and t from (2*2*8) and (2*2*9), and it is now a

matter of algebraic manipulation to establish equation

(3*5*14)*

Since

$ is

through the origin, that is, on the piston path* Thus for all N this path is given by

x -

- 56 -

and since r-s«u84|ona particle path, we have a simple differential equation for the piston path whose Integral is

x « bt2 (3.5.17)

where b is a constant. The exact value of this constant

is rather difficult to determine. In theory it may be obtained by eliminating the ratio r/s from the equations

$ (r,s) « 0 and 2bt « r - s together with the expression for t as a function of r and • •

These results may now be Interpreted as follows. Corresponding to a given value of the shook strength s a shook wave is sent into the gas on its right along a path x sb At2. This shock may be looked on as being

© caused by the motion of a piston along a path x « bt • In other words, if a piston at the origin at t = 0 beside

ft

a gas whose initial conditions are given in equations (3.5.1) and (3.5.3) is moved with constant acceleration it causes a uniform shock to move into the gas also with

constant acceleration. It should be pointed out that A

and b are not restricted to positive values. Clearly

b < A , and b is bounded below by the value

corresponding to the free expansion of the gas into a vacuum. There are two special oases. One in which a uniform retardation of the piston causes a shock whioh remains stationary at the origin, corresponding to A » 0 •

87

The other Is the case In whloh the piston Is held

stationary at the origin, that is, where b « 0 • This Is the special case whloh was discovered by Copson when H « 1 •

It Is of Interest to examine the value of b as the shook strength decreases* To avoid exoessive algebra we

i.

treat only the case N * 1 . Putting s = 1 + 6 , the expansions of a and 0 are found to be

4

,

a s - + • • • • J

4

9 “ “ 3* ( 1 “ 75*2 ~ mB*2 * ’

Since a -» ® as 8 -»0, it follows that in the

r-s plane, the piston path, £(r,s) s 0 where £ Is given by (3«5«13), must tend to simply r = 0 •

If In general we use £ = 0 In (3* 8* 13) to eliminate a from (3.5*11) and (3.5.1S) these become

4

* - o

S 9

r + 5rs + 10s

t . _______ sod--- .

r* ♦ 5r« + 10b2

Letting 6 -» 0, so that 0 - 4a» and r -» 0 on the piston path, we may eliminate s from these equations to

get the limiting value for the piston motion,

(3.5.18) x

-se-

it may also he shown that P , given by (3.5.8) also

tends to infinity as 0 ->0 • Thus the mapping of the

hack of the shook in the r-s plane also tends to r « 0 ,

while in the physical plane X -> - , and the shook

becomes

x . - £ t2

whioh is simply the equation of the r characteristic

through the origin in the original flow, as may be seen by putting r s 0 and eliminating a in equations (3.5.1) and (3.5.S). This is of oourse the expected result.

The equation (3.5.18) will now be obtained in another way. Since in the r-s plane both the piston path, and the back of the shock path are represented in the limit by the characteristic r « 0 , it follows that the region between them must be the simple wave characterised by r s 0 . Then since we have x s -

a8,

straight lines for the simple wave have the equation

x + 2 c (u - c)(t - as) • (3.5.19)

But r « 0 in the region, so -u « 3c = a and a is

found as a function of x and t from (3.5.19). Then the particle paths are = u « -a , and on integrating this, the path through the origin is found to be, as before,

The simple wave between the piston and the characteristic Is shown in figure XIV.

The situation here is different from the classical simple wave problem in which a piston is withdrawn from a gas at rest as now the straight oharaoterlstlos meet the piston Instead of originating on it* The a-characteristics tend to converge, but a simple calculation shows that their envelope is formed to the left of the piston, and so the motion is continuous*

The value of b for free expansion of the gas into a vacuum is found by eliminating r and s from (3*5* 11),

(3*5*12) and the condition for such an expansion, namely

r + s « 0 • It is b e - -zj- • For values of b between

ma

this and - the motion to the right of r = 0 will be

unaltered* To the left the solution will have the form of

(3*5*10) with "§ « • while "a takes some value between ^a (the value for expansion into a vacuum) and +oo (the value for the simple wave ease of figure VII)*

" ” i

- 59 -

§ 6*

I

where S r Lit" r +T

ix

as

. m* on 3 = PH

and i

CB.e.1)

n,

so

Under these boundary conditions, a contour Integral solution of the type of (2.2.12) may be obtained and expressed as shown In Appendix in the form

ur- »♦«)

0 qx*'4*** ,

where additional constants have been absorbed in a and

4

P •

To this corresponds the expression for $ >

[<*R"*~Flwnri*4;N+~M; JL ) -pi*4~FUMV*H;w*~n;

j

In terms of R and S equations (2.2.8) and (2.2.9)