# Ngô Quốc Anh

## April 16, 2020

### Restriction of gradient, Laplacian, etc on level sets and applications

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 18:10

This topic is devoted to proofs of several interesting identities involving derivatives on level sets. First, we start with the case of gradient. We shall prove

The first identity

$\displaystyle \partial_\nu f = \pm |\nabla f|$

on the level set

$\displaystyle \big\{ x \in \mathbf R^n : f(x) =0 \big\}.$

The above identity shows that while the right hand side involves the value of $f$ in a neighborhood, however, the left hand side indicates that only the normal direction is affected. Heuristically, any change of $f$ along the level set does not contribute to any derivative of $f$, namely, on the boundary of the level set, the norm of $\nabla f$ is actually the normal derivative $\partial_\nu f$. Therefore, the only direction taking into derivatives of $f$ is in the normal direction and this should be true for higher-order derivatives of $f$.

Next we prove the following

The second identity

$\displaystyle \partial_\nu \big(x \cdot \nabla f \big)=(\partial_\nu^2 f) (x \cdot \nu)$

on the level set

$\displaystyle \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}.$

Combining the above two identities, we can prove

The third identity

$\displaystyle \partial_\nu^2 f=\Delta f$

on the level set

$\displaystyle \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}$

which basically tells us how to compute the restriction of Laplacian on level sets. This note is devoted to a rigorous proof of the above facts together with a simple application of all these identities.

Assume that $u$ is a non-negative classical solution to the following equation

$\displaystyle \Delta^2 u =f(x, u)$

in smooth, bounded domain $\Omega \subset \mathbf R^n$ together with the following boundary conditions

$\displaystyle \partial_1 u = \cdots= \partial_n u = 0$

on $\partial \Omega$. Then we have the following Pohozaev type identity

$\displaystyle \int_\Omega (x \cdot \nabla u) f(x,u) =-\frac{n-4}2\int_\Omega (\Delta u)^2-\frac 12 \int_{\partial\Omega}(\Delta u)^2 (x\cdot\nu).$

1. Estimate of $\partial_\nu f$ on $\Omega$. Denote

$\displaystyle \Omega= \big\{ x \in \mathbf R^n : f(x) =0 \big\}.$

Since $f$ is constant along $\Omega$, the gradient $\nabla f$ and the unit outer normal $\nu$ have the same direction. Hence

$\displaystyle \nabla f = \pm |\nabla f| \nu.$

Since $\partial_\nu f = \nu \cdot \nabla f$, we deduce that

$\displaystyle \partial_\nu f=\nu \cdot \big(\pm |\nabla f| \nu \big) = \pm |\nabla f|,$

which gives the desired identity.

2. Estimate of $\partial_\nu \big( x \cdot \nabla f \big)$ on $\Omega$. We also denote

$\displaystyle\Omega = \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}.$

We shall show that the second identity actually follows from the first identity. By direction computation we have

$\displaystyle \partial_\nu \big(x \cdot \nabla f \big) = \nu \cdot \nabla \big(x \cdot \nabla f \big) = \sum_i \nu_i \cdot \partial_i \Big( \sum_j x_j \partial_j f \Big),$

which implies

$\displaystyle \partial_\nu \big(x \cdot \nabla f \big) =\sum_i \nu_i \cdot \Big( \sum_j \delta_{ij} \partial_j f + x_j \partial_i\partial_j f \Big) = \sum_i \nu_i \partial_i f + \sum_i \sum_j \nu_i x_j \partial_i\partial_j f.$

Hence

$\displaystyle \partial_\nu \big(x \cdot \nabla f \big) =\nu \cdot \nabla f + \sum_j x_j \sum_i \nu_i \partial_i (\partial_j f) = \sum_j x_j \partial_\nu (\partial_j f) = x \cdot \nabla (\partial_\nu f)$

since $\nabla f = 0$ on $\Omega$. However, because

$\displaystyle \partial_\nu f = \sum_i \nu_i \partial_i f,$

clearly $\Omega$ is the zero level set of $\partial_\nu f$. Hence

$\displaystyle \nabla(\partial_\nu f) =\pm |\nabla(\partial_\nu f)| \nu,$

which implies that

$\displaystyle x \cdot \nabla (\partial_\nu f) = \pm |\nabla(\partial_\nu f)| (x \cdot \nu).$

Thus, we have shown that

$\displaystyle\partial_\nu \big(x \cdot \nabla f \big) = \pm |\nabla(\partial_\nu f)| (x \cdot \nu).$

Note that, in the same fashion of the first identity, we further have

$\displaystyle \partial_\nu^2 f= \partial_\nu \big(\partial_\nu f \big) =\nu \cdot \nabla(\partial_\nu f)=\pm |\nabla(\partial_\nu f)| .$

Hence, we arrive at

$\displaystyle \partial_\nu \big(x \cdot \nabla f \big)= |\partial_\nu^2 f| (x \cdot \nu).$

This completes the proof.

