- THAT Q = 0 IMPLIES THE SOLUTION IS THE KING-ROSENAU SOLUTION 123
By (29.186a) and (29.188), we find that A (B - E) is constant. This and (29.191)and (29.192) imply that B = E.
By (29.189d), (29.189e), and B = E , we have A(x5 - Y5) ---+ 0. On the other
hand, (29.190) implies that A^2 x5y5---+ 0. Therefore
0 = t-t-oo lim A^2 (x 0 2 - y 0 2 )^2 = t-t-lim oo A^2 (x^2 0 + y 0 2 )^2 'so that
(29.193) A(x6 + Y5) ---+ 0.From (29.189d), we conclude that B = E ---+ l. This, (29.189g), (29.189h), and
(29.193) imply that
(29.194a) A^112 x 1 ---+ 0,
(29.194b) A^112 y 1 ---+ 0.Thus, by (29.189i), we have AC---+ 0.Let X (t) ~ xo(t) - x1(t) and Y(t) ~ Yo(t) - y 1 (t). Since F = 0 and B = E ,
by (29.186) we have
(29.195a) xb = -4BX,
(29.195b) Yb= -4BY,Thusx~ = 16AX ( X^2 + Y^2 + ~),
y 1 I = 16AY ( X^2 + Y^2 + c) B.
X' = -16AX ( X^2 + Y^2 + ~ + ~),
Y' = -16AY ( X^2 + Y^2 + ~ +
4~).
Define Z ~ X^2 + Y^2. Then
(^1) ( AC B)
z = -32Z AZ+ B + 4.
Combining this with A' = 4AB, we obtain
(AZ)'= -32AZ (AZ+ AJ + ~).
Since limt-t-oo AZ = 0 by (29.193) and (29.194), we obtain AZ = 0. Because
Z(t) = 0, by (29.195) we have that xo (t) = x1(t) and Yo(t) = y 1 (t) are independent
oft. Then (29.180) implies that xo (t) = Yo(t) = 0 and hence x1 (t) = y1(t) = 0.
We conclude that v(t) is of the form (29.181). Since v > 0, we have C > 0. This
completes Step 2.
STEP 3. If g (t) is not round, then it is the King- Rosenau solution.
Regarding the coefficients in (29.181), we have by (29.186a) and (29.186d) that