Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I'm trying to solve a system of four pdes and I know that the Newton method won't converge.

Are there other numerical methods that I can use?

Any help would be greatly appreciated!

Thanks,

Eve

Hi,

I'm an industrial engineering student who's running into problems with solving a simple system of non-linear equations.

Why do I get this error when issuing the solve command? I'm pretty sure I'm only passing two lists into the function?

Thanks a lot in advance,

Joshua

 

Hi!

Is there any way to remove the empty space that comes under images when printing the project, while using document mode?

Regards

Nicolai

I have a matrix A for which the basis of the left null space using NullSpace() is the empty set {}  while the column space is {e1,e2,e3}. By definition, we need every vector in col space . every vector in basis of left null space =0 but how would I show that in this case? Can I determine another basis for the left null space?

I've got the following matrix :

A:=[<a,a-1,-b>|<a-1,a,-b>,<b,b,2a-1>] where <> are the column elements of A, a is  a real number defined on [0,1] and b^2=2a(1-a) 

a) to show A is an orthogonal matrix, I understand that I need A.Transpose(A)=Identity(3*3) but is there a way in which I can let a take a random real numbered value between 0 and 1? The rand() only returns an integer within a range. Directly multiplying A and Transpose(A) will return an expression in a, so what's the right approach?

b) from a) we can infer that A is a matrix that describes a rotation in e1,e2,e3 where these are the standard bases vectors in R3. How can I determine the rotation axis? The hint I've been given says I need to consider the Eigenvalues and eigen vectors but I don't quite understand how.

I wish to use closed Newton-Cotes with n=2, also known as Simpson's Rule to numerically integrate an improper integral.

 

If it matters the integrand is (cos(2x))/(x^1/3), integrating between x=0..1

I've tried a few different (but similar) code but to no avail. Here is some stuff I've tried:

 

1.

 

with(Student[NumericalAnalysis]):

with(Student[Calculus 1]):

Simp1 := ApproximateInt(cos(2*x)/x^(1/3), x = 0 .. 1, method = newtoncotes[2]);

 

This gives me an output message that says "Float(infinity)".

 

2.

with(Student[NumericalAnalysis]):

with(Student[Calculus 1]):

Simp2 := int(exp(-x)/sqrt(1-x), x = 0 .. 1);

 

This doesn't have Simpson's rule as an option.

 

I think I'm on the right track with my first try, since I guess it wasn't tecnically an error message, but I'm not sure how to alter the code accordingly to get a numerical value instead. Thanks for any help.

 

 

 

 

 

 

I am trying to simplify equation 18 using equations 8 and 9. It should look a little like equation 21, but instead I get the results in equations 19 and 20.  I tried using different substituions, but algsubs gets the closest answer. A few terms are going to zero after the substitution.

When I substitute Z(X) then Zbar(X) terms vanish, and visa versa.


Initialize the metric and tetrad

 

restart; with(Physics); with(Tetrads)

0, "%1 is not a command in the %2 package", Tetrads, Physics

(1.1)

X = [zetabar, zeta, v, u]

X = [zetabar, zeta, v, u]

(1.2)

ds2 := Physics:-`*`(Physics:-`*`(2, dzeta), dzetabar)+Physics:-`*`(Physics:-`*`(2, du), dv)+Physics:-`*`(Physics:-`*`(2, H(zetabar, zeta, v, u)), (du+Physics:-`*`(Ybar(zetabar, zeta, v, u), dzeta)+Physics:-`*`(Y(zetabar, zeta, v, u), dzetabar)-Physics:-`*`(Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), dv))^2)

2*dzeta*dzetabar+2*du*dv+2*H(zetabar, zeta, v, u)*(du+Ybar(zetabar, zeta, v, u)*dzeta+Y(zetabar, zeta, v, u)*dzetabar-Y(zetabar, zeta, v, u)*Ybar(zetabar, zeta, v, u)*dv)^2

(1.3)

PDEtools:-declare(ds2)

