MaplePrimes Questions

Hi, I'm new to Maple and was trying to use it to solve 3 equations with 3 unknowns, in terms of another 2 parameters. This is what I put in, and the error that came up:

solve({(1-x)/(1+b) = b*y, (1-y)(1-a)/(-a*b+1) = b*z, (1-z)(1-b)/(-a*b+1) = a*x}, {x, y, z}); 

Error, (in SolveTools:-LinearSolvers:-Algebraic) unable to compute coeff

I want to solve the equations to get x,y,z in terms of a,b but I don't understand the error coming up - have I done something wrong or is it because not all the variables appear in all of the equations? As there are 3 equations and 3 unknowns there is a solution, but I want to check the answer I found on paper with something (the algebra got a bit messy!)

Any help greatly appreciated! :)

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_[]

`𝔢`[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!

Consider the problem of a hard-hit baseball. The air-friction drag on a baseball is approximately given by the following formula

and subsequent differential equations : 

F:=-((C_d)*rho*Pi*(r)^2*v*v)/2;
m:=0.145;
v0:=65;
g:=9.81;
v_x:=diff(x(t),t);
v_y:=diff(y(t),t);
d2v_x:=-((C_d)*rho*Pi*(r^2)*(v_x)*sqrt((v_x)^2 +(v_y)^2))/(2*m);
d2v_y:=-((C_d)*rho*Pi*(r^2)*(v_y)*sqrt((v_x)^2 +(v_y)^2))/(2*m)-g;

where

C[d] is the drag coefficient (about 0.35 for a baseball)

•
rho[air] is the density of air (about 1.2 kg/
3m
)
•
r is the radius of the ball (about 0.037 m)

v is the vector velocity of the ball

Then if given that : 

Power hitters in baseball say they would much rather play in Coors Field in Denver than in sea-level stadiums because it is so much easier to hit home runs. The air pressure in Denver is about 10% lower than it is at sea level. The field dimensions at Coors Field are:

Left Field - 347 feet (106 m)
Left-Center - 390 feet (119 m)
Center Field - 415 feet (126 m)
Right-Center - 375 feet (114 m)
Right Field - 350 feet (107 m)

 1. Overlay two plots: one at sea level and one in Denver to show why power hitters prefer Coors field.

2. Find the initial magnitude of velocity, v0

needed to hit a home run to Right-Center, where v_x(0)=v0/sqrt(2) and v_y(0)=v0/sqrt(2)

I don't quite understand how to use the field dimensions for both 1 and 2 and am pretty clueless as to how to approach this question using the ordinary differential equations mentioned above.

 

 

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 

 

Is it possible to solve DDE with dsolve?

restart:

Eq1 := diff(x(t), t) = 1-.1*x(t)-0.5e-3*x(t)*v(t)/(1+0.1e-5*v(t));

Eq2 := diff(y(t), t) = 0.5e-3*x(t-10)*v(t-10)/(1+0.1e-5*v(t-10))-.3*y(t)-y(t)*z(t);

Eq3 := diff(v(t), t) = 200*y(t-10)-8*v(t);

Eq4 := diff(z(t), t) = 2*y(t)*z(t)-.15*z(t);

ics := x(0) = 1, y(0) = 1, v(0) = 5, z(0) = 1;

Thanks

I am trying to solve the Morrison equation for a normal force acting on a cylinder in a viscous fluid by using the potentional theory. With my limited experience in Maple I do not get why my dubble integral cannot be computed. Any help would be appreciated, tips are also very welkom as I am trying to expand my knowledge of Maple. 

Hi all,

I have lots of contstraint equations group and I want to fund a group of parameters which can fit them. 

For example, these are a simple constraint eqqations group:
eqs:{x1>0, x2>0 x1<1000, x2<1000, x1+x2>300,x1+x2<700}

Through SolveTools library, I can determine whether there is a group of parameters.


with(SolveTools[Inequality]);

LinearMultivariateSystem({x1 > 0, x1+x2 > 300, x2 < 1000, x1+x2 < 700, x2 > 0*x1 and 0*x1 < 1000}, [x1, x2]);

{[{x1 <= 300, 0 < x1}, {x2 < -x1 + 700, 300 - x1 < x2}],[{300 < x1, x1 < 700}, {0 < x2, x2 < -x1 + 700}]}

 

Then, if I want to find a group of parameter a group of parameters (ex, x1=300, x2=200 in this case), how should I do?

