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).

See also:

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