AN-04 proof edition · CC0; exact source credit and separately licensed prerequisites

Hyperbolic Cauchy problems at a principal-type boundary

Original independent programme exposition and all twenty complete original solutions are retained. Current source and proof review: GPT-6 Astra (OpenAI), Ultra, 5 October 2026. This lesson, its added proof completions and original figure are dedicated under CC0. Linked components retain their own licences.

Begin with higher-order Cauchy roots, jets and propagation and the necessity of hyperbolicity. We now prove energy estimates, supported existence and boundary propagation when two real normal roots merge at the initial surface while the characteristic remains of principal type. The ordinary symbol calculus supplies the composition and Sobolev mapping conventions used below.

Use coordinates $(x',t)$, $D=-i\partial$, the Hermitian pairing linear in its first entry, and the tangential Sobolev norm $\|\cdot\|_s$. All symbols are ordinary $S^r_{1,0}$, with $t$ as a smooth parameter on a fixed compact interval. The mixed half-space proofs in the higher-order Cauchy lesson supply restricted/supported duality, the complete normal jet norm and the all-real normal step. Boundary distributions and intrinsic derivatives retain their exact AN-03 interfaces. No raw ambient derivative is substituted for an intrinsic derivative.

0. Exact programme inputs and the weak-limit step

The complete scalar sharp bound, finite-matrix sharp bound and scalar Fefferman–Phong proof supply the distinct positivity estimates used below. Metric Weyl products, operator domains and action and quantization conversion fix the signs, domains and remainders. Finite matrices are treated entrywise where their fixed dimension permits it; the scalar Fefferman–Phong conclusion is never asserted for matrices.

Mixed spaces and half-space duality, together with the two-weight Hilbert-space proof, prove the actual restricted/support spaces, all-real normal step, traces, tangential family maps and differential recovery. Their exact receivers are also collected in Section 4 and HC45 of the earlier higher-order Cauchy lesson. The boundary wavefront and tangential companion supplies the intrinsic extension, compressed cotangent and proper tangential action; its exact file and locators are recorded in the proof map. The normal-extension proof supplies the boundary jet calculus. These are complete earlier programme arguments, including the pure-normal exception to tangential smoothing.

Weak subsequences in the actual Hilbert space. If $(u_j)$ is bounded by $C$ in a Hilbert space $H$, apply finite Gram–Schmidt successively to the sequence, omitting each vector already in the span. This gives a finite or countable orthonormal family $e_k$ whose closed span contains every $u_j$. Each coordinate $(u_j,e_k)$ lies in the compact complex disk of radius $C$. Successive subsequences and a diagonal extraction make every coordinate converge to $c_k$. For any finite $N$, orthogonality gives

The finite square sums increase to a finite limit. Hence $\sum c_ke_k$ is Cauchy by orthogonality; completeness gives $u\in H$, with $\|u\|\le C$. For $v\in H$, the same orthogonality inequality shows that $\sum(v,e_k)e_k$ converges to $v_0$, and $v-v_0$ is orthogonal to all $e_k$, hence to every $u_j$ and $u$. Testing a finite partial sum of $v_0$ passes to the coordinate limit. Its remaining norm tail has pairing bounded by $2C$ times that tail, uniformly in $j_\ell$, proving $(u_{j_\ell},v)\to(u,v)$. A finite family uses the same argument without an infinite tail; the zero-span case is immediate.

Also $\|u\|\le\liminf_\ell\|u_{j_\ell}\|$: if $u\ne0$, test against $u/\|u\|$ and apply Cauchy–Schwarz; if $u=0$, the assertion follows from nonnegativity. A continuous linear functional vanishing on every $u_{j_\ell}$ vanishes on $u$. Consequently distributional equations and support restrictions pass through this weak limit, because each compact smooth test pairing is continuous in the mixed Hilbert norm. For supported representatives this continuity follows from their Fourier norm and Cauchy–Schwarz; for restricted spaces it follows by testing inside the domain and taking the infimum over extensions. The Hilbert representation and adjoint proof T1 and the Bochner construction supply all completeness and pairing facts used here. No separability of the ambient Hilbert space is assumed.

