# Ngô Quốc Anh

## April 30, 2010

### The Pohozaev identity: Toda systems and a priori estimates

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 10:15

In this topic we consider the analysis of solutions of the following system entitled Toda system

$\displaystyle\left\{ \begin{gathered} - \Delta {u_1} = 2{e^{{u_1}}} - {e^{{u_2}}}, \hfill \\ - \Delta {u_2} = - {e^{{u_1}}} + 2{e^{{u_2}}}. \hfill \\ \end{gathered} \right.$

Following is our main result

Lemma 1. The following identities

$\displaystyle\begin{gathered} \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla {u_1}{|^2}dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_1}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_1}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_1}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_1}}}dx} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} , \hfill \\ \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla {u_2}{|^2}dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_2}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_2}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_2}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_2}}}dx} - \int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} , \hfill \\ \end{gathered}$

hold.

## April 29, 2010

### Surface integrals over a sphere when its radius varies

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Giải Tích Cổ Điển — Tags: — Ngô Quốc Anh @ 1:53

Let us consider the following integral in $\mathbb R^3$

$\displaystyle\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}}$

when $t$ varies.

Obviously, the sphere $|{\mathbf{x}}| = t$ can be parametrized as the following

$\displaystyle {\mathbf{x}} = t\left( {\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta } \right)$

so

$\displaystyle\begin{gathered} \iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi )){t^2}\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi ))\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{|{\mathbf{x}}| = 1} {f \left(t\mathbf{x} \right)d{\sigma _{\mathbf{x}}}}. \hfill \\ \end{gathered}$

If we wish to work on the average, the formula is much simpler than that, precisely

$\displaystyle\frac{1}{{4\pi {t^2}}}\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \frac{1}{{4\pi }}\iint\limits_{|{\mathbf{x}}| = 1} {f(t{\mathbf{x}})d{\sigma _{\mathbf{x}}}}$

that means

$\displaystyle \overline {\iint\limits_{|{\mathbf{x}}| = t} } f({\mathbf{x}})d{\sigma _{\mathbf{x}}}=\overline {\iint\limits_{|{\mathbf{x}}| = 1} } f({t\mathbf{x}})d{\sigma _{\mathbf{x}}}$

where the bar means the average.

More general, we get

$\displaystyle\iint\limits_{|{\mathbf{x}}| = {t_1}} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = {\left( {\frac{{{t_1}}}{{{t_2}}}} \right)^2}\iint\limits_{|{\mathbf{x}}| = {t_2}} {f\left( {\frac{{{t_1}}}{{{t_2}}}{\mathbf{x}}} \right)d{\sigma _{\mathbf{x}}}}$.

## April 27, 2010

### Surface integral: The symmetric property, 2

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Tags: — Ngô Quốc Anh @ 16:07

Today, we try to evaluate the following surface integral in $\mathbb R^3$. I found this result in a paper published in Math. Z. 198, 277-289 (1988). This topic can be considered as a continued part to the following topic.

Proposition. Let $f(x)$ be a continuous function. Then we have

$\displaystyle\frac{1}{t}\iint\limits_{|{\mathbf{y}} - {\mathbf{x}}| = t} {f(|{\mathbf{y}}|)d{\sigma _{\mathbf{y}}}} = \frac{{2\pi }}{r}\int\limits_{|r - t|}^{r + t} {sf(s)ds}$

here we denote $r=|\mathbf{x}|$.

Proof. Remark that the integral only depends on $|\mathbf{x}|$, so that we may assume $\mathbf{x} = (0, 0, r)$ and introduce the following spherical coordinate

$\mathbf{y}=\mathbf{x}+t\omega$

where

$\displaystyle \omega = (\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$.

This system of coordinates can be seen via the picture below

## April 26, 2010

### Conformal invariant operators: Laplacian operators, 2

Filed under: Riemannian geometry — Ngô Quốc Anh @ 19:23

Let me provide another proof of Example 2 in this topic.

Example 2. The conformal Laplacian operator acting on a smooth function $u$ as the following

$\displaystyle L_g u= - {\Delta _g}u + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_gu$

is conformal invariant.

The proof relies on the following fact

Proposition. Under a conformal change $g_\omega = e^{2\omega}g$, we have

$\displaystyle d{v_{{g_\omega }}} = {e^{n\omega }}d{v_g}$

where $dv_{g_\omega}$ and $dv_g$ are the volume elements with respect to metrics $g_\omega$ and $g$ respectively.

We are now in a position to prove example 2. For simplicity, let us consider the following conformal change $g_\varphi = \varphi^\frac{4}{n-2}g$. The proof is divided into three steps.

Step 1. Showing

$\displaystyle {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{\Delta _{{g_\varphi}}}u = {\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u$.

Proof. Indeed, for any test function $v\in C_0^\infty(M)$ we have

$\displaystyle \int_M {{\varphi ^{ - \frac{{n + 2}}{{n - 2}}}}\left( {{\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u} \right)vd{v_{{g_\omega }}}} = \int_M {\left( {{\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u} \right){d{v_g}}}$.

