Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 / 3

Write $P=P_N+P_{N-1}+\cdots +P_0$ as a sum of homogeneous parts. The operator is elliptic whenContinuity on the unit sphere gives $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Uniformly in $\omega$,so for sufficiently large $t$, $|P(t\omega )|\ge (c/2)t^N$. Since $t^N\asymp ⟨t\omega {⟩}^N$ at large $t$,for sufficiently large $|\lambda |$.

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A parametrix for $P(D)$ is a distribution $E$ for whichwith $r\in C^{\infty }$. Choose a smooth cutoff $\chi$ that vanishes on a large ball containing every real zero of $P$ and equals one outside a slightly larger ball. Ellipticity makesa symbol of order $-N$. For $E={\mathcal{F}}^{-1}q$,Since $1-\chi$ is smooth and compactly supported, its inverse Fourier transform is smooth. Thus $E$ is a parametrix.

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The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^n;N)=S_{1,0}^N$ consists of $c\in C^{\infty }(X\times {\mathbb{R}}^n)$ such that, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,The defining estimate givesThe Leibniz rule givesand the triangle inequality givesThese are the basic rules of symbol calculus.

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The assumed lower bound gives $|a|=|Q{|}^{-1}{\lesssim }_K⟨\lambda {⟩}^{-M}$ at large frequency. Differentiating $1/Q$ repeatedly expresses every derivative as a finite sum of products of derivatives of $Q$ divided by powers of $Q$. Since $D_{\lambda }^{\beta }Q$ has polynomial order at most $M-|\beta |$, induction and symbol calculus giveThe interpolation region $R\le |\lambda |\le R+1$ is compact in frequency and causes no problem. Hence

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For any smooth amplitude $v(x,\lambda )$,because $D_x=-i{\partial }_x$. Applying each term of the differential operator under the oscillatory integral givesThe identity is justified distributionally by regularizing the frequency integral and integrating by parts.

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Since $Q(x,\lambda )$ is a polynomial of degree at most $M$ in $\lambda$, its Taylor formula is exact:The $\alpha =0$ term is $Q{a}_1=1$ outside the cutoff region; its difference from one is a symbol of order $-\infty$. For $|\alpha |\ge 1$,so their product lies in $S^{-|\alpha |}\subset S^{-1}$. Defining $b_1$ as the negative of the cutoff remainder and these lower-order terms yields

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TakeBy symbol calculus, $a_2\in S^{-M-1}=S^{N_Q-1}$. Its zeroth-order contribution is $Q{a}_2=b_1$ outside the compact transition region and therefore cancels the order $-1$ remainder. Every term incontains at least one $\lambda$ derivative of $Q$ and has order at most $(M-1)+(-M-1)=-2$. Absorbing these terms and the smoothing cutoff contribution into $b_2$ gives

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Iterate the correction: after constructing $b_{j-1}\in S^{-(j-1)}$, setThe same calculation improves the remainder by one order:Let $A_j$ and $B_N$ be the corresponding inverse oscillatory integrals. ThenWhen $N>n$, the frequency integral defining $B_N$ converges absolutely and depends continuously on $x$, so $B_N\in C(X)$. This completes the finite-order parametrix construction.

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