Ybar(zetabar, zeta, v, u)*`will now be displayed as`*Ybar

(1.4)

NULL

vierbien = Matrix([[1, 0, -Ybar(zetabar, zeta, v, u), 0], [0, 1, -Y(zetabar, zeta, v, u), 0], [Physics:-`*`(H(zetabar, zeta, v, u), Y(zetabar, zeta, v, u)), Physics:-`*`(H(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), 1-Physics:-`*`(Physics:-`*`(H(zetabar, zeta, v, u), Y(zetabar, zeta, v, u)), Ybar(zetabar, zeta, v, u)), H(zetabar, zeta, v, u)], [Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u), -Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), 1]])

vierbien = (Matrix(4, 4, {(1, 1) = 1, (1, 2) = 0, (1, 3) = -Ybar(zetabar, Zeta, v, u), (1, 4) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = -Y(zetabar, Zeta, v, u), (2, 4) = 0, (3, 1) = H(zetabar, Zeta, v, u)*Y(zetabar, Zeta, v, u), (3, 2) = H(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (3, 3) = 1-H(zetabar, Zeta, v, u)*Y(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (3, 4) = H(zetabar, Zeta, v, u), (4, 1) = Y(zetabar, Zeta, v, u), (4, 2) = Ybar(zetabar, Zeta, v, u), (4, 3) = -Y(zetabar, Zeta, v, u)*Ybar(zetabar, Zeta, v, u), (4, 4) = 1}))

(1.5)

``

NULL

Setup(tetrad = rhs(vierbien = Matrix(%id = 18446744078408794830)), metric = ds2, mathematicalnotation = true, automaticsimplification = true, coordinatesystems = (X = [zetabar, zeta, v, u]), signature = "+++-")

[automaticsimplification = true, coordinatesystems = {X}, mathematicalnotation = true, metric = {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}, signature = `+ + + -`, tetrad = {(1, 1) = 1, (1, 3) = -Ybar(X), (2, 2) = 1, (2, 3) = -Y(X), (3, 1) = H(X)*Y(X), (3, 2) = H(X)*Ybar(X), (3, 3) = 1-H(X)*Y(X)*Ybar(X), (3, 4) = H(X), (4, 1) = Y(X), (4, 2) = Ybar(X), (4, 3) = -Y(X)*Ybar(X), (4, 4) = 1}]

(1.6)

``

Verification of Tetrad

 

I will try to verify the tetrad from (Kerr and Schild (1965)). However, the tetrad given in the paper seems to have the third tetrad with the wrong sign. I changed the sign and get the correct verification,

    e_[]

`&efr;`[a, mu] = (Matrix(4, 4, {(1, 1) = 1, (1, 2) = 0, (1, 3) = -Ybar(X), (1, 4) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = -Y(X), (2, 4) = 0, (3, 1) = H(X)*Y(X), (3, 2) = H(X)*Ybar(X), (3, 3) = 1-H(X)*Y(X)*Ybar(X), (3, 4) = H(X), (4, 1) = Y(X), (4, 2) = Ybar(X), (4, 3) = -Y(X)*Ybar(X), (4, 4) = 1}))

(2.1)

g_[]

g[mu, nu] = (Matrix(4, 4, {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 1) = 1+2*H(X)*Y(X)*Ybar(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 1) = -2*H(X)*Y(X)^2*Ybar(X), (3, 2) = -2*H(X)*Ybar(X)^2*Y(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 1) = 2*H(X)*Y(X), (4, 2) = 2*H(X)*Ybar(X), (4, 3) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}))

(2.2)

Physics:-`*`(e_[a, mu], e_[a, nu]) = g_[mu, nu]

Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[`~a`, nu] = Physics:-g_[mu, nu]

(2.3)

TensorArray(Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[`~a`, nu] = Physics:-g_[mu, nu])

