Prakash J

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These are questions asked by Prakash J

eq1 := diff(f(x), x, x, x)+(1/2)*x*cos(alpha)*(diff(f(x), x, x))+(1/2)*sin(alpha)*f(x)*(diff(f(x), x, x))+G[r]*theta(x)+G[m]*phi(x) = 0;

eq2 := diff(theta(x), x, x)+(1/2)*Pr*cos(alpha)*x*(diff(theta(x), x))+sin(alpha)*f(x)*(diff(theta(x), x))+N[b]*(diff(theta(x), x))*(diff(phi(x), x))+N[t]*(diff(theta(x), x))^2 = 0;

eq3 := diff(phi(x), x, x)+(1/2)*Le*cos(alpha)*x*(diff(phi(x), x))+sin(alpha)*f(x)*(diff(phi(x), x))+N[t]*(diff(theta(x), x, x))/N[b] = 0;

ics := f(0) = 0, (D(f))(0) = gamma*((D@@2)(f))(0), theta(0) = 1+tau*(D(theta))(0), phi(0) = 1;

bcs := (D(f))(infty) = 0, theta(infty) = 0, phi(infty) = 0;

Parameters := G[r] = 5, G[m] = 3, Pr = 7, N[b] = .1, N[t] = .1, Le = 1, gamma = .2, tau = .1, alpha = 30*degree;

I have a equation

((D@@2)(theta))(eta) = -(1/2)*(D(theta))(eta)*(-2*(D(phi))(eta)*beta*epsilon*lambda*D[B]+2*(D(phi))(eta)*beta*epsilon*mu*D[B]+f(eta)*sin(alpha)*beta*nu+2*(D(theta))(eta)*gamma*epsilon*D[t]-2*(D(theta))(eta)*beta*epsilon*D[t]+cos(alpha)*beta*eta*nu)/(beta*sigma)

and a parameters expression

Pr:=nu/sigma; N[b] := epsilon*D[B](mu-lambda)/sigma; N[t] := epsilon*D[t](gamma-beta)/(gamma*sigma); Le := nu/D[B]

How can I seperate common terms and substitute this parameters and got this following expression

((D@@2)(theta))(eta) = -(1/2)*Pr*(D(theta))(eta)*eta*cos(alpha)-(1/2)*Pr*(D(theta))(eta)*sin(alpha)*f(eta)-N[b]*(D(theta))(eta)*(D(phi))(eta)-N[t]*(D(theta))(eta)

How to conver a patial differetial equation to ordinary differential equation with or without dchange?
 

restart

declare(u(x, y, t), v(x, y, t), T(x, y, t), C(x, y, t), eta(x, y, t), psi(x, y, t), f(eta), theta(eta), phi(eta));

declare(u(x, y, t), v(x, y, t), T(x, y, t), C(x, y, t), eta(x, y, t), psi(x, y, t), f(eta), theta(eta), phi(eta))

(1)

eta := proc (x, y, t) options operator, arrow; y/(nu*t+nu*x/U[w])^(1/2) end proc:

eq1 := diff(T(x, y, t), t)+u*(diff(T(x, y, t), x))+v*(diff(T(x, y, t), y))-sigma*(diff(T(x, y, t), y, y))-epsilon*D[B]*(diff(T(x, y, t), y))*(diff(C(x, y, t), y)) = 0

diff(T(x, y, t), t)+U[w]*(D(f))(y/(nu*t+nu*x/U[w])^(1/2))*(diff(T(x, y, t), x))+(-(1/2)*f(y/(nu*t+nu*x/U[w])^(1/2))*nu/(nu*t+nu*x/U[w])^(1/2)+(1/2)*(D(f))(y/(nu*t+nu*x/U[w])^(1/2))*y*nu/(nu*t+nu*x/U[w]))*(diff(T(x, y, t), y))-sigma*(diff(diff(T(x, y, t), y), y))-epsilon*D[B]*(diff(T(x, y, t), y))*(diff(C(x, y, t), y)) = 0

(2)

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I have a PDE

eq1 := du/dx+dv/dy = 0; eq2 := du/dt+u*du/dx+v*du/dy-nu*d^2*u/dy^2 = 0

where u(x, y, t), v(x, y, t) and

eta(x, y, t):=y/((nu*t*cos(alpha)+(nu*x)/(U[w])*sin(alpha))^(1/(2)));

psi(x, y, t):=U[w]*(nu*t*cos(alpha)+(nu*x)/(U[w])*sin(alpha))^(1/(2))*f(eta(x,y,t));

u = diff(psi, y); v= -diff(psi, x).

How to substitiute u = diff(psi, y); v= -diff(psi, x) in eq1 and eq2, Also find the value of nu.

How to solve ordinary differemtial equation system with initial conditions and boundary conditions. Here, some initial conditions are unknown variables. So how to find these  values of parameters.

eq1 := diff(f(x), x, x, x)+(1/2)*cos(alpha)*x*(diff(f(x), x, x))+(1/2)*sin(alpha)*f(x)*(diff(f(x), x, x)) = 0;

eq2 := diff(g(x), x, x)+diff(g(x), x)+(diff(g(x), x))*(diff(h(x), x))+cos(alpha)*x*(diff(g(x), x))+sin(alpha)*f(x)*g(x) = 0;

eq3 := diff(g(x), x, x)+diff(h(x), x, x)+1/2*(cos(alpha)*x+sin(alpha)*f(x)) = 0

ics:=f(0)=0, f'(0)=1, f''(0)=a[1], g(0)=1, g'(0)=a[2], h(0)=1, h'(0)=a[3];

bcs:=f(x) , g(x), h(x) tends to 0 ad x tends to infinity

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