# Prescribing a gradient direction

Let $$\Omega= \{(x,y) : \frac{1}{2} \leq x^2+y^2 \leq 1\}$$ and $$S = \{(x,y) : x^2+y^2 = 1\}$$ the unit circle, and $$X=w^{1.\infty}(\Omega;\mathbb{R})$$ the space of Lipschitz valued functions. We denote by $$\left| \cdot \right|$$ the euclidean norm in $$\mathbb{R}^2$$ or the modulus in $$\mathbb{R}$$.

Given $$\phi \in C^\infty(\Omega;S)$$, I am looking for a continuous linear operator $$P$$ acting on $$X$$ such that

For all $$f\in X$$, $$|D(Pf)\cdot \phi| = |D(Pf)| ,$$ and

$$\text{ If } |Df \cdot \phi| = |Df| \mbox{ in } \Omega, \text{ then }P(f)=f.$$

In other words, if the gradient of $$f$$ is parallel to $$\phi$$ then $$f$$ is unchanged by $$P$$, whereas if it isn't, $$P$$ "projects" it on that space.

There is one simple case when one can do it naturally, it is when $$\phi = (x,y)/|(x,y)|$$, as in this case one can choose $$Pf(x,y) = \frac{1}{2\pi}\int_{0}^{2\pi} f\left(|(x,y)|\cos \theta, |(x,y)| \sin \theta \right) d\theta.$$ As $$Pf$$ is radial, $$D(Pf)$$ is parallel to $$\phi$$.

If there exist two sufficiently smooth functions such that $$u,v$$ such that

a. $$\phi= Du$$

b. $$\min \det (Du,Dv)>0$$

c. $$v$$ is such that for all $$u$$ $$\int_{\{u=c\}\cap\{x\in \Omega\}} dv=1$$ then, writing $$x=s(u,v) \text{ and } y=t(u,v)$$ we can generalize the radial case to $$Pf(x,y) = \int_{\{v : (s(u,v),t(u,v))\in \Omega \}} f\left(s(u(x,y),v),t(u(x,y),v)\right) dv.$$ As pointed out by Mikhail Skopenkov in his remark, this is easier said than done. Suppose $$\phi=(1,0)$$, then $$u=x$$; but the natural choice of $$v=y/l(x)$$, where $$l(x)=\begin{cases} 2\sqrt{1-x^2} & \text{ when } \frac{1}{\sqrt2}\leq |x| \leq 1 \\ 2\sqrt{1-x^2} - 2\sqrt{\frac{1}{2} - x^2}& \text{ when } 0 \leq |x| < \frac{1}{\sqrt2} \end{cases}$$ does not work, as $$l$$ isn't Lipschitz at $$x=1/\sqrt{2}$$, in fact only in $$W^{1,s}$$ for $$s<2$$, thus $$Pf(x,y) = \int_{v : (x,v l(x)) \in \Omega} f\left(x,v l(x)\right) dv,$$ which is indeed a function of $$x$$, and corresponds to averaging $$f$$ in the direction $$(0,1)$$ orthogonal to $$\phi$$, isn't regular.

Any hint on how to go for general directions $$\phi$$ would be great.

• Could you perhaps make the question a little clearer? When you say "this works", you have not specified what you mean by "a projection", so it is not clear what you mean by "works". I think you want $Pf$ to also be a function on the annulus, but constant along radial directions, so really a function on the circle. But what else do you want? The phrase "is what I am looking for ... but I am not sure this is the case" is unclear to me. – Ben McKay Nov 10 '18 at 8:53
• @BenMcKay $Pf$ is constant on concentric cirles in the example, if I read it correctly. "Projection in the direction of $\phi$" seems to mean "the gradient at each point $(x,y)$ is contained in the line generated by $\phi(x,y)$". But I agree the question should be formulated more clearly. Also, one should probably avoid solutions of the type "fix a function $g$ with $Dg\in\phi\times\mathbb R$ and set $Pf=<f,g>_{L^2}g$". – Sebastian Goette Nov 10 '18 at 10:51
• Thank you both very much, in particular @SebastianGoette as your comment allowed me to define what I wanted. – username Nov 12 '18 at 17:25
• The function $l(x)$ does not seems to be $C^\infty$, hence the resulting $P$ does not seem to an operator on $C^\infty(\Omega;\mathbb{R})$. – Mikhail Skopenkov Nov 14 '18 at 13:09
• Since you are working in two dimensions, the orthogonal distribution to $\phi$ is integrable, and $Pf$ has to be constant along the leaves of this distribution. Does it not suffice to just average $f$ over these leaves? – Willie Wong Nov 14 '18 at 22:12

Here is a comment that shows possible obstructions to the existence of such an operator. I'm assuming a bit more regularity here.

