Does unique continuation also hold for derivatives of solutions?

Question

Let $D \subset \mathbf{R}^n$ be the unit ball, and $u \in C^2(D)$ be a solution of the linear elliptic PDE \begin{equation} a^{ij} D_{ij} u + b^i D_i u + cu = 0 \quad \text{in $D$}, \end{equation} where the coefficients are regular, say of class $C^d$ for some integer $d \geq 1$.

Assume that $u(0) = 0$, $Du(0) = 0$. Strong unique continuation means that $u$ has finite order of vanishing at the origin unless it vanishes identically: there exists $N > 0$ so that $r^{-n} \int_{D_r} u^2 \notin O(r^N)$ as $r \to 0$ if $u \not \equiv 0$.

Question. Under the given hypotheses, do the partial derivatives $D_k u$ of $u$ also satisfy the strong unique continuation property?

Answer 1

In general there's no way what you want (as clarified in this comment) can be true.

Let $n = 3$, and fix $k = 3$. Consider the function $u = (x_1)^2 - (x_2)^2$ which is harmonic.

Let $\Phi: D \to D$ be a diffeomorphism that is the identity at a neighborhood of the origin. Then the Laplacian pulls back to a Laplace-Beltrami operator $L$ on $D$, and the pullback $\tilde{u} = \Phi^*u$ is a solution to $L\tilde{u} = 0$. Note that $L$ has no 0-th order terms. Note that $\tilde{u}$ satisfies $\tilde{u}(0) = \nabla \tilde{u}(0) = 0$, and that $\partial_3 \tilde{u}$ vanishes to infinite order at origin (since this holds of $u$ and $\Phi$ is identity near the origin).

However, for general $\Phi$, the function $\partial_3\tilde{u}$ does not vanish identically in $D$.