))+2*epsilon(X)*(k1(X))(X)*(a(t))(X)*(diff((a(t))(X), r)))*a(t)+epsilon*psi[r]*(epsilon*k1(X)*(a(t)^2)-1)*a(t)-epsilon^2*psi[theta]*(a(t)^3)*k2(X)-epsilon^2*psi[phi]*(a(t)^3)*k3(X)-(2*(diff(a(t), t))*psi(X)*epsilon+psi[t]*a(t)*epsilon-(diff(a(t), t)))*(epsilon*k0(X)*(a(t)^2)+1))*(1/a(t))]])

where epsilon is now a function!

 

 

 

This may be a silly question, but does there exist some simple way of (Taylor) expanding an expression of 'small' functions in terms of these functions.

A simple example: Assume that diff(f(x),x) and g(x) are two functions both with range, say, in [-a,+a], where a << 1, and consider the following expression:

sqrt(1 + diff(f(x),x)) * (2 + g(x));

Its expansion to first order in terms of diff(f(x),x) and g(x) should be 2 + diff(f(x),x) + g(x). My problem is that mtaylor does not accept functions as variables to expand on, and I would prefer not to have to substitute back and forth with some 'placeholders'.

The help text for dsolve,numeric,events describes many kinds of triggers and actions and other event specifications, but shows only a very limited selection of examples of these possibilities.

Please tell me of any sources containing extensive examples of the wide variety of available specifications and the problems they are useful in solving.

> restart:with(plots):blt:=7:

 

 

> lambda:=2:m:=3:s:=1:

> Eq1:=(diff(f(eta),eta,eta,eta))+(f(eta)*(diff(f(eta),eta,eta)))-((diff(f(eta),eta))^2)+lambda*(((f(eta)*(diff(f(eta),eta,eta,eta))))-2*(diff(f(eta),eta))*(diff(f(eta),eta,eta,eta))^2)-(M/(1+m^2))*((diff(f(eta),eta)+ms))=0;

 

>

> Eq2:=(diff(h(eta),eta,eta))+(f(eta)*(diff(h(eta),eta)))-((diff(f(eta),eta))*(h(eta)))+lambda*(((f(eta)*(diff(h(eta),eta,eta,eta))))+(h(eta)*(diff(f(eta),eta,eta,eta)))+(diff(f(eta),eta,eta))*(diff(h(eta),eta))-2*(diff(f(eta),eta))*(diff(h(eta),eta,eta))+(M/(1+m^2)))*(m*(diff(f(eta),eta)-h))=0;

 

>

> Eq3:=((f(eta))*(diff(theta(eta),eta)))+Pr*((diff(theta(eta),eta,eta)))=0;

 

>

> bcs1 := f(0) = 0, (D(f))(0) = 1, (h)(0) = 0, (theta)(0) = 1, (D(f))(blt) = 0,  h(blt) = 0, theta(blt) = 0, D(D(f))(blt) = 0, (D(h))(blt)=0;

 

>

> L := [1,2,3];

 

> for k to 3 do R := dsolve(eval({Eq1, Eq2, Eq3, bcs1}, M = L[k]), [f(eta),h(eta),theta(eta)], numeric, maxmesh=10000, output = listprocedure);;Y || k := rhs(R[3]); YA || k := rhs(R[6]);YB || k := rhs(R[5]);YC || k := rhs(R[4]);YD || k := rhs(R[7]);end do:

>

 

>

>

Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

for example is there an existing package for reducing groups of matrices like the one below to only its unique elements, or do i basically need to use  linear algebra matrix operations ie finding the basis of the set via echeleon reduction blah blah

 

{Matrix(2, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = 0, (2, 2) = -673/2880}), Matrix(2, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = 0, (2, 2) = -5/96}), Matrix(2, 2, {(1, 1) = 0, (1, 2) = -(2521/17920)*Zeta(5), (2, 1) = 0, (2, 2) = -(2521/17920)*Zeta(3)-2087/1920}), Matrix(2, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = 0, (2, 2) = 0}), Matrix(2, 2, {(1, 1) = 0, (1, 2) = -(7/320)*Zeta(5), (2, 1) = 0, (2, 2) = -(7/320)*Zeta(3)-499/840})}

i want a scheme of fractional differential equation so that i solve my questions and make a code of it.

please provide me the scheme

I was learning about Maple, but I do not understand this code below. This does not look correct.

http://www.maplesoft.com/support/help/Maple/view.aspx?path=Task/ConvertExpressionToFunction

When I type it on my worksheet, I get these errors

 

 

Sometimes I wonder if any one at Maplesoft actually looks at their own help web pages and try to use them, or is just the poor users who do that.