Matrix(4, 4, {(1, 1) = 2*H(X)*Y(X)^2 = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X) = 2*H(X)*Y(X), (2, 1) = 1+2*H(X)*Y(X)*Ybar(X) = 1+2*H(X)*Y(X)*Ybar(X), (2, 2) = 2*H(X)*Ybar(X)^2 = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X) = 2*H(X)*Ybar(X), (3, 1) = -2*H(X)*Y(X)^2*Ybar(X) = -2*H(X)*Y(X)^2*Ybar(X), (3, 2) = -2*H(X)*Ybar(X)^2*Y(X) = -2*H(X)*Ybar(X)^2*Y(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2 = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X) = 1-2*H(X)*Y(X)*Ybar(X), (4, 1) = 2*H(X)*Y(X) = 2*H(X)*Y(X), (4, 2) = 2*H(X)*Ybar(X) = 2*H(X)*Ybar(X), (4, 3) = 1-2*H(X)*Y(X)*Ybar(X) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X) = 2*H(X)})

(2.4)

Physics:-`*`(e_[a, mu], e_[b, mu]) = eta_[a, b]

Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[b, `~mu`] = Physics:-Tetrads:-eta_[a, b]

(2.5)

NULL

TensorArray(Physics:-Tetrads:-e_[a, mu]*Physics:-Tetrads:-e_[b, `~mu`] = Physics:-Tetrads:-eta_[a, b])

Matrix(4, 4, {(1, 1) = 0 = 0, (1, 2) = 1 = 1, (1, 3) = 0 = 0, (1, 4) = 0 = 0, (2, 1) = 1 = 1, (2, 2) = 0 = 0, (2, 3) = 0 = 0, (2, 4) = 0 = 0, (3, 1) = 0 = 0, (3, 2) = 0 = 0, (3, 3) = 0 = 0, (3, 4) = 1 = 1, (4, 1) = 0 = 0, (4, 2) = 0 = 0, (4, 3) = 1 = 1, (4, 4) = 0 = 0})

(2.6)

``

gamma_[4, 2, 1]

diff(Y(X), zeta)-(diff(Y(X), u))*Ybar(X)

(2.7)

SumOverRepeatedIndices(Physics:-`*`(Physics:-`*`(D_[nu](e_[4, mu]), e_[2, mu]), e_[1, `~nu`]))

diff(Y(X), zeta)-(diff(Y(X), u))*Ybar(X)

(2.8)

NULL

``

For equation 2.8 we get the following:

SumOverRepeatedIndices(Physics:-`*`(Physics:-`*`(Riemann[`~sigma`, rho, mu, nu], e_[4, `~rho`]), e_[4, `~nu`]))

(-Physics:-Riemann[`~sigma`, 4, 4, mu]*Y(X)^2+(Physics:-Riemann[`~sigma`, 4, 1, mu]+Physics:-Riemann[`~sigma`, 1, 4, mu])*Y(X)-Physics:-Riemann[`~sigma`, 1, 1, mu])*Ybar(X)^2+((Physics:-Riemann[`~sigma`, 2, 4, mu]+Physics:-Riemann[`~sigma`, 4, 2, mu])*Y(X)^2+(-Physics:-Riemann[`~sigma`, 2, 1, mu]+Physics:-Riemann[`~sigma`, 3, 4, mu]+Physics:-Riemann[`~sigma`, 4, 3, mu]-Physics:-Riemann[`~sigma`, 1, 2, mu])*Y(X)-Physics:-Riemann[`~sigma`, 3, 1, mu]-Physics:-Riemann[`~sigma`, 1, 3, mu])*Ybar(X)-Physics:-Riemann[`~sigma`, 2, 2, mu]*Y(X)^2+(-Physics:-Riemann[`~sigma`, 2, 3, mu]-Physics:-Riemann[`~sigma`, 3, 2, mu])*Y(X)-Physics:-Riemann[`~sigma`, 3, 3, mu]

(1)

 

Now we replicate eqn 2.16. These are the conditions for e[4,mu] to be geodesic and shear-free. The outputs are eqn 3.5.