3. Estimate of $\partial_\nu^2 f$ on $\Omega$. As routine, we denote

$\displaystyle\Omega = \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}.$

On one hand, because $\partial_\nu f = \nu \cdot \nabla f$, we easily get

$\displaystyle \Delta f = {\rm div} (\nabla f) = {\rm div} \big( \nu \cdot \nabla f \big) = \nu \cdot \nabla \big( \partial_\nu f \big).$

Now because $\Omega$ is the zero level set of $\partial_\nu f$, we know that

$\displaystyle \nabla \big( \partial_\nu f \big)=\pm \big|\nabla \big( \partial_\nu f \big)\big| \nu$

on $\Omega$; hence yielding

$\displaystyle \Delta f =\pm \big|\nabla \big( \partial_\nu f \big)\big|$

on $\Omega$. Now as in the proof of the second identity (or by the first identity for $\partial_\nu f$) we obtain

$\displaystyle \pm \big|\nabla \big( \partial_\nu f \big)\big| =\partial_\nu \big( \partial_\nu f \big) = \partial_\nu^2 f.$

This gives us the desired identity.

4. A Pohozaev type identity. Multiplying both sides of the equation by $x \cdot \nabla u$ and integrating the resulting equation over $\Omega$ give

$\displaystyle \int_\Omega f(x, u)= \int_\Omega (x \cdot \nabla u) \Delta^2 u.$

Integration by parts gives

$\displaystyle \int_\Omega (x \cdot \nabla u) \Delta^2 u =\int_\Omega \Delta u \Delta (x \cdot \nabla u) +\int_{\partial \Omega}(x \cdot \nabla u) \partial_\nu \Delta u- \int_{\partial\Omega} \Delta u \partial_\nu (x \cdot \nabla u).$

Using the boundary condition, we know that

$\displaystyle\int_{\partial \Omega}(x \cdot \nabla u) \partial_\nu \Delta u.$

For the remaining boundary term we use the second identity to get

$\displaystyle \int_{\partial\Omega} \Delta u \partial_\nu (x \cdot \nabla u)=\int_{\partial\Omega} \Delta u (\partial_\nu^2 u) (x \cdot \nu)=\int_{\partial\Omega} \big(\Delta u\big)^2 (x \cdot \nu).$

Of course, in view of the third identity, we do not distinguish the two terms $\Delta u$ and $\partial_\nu^2 u$ on $\Omega$. Finally, the last term can be estimated by using $\Delta (x \cdot \nabla u)=2\Delta u + (x \cdot \nabla \Delta u)$ to get

$\displaystyle\int_\Omega \Delta u \Delta (x \cdot \nabla u)=2\int_\Omega (\Delta u)^2+\int_\Omega \Delta u (x \cdot \nabla \Delta u).$

However, integration by parts gives

$\displaystyle\int_\Omega \Delta u (x \cdot \nabla \Delta u)=\frac 12\int_\Omega (x \cdot \nabla \big[ (\Delta u)^2\big] ) = -\frac n2 \int_\Omega (\Delta u)^2 +\frac 12 \int_{\partial\Omega}(\Delta u)^2 (x\cdot\nu).$

Hence

$\displaystyle\int_\Omega \Delta u \Delta (x \cdot \nabla u)=-\frac{n-4}2\int_\Omega (\Delta u)^2+\frac 12 \int_{\partial\Omega}(\Delta u)^2 (x\cdot\nu).$

Thus

$\displaystyle \int_\Omega (x \cdot \nabla u) \Delta^2 u dx=-\frac{n-4}2\int_\Omega (\Delta u)^2-\frac 12 \int_{\partial\Omega}(\Delta u)^2 (x\cdot\nu).$

Hence, we obtain the following Pohozaev type identity

$\displaystyle \int_\Omega (x \cdot \nabla u) f(x,u) =-\frac{n-4}2\int_\Omega (\Delta u)^2-\frac 12 \int_{\partial\Omega}(\Delta u)^2 (x\cdot\nu).$