A basic stability question

Question

Let $u_i \in C^1(\Omega)$ with $|\nabla u_i|>0$ in a simply connected region $\Omega$ with connected boundary, and $u_1=u_2$ on $\partial \Omega$. Assume $$ \nabla u_i(x) \cdot V_i (x)=|\nabla u_i(x)||V_i(x)|, \ \ \forall x \in \Omega$$ for two vector fields $V_i\in L^{\infty}(\Omega)$ with $|V_i|>0$, $i=1,2$. I wonder if $||V_2-V_1||_{L^2}$ being small would imply $u_2-u_1$ is also small in some norm (ideally in $H^1_0$ norm).

Answer 1

Let $\Omega$ be the unit ball in $\mathbb{R}^2$.

Let $u_k(x,y) = \tan^{-1}(k^3 x)$.

Let $v_k(x,y)$ be a function that agrees with $u_k$ on $\partial\Omega$, and is constant on the level sets of $\{ (x- k)^2 + y^2\}$.

So $\nabla u_k / |\nabla u_k| = \partial_x$, and $ \nabla v_k / |\nabla v_k| = \partial_x + O(1/k) $, so the directions of their gradients differ only by a little.

Let $B_k$ be the ball of radius $1/k^3$ centered at $(k - \sqrt{k^2 + 1} \approx -1/(2k), 0)$.

For large $k$, we have that $\nabla u_k \approx 1/k$ on $B_k$.

On the other hand, $\nabla v_k \approx k^3$ on $B_k$.

This implies that $\|u_k - v_k\|_{H^1_0(\Omega)} \geq \|\nabla u_k - \nabla v_k\|_{L^2(B_k)} \approx 1$ is bounded uniformly away from zero.

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This shows that, under the assumptions that

• $\Omega$ is simply connected with connected boundary, and bounded
• $u,v\in C^1(\Omega)$ are both functions bounded by $M$
• $u = v$ on $\partial\Omega$.
• and that $\nabla u, \nabla v \neq 0$ on $\Omega$

there does not exist a constant $C$ such that $$ \|u - v\|_{H^1_0(\Omega)} \leq C \| \frac{\nabla u}{|\nabla u|} - \frac{\nabla v}{|\nabla v|} \|_{L^2(\Omega)}$$

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Edit: let me give a slightly easier to check counterexample.

Let $u_k = \tan^{-1}( k^3 (\sqrt{(x-k)^2 + y^2} - k) )$

Let $v_k = \tan^{-1}( k^3 (\sqrt{(x-k)^2 + 1 - x^2}-k))$

Along the set $\{x^2 + y^2 = 1\}$ to two functions obviously agree.

Now let $B_k$ be the ball of radius $k^{-3}$ centered at the origin.

The gradients of the functions can be computed entirely explicitly

$$ \nabla u_k = \frac{k^3}{1 + k^6\left( \sqrt{(x-k)^2 + y^2} - k \right)^2 } \frac{1}{\sqrt{(x-k)^2 + y^2}} \cdot (x-k, y) $$

$$ \nabla v_k = \frac{k^3}{1 + k^6 \left( \sqrt{(x-k)^2 + 1 - x^2} - k \right)^2} \frac{1}{\sqrt{(x-k)^2 + 1 - x^2}} (-k,0) $$

Evaluating at the origin one finds

$$ \nabla u_k(0,0) = (-k^3, 0) $$

and

$$ \nabla v_k(0,0) = \frac{k^3}{1 + k^6 (\underbrace{\sqrt{1 + k^2} - k}_{\approx k^{-1}})^2} \frac{1}{\sqrt{1+k^2}} (-k,0) = O(k^{-1})$$

and the argument proceeds similarly to above.

Answer 2

Let $\Omega$ be the unit ball. Let $f$ be any non-positive radially symmetric smooth function and let $V_1 = V_2 = r f\partial_r$, hence smooth vector fields on $\Omega$.

Any and all radially decreasing function $u$ solve $V_i \cdot \nabla u = |\nabla u| |V_i|$. Clearly you can arrange for two of them to have the same boundary values while not identically the same.

If you don't like that $V$ vanish at the origin: instead of the unit ball, use the annulus $B_1(0) \setminus B_{1/2}(0)$.