33D (Integrable Systems)

The system prescribes both first derivatives of the same vector

ψ

, so it is overdetermined: the two mixed derivatives must agree. Differentiating gives

\partial_{\overset{z}{ˉ}} \partial_{z} ψ \partial_{z} \partial_{\overset{z}{ˉ}} ψ = (\partial_{\overset{z}{ˉ}} U) ψ + U \partial_{\overset{z}{ˉ}} ψ = (\partial_{\overset{z}{ˉ}} U + U V) ψ, = (\partial_{z} V) ψ + V \partial_{z} ψ = (\partial_{z} V + V U) ψ .

(1)

Thus every solution satisfies

(\partial_{\overset{z}{ˉ}} U - \partial_{z} V + [U, V]) ψ = 0.

(2)

For the system to possess arbitrary nontrivial local solution data, the coefficient matrix must vanish:

\partial_{\overset{z}{ˉ}} U - \partial_{z} V + [U, V] = 0 .

(3)

This is the zero-curvature condition, or the compatibility condition for an overdetermined linear system.

b

 0  0

Solution

 0  0

Substitute

U = \frac{1}{2} (u_{z} λ^{- 1} e^{u} λ^{- 1} e^{- u} - u_{z}), V = \frac{1}{2} (- u_{\overset{z}{ˉ}} λ e^{- u} λ e^{u} u_{\overset{z}{ˉ}}) .

(1)

The off-diagonal entries of

U_{\overset{z}{ˉ}} - V_{z} + [U, V]

(2)

cancel identically. Its upper-left entry is

u_{z \overset{z}{ˉ}} + \frac{1}{4} (e^{- 2 u} - e^{2 u}) = u_{z \overset{z}{ˉ}} - \frac{1}{2} sinh (2 u),

(3)

and its lower-right entry is the negative of this. Hence compatibility is equivalent to

u_{z \overset{z}{ˉ}} = \frac{1}{2} sinh (2 u) .

(4)

This is the Sinh-Gordon equation in complex light-cone coordinates.

c

 0  0

Solution

 0  0

When

u

is independent of

y

and

z = x + i y

\partial_{z} = \partial_{\overset{z}{ˉ}} = \frac{1}{2} \frac{d}{d x} .

(1)

Put

L = U - V, M = U + V .

(2)

The zero-curvature equation becomes

\frac{1}{2} L_{x} + [U, V] = 0.

(3)

Since

[M, L] = [U + V, U - V] = - 2 [U, V],

(4)

we obtain the Isospectral Lax equation

L_{x} = [M, L] .

(5)

Therefore, for every positive integer

n

, the cyclic property of the matrix trace gives

\frac{d}{d x} tr (L^{n}) = n tr (L^{n - 1} [M, L]) = n tr (L^{n - 1} M L - M L^{n}) = 0.

(6)

These are the trace invariants of a Lax equation.

The PDE from part (b) reduces to

\frac{1}{4} u_{xx} = \frac{1}{2} sinh (2 u) .

(7)

With

ϕ = 2 u

, this is

ϕ^{''} = 4 sinh ϕ .

(8)

To extract a nontrivial invariant, use

n = 2

. Writing

p = u_{x}

, direct multiplication gives

tr (L^{2}) = \frac{1}{2} [p^{2} + λ^{2} + λ^{- 2} - 2 cosh (2 u)] .

(9)

The terms involving

λ

are constant, and

p = ϕ^{'} /2

. Hence

\frac{1}{2} (ϕ^{'})^{2} - 4 cosh ϕ = C

(10)

is a first integral. Direct differentiation verifies it:

\frac{d}{d x} [\frac{1}{2} (ϕ^{'})^{2} - 4 cosh ϕ] = ϕ^{'} (ϕ^{''} - 4 sinh ϕ) = 0.

(11)

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 33D (Integrable Systems)

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