2[zeta]/y = -y^(-2)[nabla]²[psi] = dH/d[psi] (4)

is a function of [psi], say [f]´([psi]), and constant along a stream line;

dH/dv = 2q[zeta], H - [f]([psi]) = constant, (5)

throughout the liquid.

When the motion is irrotational,

d[phi] 1 d[psi] d[phi] 1 d[psi]

[zeta] = 0, u = - ------ = - --- ------, v = - ------ = --- ------, (6)

dx y dy dy y dx

d²[psi] d²[psi] 1 d[psi]

[nabla]²[psi] = 0, or ------- + ------- - --- ------ = 0. (7)

dx² dy² y dy

Changing to polar coordinates, x = r cos[theta], y = r sin[theta], the

equation (2) becomes, with cos[theta] = [mu],

d²[psi] d²[psi]

r²------- + (1 - [mu]²) ------- = 2[zeta]r³ sin [theta], (8)

dr² d[mu]²

of which a solution, when [zeta] = 0, is

/ B \ dPn / B \ dPn

[psi] = ( Ar^(n+1) + --- ) (1 - [mu]²) ----- = ( Ar^(n-1) + ------- ) y²-----, (9)

\ r^n / d[mu] \ r^(n+2) / d[mu]

[phi] = {(n + 1)Ar^n - nBr^(-n-1)} Pn, (10)

where Pn denotes the zonal harmonic of the nth order; also, in the

exceptional case of

[psi] = A0 cos[theta], [phi] = A0/r;

[psi] = B0r, [phi] = -B0 log tan ½[theta]

= -½B0 sh(-1) x/y. (11)

Thus cos[theta] is the Stokes' function of a point source at O, and PA

- PB of a line source AB.

The stream function [psi] of the liquid motion set up by the passage

of a solid of revolution, moving with axial velocity U, is such that

1 d[psi] dy

--- ------ = -U --, [psi] + ½Uy² = constant, (12)

y ds ds

over the surface of the solid; and [psi] must be replaced by [psi]´ =

[psi] + ½Uy² in the general equations of steady motion above to obtain

the steady relative motion of the liquid past the solid.

For instance, with n = 1 in equation (9), the relative stream function

is obtained for a sphere of radius a, by making it

[psi]´ = [psi] + ½Uy² = ½U(r² - a³/r) sin² [theta],

[psi] = -½Ua³ sin² [theta]/r; (13)

and then

[phi]´ = Ux(1 + ½a³/r²), [phi] = ½Ua³ cos [theta]/r², (14)

d[phi] a³ d[phi] a³

- ------ = U -- cos [theta], - --------- = ½U -- sin [theta], (15)

dr r³ rd[theta] r³

so that, if the direction of motion makes an angle [psi] with Ox,

tan ([psi] - [theta]) = ½ tan [theta],

tan [psi] = 3 tan [theta]/(2 - tan² [theta]), (16)

Along the path of a liquid particle [psi]´ is constant, and putting it

equal to ½Uc²,

(r² - a³/r) sin² [theta] = c², sin² [theta] = c²r/(r³ - a³), (17)

the polar equation; or

y² = c²r³/(r³ - a³), r³ = a³y²/(y² - c²), (18)

a curve of the 10th degree (C10).

Encyclopaedia Britannica, 11th Edition, "Hydromechanics" to "Ichnography" Volume 14, Slice 2 · 第 39 / 120 题