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). 



A basic stability question
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. 


