Worksheet 7
Toda equation
Definition. Consider Painlev´e-II equation. Since momentum and Hamiltonian are expressed in terms of
q(x), the action of B¨acklund transformation B+ can be extended from pairs (q(x), α) to pairs (p(x), α) and
(H(x), α). We denote
(qn(x), α + n) = B
n
+(q(x), α) (1)
(pn(x), α + n) = B
n
+(p(x), α) (2)
(Hn(x), α + n) = B
n
+(H(x), α) (3)
Definition. Define tau function for Painlev´e-II equation using formula
d
dx ln(τn(x)) = Hn(x) (4)
1. Show that Hn+1(x) = Hn(x) − qn+1(x).
2. Show that the Toda equation holds
d
2
dx2
ln(τn(x)) = C
τn+1(x)τn−1(x)
τ
2
n
(x)
(5)
for some constant C.
Definition. Consider Painlev´e-III equation. Fix ε = ±1. Define auxiliary Hamiltonian using formula
h(x) = 1
2
H(x) + q(x)p(x) − εx2 +
1
4
(β − 4ε)(β + ε(α − 2))
(6)
Definition. Consider Painlev´e-III equation. Choose ε = 1. Since momentum and Hamiltonian are expressed
in terms of q(x), the action of B¨acklund transformation B1 can be extended from pairs (q(x), α, β) to pairs
(p(x), α, β) and (H(x), α, β). We denote
(qn(x), α + 2n, β + 2n) = B
n
1
(q(x), α, β) (7)
(pn(x), α + 2n, β + 2n) = B
n
1
(p(x), α, β) (8)
(Hn(x), α + 2n, β + 2n) = B
n
1
(H(x), α, β) (9)
(hn(x), α + 2n, β + 2n) = B
n
1
(h(x), α, β) (10)
Definition. Define tau function for Painlev´e-III equation using formula
x
d
dx ln(τn(x)) = hn(x) (11)
3. Show that hn+1(x) = hn(x) − pnqn(x) −
3
2 +
α
4 +
3β
4 + 2n.
4. Show that the Toda equation holds
x
d
dxx
d
dx ln(τn(x)) = C
τn+1(x)τn−1(x)
τ
2
n
(x)
(12)
for some constant C.
Definition. Consider Painlev´e-III equation. Choose ε = −1. Since momentum and Hamiltonian are expressed in terms of q(x), the action of B¨acklund transformation B3 can be extended from pairs (q(x), α, β)
to pairs (p(x), α, β) and (H(x), α, β). We denote
(qn(x), α + 2n, β − 2n) = B
n
3
(q(x), α, β) (13)
(pn(x), α + 2n, β − 2n) = B
n
3
(p(x), α, β) (14)
(Hn(x), α + 2n, β − 2n) = B
n
3
(H(x), α, β) (15)
(hn(x), α + 2n, β − 2n) = B
n
3
(h(x), α, β) (16)
Definition. Define tau function for Painlev´e-III equation using formula
x
d
dx ln(τn(x)) = hn(x) (17)
5. Show that hn+1(x) = hn(x) − pnqn(x) −
3
2 +
α
4 −
3β
4 + 2n.
6. Show that the Toda equation holds
x
d
dxx
d
dx ln(τn(x)) = C
τn+1(x)τn−1(x)
τ
2
n
(x)
(18)
for some constant C.