1. A pair of merging roots has a smooth quadratic factor

Normalize the noncharacteristic principal polynomial as

This is the index convention on printed page 404, PDF page 419 of the 2007 edition. Assume every normal root is simple and real for $t>0,\xi'\ne0$. At $t=0$ the roots are real; at every double root require

The necessity lesson supplies this alternative under the real principal-type hypotheses. A multiple normal root here is not a genuine double characteristic: $dp\ne0$ on the punctured characteristic set. Conversely the quadratic inequality forces exactly that distinction at the root in question.

Fix a double root $\tau_0$ and a small complex circle around it, disjoint from all other roots. Polynomial coefficients depend smoothly on the real parameters. On a smaller parameter patch the circle remains free of zeros and encloses exactly two roots, with multiplicity. Their power sums are smooth, even where the two individual roots fail to be smooth:

The complete polynomial circle-count identity and homotopy proof, together with the weighted circle moments A5, counts the multiplicities and gives these power sums. Differentiation under the fixed compact contour, whose denominator is bounded away from zero, proves every parameter derivative. Set

The two-root factor is $(\tau-a)^2-b$. Polynomial division gives

where $q$ is monic of degree $m-2$ in $\tau$, has smooth coefficients and is nonzero at the chosen double root. Reality of the coefficients makes $a,b$ real. For $t\ge0$ both enclosed roots are real, so $b=(\tau_+-\tau_-)^2/4\ge0$. Construct the patch first on the unit tangential sphere and extend conically; homogeneity gives degrees one and two for $a,b$. This contour argument proves the specific polynomial factor used here without importing a smooth preparation theorem for arbitrary functions.

At the double root, $b=0$, $a=\tau_0$, and

Thus $b_t>0$. On a fixed smaller conic patch, continuity and homogeneity give $b_t\ge c|\xi'|^2$ for $0\le t\le T$. Since $b(x',0,\xi')\ge0$, integration in $t$ gives

All constants belong to that fixed patch. No uniform root-gap bound persists at the boundary, and taking separate smooth square roots of $b$ there would lose the smoothness that the quadratic factor retains.

The Tricomi model is $P=D_t^2-t|D'|^2$. Its principal roots are $\pm\sqrt t\,|\xi'|$ for $t>0$, they merge at $t=0$, and the operator is elliptic for $t<0$ at nonzero covectors. At the boundary root $p_t=-|\xi'|^2$, $p_{\tau\tau}=2$, hence the required product is negative. In particular the initial surface remains noncharacteristic because its pure normal coefficient is one.

2. Exact positivity interfaces and the second-order operator

Consider

with $a\in S^1$, $b\in S^2$, $\operatorname{Im}a\in S^0$, $\operatorname{Im}b\in S^1$, and

The actual source writes a lower order $C_2(1+|\xi'|)$; this is equivalent to the displayed form with an adjusted constant.

The exact earlier scalar sharp Gårding proof G1–G10 and finite-matrix extension M1–M4 give the classical lower bound for a nonnegative order $2r+1$ symbol, $-C\|u\|_{H^r}^2$, uniformly on bounded finite-seminorm families. Here the matrix has fixed size two. We use $r=1/2$ for scalar order two and $r=-1/2$ for matrix order zero.

The complete scalar Fefferman–Phong induction, Sections 5 and 7–9 gives the stronger scalar endpoint. On $g=|dx'|^2+\langle\xi'\rangle^{-2}|d\xi'|^2$, $h=\langle\xi'\rangle^{-1}$, a nonnegative scalar $S^2=S(h^{-2},g)$ symbol has Weyl lower bound $-C\|u\|_0^2$. Its real left quantization has the same lower bound for its symmetric part: the first left-to-Weyl correction is purely imaginary and the real remainder has order zero. Constants use only finitely many seminorms; the cited programme proof includes the complete metric, coefficient norm, partition, finite-derivative and quantization prerequisites.