 

gamma_[4, 1, 1] = 0

diff(Ybar(X), zeta)-(diff(Ybar(X), u))*Ybar(X) = 0

(2)

gamma_[4, 2, 2] = 0

diff(Y(X), zetabar)-(diff(Y(X), u))*Y(X) = 0

(3)

gamma_[1, 4, 4] = 0

(diff(Ybar(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Ybar(X), zeta))-Ybar(X)*(diff(Ybar(X), zetabar))-(diff(Ybar(X), v)) = 0

(4)

gamma_[2, 4, 4] = 0

(diff(Y(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Y(X), zeta))-(diff(Y(X), zetabar))*Ybar(X)-(diff(Y(X), v)) = 0

(5)

gamma_[3, 4, 4] = 0

0 = 0

(6)

gamma_[4, 4, 4] = 0

0 = 0

(7)

shearconditions := {diff(Y(X), zetabar)-(diff(Y(X), u))*Y(X) = 0, diff(Ybar(X), zeta)-(diff(Ybar(X), u))*Ybar(X) = 0, (diff(Y(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Y(X), zeta))-(diff(Y(X), zetabar))*Ybar(X)-(diff(Y(X), v)) = 0, (diff(Ybar(X), u))*Y(X)*Ybar(X)-Y(X)*(diff(Ybar(X), zeta))-Ybar(X)*(diff(Ybar(X), zetabar))-(diff(Ybar(X), v)) = 0}:

 

Now we can define the rotation coefficients associated with rotation and expansion z = theta - i omega

 

gamma_[2, 4, 1] = Z(X)

-(diff(Y(X), zeta))+(diff(Y(X), u))*Ybar(X) = Z(X)

(8)

gamma_[1, 4, 2] = Zbar(X)

-(diff(Ybar(X), zetabar))+(diff(Ybar(X), u))*Y(X) = Zbar(X)

(9)

PDEtools:-declare(Z(X), Zbar(X))

Zbar(zetabar, zeta, v, u)*`will now be displayed as`*Zbar

(10)

Zdefinitions := {-(diff(Y(X), zeta))+(diff(Y(X), u))*Ybar(X) = Z(X), -(diff(Ybar(X), zetabar))+(diff(Ybar(X), u))*Y(X) = Zbar(X)}

{-(diff(Y(X), zeta))+(diff(Y(X), u))*Ybar(X) = Z(X), -(diff(Ybar(X), zetabar))+(diff(Ybar(X), u))*Y(X) = Zbar(X)}

(11)

We now show that the tetrad vectors are propogated parallel along each curve of the congruence of null geodesics which have e[4,~mu] as tangents.

 

   

We now use the tetrad form of the Ricci tensor. In order to use this in Maple we need to create a Ricci Tensor Tetrad function.

 

RicciT := proc (a, b) options operator, arrow; SumOverRepeatedIndices(Ricci[mu, nu]*e_[a, `~mu`]*e_[b, `~nu`]) end proc

proc (a, b) options operator, arrow; Physics:-SumOverRepeatedIndices(Physics:-`*`(Physics:-`*`(Physics:-Ricci[mu, nu], Physics:-Tetrads:-e_[a, `~mu`]), Physics:-Tetrads:-e_[b, `~nu`])) end proc

(12)

SlashD := proc (f, a) options operator, arrow; SumOverRepeatedIndices(D_[b](f)*e_[a, `~b`]) end proc

proc (f, a) options operator, arrow; Physics:-SumOverRepeatedIndices(Physics:-`*`(Physics:-D_[b](f), Physics:-Tetrads:-e_[a, `~b`])) end proc

(13)

SlashD(f(X), 1)

diff(f(X), zeta)-Ybar(X)*(diff(f(X), u))

(14)

SlashD(f(X), 2)

diff(f(X), zetabar)-Y(X)*(diff(f(X), u))

(15)

SlashD(f(X), 3)