Let $$(M,g)$$ be a compact Riemannian manifold with boundary, and let $$\text{vol}_g$$ be its Riemannian volume form. Equip the vector space $$C^{\infty}(M)/\mathbb{R}$$ of smooth functions integrating to zero with the inner product $$\langle \alpha,\beta\rangle=\int_{M}g(\text{grad}\,\alpha,\text{grad}\,\beta)\text{vol}_g.$$ For a fixed vector field $$\phi$$ on M, define the subspace $$V_{\phi}=\{f\in C^{\infty}(M)/\mathbb{R}\;\vert\;\text{grad}\,f=\lambda\cdot\phi\text{ for some }\lambda\in C^{\infty}\}.$$ Then I think your problem (correct me if I'm misinterpreting your question) can be reduced to computing for a given $$\phi\in\text{Vect}(M)$$ and $$f\in C^{\infty}(M)/\mathbb{R}$$ its orthogonal projection onto $$V_{\phi}$$ using the inner product above.

Suppose that the projection $$Pf$$ of $$f$$ in the above sense exists. We have $$0=\int_{M}\lambda g(\text{grad}\,Pf-\text{grad}\,f,\phi)\text{vol}_g,\quad \forall \lambda\in C^{\infty}(M),$$ which implies that $$\text{grad}\,Pf-\text{grad}\,f$$ has to be orthogonal to $$\phi$$ at every point in $$M$$. Thus, in order to find $$Pf$$, we can perform orthogonal projection of $$\text{grad}\,f$$ onto the subspace generated by $$\phi$$ in each tangent space $$T_p M$$ using the inner product $$g_p$$, which leads to a vector field $$X_{\phi}$$. We then have to find a potential, i.e. a function $$Pf\in C^{\infty}(M)/\mathbb{R}$$ such that $$\text{grad}\,Pf=X_{\phi}$$. However, the existence of such a function leads to a contradiction if $$X_{\phi}=\lambda\cdot\phi$$ with some $$\lambda\in C^{\infty}$$, for which $$d\lambda\wedge\phi^{\flat}+\lambda d\phi^{\flat}\neq0,$$ where $$\phi^{\flat}$$ is the one-form dual to $$\phi$$.

For a counterexample when $$M$$ is the annulus in $$\mathbb{R}^2$$ and $$g$$ is the Euclidean metric on $$\mathbb{R}^2$$ restricted onto $$M$$, choose $$f=xy$$ and $$\phi=\partial_x$$ (i.e. $$\phi=(1,0)$$). Then $$df=y dx+x dy$$ and thus the orthogonal projection onto $$\phi$$ is given by $$X_{\phi}^{\flat}=y dx$$. But we have $$dX_{\phi}^{\flat}=-\text{vol}_g\neq 0$$, so there cannot be a function $$Pf$$ such that $$d(Pf)=X_{\phi}^{\flat}$$.

• $g_p$ is just the Riemannian metric at $p$ (as a quadratic form on $T_pM$). – S.Surace Nov 14 '18 at 22:15
• You are probably answering my question precisely. Unfortunately I am not a geometer, and while I am happy with the Hilbert part of your argument, I am less comfortable with the rest (what is g_p?). Would a worked example on B2\B1 with grad$\phi =(1,0)$ for example be possible? – username Nov 14 '18 at 22:29
• I changed the bit about $\phi$ (it is now an arbitrary vector field). I will think a bit about sufficient conditions to be put on $\phi$ and $f$ such that $Pf$ exists. – S.Surace Nov 14 '18 at 22:33
• Even better, if you could find explicit examples when it doesn't exist... – username Nov 14 '18 at 22:56
• @username I added a counterexample. – S.Surace Nov 14 '18 at 23:15