Replacing $A$ by its self-adjoint part changes it by order zero. Replacing $B$ by its self-adjoint part changes it by order one. The resulting change in $P$ has the form of a tangential order-one term on $u$, or order-zero term on $D_tu$. Its squared norm is bounded by $C(\|u\|_1^2+\|D_tu\|_0^2)$. In a forward estimate with $\lambda$ times these two norms, it is absorbed once $\lambda$ is sufficiently large. In the backward estimate, multiplying its squared norm by $t^2$ gives precisely the weighted norms controlled there, and is absorbed in the same way.

Scalar sharp Gårding permits adding an order-one correction to $B$ so that $B(t)\ge0$. A further order-one positive correction gives

For precision, the strengthened symbol inequality follows by adding $C\langle\xi'\rangle$ with a large fixed $C$, since $t\le1$ and the original defect has order one. Scalar sharp Gårding then bounds the real Weyl operator from below by $-C'\|u\|_{1/2}^2=-C'(\langle D'\rangle u,u)$. Adding a further $C'\langle D'\rangle$ makes the operator nonnegative while preserving the symbol inequality. These changes are among the already controlled lower terms. Hence the proof may use self-adjoint $A,B$, nonnegative $B$, and this stronger symbol inequality. For the symbol inequalities choose their real Weyl symbols; passing between Weyl and left quantization retains the order-zero correction to $A$ and the order-one correction to $B$ just described. A self-adjoint operator need not have a real full left symbol. The complete operator, including these changes, must be retained until the absorption step.

Conjugating by $\langle D'\rangle^s$ changes $A$ by order zero and $B$ by order one, because the ordinary commutator loses one order. Therefore prove the estimates at $s=0$, then absorb the conjugation remainders by choosing $\lambda>\lambda_s$. This is the reason the source can take the principal constant independent of $s$ while allowing the threshold to depend on $s$.

3. Forward energy and the singular weight at the initial surface

Let $0\le t_0<T\le1$, with $u=D_tu=0$ at $t=t_0$. Initially take $t_0>0$, so every integration is ordinary. Write $v=(D_t-A)u$, $w=e^{-2\lambda t}$, $\psi=w/t$. For a self-adjoint $A$,

The initial boundary term vanishes. The terminal term is nonnegative. Moreover

The last term is also nonnegative. In the tangential principal symbol calculus, with $\dot b=b_t-\{a,b\}$, the commutator symbol is

up to an order-one remainder times $\psi$. Here brackets are tangential brackets. Taylor expansion along the space-frequency vector field $\partial_t-H_a$, with the initial nonnegativity of $b$, gives

The order-one term includes the already permitted lower symbol terms. The expansion follows by integrating $\dot b$ along that flow and comparing its value at the endpoint: the difference is a double time integral of the second flow derivative, bounded by $Ct^2\langle\xi'\rangle^2$. Lower order corrections retain $Ct\langle\xi'\rangle$. A fixed smooth flow patch and finite symbol bounds make the constants uniform in $t_0,T$.

Combining this with $b\ge t\langle\xi'\rangle^2/C_1$, for large $\lambda$ the scalar symbol lower bound contains $(\lambda/C_1)w\langle\xi'\rangle^2$. Apply scalar sharp Gårding to the remaining nonnegative order-two symbol after dividing by $|\psi'|=w(t^{-2}+2\lambda/t)$. Its symbol seminorms on the fixed interval are uniformly bounded for large $\lambda$; the time parameter is not differentiated in this tangential application. The resulting error is bounded by

The order-one remainder has the same form. Young's inequality absorbs a fixed portion of $\lambda\int w\|u\|_1^2$, leaving

All integrals in this section are over $[t_0,T]$. The forcing term is absorbed into half the first left term plus $C\int w\|Pu\|_0^2$, since

For the remaining singular error, define

The exact identity is $-(wK)'=wW_u$. Set $I^2=\int wW_u\|u\|_0^2$. Integration by parts, the zero initial trace and the nonnegative terminal term give

The contribution of $A$ has zero real part in the derivative calculation, since $A$ is self-adjoint. Weighted Cauchy–Schwarz gives

For $z=\lambda t\ge0$,

so $K^2/W_u\le t^{-2}+\lambda/t$. It follows that

Also $(\lambda/t+t^{-2})^2\le C W_u$, by the bounded elementary ratio $(z+1)^2/(2z^3+z^2+2z+3)$. Hence its coefficient $C/\lambda$ can be absorbed by taking $\lambda$ large. Keeping fixed portions of all the positive terms proves