(1+H(X)*Y(X)*Ybar(X))*(diff(f(X), u))-H(X)*((diff(f(X), zeta))*Y(X)+Ybar(X)*(diff(f(X), zetabar))+diff(f(X), v))

(16)

SlashD(f(X), 4)

-Y(X)*Ybar(X)*(diff(f(X), u))+(diff(f(X), zeta))*Y(X)+Ybar(X)*(diff(f(X), zetabar))+diff(f(X), v)

(17)

NULL

The geodesic and shear free condition given by Lemma 1 in (Goldberg and Sachs (1962)). Kerr uses the fourth tetrad instead of the third so we need to modify the Ricci tensor conditions. The equations (2) - (5) enforce the first Lemma.

 

   

 

Notice that none of the previous Ricci conditions can be used to solve for H.  We can use the remaining field equations to find the partial differential equations necessary to derive the metric.

 

  simplify(RicciT(1, 2), shearconditions) = 0

H(X)*(diff(diff(Y(X), zeta), zetabar))*Ybar(X)-H(X)*Ybar(X)*Y(X)*(diff(diff(Ybar(X), u), zetabar))-H(X)*Ybar(X)^2*(diff(diff(Y(X), u), zetabar))-H(X)*Y(X)^2*(diff(diff(Ybar(X), u), zeta))-2*H(X)*Y(X)*Ybar(X)*(diff(diff(Y(X), u), zeta))+H(X)*Y(X)^2*Ybar(X)*(diff(diff(Ybar(X), u), u))-H(X)*Y(X)*(diff(diff(Ybar(X), u), v))+H(X)*Y(X)*Ybar(X)^2*(diff(diff(Y(X), u), u))-H(X)*(diff(diff(Y(X), u), v))*Ybar(X)+H(X)*(diff(Ybar(X), zetabar))^2+(-3*H(X)*Y(X)*(diff(Ybar(X), u))-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Ybar(X), zetabar))+H(X)*(diff(Y(X), zeta))^2+(-4*H(X)*(diff(Y(X), u))*Ybar(X)-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Y(X), zeta))+2*H(X)*Y(X)^2*(diff(Ybar(X), u))^2-Y(X)*(-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)+diff(H(X), v))*(diff(Ybar(X), u))+2*(H(X)*(diff(Y(X), u))*Ybar(X)+(1/2)*(diff(H(X), u))*Y(X)*Ybar(X)-(1/2)*(diff(H(X), zeta))*Y(X)-(1/2)*(diff(H(X), zetabar))*Ybar(X)-(1/2)*(diff(H(X), v)))*(diff(Y(X), u))*Ybar(X) = 0

(18)

-(diff(H(X), zetabar))*Ybar(X)*Z(X)-Y(X)*(diff(H(X), zeta))*Z(X)-H(X)*(diff(Y(X), zeta))^2+Z(X)*((diff(H(X), u))*Y(X)*Ybar(X)+2*H(X)*Z(X)-(diff(H(X), v))) = 0

-(diff(H(X), zetabar))*Ybar(X)*Z(X)-(diff(H(X), zeta))*Y(X)*Z(X)-H(X)*(diff(Y(X), zeta))^2-Z(X)*(-(diff(H(X), u))*Y(X)*Ybar(X)-2*H(X)*Z(X)+diff(H(X), v)) = 0

(19)

Zbar(X)*(-(diff(H(X), v))-(diff(H(X), zetabar))*Ybar(X)-(diff(H(X), zeta))*Y(X)+(diff(H(X), u))*Y(X)*Ybar(X)+H(X)*(diff(Ybar(X), zetabar)+2*Zbar(X))) = 0

-Zbar(X)*(-(diff(H(X), u))*Y(X)*Ybar(X)+(diff(H(X), zeta))*Y(X)+(diff(H(X), zetabar))*Ybar(X)-H(X)*(diff(Ybar(X), zetabar))-2*H(X)*Zbar(X)+diff(H(X), v)) = 0

