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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 1

When $u$ is independent of $y$ and $z=x+iy$,
$$
\partial_z=\partial_{\bar z}=\frac12\frac d{dx}.
$$
Put
$$
L=U-V,\qquad M=U+V.
$$
The zero-curvature equation becomes
$$
\frac12L_x+[U,V]=0.
$$
Since
$$
[M,L]=[U+V,U-V]=-2[U,V],
$$
we obtain the <Isospectral Lax equation>
$$
\boxed{L_x=[M,L]}.
$$
Therefore, for every positive integer $n$, the cyclic property of the <matrix trace> gives
$$
\begin{aligned}
\frac d{dx}\operatorname{tr}(L^n)
&=n\operatorname{tr}(L^{n-1}[M,L])\\
&=n\operatorname{tr}(L^{n-1}ML-ML^n)=0.
\end{aligned}
$$
These are the <trace invariants of a Lax equation>.

The PDE from part (b) reduces to
$$
\frac14u_{xx}=\frac12\sinh(2u).
$$
With $\phi=2u$, this is
$$
\phi''=4\sinh\phi.
$$
To extract a nontrivial invariant, use $n=2$. Writing $p=u_x$, direct multiplication gives
$$
\operatorname{tr}(L^2)
=\frac12\left[
p^2+\lambda^2+\lambda^{-2}-2\cosh(2u)
\right].
$$
The terms involving $\lambda$ are constant, and $p=\phi'/2$. Hence
$$
\boxed{
\frac12(\phi')^2-4\cosh\phi=C
}
$$
is a <first integral>. Direct differentiation verifies it:
$$
\boxed{\frac d{dx}
\left[\frac12(\phi')^2-4\cosh\phi\right]
=\phi'(\phi''-4\sinh\phi)=0.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 33D / c / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 33D / c / Solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 33D / c / Solution

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