This is the stronger estimate represented by source (23.4.5); constants absorb the harmless numerical factors. Since $0<t\le1$, it implies

The term $Au$ is controlled by $\|u\|_1$. Conjugation and the lower term absorption of Section2 give the stated estimate in every real tangential Sobolev order. For $t_0=0$, smooth $u$ with $u=D_tu=0$ has $u=O(t^2)$, $v=O(t)$. Consequently the singular integrals above are finite and their boundary terms tend to zero. First integrate over $[\epsilon,T]$, retain the actual boundary terms and let $\epsilon\downarrow0$; this justifies the endpoint, rather than assigning $\psi(0)$ a value.

We shall also need the terminal terms that this argument permits us to retain. At $s=0$, the energy identity controls $\psi(T)\|v(T)\|_0^2$ together with the integrated positive terms, because $\psi(T)(Bu(T),u(T))\ge0$. The Hardy identity before dropping its endpoint is

Young's inequality and the already controlled kinetic integral therefore bound both terminal expressions by the same forcing norm. At order $s=-1/2$ this gives

Here one may take $v$ with the self-adjoint part of $A$. The order-zero conjugation defect and the order-one defect of $B$ are absorbed in the integrated terms as above. Replacing that $v$ by the original shifted derivative adds an order-zero image of $u(T)$, controlled by the second terminal expression since $\psi\le K w$ for large $\lambda$. No forcing beyond $T$ is used. Explicitly, for $z=\langle D'\rangle^{-1/2}u$, first replace the conjugated coefficients by self-adjoint coefficients with a nonnegative corrected $B_s$, using only the already permitted order-zero and order-one changes. Apply the $s=0$ argument, including its terminal terms, to this modified equation for $z$. Its extra forcing is $C_0D_tz+C_1z$, bounded by the integrated positive terms with a coefficient absorbed for large $\lambda$. Thus positivity at the terminal point is used for this corrected operator, not assumed for the raw conjugate of $B$. The shifted derivatives differ by order zero, and the retained Hardy endpoint controls that difference as stated.

4. Backward energy has different weights

Take instead $u=D_tu=0$ at $t=T$. Use $w_+=e^{2\lambda t}$, $\psi_-=-tw_+$, so

The initial endpoint contributes $-\psi_-(t_0)\|v(t_0)\|_0^2\ge0$, while the terminal term vanishes. The scalar commutator principal symbol is

For the backward sign one needs a different Taylor comparison. Follow the same flow from time $t$ to time $2t$. Nonnegativity at the latter point and the second flow derivative bound give $b(t)+t\dot b(t)\ge-Ct^2\langle\xi'\rangle^2$, modulo the order-one correction $-Ct\langle\xi'\rangle$. If $2t>1$, the assertion follows from bounded symbol seminorms and $t\ge1/2$; near zero the flow stays in the fixed extension patch. Combining this comparison with $b\ge t\langle\xi'\rangle^2/C_1$ gives, after retaining a smaller positive principal term,

The last term is controlled by Young's inequality in the variable $t\langle\xi'\rangle$. Apply the scalar Fefferman–Phong input to the resulting nonnegative order-two symbol, normalized on a bounded finite-seminorm family; include its order-zero error and the real lower symbol terms. This yields the source's backward energy inequality

The zero-order error matters: applying only the order-two matrix sharp bound would leave the wrong unweighted $H^{1/2}$ error at $t=0$. Scalarity is used exactly here.

The terminal zero trace gives the second elementary identity

It includes a nonnegative initial trace term in its first inequality. Hence $\lambda^2\int w_+\|u\|_0^2\le\int w_+\|v\|_0^2$. Use one fixed portion of the $v$ term for this control, absorb the forcing term by Young, and absorb $C\lambda\|u\|_0^2$ when $\lambda$ is large. Recovering $D_tu=v+Au$, then conjugating and absorbing lower terms, proves

The estimate is uniform in $t_0\ge0$ at fixed $T\le1$. It does not say that the top tangential and normal energies are unweighted near the initial surface. Their $t^2$ weights must survive the higher-order factorization and the adjoint existence argument.

