Find the equilibrium (time independent) solution rho(x), u(x), in 0 lessthanoreq

Question

Find the equilibrium (time independent) solution rho(x), u(x), in 0 lessthanorequalto x lessthanorequalto l, to the equations of gas dynamics in one space dimension for a gas subject to a constant vertical gravitational force: partial differential rho/partial differential t + partial differential(rho u)/partial differential x = 0 partial differential rho u/partial differential t + partial differential(rho u^2 + p(partial differential))/partial differential x = -g p(partial differential) = a partial differential^gamma: gamma > 1, with boundary conditions u(0, t) = 0 and rho(l, t) = rho_0.

Explanation / Answer

given equations
d(rho)/dt + d(rho*u)/dx = 0
d(rho*u)/dt + d(rho*u^2 + p(rho))/dx = -g
p(rho) = a*rho^(gamma)

so for time independent solutions
d(rho)/dt = 0, so rho is not a fucntion of time, t
d(rho*u)/dt = 0, so rho*u is not a function of time too,
now since rho is not a functgion of time, so u is also not a function of time

hence
d(rho*u)/dx = 0
hence rho*u = k [ constant]

d(rho*u^2 + p(rho))/dx = -g
=> rho*u^2 + a*rho^(gamma) = -gx + C [ C is the constant of integration ]
at x = 0, u = 0
a*rho(0,t)^(gamma) = C

hence, rho = k/u
u*k + a*(k/u)^(gamma) = -gx + a*rho(0,t)^(gamma)

the solution to this equation in u gives u
rho = k/u

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