Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 5 / 2

The total differential order of ${\partial }_t-{\partial }_x^2$ is two, so its principal symbol is $p(\tau ,\xi )=-{\xi }^2$. The conormal to $t=0$ is $(1,0)$, on which $p$ vanishes. The initial line is a characteristic hypersurface for this total-order symbol; the first-order time derivative does not enter it.

Suppose a real analytic solution existed near $(0,0)$. Repeated use of the heat equation gives ${\partial }_t^{k}u={\partial }_x^{2k}u$. The initial power series is ${\sum }_{j\ge 0}(-1{)}^j{x}^{2j}$ near zero, henceThe time Taylor series at $x=0$ would therefore have coefficients $(-1{)}^k(2k)!/k!$. The ratio of successive absolute coefficients is $2(2k+1)\to \infty$, giving radius of convergence zero. This contradicts the assumed real analytic regularity. No such analytic local solution exists, although the initial function itself is real analytic.

For a defining function $\varphi$ with $d\varphi \ne 0$, the characteristic hypersurface test is that the principal symbol vanish at $d\varphi$.

For the wave equation with speed $c>0$,Thus its characteristic hypersurfaces satisfy ${\varphi }_t^2=c^2|{\nabla }_x\varphi {|}^2$. In one space dimension the two families are $x\pm ct=\text{constant}$; cones are characteristic away from their vertices.

For the free Schrodinger equation, in normalized units,Its total-order characteristic hypersurfaces satisfy ${\nabla }_x\varphi =0$. Their normal is purely temporal, so locally they are constant-time hypersurfaces. Multiplying the equation by a nonzero constant or choosing the opposite sign convention does not change this test.

For the Laplace equation,There are no real characteristic hypersurfaces for the Laplace equation, since their normal cannot be zero. These statements concern the ordinary total-order principal symbol, not a weighted space-time grading.

The interior elliptic regularity assertion for the Laplace equation is that a harmonic function is smooth, in fact real analytic, throughout $U$. No boundary regularity of its unspecified boundary values is implied.

First let $u\in C^2(U)$ and $\Delta u=0$. On a ball compactly contained in $U$, differentiating its spherical average and applying the divergence theorem expresses that derivative as a constant factor times ${\int }_{B_r}\Delta u$, which is zero. The spherical average tends to $u$ at the center as $r\to 0$. This proves the mean value property for harmonic functions.

Choose a radially symmetric smooth mollifier ${\eta }_r$, supported in $B_r$, with integral one. By integrating the spherical mean value property,whenever $B_r(x)⋐U$. For fixed $r$ the right side is a smooth convolution, since all derivatives can be placed on ${\eta }_r$. Thus $u$ is smooth. The same argument proves the Weyl lemma for a distributionally harmonic $u\in L_{\mathrm{loc}}^1$: first mollify $u$, apply the fixed-radius identity, and let the mollification radius tend to zero in distributions to obtain the same smooth representative.

To prove real analytic regularity, differentiating the fixed-radius convolution gives the interior derivative estimate for a harmonic functionfor any harmonic $v$. All derivatives of $u$ are harmonic. On nested balls between $B_R(a)$ and $B_{R/2}(a)$, apply this estimate $k$ times, decreasing the radius by $R/(2k)$ each time. For $|\alpha |=k$,The inequality $k^k\le e^{k}k!$ follows by integrating $\log s$ below the sum defining $\log (k!)$. Apply the one-dimensional Taylor theorem along each segment, expanding directional derivatives by the multinomial formula. The remainder is bounded by $M(A{\sum }_j|h_j|{)}^k$, so it tends to zero for sufficiently small $h$. This gives a locally convergent multivariate Taylor series. A harmonic function is real analytic in the interior.

The usual inhomogeneous elliptic regularity statement also follows: if $\Delta u=f$ and $f$ is smooth, take a cutoff $\chi$ equal to one near a given point and set $w=\Phi \ast (\chi f)$, where $\Phi$ is a fundamental solution of the Laplace equation with $\Delta \Phi =\delta$. Moving every derivative to the compactly supported smooth function $\chi f$ shows $w$ is smooth. Locally $u-w$ is harmonic, so $u$ is smooth there too.