For precision in the positivity step, subtract a fixed fraction of $\lambda t^2\langle\xi'\rangle^2$. Young's inequality bounds the term $-Ct\langle\xi'\rangle$ below by a further small fraction of that positive term and $-C/\lambda$. After adding this constant, divide the remaining scalar nonnegative symbol by $1+\lambda$. Its tangential symbol seminorms are uniformly bounded: every occurrence of $\lambda t b$ has been divided by $1+\lambda$, and no normal derivative is part of those seminorms. Scalar Fefferman–Phong gives a lower error $-C(1+\lambda)\|u\|_0^2$. Real quantization corrections and the permitted imaginary lower symbols have the same bound or are absorbed into the weighted positive term. This explains both the scalar hypothesis and the exact unweighted error used above.

5. Reconstructing all normal jets without separating a double pair

Let $P$ now be a differential operator of order $m$, normalized with coefficient one on $D_t^m$. Work on a compact coordinate patch and a small cone in $\xi'\ne0$. Its principal polynomial factors into monic smooth polynomials $q_\mu$ of degrees $d_\mu\in\{1,2\}$, with $\sum_\mu d_\mu=m$. Different factors have no common root. A quadratic factor may have its two roots equal at $t=0$.

The concrete algebra needed for the estimates is

These $m$ polynomials form a basis of the polynomials of degree at most $m-1$. Indeed, reduce a linear relation modulo $q_\mu$. All other terms vanish. The remaining product is invertible in $\mathbb C[\tau]/(q_\mu)$, by coprimality between factors, so its coefficient polynomial of degree below $d_\mu$ is zero. Repeat for every $\mu$. Dimension proves the assertion. The quotient algebra still has dimension two at an internal double root; no division by that internal root gap has occurred.

The quotient inverse, including a double root. Put $R_\mu=\prod_{\nu\ne\mu}q_\nu$. For a linear $q_\mu=\tau-r$, the constant $1/R_\mu(r)$ is its inverse modulo $q_\mu$. If $q_\mu$ has two distinct roots $r,s$, the polynomial

has $R_\mu V-1$ zero at both roots, so division by the two linear factors proves divisibility by $q_\mu$. If $q_\mu=(\tau-r)^2$, use instead

The value and first derivative of $R_\mu V-1$ at $r$ vanish, so its polynomial Taylor formula shows divisibility by $(\tau-r)^2$. The denominators $R_\mu(r)$ never vanish because different factors share no root. These pointwise formulas prove invertibility of the coefficient matrix at every parameter, without choosing smooth individual roots. The matrix entries themselves are smooth factor coefficients; its inverse is the adjugate divided by its nonzero determinant and is therefore smooth, with all derivatives bounded on a smaller compact cone. This proves precisely the algebraic and parameter assertions used in PT1 and PT5.

Put $L=\langle D'\rangle$, and define

The second expression includes the highest normal derivative in tangential order $s-1$. It is an error norm, not the lower jet in the backward conclusion.

Quantize $q_\mu$ as a first order operator with real principal symbol or as $(D_t-A_\mu)^2-B_\mu$. Quantize the monic quotient $\prod_{\nu\ne\mu}q_\nu$, in normal powers on the right, as $Q_\mu$. Then

Here $R_\mu$ has total order at most $m-1$ and normal degree at most $m-1$. The order $m$ terms agree by construction, and the two monic $D_t^m$ terms cancel exactly. Differentiating a coefficient in time and each ordinary tangential composition correction lowers the total order by one. Thus all coefficient time derivatives belong to $R_\mu$; none is silently omitted.

Choose a finite tangential cone partition $\chi_\alpha$, with a larger cutoff for each cone. Its base cutoffs cover the compact time interval too: near zero use the simple/quadratic clusters, and away from zero use the strictly separated simple roots. Replace $Q_\mu u$ in that cone by $Q_\mu\chi_\alpha u$. The added $[P,\chi_\alpha]$, including derivatives of its time cutoffs, has the same total and normal order bound as $R_\mu$. Extend each principal factor off the smaller cone with bounded symbol seminorms and, for a quadratic factor, its positive $b\ge ct\langle\xi'\rangle^2$ property. This can be done by blending with $ct\langle\xi'\rangle^2$ where the local lower bound still holds. On the complement of the larger cone the proper composition error is tangentially smoothing with finite normal degree. These errors obey the same lower order bounds used below. The finite time cover accounts for the fixed-interval uniformity asserted in (PT9).