 

 

I noticed Maple gives different order of term (but still correct) when variable y or z is used, vs. other letters in this computation.

In this example below, integral results are given, using one letter of the alphabet at a time in the integrand. 

Only when the variable is y or z, Maple shows expression in different form from all the others (negative sign pulled out). The result are correct ofcourse, but was wondering why this hapens only for these two variables? I would have expected that variable name (letter used) should not make any difference to the final form.

restart;
interface(rtablesize=infinity):
chars:=[seq(parse(StringTools:-Char(i)),i=97..122)]: #generate a..z variables, may be there is better way?
Matrix([seq([chars[i0],simplify(int((c0^2 - c1^2)/(chars[i0] - x0+I*c3), x0))],i0=1..nops(chars))]);


Gives (this below. Notice the very last 2 entries below)

Is this something common in Maple? Does it depend on the computation being performed? Can a user do something to tell Maple not to change order/form of expression depending on what variable letter is used?

 

hi

please help me for dsolve differential equations...after much time dont answer!!!

thanks..

dsolve.mw

Digits := 15; SYS := [.16783*h1(theta)-0.96238e-3*(diff(h1(theta), theta, theta))+0.61603e-1*(diff(h2(theta), theta))+0.14870e-4*(diff(h2(theta), theta, theta, theta))-.23703*h3(theta)-0.84431e-3*(diff(h3(theta), theta, theta))+3.4919*10^(-7)*(diff(h1(theta), theta, theta, theta, theta)) = 0, 2.3940*h2(theta)-.35329*(diff(h2(theta), theta, theta))-0.68260e-1*(diff(h1(theta), theta))-0.16526e-4*(diff(h1(theta), theta, theta, theta))+3.0808*(diff(h3(theta), theta))-0.17833e-2*(diff(h3(theta), theta, theta, theta)) = 0, 9.4813*10^(-7)/((1.+1.5802*10^(-8)*h3(theta))*ln(10.+1.5802*10^(-7)*h3(theta))^2)-3.1867/((1.-0.26556e-1*h3(theta))*ln(10.-.26556*h3(theta))^2)-7.6530/((1.-0.31888e-1*h3(theta))*ln(10.-.31888*h3(theta))^2)-4.2551/((1.-0.35459e-1*h3(theta))*ln(10.-.35459*h3(theta))^2)-9.0315/((1.-0.37632e-1*h3(theta))*ln(10.-.37632*h3(theta))^2)-4.6587/((1.-0.38822e-1*h3(theta))*ln(10.-.38822*h3(theta))^2)-9.4520/((1.-0.39384e-1*h3(theta))*ln(10.-.39384*h3(theta))^2)+0.74999e-1/((1.+0.12500e-2*h3(theta))*ln(10.+0.12500e-1*h3(theta))^2)-.69143/((1.-0.28810e-2*h3(theta))*ln(10.-0.28810e-1*h3(theta))^2)-.12945/(1.-0.38836e-1*h3(theta))^4-0.38260e-1/(1.-0.11478e-1*h3(theta))^4-0.24826e-1/(1.-0.37240e-2*h3(theta))^4+0.74712e-3*(diff(h3(theta), theta, theta, theta, theta))+2.6337*10^(-8)/(1.+1.5802*10^(-8)*h3(theta))^4-0.27242e-1*(diff(h3(theta), theta, theta))-3.0707*(diff(h2(theta), theta))+0.17833e-2*(diff(h2(theta), theta, theta, theta))-.23618*h1(theta)-0.84126e-3*(diff(h1(theta), theta, theta))-0.89222e-1/(1.-0.26767e-1*h3(theta))^4-.21340/(1.-0.32010e-1*h3(theta))^4-.13154/(1.-0.19732e-1*h3(theta))^4-0.88519e-1/(1.-0.26556e-1*h3(theta))^4-4.7454/((1.-0.39545e-1*h3(theta))*ln(10.-.39545*h3(theta))^2)-0.36390e-1/(1.-0.10917e-1*h3(theta))^4-1.3100/((1.-0.10917e-1*h3(theta))*ln(10.-.10917*h3(theta))^2)-.89374/((1.-0.37240e-2*h3(theta))*ln(10.-0.37240e-1*h3(theta))^2)-1.3773/((1.-0.11478e-1*h3(theta))*ln(10.-.11478*h3(theta))^2)-.11820/(1.-0.35459e-1*h3(theta))^4-.26259/(1.-0.39388e-1*h3(theta))^4-4.7356/((1.-0.19732e-1*h3(theta))*ln(10.-.19732*h3(theta))^2)+28.586*h3(theta)-.21258/(1.-0.31888e-1*h3(theta))^4-9.4531/((1.-0.39388e-1*h3(theta))*ln(10.-.39388*h3(theta))^2)-4.6603/((1.-0.38836e-1*h3(theta))*ln(10.-.38836*h3(theta))^2)-9.0393/((1.-0.37664e-1*h3(theta))*ln(10.-.37664*h3(theta))^2)-4.2631/((1.-0.35526e-1*h3(theta))*ln(10.-.35526*h3(theta))^2)-7.6824/((1.-0.32010e-1*h3(theta))*ln(10.-.32010*h3(theta))^2)-.25088/(1.-0.37632e-1*h3(theta))^4-.12919/(1.-0.19378e-1*h3(theta))^4-.11842/(1.-0.35526e-1*h3(theta))^4-.12941/(1.-0.38822e-1*h3(theta))^4+0.20833e-2/(1.+0.12500e-2*h3(theta))^4-0.19206e-1/(1.-0.28810e-2*h3(theta))^4-4.6507/((1.-0.19378e-1*h3(theta))*ln(10.-.19378*h3(theta))^2)-.26255/(1.-0.39384e-1*h3(theta))^4-.13182/(1.-0.39545e-1*h3(theta))^4-.25109/(1.-0.37664e-1*h3(theta))^4-3.2120/((1.-0.26767e-1*h3(theta))*ln(10.-.26767*h3(theta))^2) = 0]