The Cauchy problem for a partial differential equation here prescribes both the value $u{|}_{\Gamma }=g$ and the normal derivative ${\partial }_{n}u{|}_{\Gamma }=h$, together with $\Delta u=0$. Only one of these traces would be boundary data for a usual elliptic boundary problem, rather than full Cauchy data.

Every real hypersurface is a non-characteristic hypersurface for the Laplace equation. In local real analytic coordinates flattening the real analytic hypersurface $\Gamma$, the coefficient of the second transverse derivative is nonzero: its principal coefficient is the squared length of the conormal. The equation can therefore be solved for that second derivative. The normal derivative data determine the transverse first derivative, because the coefficient relating them is nonzero and the tangential first derivatives are already determined by $g$.

The coefficients, flattened Cauchy data, and coordinate change are all real analytic. The Cauchy-Kovalevskaya theorem applies, giving a unique local real analytic solution around each point of $\Gamma$. This is a local existence assertion, not a claim of stable dependence in arbitrary Sobolev space norms.

The Cauchy-Kovalevskaya theorem cannot be applied to merely $C^2$, non-$C^3$ Cauchy data. It requires real analytic data.

There is also no $C^2$ solution of the Laplace equation on a neighborhood of a point where one of these prescribed traces fails to be $C^3$. By interior elliptic regularity, any such solution would be smooth and real analytic. On the real analytic hypersurface retained from the preceding part, both its restriction and its normal derivative would then be real analytic, hence $C^3$. This contradicts the prescribed trace. There is no solution on a neighborhood of all of $\Gamma$ with the stated non-$C^3$ data. This does not exclude solutions near other points where the data happen to be real analytic.

Use unit speed and write the Cauchy data as $u(0)=f$, $u_t(0)=g$. For finite-energy data define the wave energy estimate quantityMultiply $u_{tt}-{\Delta }_{x}u=0$ by $u_t$ and use integration by parts. With compact support or sufficient decay the boundary flux is zero, soFor general finite-energy solutions, cutoff or approximation arguments justify this identity; equivalently the local estimate below, applied in both time directions and with radii tending to infinity, gives the same equality. Arbitrary smooth data need not have finite global energy; then the global bound with an infinite right side is uninformative, while the local estimate remains useful.

If $f\in H^1$ and $g\in L^2$, the fundamental theorem of calculus and the wave energy estimate further giveTogether these yield an a priori bound for $\Vert u(t){\Vert }_{H^1}+\Vert u_t(t){\Vert }_2$ on each bounded time interval. No existence assumption is proved by the estimate itself; it controls any sufficiently regular solution.

The local wave energy estimate is, for $T\ge 0$, $r>0$, and any center $x_0$,To prove it, let $R=r+T$ and integrate the local energy estimate identity ${\partial }_{t}e=\mathrm{div}(u_t\nabla u)$, where $e=(u_t^2+|\nabla u{|}^2)/2$, over the shrinking ball $B_{R-t}(x_0)$. Differentiation of this moving-domain integral yieldsIntegrating in time proves the local wave energy estimate, without assumptions at spatial infinity.

Let the union of the initial supports be a compact set $K$. If $\mathrm{dist}(x_0,K)>T$, choose $r>0$ with $r+T<\mathrm{dist}(x_0,K)$. The initial energy on $B_{r+T}(x_0)$ vanishes. Applying the shrinking-ball identity up to every intermediate time shows $u_t$ and $\nabla u$ vanish throughout that cone. In particular, along the vertical segment through $x_0$, $u_t=0$; its initial value is also zero, so $u(T,x_0)=0$. This last value check removes the constant ambiguity invisible to gradient energy.

Consequently the finite propagation speed conclusion isThis set is compact for each finite $t$. The speed is at most one in these units, or $c$ for $u_{tt}-c^2\Delta u=0$. Time reversal gives the same conclusion for negative time.