Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 61 / 2

Write $E_k=e^{i(kx+k^{3}t)}$, $I={\int }_L\phi (l)d\lambda (l)$ and $A_k={\int }_L\phi (l)/(k+l)d\lambda (l)$. Introduce the dressing operatorThe given linear integral equation is $\mathcal{F}\phi =E$, and $q=-I_x$. The appropriate contour and measure must permit differentiation under these integrals and avoid untreated kernel singularities.

For the specified spatial operator, direct differentiation givesInside the integral, replace ${\phi }_{xx}(l)+q\phi (l)$ by $M_l\phi (l)+il{\phi }_x(l)$. It follows thatApply $M_k$ to $\phi +i{E}_k{A}_k=E_k$. The last term contributes $-E_k{I}_x=q{E}_k$, canceling the right side, soThis is precisely the homogeneous integral equation. Uniqueness of the original inhomogeneous solution implies a trivial homogeneous null space, hence $M\phi =0$. This calculation is the spatial identity in Cauchy-kernel dressing for the KdV equation.

For the specified time operator, $N_k{E}_k=3ikq{E}_k$, because the derivatives of $e^{i(kx+k^{3}t)}$ cancel between ${\partial }_t$ and ${\partial }_x^3$. Expanding the derivatives of $E_k{A}_k$ givesFor completeness, the extra derivative terms before simplification are $3ik{A}_{k,xx}-3k^2{A}_{k,x}+3ikq{A}_k$. Substituting ${\phi }_{xx}+q\phi =M_l\phi +il{\phi }_x$ turns their derivative contribution into $-3k{\int }_L{\phi }_{x}d\lambda =-3k{I}_x$.

Applying $N_k$ to the original linear integral equation and using $I_x=-q$ cancels its right side. ThusPart (a) gives $E_k{\int }_L{M}_l\phi /(k+l)d\lambda =i{M}_k\phi$. Therefore the exact requested identity isSince $M\phi =0$, this becomes homogeneous too. The same uniqueness argument gives $N\phi =0$; it is not necessary to assume a separate uniqueness theorem for the differential time equation.

The two differentiated linear integral equations have now supplied both equations of the Lax pair. To verify the nonlinear consequence explicitly, put $D={\partial }_x$ and take the operator commutator in the order $[N,M]=NM-MN$. The product rule givesFor example, $[D^3,q]=3q_x{D}^2+3q_{xx}D+q_{xxx}$ and $[3qD,D^2]=-6q_x{D}^2-3q_{xx}D$, which explain the cancellations. Acting on $\phi$ yieldsThe integral equation has the large-real-$k$ normalization $\phi /E_k=1+O(k^{-1})$ under the stated contour regularity, so its eigenfunction is not zero for every $k$ at a fixed $(x,t)$. Consequently $q_t+q_{xxx}+6q{q}_x=0$.

Thus a time-independent contour and measure determine $\phi$ through a linear spectral integral equation; $q=-{\partial }_x{\int }_L\phi d\lambda$ then reconstructs the nonlinear solution. This is the asserted linearisation. It does not assert that sums of reconstructed nonlinear solutions are again solutions, and it depends on the given uniqueness and differentiability hypotheses.

Take the spectral measure to have $n$ point masses at $k_j=i{\kappa }_j$, with distinct ${\kappa }_j>0$ and positive weights $c_j$. SetEvaluating the linear integral equation at these points gives the explicit algebraic system and reconstructionThe sign is fixed by $i/(i{\kappa }_i+i{\kappa }_j)=1/({\kappa }_i+{\kappa }_j)$. No continuum integration remains.

There is also a useful determinant form. Define $A_{ij}=c_j{E}_i/({\kappa }_i+{\kappa }_j)$, $B=I+A$, $\tau =\det B$ and $\Lambda =\mathrm{diag}({\kappa }_1,\dots ,{\kappa }_n)$. Then $A_x=-\Lambda A$ and $\Lambda A+A\Lambda =E{c}^T$. Using the derivative of a determinant and cyclicity of the trace,Hence the finite-rank KdV dressing determinant givesPositivity ensures a regular real solution: the Cauchy matrix $C_{ij}=1/({\kappa }_i+{\kappa }_j)$ is the Gram matrix of the independent functions $e^{-{\kappa }_{i}s}$ on $s>0$, so it is positive definite. The matrix $A=\mathrm{diag}(E)C\mathrm{diag}(c)$ is diagonally similar to $\mathrm{diag}(\sqrt{cE})C\mathrm{diag}(\sqrt{cE})$. Its eigenvalues are positive, and $\tau >0$. The resulting reflectionless multisoliton solution has $n$ distinct scales and velocities; the weights fix their positions. Arbitrary signed or complex weights need not give a nonsingular real solution.

For a single spectral mass, the algebraic equation givesDifferentiate the reconstruction, writing $r=cE/(2\kappa )$:With $x_0={\kappa }^{-1}\log [c/(2\kappa )]$, the concise one-soliton answer isIts speed is ${\kappa }^2$ and its maximum is ${\kappa }^2/2$. Equivalently, setting $\eta =\kappa /2$ gives $q=2{\eta }^2{\mathrm{sech}}^2[\eta (x-4{\eta }^{2}t-x_0)]$. To check the sign and speed independently, this profile obeys $q_{xx}={\kappa }^{2}q-3q^2$ and $q_t=-{\kappa }^2{q}_x$; differentiation gives the stated KdV equation exactly.