- #1
aqualonebear
- 5
- 0
Hi...
Suppose we consider a circular cylinder moving with constant velocity U in x-direction in a two-dimensional unbounded, irrotional, incompressible, inviscid fluid. If the motion of the fluid is completely resulted from the motion of the body, we know the velocity field of fluid can be described by ( \psi_y, -\psi_x), where \psi is the stream function which satisfys the boundary condition
\psi = y + constant, on the cirlce
and also
( \psi_y, -\psi_x) goes to zero as (x,y) goes to infinity
and it satisfies the laplace equation as well.
My question is , is this stream function unique up to a constant?
Actually if the radius of the circle is R, one solution of \psi is
\psi = y U R^2/(x^2+y^2).
But I think
\psi = y U R^2/(x^2+y^2) + log ((x^2+y^2)/R^2)
is also a solution since it satisfies all the conditions.
Thanks a lot.
Suppose we consider a circular cylinder moving with constant velocity U in x-direction in a two-dimensional unbounded, irrotional, incompressible, inviscid fluid. If the motion of the fluid is completely resulted from the motion of the body, we know the velocity field of fluid can be described by ( \psi_y, -\psi_x), where \psi is the stream function which satisfys the boundary condition
\psi = y + constant, on the cirlce
and also
( \psi_y, -\psi_x) goes to zero as (x,y) goes to infinity
and it satisfies the laplace equation as well.
My question is , is this stream function unique up to a constant?
Actually if the radius of the circle is R, one solution of \psi is
\psi = y U R^2/(x^2+y^2).
But I think
\psi = y U R^2/(x^2+y^2) + log ((x^2+y^2)/R^2)
is also a solution since it satisfies all the conditions.
Thanks a lot.