(20)

Physics:-`*`(SlashD(H(X), 4), Z(X)+Zbar(X)) = Physics:-`*`(H(X), SlashD(Z(X), 4)+SlashD(Zbar(X), 4))

-(-(diff(H(X), v))-(diff(H(X), zeta))*Y(X)+Ybar(X)*((diff(H(X), u))*Y(X)-(diff(H(X), zetabar))))*(Z(X)+Zbar(X)) = H(X)*(-Y(X)*Ybar(X)*(diff(Z(X), u))+Ybar(X)*(diff(Z(X), zetabar))+Y(X)*(diff(Z(X), zeta))+diff(Z(X), v)-Y(X)*Ybar(X)*(diff(Zbar(X), u))+Ybar(X)*(diff(Zbar(X), zetabar))+Y(X)*(diff(Zbar(X), zeta))+diff(Zbar(X), v))

(21)

``

NULL

NULL


Download Deriving_the_Kerr_Metric.mw

Hi,

 

I'm trying to create interactive plots by using Explore to help demonstrate the effects parameters have on functions. I created one successfully to illustrate shifts and stretches of a polynomial:

 

transform(A,B,X,H,P,K):=Explore(plot(a*(b*x+h)^(p)+k,x=X),parameters=[a=A, b= B,h=H,p=P,k=K],placement=right)

 

However when I try to do the same with a solved ODE it returns an error message:

 

Explore(plot(1/(-p*x+x+1)^(1/(p-1)), x = -5 .. 5), parameters = [p = -20 .. 20], placement = right);

 

Executing this gives the error message: 

Warning, expecting only range variable x in expression 1/((-p*x+x+1)^(1/(p-1))) to be plotted but found name p
INTERFACE_PLOT(AXESLABELS(x, ""),

VIEW(-5. .. 5., DEFAULT, _ATTRIBUTE("source" = "mathdefault"))),

parameters = [p = -20 .. 20], placement = right

 

I'm not sure why it is having difficulty dealing with "p" when it had no difficulty with the first. Any help would be appreciated!

Hey guys, I'm new to the forum, so please tell if I need to set up the question in a different way :) I've tried to find an answer for this, but have struggled since our learning book is in danish, so the used terms may not be technically correct,

 

Anyways, how do you solve this problem in maple? 

 

Find the complete real solution for the differential equation system:

 

 

For the homogenous part I've found. (Is this correct?)

  , c1,c2 € R

 

 

I've tried to find an answer for the inhomogenous part but I get a really complicated result, so I doubt it's correct.

 

Thanks for the help :)

-Alex 

 

I am considering options to access Maple features from within a Fortran program using the OpenMaple API. I do not find any examples illustrating this (or even precise statements if this is possible in principle). (Quite surprising to me considering the presence of Fortran in scientific computing).

Is the only way to go to write a C wrapper for the API? If yes, is there an example for such a wrapper which performs common conversions between Fortran and C data types.

Thanks!

I have been Maple  18 with no problem. Then, today, the  "=" symbol and  "+" symbol don't work.  Does anyone know what to do about this?

 

hai everyone. i am currently trying to solve an integration of the following ∫g(η)dη . integrate from 0 to 10.

from the following odes.

f ''' +1-(f ')2 +ff ''=0,

g''-gf'+fg'=0,

with boundary conditions f(0)=0, f'(0)=λ, f'(∞)=1, g(0)=1,g(∞)=0

First, i solve the odes using the shooting method. then i used the trapezoidal rule to solve for the integration of g(eta) using the following codes

> with(student);
> trapezoid(g(eta), eta = 0 .. 10, 10);
> evalf(%);

it seems that it can not read the data from the shooting method. can anyone suggest why it is happening?

thank you verymuch for your concern :)

Hello

I wantt to assigne value to a variable that doesn't exist before the radio button. How can i do that?

Could you please help me?

First 205 206 207 208 209 210 211 Last Page 207 of 2218