To see the inverse estimate explicitly, form the normalized jet columns

and the $m$ rows

The principal coefficient matrix taking $Z$ to $W$ is order zero. On the unit tangential sphere it is the basis matrix (PT1), with its columns multiplied by nonzero scalar powers of $|\xi'|$. It is invertible on a smaller compact cone. The exact ordinary AN03 matrix parametrix gives an order-zero inverse there, with an order-minus-one error after one step. Hence, summing the finite cover and adding the bounded-frequency piece,

Time derivatives in (PT4) differentiate the coefficients of $Q_\mu$ as well; these terms again have total order $m-2$ in this reconstruction and contribute to $E_s^-$. At bounded tangential frequency the Sobolev weights are comparable, so that piece is also controlled by $E_s^-$. This is an ordinary square matrix calculation after column normalization, with no general boundary system theorem being assumed.

For a quadratic factor the forward estimate of Section 3 controls precisely the two rows $LQ_\mu\chi_\alpha u$ and $D_tQ_\mu\chi_\alpha u$. The standard first order energy controls the one row for a linear factor: integration with $e^{-2\lambda t}$ gives $\lambda^2\int e^{-2\lambda t}\|w\|_s^2\le C_s\int e^{-2\lambda t}\|q_\mu w\|_s^2$ for its zero initial trace. Use the weaker coefficient $\lambda$ so that all factors have one common inequality. Every required initial jet of $Q_\mu\chi_\alpha u$ vanishes if $D_t^j u(t_0)=0$ for $j<m$. Equations (PT3)–(PT5) give

The $R_\mu$ terms map the top jet to $H^s$; their coefficient is the fixed $C_s$ on the left. It remains to absorb the lower reconstruction error.

For a Hilbert space valued function $g$ with $g(t_0)=0$, integration of $(w\|g\|^2)'$, including the nonnegative terminal term, proves

Apply this to $D_t^j u$ in $H^{s+m-2-j}$. For $j<m-1$, its derivative is a top jet in (PT2). For $j=m-1$, use the monic equation

Thus $\int w E_s^-\le C_s\lambda^{-2}\int w(E_s+\|Pu\|_s^2)$. Inserting this into (PT6) and taking $\lambda$ large proves $\lambda\int wE_s\le C_s\int w\|Pu\|_s^2$. Fix one such $\lambda_s$. Comparing the exponential weights on $[0,1]$ proves

All constants are uniform for $0\le t_0<T\le1$ in the fixed chart.

6. Weighted backward reconstruction and the unweighted lower jet

For terminal zero jets use $w_+=e^{2\lambda t}$. Section 4 controls the quadratic rows with $\lambda t^2$. For a first order factor integrate with $\psi=-t^2w_+$. Its positive derivative is $-\psi'=(2t+2\lambda t^2)w_+$; Young's inequality gives

Order-zero imaginary terms are absorbed for large $\lambda$. The inverse calculation (PT5) therefore gives

For $g(T)=0$, integration of $(t^2w_+\|g\|^2)'$ and the nonnegative initial term proves the counterpart of (PT7) with weight $t^2w_+$. Applying (PT8) again absorbs the last term of (PT10). Hence $\lambda\int t^2w_+E_s\le C_s\int t^2w_+\|Pu\|_s^2$.

One more Hardy calculation supplies every lower jet, rather than trying to reconstruct it from only the quadratic rows. For any terminal-zero $g$,

The first inequality keeps the initial boundary term, which has the correct sign; the second follows by Cauchy–Schwarz. Take $g=D_t^ju$ in $H^{s+m-2-j}$, for $0\le j\le m-2$. Its derivative is a top jet already controlled by (PT10). Fixing $\lambda_s$ and comparing weights proves

For $m=1$ the second sum is empty. First order problems also have the ordinary unweighted backward estimate of the strict Cauchy lesson. The quadratic case and (PT12) do not give an unweighted top jet at $t=0$.