We now use Laplace-de Rham operator for the sake of simplicity. Note that Laplace–de Rham operator is the Laplacian differential operator on sections of the bundle of differential forms on a pseudo-Riemannian manifold. However, the Laplace-de Rham operator is equivalent to the definition of the Laplace–Beltrami operator when acting on a scalar function. Precisely,

$\displaystyle \int_M {v{\Delta _g}ud{v_g}} = \int_M {\langle dv,du\rangle d{v_g}}$.

Therefore

$\displaystyle\begin{gathered} \int_M {\left( {{\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u} \right)\varphi vd{v_g}} = \int_M {\left( {\left\langle {d(\varphi u),d(\varphi v)} \right\rangle - \left\langle {d(\varphi ),d(\varphi uv)} \right\rangle } \right)d{v_g}} \hfill \\ \qquad= \int_M {{\varphi ^2}\left\langle {d(u),d(v)} \right\rangle d{v_g}} \hfill \\ \qquad= \int_M {{\varphi ^{ - \frac{4}{{n - 2}}}}\left\langle {d(u),d(v)} \right\rangle {\varphi ^{\frac{{2n}}{{n - 2}}}}d{v_g}} \hfill \\\qquad = \int_M {v{\Delta _{{g_\varphi }}}ud{v_g}} . \hfill \\ \end{gathered}$

Step 2. Showing the scalar curvature equation

$\displaystyle \Delta_g \varphi + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_{{g_\varphi }}{\varphi ^{\frac{{n + 2}}{{n - 2}}}} = \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi$.

Equivalently,

$\displaystyle {L_g}(\varphi ) = \frac{n - 2}{4(n - 1)}{\rm Scal}_{{g_\varphi }}\varphi ^{\frac{{n + 2}}{{n - 2}}}$.

Step 3. Showing

$\displaystyle {L_g}(\varphi u) = {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{L_{{g_\varphi }}}(u)$.

Proof. Clearly

$\displaystyle\begin{gathered} {L_g}(\varphi u) = - {\Delta _g}(\varphi u) + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi u \hfill \\ \qquad= - {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{\Delta _{{g_\varphi }}}u - {\Delta _g}(\varphi )u + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi u \hfill \\ \qquad= - {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{\Delta _{{g_\varphi }}}u + {L_g}(\varphi )u \hfill \\ \qquad= {\varphi ^{\frac{{n + 2}}{{n - 2}}}}\left[ { - {\Delta _{{g_\varphi }}}u + \frac{n - 2}{4(n - 1)}{\rm Scal}_{{g_\varphi }}u} \right] \hfill \\ \qquad= {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{L_{{g_\varphi }}}(u). \hfill \\ \end{gathered}$

## April 25, 2010

### Explicit One-Step Schemes for the Advection Equation: The upwind scheme

Filed under: Giải tích 9 (MA5265), PDEs — Ngô Quốc Anh @ 1:56

Let us consider a very simple class of schemes solving the convection equations entitled the upwind schemes. Precisely, let us consider

$\displaystyle u_t+cu_x=0, \quad x \in \mathbb R, t>0$

together with the following initial data

$\displaystyle u(x,0)=u_0(x), \quad x\in \mathbb R$.

It is well-known that the above problem has a unique solution given by $u(x, t ) = u_0(x - c t )$, which is a right traveling wave of speed $c$.

Since the upwind schemes are well-known and appear in most of numerical analysis’s textbooks, what I am trying to do here is to give some basic ideas together with their motivation.

First of all, the original upwind schemes are just explicit one-step schemes. Upwind schemes use an adaptive or solution-sensitive finite difference stencil to numerically simulate more properly the direction of propagation of information in a flow field. The upwind schemes attempt to discretize hyperbolic partial differential equations by using differencing biased in the direction determined by the sign of the characteristic speeds. Historically, the origin of upwind methods can be traced back to the work of Courant, Isaacson, and Reeves who proposed the CIR method.

Discretization. Like the finite difference methods, we need to use a a grid with points $(x_j , t_n)$ defined by

$x_j=\pm j\Delta x, j \geqslant 0, \quad t^n = \pm n\Delta t, n \geqslant 0$.

for some spatial grid size $\Delta x$ and time step $\Delta t$.

## April 24, 2010

### Conformal invariant operators: Laplacian operators

Filed under: Riemannian geometry — Ngô Quốc Anh @ 9:23

Let $M$ be an $n$-dimensional differentiable connected Riemannian manifold with metric tensor $g$. In order to distinguish $g$ from other metrics on $M$, we shall denote the manifold $M$ with $g$ by $(M, g)$.

Definition. If $g'$ is another metric on $M$, and there is a function $\omega$ on $M$ such that $g' = e^{2\omega}g$, then $g$ and $g'$ are said to be conformally related or conformal to each other, and such a change of metric $g \to g'$ is called a conformal change of Riemannian metric.

