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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).

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    $\begingroup$ Please be more thoughtful about tags going forward. Your PDE is obviously not linear (absolute value function is not linear). Your PDE is not elliptic (a fortiori since my answer below; but I don't see any reason for you to believe that it is elliptic to start with). There's no hard dependencies on function spaces, and I don't see any special function entering the discussion. $\endgroup$ Aug 15, 2018 at 13:44
  • $\begingroup$ I added the assumption $|V_i|>0$. $\endgroup$
    – User4966
    Aug 15, 2018 at 15:51
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    $\begingroup$ If you are so happy to move goal posts, you may as well add that $\Omega$ is bounded (to rule out exterior of a ball), and that $|V_i| = 1$ (in fact, a fortiori if $V_i$ is the direction of the gradient of $\nabla u_i$, and $u_i$ is $C^1$, then $V_i \in C^0$). Fixing the size of $V_i$ gets rid of the scaling issue also. In fact, you might as well re-phrase your question as: let $u, v\in C^1(\Omega)$ be two functions that agree on $\partial\Omega$ with non-vanishing gradient. Can $\|u-v\|_{H^1}$ be bounded by $\| \nabla u / |\nabla u| - \nabla v / |\nabla v|\|_{L^2}$? $\endgroup$ Aug 15, 2018 at 17:02
  • $\begingroup$ The answer to which would still be "no" because you can replace $u, v$ by $\lambda u, \lambda v$. So you still need to fix a scale. $\endgroup$ Aug 15, 2018 at 17:11

2 Answers 2

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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.


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 $$ \|u - v\|_{H^1_0(\Omega)} \leq C \| \frac{\nabla u}{|\nabla u|} - \frac{\nabla v}{|\nabla v|} \|_{L^2(\Omega)}$$


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.

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  • $\begingroup$ I doubled checked the computations. It seems to me that $\nabla v_k \approx \frac{1}{k^3}$ and $\nabla u_k \approx \frac{1}{k^3}$. So this doesn't seem to provide a counterexample. $\endgroup$
    – User4966
    Aug 16, 2018 at 20:29
  • $\begingroup$ @MathStudent: your computation for $\nabla u_k$ cannot be right. By definition you have that $\nabla u_k(x,y) = k^3 / (1 + k^6 x^2) \partial_x$. Evaluating it at $(-1/2k,0)$ you should get something of the order $k^3 / (1 + k^4)$ which is approximately $1/k$. The value of $\nabla v_k(x,y)$ is hard to compute exactly, but the key idea is that at $(-1/2k,0)$, $v_k = 0$, and hence its derivative is expected to be of order $k^3$. You should double check your computations again. $\endgroup$ Aug 16, 2018 at 21:28
  • $\begingroup$ You are absolutely right. Thank you, Willie. $\endgroup$
    – User4966
    Aug 16, 2018 at 21:48
  • $\begingroup$ @ Willie Wong: Do you think the result may hold if we additionally assume that $|\nabla u_i|$ are bounded( i.e. $\lambda \leq |\nabla u_i| \leq \lambda$)? $\endgroup$
    – User4966
    Aug 25, 2018 at 16:24
  • $\begingroup$ @MathStudent: Not sure. If your assumed bound on $V_2 - V_1$ were in $L^\infty$ and not $L^2$, then ODE arguments will tell you that the level sets of $u_1$ and $u_2$ are similar, which coupled with $L^\infty$ bounds on $\nabla u_i$ would (probably) imply using a simple triangle inequality argument that $\nabla u_1 - \nabla u_2$ is controlled (even in $L^\infty$). It is not immediately clear to me whether in $L^2$ (and in higher dimensions) there may be more delicate counterexamples. $\endgroup$ Aug 26, 2018 at 1:23
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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 $B_1(0) \setminus B_{1/2}(0)$.

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  • $\begingroup$ Basically: your equation fixes the direction but not the magnitude of the gradient, and the magnitude does matter. $\endgroup$ Aug 15, 2018 at 13:51
  • $\begingroup$ Thank you, Willie. Actually I forgot to mention that $\Omega$ is simply connected. Will there be a counter example under this assumption? $\endgroup$
    – User4966
    Aug 15, 2018 at 15:53
  • $\begingroup$ The unit ball is simply connected. The annulus in $\mathbb{R}^d$ for $d \geq 3$ is also simply connected. The interval in $\mathbb{R}$ is also simply connected. $\endgroup$ Aug 15, 2018 at 16:38
  • $\begingroup$ Of course, but we also require $|\nabla u_i|>0$. So your example on unit ball will not be a counter example. What if we require $\partial \Omega$ to be connected? Then I guess there shouldn't be a trivial counter example. $\endgroup$
    – User4966
    Aug 15, 2018 at 16:45

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