Source comparison note: the two references to (23.4.6) in the forward higher order proof and its forward existence application, on printed 409–410/PDF 424–425 of the 2007 edition, have incompatible backward weights. The estimate used at those steps is (23.4.8), represented here by (PT9). The source is preserved; the teaching argument uses its actual forward orders and direction.

7. Supported existence, arbitrary orders and the extra half derivative in the data

Let $X_0=\{\varphi=0\}$, $X_+=\{\varphi>0\}$, and let $P$ be a smooth differential operator of order $m$. Assume $X_0$ is noncharacteristic, $P$ is strictly hyperbolic for the positive levels and its real principal symbol remains of principal type at $X_0$. The necessity result and Section 1 give exactly the simple/quadratic factors needed above. For any relatively compact equation region $Y$, supported $f\in H_s^{\mathrm{loc}}(X)$ has a supported solution

Uniqueness holds in a neighborhood of $X_0\cap Y$. A semiglobal extension through the positive interior uses the controlled strict Cauchy regions of the preceding lesson; uncontrolled incoming lateral boundaries are not uniqueness regions.

Here are the local steps and their orders. Straighten $\varphi=t$ and normalize the leading normal coefficient. On a compact chart choose $M$ larger than every characteristic normal-to-tangential ratio. The corresponding short forward and backward cones have noncharacteristic supporting planes and lie in the equation domain after shrinking time. The strict Cauchy existence and uniqueness result then gives, for smooth forcing and $\epsilon>0$, a solution with all initial jets zero at $t=\epsilon$, supported in the forward cone. Extend it by zero below $\epsilon$. No boundary delta occurs in $Pu_\epsilon$, because all $m$ initial jets vanish; its forcing is $1_{t>\epsilon}f$.

Estimate (PT9) is uniform in $\epsilon$. It bounds $u_\epsilon$ in the mixed space $H_{(m-1,q)}$ when the forcing is in $H_{(0,q)}=L^2_tH^q_{x'}$. Here and below supported spaces mean the actual ambient supported representative, with local cutoffs. Weak compactness gives a supported limit satisfying the ambient equation. For smooth forcing all tangential orders are available; the monic differential equation and the exact normal recovery theorem give all normal orders as well. The limit is smooth up to the initial surface and its Cauchy jets vanish. Density in $L^2_tH^q_{x'}$, (PT9) for differences and the same support cone produce the solution at every real $q$. Local uniqueness, proved just below, identifies all these limits.

For forcing in $H_{(r,q)}$, $r\ge0$, the mixed embedding first gives $f\in H_{(0,r+q)}$, so the construction yields $u\in H_{(m-1,r+q)}$. Differential normal recovery B.2.9, with forcing $Pu\in H_{(r,q)}$, now gives

The total order $r+q+m-1$ is preserved, and the allowed normal order comes from the equation. The theorem is being used here for a differential $P$; Section 8 supplies a separate proof for tangential pseudodifferential models.

Negative integer normal orders follow by the exact one-sided supported decomposition already proved in the strict Cauchy lesson. At order $(r,q)$ write

with both representatives supported on the positive side. This is the forward supported decomposition, distinct from the backward kernel used to extend a restricted distribution. Suppose existence is established at normal order $r+1$. Solve for $u_0,u_1$ with these two forcings. Then

The commutator has total order $m$ and normal degree at most $m-1$, because the leading normal coefficient is one. It maps $u_1$ to $H_{(r+1,q-1)}$. Solve once more at that already established normal order and subtract its solution. The final result lies in $H_{(r+m-1,q)}$, by the mixed embedding for $u_0$ and the correction, and differentiation for $u_1$. Every term preserves support. This is finite induction from $r=0$, not an infinite iteration.

