1547845447-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_IV__Chow_

13. THAT Q = 0 IMPLIES THE SOLUTION IS THE KING-ROSENAU SOLUTION 119

We also have that

V1112 = 3v1222 and V2221 = 3v1121,

which imply that v 1112 = V2221 = 0. Hence we have that v 111 is a function of x
only and that V222 is a function of y only. Thus, the V1111 = V2222 in (29.178) is a
constant. We obtain that
vrn = 2(C1x + C2) and v222 = 2(C1y + C3)

for some constants C 1 (t), C2(t), and C3(t).
Integrating this yields

v11 = C1x^2 + 2C2x +Ji,

V22 = C1y^2 + 2C3y + h

for some functions Ji (y, t) and h (x , t). Then (29.177) implies

2C1x + 2C2 = 38xh and 2C1y + 2C3 = 38yfi·

We conclude that

where C 4 (t) and C 5 (t) are constants.

2

v112 = 3(C1y + C3)

Taking the derivative of this, we have that

and

Integrating this, we obtain

2

V12 = 3(C1xy + C3x + C2y) + C5,

where C 6 (t) is a constant.
Summarizing, we have with

( )

_. C2

29.179 x =;= x + Ci and

that

• -2 1 -2
  v11 = C1x + 3C1y + C1,

V22 - = 3 le ix -2 + c iY - 2 + c 3,

(^2) c -- c
V12 = 3 iXY + g,
for some constants C 7 (t), Cs(t), and Cg(t). From this we co nclude that
q=;=v-. _ C1 (-2 -2)2
12
x +y
satisfies
q11 = C1, q22 =Cs, q12 =Cg.
Therefore q is a quadratic polynomial in x and y. This completes Step 1.