I've got the following four differential equations :

v_x:=diff(x(t),t);
v_y:=diff(y(t),t);
d2v_x:=-((C_d)*rho*Pi*(r^2)*(v_x)*sqrt((v_x)^2 +(v_y)^2))/(2*m);
d2v_y:=-((C_d)*rho*Pi*(r^2)*(v_y)*sqrt((v_x)^2 +(v_y)^2))/(2*m)-g;


and the following initial value conditions:

x(0)=0,y(0)=0,v_x(0)=v0/sqrt(2),v_y(0)=v0/sqrt(2) given v0=65 

I need to solve these using the numeric type and then draw overlaid plots

(i) setting C_d=0

(ii) leaving C_d as a variable

before plotting y(t) vs x(t). The hint for this last part is that the path can be seeing using [x(t),y(t)] instead of [t,y(t)]

I've tried to do it but seemed to have several syntax errors.

 

 

I've got the following diff.eq

y'(x)=sin(x*y(x)) given y(0)=1 

and need to solve it numerically which is why I've used:

dy4:=diff(y(x),x);
eqn4:=dy4=sin(x*y(x));
ic1:=y(0)=1;
ans3:=dsolve({eqn4,ic1},y(x),type=numeric);

This code doesn't return a value though and in fact, ans3 is being displayed as a procedure

"ans3:=proc(x_rkf45) ... end proc"

I don't quite understand why and what I need to do to get the required numerical solution

 

Hello

Could you plese help me to plot my function. i want to plot a function in which variables are have able formatting. I have attache my file.

Thank you.


restart

L[b] := 400:

L[c] := 400:

q := 5:

M[ab] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), theta[b]):``

M[ba] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), VectorCalculus:-`*`(2, theta[b])):

M[bc] := VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[b]), 1/L[b]), VectorCalculus:-`+`(VectorCalculus:-`*`(2, theta[b]), theta[c])), VectorCalculus:-`*`(VectorCalculus:-`*`(q, L[b]^2), 1/12)):

M[cb] := VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[b]), 1/L[b]), VectorCalculus:-`+`(theta[b], VectorCalculus:-`*`(2, theta[c]))), VectorCalculus:-`-`(VectorCalculus:-`*`(VectorCalculus:-`*`(q, L[b]^2), 1/12))):

M[cd] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), VectorCalculus:-`*`(2, theta[c])):

M[dc] := VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, E), I[c]), 1/L[c]), theta[c]):

eq1 := VectorCalculus:-`+`(VectorCalculus:-`*`(VectorCalculus:-`+`(M[ab], M[ba]), 1/L[c]), VectorCalculus:-`*`(VectorCalculus:-`+`(M[cd], M[dc]), 1/L[c])) = 0:

eq2 := VectorCalculus:-`+`(M[ba], M[bc]) = 0:

solve([eq1, eq2], {theta[b], theta[c]})

{theta[b] = -(40000000/3)/(E*(I[b]+2*I[c])), theta[c] = (40000000/3)/(E*(I[b]+2*I[c]))}

(1)

theta[b] := rhs({theta[b] = -(40000000/3)/(E*(I[b]+2*I[c])), theta[c] = (40000000/3)/(E*(I[b]+2*I[c]))}[1]):

theta[c] := rhs({theta[b] = -(40000000/3)/(E*(I[b]+2*I[c])), theta[c] = (40000000/3)/(E*(I[b]+2*I[c]))}[2]):

subs(theta[b] = theta[b], M[ba])

-(400000/3)*I[c]/(I[b]+2*I[c])

(2)

subs(theta[b] = theta[b], M[bc]):

``

EXplore(plot(M[ba], I[b] = 1 .. 10), parameters = [I[c] = 1 .. 20])

Error, (in plot) unexpected option: I[b] = 1 .. 10

 

``


Download shiboft2.mw

Matlab seems to be pretty strong at doing color plots with separate color bars, e.g.

Is this also possible in Maple and somehow in combination with `plots[surfdata](...,color=zhue,...)`?

In a 3d coordinate system I have a circular spacecurve with z-minimum -4 and z-maximum +4. In the same 3d coordinate system I have a 3d surface plot with z-minimum -0.5 and z-maximum +1.3 . When I choose the color option "Z(Hue)" in order to color-code the z-values on the 3d surface and make the topography more clear, I mostly get a totally green 3d surface. It seems that the color scaling is coupled with the spacecurve with z-values of +-4 . How can I uncouple the color scaling from the spacecurve and couple it with the z-range of the surface, while the color-limits shall be at +-1.3 ?

I have a nice family of functions of the form:

W:=(p,n,mu,w)->sum(w[k] * (n-k)* mu(n-k),k=1..n)

which can be evaluated for different p's using the operator mu*diff(...,mu)

The recursion begins with p=0 and proceeds using mu*diff(W(p,n,mu,w),mu) = W(p+1,n,mu,w).

Can anybody implement this procedure in Maple

Thank you 

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