For arbitrary isotropic $s$, choose an integer $r\le\min(0,s)$ and $q=s-r$. Then $H_s\subset H_{(r,q)}$. The preceding construction gives a solution with total order $s+m-1$, possibly with too little normal order. Apply the whole-space differential normal-recovery argument on the ambient equation $Pu=f$, including across zero. This is the whole-space counterpart of B.2.9, with the same normal and total-order inequalities, used and checked in Section6 of the strict Cauchy lesson. It raises the normal order to $s+m-1$ without changing the total, yielding the actual isotropic conclusion (PT13). It changes the regularity conclusion for the same distribution and therefore preserves its support. Using only restricted recovery would not establish ambient regularity across the initial surface. This last recovery is essential; the tangential energy alone does not imply the isotropic conclusion when $s>0$.

To prove uniqueness, take a compact smooth $g$ in a smaller backward cone and solve $P^*v=g$ with terminal zero data in the strict positive interior. The adjoint has the same real principal factors and permitted lower terms. Estimate (PT12) bounds its lower jet without a $t^2$ weight in every tangential order, uniformly down to $t=0$. For $m\ge2$ this supplies an extendible mixed seed of normal order $m-2$; all tangential orders and differential normal recovery then give all normal orders up to the boundary on a fixed smaller cone. For $m=1$ use the ordinary first order backward estimate. Thus $v$ is actually smooth up to $X_0$, rather than merely having a weighted top jet.

Extend $v$ smoothly across zero and cut it off within the controlled cone. Its terminal and lateral support properties make this cutoff legitimate. For any ambient supported homogeneous distribution $u$,

The first equality uses equality on the positive cone and matching all boundary jets; an arbitrary smooth extension of $v$ has the same jets there. Terms flat on the closed positive side pair to zero with a finite-order supported distribution. Hence every such $g$ annihilates $u$. This proves the claimed local uniqueness, including possible boundary-supported distributions. It also permits the finite cone patching and strict interior continuation used for (PT13).

For nonzero Cauchy data and isotropic forcing of nonnegative order $s$, choose a trace extension

Here is a continuous trace extension with all the required orders. Take a smooth compact function $\chi$ equal to one near zero, put $\rho_\xi=\langle\xi'\rangle$, and set

Every derivative of $\chi$ at zero except its value vanishes, so $D_t^ku_0(0)=\psi_k$ for $k<m$. For one summand its full time Fourier transform has the form $\rho_\xi^{-j-1}F_j(\tau/\rho_\xi)\widehat\psi_j$, with $F_j$ Schwartz. For any real $R,q$, direct integration of the mixed Fourier weight gives

Apply this with $(R,q)=(s+m,0)$, or $(r+m,q)$ in the mixed statement. Proper base cutoffs retain the norm bounds. The strict trace threshold in B.2.7 is satisfied, so these are the actual continuous traces, not merely formal Fourier values.

The ordinary differential image of this extension lies in $\overline H^s$. The remainder $F=f-Pu_0$ embeds in $L^2_tH^s_{x'}$ on the positive half. Extend it by zero in that space, where the extension is bounded, and use the supported $r=0$ construction above. This gives a correction in ambient $H_{(m-1,s)}$. Restricted differential normal recovery then gives $\overline H^{s+m-1}$. No bounded zero extension in isotropic $H^s$ has been assumed. Equivalently, in mixed spaces with $r\ge0$, use the displayed extension in $\overline H_{(r+m,q)}$, data $H^{r+q+m-1/2-j}$, zero extension of the remainder in $L^2_tH^{r+q}_{x'}$, and recovery to $\overline H_{(r+m-1,q)}$.

To justify all $m$ jets of the correction, the same restricted recovery also gives $\overline H_{(s+m,-1)}$, or $\overline H_{(r+m,q-1)}$ in the mixed case: its normal order is allowed by the equation and its total order is exactly the seed total. Since $s\ge0$, or $r\ge0$, all these traces exist. They vanish because the ambient monic equation has no boundary delta derivatives; the coefficient of the highest delta derivative first forces the zeroth jet to vanish, and induction forces each subsequent jet. Compared with the strict Cauchy proposition, which uses $H^{s+m-1-j}$, (PT17) requires exactly half a derivative more. This sufficient trace-extension reduction is stated where its traces exist; it does not silently extend a positive-normal-order trace theorem to arbitrary negative normal orders.