Holmgren's uniqueness theorem

We will use the multi-index notation: Let ${\displaystyle \alpha =\{\alpha _{1},\dots ,\alpha _{n}\}\in \mathbb {N} _{0}^{n},}$, with ${\displaystyle \mathbb {N} _{0}}$ standing for the nonnegative integers; denote ${\displaystyle |\alpha |=\alpha _{1}+\cdots +\alpha _{n}}$ and

${\displaystyle \partial _{x}^{\alpha }=\left({\frac {\partial }{\partial x_{1}}}\right)^{\alpha _{1}}\cdots \left({\frac {\partial }{\partial x_{n}}}\right)^{\alpha _{n}}}$.

Holmgren's theorem in its simpler form could be stated as follows:

Assume that P = ∑_|α| ≤m A_α(x)∂^α
_x is an elliptic partial differential operator with real-analytic coefficients. If Pu is real-analytic in a connected open neighborhood Ω ⊂ Rⁿ, then u is also real-analytic.

This statement, with "analytic" replaced by "smooth", is Hermann Weyl's classical lemma on elliptic regularity:[2]

If P is an elliptic differential operator and Pu is smooth in Ω, then u is also smooth in Ω.

This statement can be proved using Sobolev spaces.

Let ${\displaystyle \Omega }$ be a connected open neighborhood in ${\displaystyle \mathbb {R} ^{n}}$, and let ${\displaystyle \Sigma }$ be an analytic hypersurface in ${\displaystyle \Omega }$, such that there are two open subsets ${\displaystyle \Omega _{+}}$ and ${\displaystyle \Omega _{-}}$ in ${\displaystyle \Omega }$, nonempty and connected, not intersecting ${\displaystyle \Sigma }$ nor each other, such that ${\displaystyle \Omega =\Omega _{-}\cup \Sigma \cup \Omega _{+}}$.

Let ${\displaystyle P=\sum _{|\alpha |\leq m}A_{\alpha }(x)\partial _{x}^{\alpha }}$ be a differential operator with real-analytic coefficients.

Assume that the hypersurface ${\displaystyle \Sigma }$ is noncharacteristic with respect to ${\displaystyle P}$ at every one of its points:

${\displaystyle \mathop {\rm {Char}} P\cap N^{*}\Sigma =\emptyset }$.

Above,

${\displaystyle \mathop {\rm {Char}} P=\{(x,\xi )\subset T^{*}\mathbb {R} ^{n}\backslash 0:\sigma _{p}(P)(x,\xi )=0\},{\text{ with }}\sigma _{p}(x,\xi )=\sum _{|\alpha |=m}i^{|\alpha |}A_{\alpha }(x)\xi ^{\alpha }}$

the principal symbol of ${\displaystyle P}$. ${\displaystyle N^{*}\Sigma }$ is a conormal bundle to ${\displaystyle \Sigma }$, defined as ${\displaystyle N^{*}\Sigma =\{(x,\xi )\in T^{*}\mathbb {R} ^{n}:x\in \Sigma ,\,\xi |_{T_{x}\Sigma }=0\}}$.

The classical formulation of Holmgren's theorem is as follows:

Holmgren's theorem

Let ${\displaystyle u}$ be a distribution in ${\displaystyle \Omega }$ such that ${\displaystyle Pu=0}$ in ${\displaystyle \Omega }$. If ${\displaystyle u}$ vanishes in ${\displaystyle \Omega _{-}}$, then it vanishes in an open neighborhood of ${\displaystyle \Sigma }$.[3]

Consider the problem

${\displaystyle \partial _{t}^{m}u=F(t,x,\partial _{x}^{\alpha }\,\partial _{t}^{k}u),\quad \alpha \in \mathbb {N} _{0}^{n},\quad k\in \mathbb {N} _{0},\quad |\alpha |+k\leq m,\quad k\leq m-1,}$

with the Cauchy data

${\displaystyle \partial _{t}^{k}u|_{t=0}=\phi _{k}(x),\qquad 0\leq k\leq m-1,}$

Assume that ${\displaystyle F(t,x,z)}$ is real-analytic with respect to all its arguments in the neighborhood of ${\displaystyle t=0,x=0,z=0}$ and that ${\displaystyle \phi _{k}(x)}$ are real-analytic in the neighborhood of ${\displaystyle x=0}$.

Theorem (Cauchy–Kowalevski)

There is a unique real-analytic solution ${\displaystyle u(t,x)}$ in the neighborhood of ${\displaystyle (t,x)=(0,0)\in (\mathbb {R} \times \mathbb {R} ^{n})}$.

Note that the Cauchy–Kowalevski theorem does not exclude the existence of solutions which are not real-analytic.

On the other hand, in the case when ${\displaystyle F(t,x,z)}$ is polynomial of order one in ${\displaystyle z}$, so that

${\displaystyle \partial _{t}^{m}u=F(t,x,\partial _{x}^{\alpha }\,\partial _{t}^{k}u)=\sum _{\alpha \in \mathbb {N} _{0}^{n},0\leq k\leq m-1,|\alpha |+k\leq m}A_{\alpha ,k}(t,x)\,\partial _{x}^{\alpha }\,\partial _{t}^{k}u,}$

Holmgren's theorem states that the solution ${\displaystyle u}$ is real-analytic and hence, by the Cauchy–Kowalevski theorem, is unique.

• Cauchy–Kowalevski theorem
• FBI transform