- THAT Q = 0 IMPLIES THE SOLUTION IS THE KING-ROSENAU SOLUTION 121
Second, by substituting (29.182) into the LHS of (29.170), we have that (where'
denotes a time derivative)(29.184)~~ = A'((x - xo)2
- (y - Yo)
2
)^2 + B' (x - x1)
2
- E' (y - Y1)
2
- F'xy + C'
- 4Axb (x - xo) ((x - xo)^2 + (y - y 0 )^2 )
- 4Ayb (y - Yo) ((x - xo)
2
- (y - Yo)
2
)
- 2B (x - x1) x~ - 2E (y - Y1) y~.
We rewrite this in a form analogous to (29.183):(29.185)
av , 2 )2 2
at =A ((x - xo) + (y - Yo )- 4Axb (x - xo) ((x - x 0 )^2 + (y -y 0 )^2 )
- 4Ayb (y - Yo) ((x - xo)^2 + (y - Yo)^2 )
- B' (x - xo)^2 + E' (y - Yo)
2
+ F'xy
+ 2 (B' (xo - x1) - Bx~) (x - xo) + 2 (E' (Yo - Y1) - Ey~) (y - Yo)
+ B' (xo - x1)2
- E' (yo - Y1)
2
- 2Bx~ (xo - x1) - 2Ey~ (Yo - Y1) + C'.
By (29.170) and by equating the coefficients in (29.183) and (29.185), we obtain(29.186a)
(29.186b)
(29.186c)(29.186d)(29.186e)(29.186f)(29.186g)(29.186h)and
A' = 2A (B + E),
xb = -2Fyo + 4B (x1 - xo),
Yb = -2Fxo + 4E (Y1 - Yo),B' = 16A (B (xo - x1)
2
- E (Yo - Y1)
2
+ Fxoyo + c)
+2B(E-B)-F^2 ,E' = 16A ( B (xo - x1)
2
- E (Yo -y1)
2
- Fxoyo + c)
+ 2E ( B - E) - F^2 ,
F' = -2F (B + E),
, B' 2EFy 1 - F^2 x 0x 1 = B (xo - x1) - 2 (E - B) (xo - x1) - B ,
1 E' 2BFx1 - F(^2) yo
y 1 = E (Yo - Y1) - 2 (B - E)(yo - Y1) - E ,
(29.187) C' = -B' (xo - x1)^2 - E' (Yo - Y1)^2 + 2Bx~ (xo - x1) + 2Ey~ (yo - Y1)
- 2 (E - B) B (xo - x1)
2
- 2 (B - E) E (Yo - Y1)
2
- 2 (B + E) C
- F^2 (x6 + Y5) + 4BFx1Yo + 4EFxoY1·