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These are questions asked by samantasubho

I want to evaluate - del(psi)/del(x) analytically in MAPLE. I wrote the following lines for the purpose :


deltat := proc (x) options operator, arrow; lambda*x^m end proc

Gr := proc (x) options operator, arrow; g*beta*deltat*x^3/nu^2 end proc

xi := proc (x) options operator, arrow; g*beta*x/c[p] end proc

eta := proc (x) options operator, arrow; (1/4)*y*4^(3/4)*Gr^(1/4)/x end proc

psi := proc (x, y) options operator, arrow; nu*4^(3/4...

I am trying to solve three non-linear coupled initial value problems. The equations are a bit complicated. The MAPLE code is 



eq[1] := (1/6)*Ra*v(x)^3*((2*v(x)*w(x)^2*(v(x)^2/w(x)^2-3)*(diff(w(x), x))-6*w(x)^3*(1-v(x)/w(x))^2*(diff(v(x), x)))*A+v(x)^2*w(x)^2*(2*v(x)/w(x)-3)*(diff(w(x), x))-3*w(x)^4*(1-v(x)/w(x))^2*(diff(v(x), x)))/(w(x)^2*(w(x)+2*A)^2*(2*v(x)+3*B))+u(x)

eq[2] := (1/105)*(diff(u(x)^2*(v(x)^2+11*B*v(x)+39*B^2...

I tried to solve a non linear coupled boundary value problem in MAPLE using DSolve command. The code is :

alias(eta = e, theta = t)

Eq[1] := 5*(diff(F(e), `$`(e, 3)))+(m+3)*F(e)*(diff(F(e), `$`(e, 2)))-(2*m+1)*(diff(F(e), e))^2-(4*m+2)*H(e)-(m-2)*e*(diff(H(e), e)) = 0

Eq[2] := diff(H(e), e) = t(e)

Eq[3] := 5*(diff(t(e), `$`(e, 2)))/Pr-(m+3)*F(e)*(diff(t(e), e))-5*m*(diff(F(e), e))*t(e) = 0

BCs := [F(0), (D(F))(0), (D(F))(infinity), t(0)-1, t(infinity), H(infinity)]

MAPLE is not able to show me the output of nested loop. How can I get it?

for i from 0 to N do for j to 3 do

eqn[i] := sort(coeff(lhs(dequn[j]), epsilon, i)) = coeff(rhs(dequn[j]), epsilon, i)

end do

nested_loops.mwend do

I want to get the zeroth order and first order perturbation equations of the following nonlinear coupled ODE.

restart; N := 1

alias(eta = e, theta = t)

eq[1] := 5*(diff(F(e), `$`(e, 3)))+(1+3/m)*F(e)*(diff(F(e), `$`(e, 2)))-(2+1/m)*(diff(F(e), e))^2-(4+2/m)*H(e)-(1-2/m)*e*(diff(H(e), e)) = 0

eq[2] := diff(H(e), e) = t(e)

eq[3] := 5*(diff(t(e), `$`(e, 2)))/Pr+(1+3/m)*F(e)*(diff(t(e), e))-5*(diff(F(e), e))*t(e)

eqs := eq[1], eq[2], eq[3]

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