ode2 := diff(SYS[2], theta); SYS2 := {ode2, SYS[1], SYS[3]}; bcs2 := {h1(0) = 0, h1(1) = 0, h2(0) = 0, h2(1) = 0, h3(0) = 0, h3(1) = 0, ((D@@1)(h1))(0) = 0, ((D@@1)(h1))(1) = 0, ((D@@1)(h3))(0) = 0, ((D@@1)(h3))(1) = 0}; bcs22 := eval[recurse](convert(SYS[2], D), `union`({theta = 1}, bcs2)); res2 := dsolve(`union`(`union`(SYS2, bcs2), {bcs22}), 'maxmesh' = 2024, numeric, method = bvp[middefer], range = 0 .. 1, abserr = 0.1e-3, output = listprocedure)

NULL



Download dsolve.mw

 

Can you tell me the signification of the number 5 in output below.

Input: maximize(x^2+3,x=-1..2,location)

Output: 7,{[{x=2},5]}

 

Thanks

 

how can solve using assume option ?

restart:

F(z):=m*z^4-4*m*z^3+(3*m+3)*z^2-6*u;

m*z^4-4*m*z^3+(3*m+3)*z^2-6*u

(1)

solve(F(z),z assuming -0.5<m<0.5, 0<u<0.5)

Error, `<` unexpected

 

 

 

Download assume.mw

This sequence has a special name does anyone know what it is? i mean id look it up on oeis but unfortunately the asthetics of their presentation dont meet my strict standards, so yes i did work out an expression for it but if anyone knows the easiest in a singular expression.. very helpful right now.

 

1, 1, 2, 2, 3, 3, 4, 4, 5, 5

Dear all

Let q be  a real  different to one and for a fixed positive integer  n  given also 

 

Let x and y satisfies the condition

x*y -q*y*x=1

I will assume that the product is not commutative in all my computation

We would like to write the following function using only y

f(x,y)= x*y^n-q^n*y^n*x

 

all computation done I get f(x,y)=(q^n-1)/(q-1)*y^(n-1)

But how can I get the same result using Maple

Thank you very much for any help

 

 

 

 

 

Hi

 

Im trying to copy my collegues work, he uses mathcad and I use maple. I have a problem when trying to solve a system of 7 equations and 7 unknows. For some reason he dosent seem to get the same problem of "Error, recursive assignment".

You can see a screen shot of his solution here:
http://i.imgur.com/U7nE7sU.png

 

And you can see my solution here:

http://i.imgur.com/vrBTLIY.jpg

 

As you can see the code dosent work, is there anyone that could help me understand why eactly this happens and how to fix this? 

 

Best Regards

Ivan N

 

 

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