We are now in a position to define the so-called “conformal invariant operators“.

Definition. We call a metrically de fined operator $A$ conformally covariant of bi-degree $(a,b)$ if under the conformal change of metric $g_\omega = e^{2\omega}g$, the pair of  corresponding operators $A_{g_\omega}$ and $A$ are related by

$\displaystyle {A_{{g_\omega }}}(\varphi ) = {e^{ - b\omega }}A({e^{a\omega }}\varphi ), \quad \forall \varphi \in {C^0}(M)$.

Example 1. The Laplace-Beltrami operator is conformal invariant.

Proof. Let us recall from this topic that on $(M,g)$, the Laplace-Beltrami operator is defined to be

$\displaystyle \Delta_g ={\rm div}({\rm grad \left(\cdot\right)})$

and in local coordinates it is given as follows

$\displaystyle\Delta_g = \frac{1}{{\sqrt {\det \left( g\right)} }}\frac{\partial }{{\partial {x^j}}}\left( {\sqrt {\det \left( g\right)} {g^{ij}}\frac{{\partial}}{{\partial {x^i}}}} \right)$.

By a change of metric $g_\omega = e^{2\omega}g$ we need to calculate $\Delta_{g_\omega}$ and then compare with $\Delta_g$.

Indeed, in local coordinates

$\displaystyle {g_\omega } = {e^{2\omega }}{g_{ij}}d{x^i} \otimes d{x^j}$

which implies

$\displaystyle \det ({g_\omega }) = \det ({e^{2\omega }}{g_{ij}}) = {e^{2\omega }}\det (g)$.

Besides,

$\displaystyle {({g_\omega })^{ij}} = {({g_\omega })^{ - 1}} = {({e^{2\omega }}{g_{ij}})^{ - 1}} = \frac{1}{{{e^{2w}}}}{g^{ij}}=e^{-2\omega}g^{ij}$.

## April 23, 2010

### Comparison of some subspaces associated to A and A^TA

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:10

Let us start with a given $n\times m$ matrix $A$

$\displaystyle {A_{n \times m}} = \left( {\begin{array}{*{20}{c}} {{a_{11}}} & \cdots & \cdots & {{a_{1m}}} \\ \vdots & \ddots & \ddots & \vdots \\ {{a_{n1}}} & \cdots & \cdots & {{a_{nm}}} \\ \end{array} } \right)$.

The aim of this entry is to compare nullspace, column space and row space between $A$, $A^TA$ and $AA^T$. Obviously, $A^TA$ and $AA^T$ are $m\times m$ and $n \times n$ matrices respectively.

Proposition 1. The following

$\displaystyle N(A)=N(A^TA)$

holds.

Proof. Pick an arbitrary element $x \in N(A)$, i.e. $Ax=0$, we can see that

$\displaystyle A^TAx=A^T(Ax)=A^T0=0$

so

$\displaystyle N(A) \subset N(A^TA)$.

Conversely, assume $x$ is such that $A^TAx=0$, as a consequence, $x^TA^TAx=0$. This gives us the fact

$\displaystyle (Ax) \cdot (Ax)= (Ax)^T(Ax)=x^TA^TAx=0$.

Consequently, $Ax=0$ which proves

$\displaystyle N(A^TA)\subset N(A)$.

In other words, $\displaystyle N(A)=N(A^TA)$.

Remark$\displaystyle N(A)=N(AA^T)$ is no longer true since these two matrices have different dimension.

Regarding to matrix $A$, one has

$\displaystyle {\rm rank}(A)+\dim N(A)=m$.

Similarly, involving matrix $A^TA$, one gets

$\displaystyle {\rm rank}(A^TA)+\dim N(A^TA)=m$.

It now follows from Proposition 1 that

$\displaystyle {\rm rank}(A)={\rm rank}(A^TA)$.

In other words, column and row spaces associated to $A$ and $A^TA$ have the same dimension respectively.

Row spaces. We prove the following

Proposition 2. The following

$\displaystyle RS(A)=RS(A^TA)$

holds.

Proof. The way to compare column spaces is to use the following facts

$\displaystyle RS(A)=CS(A^T)$

and

$\displaystyle CS(A^T)=A^T(\mathbb R^n)$.

Equivalently, from the first fact we need to show that

$\displaystyle CS(A^T)=CS(A^TA)$.

In term of the second fact, once you have a suitable matrix $Q$ the column space of $AQ$ is indeed contained in the column space of $A$. Therefore

$\displaystyle CS(A^TA) \subset CS(A^T)$

which turns out to be

$\displaystyle RS(A^TA) \subset RS(A)$

since they have the same dimension, equality occurs.

Remark. This comes from the proof above. If you have a good matrix $Q$, the following is true

$\displaystyle CS(AQ) \subset CS(A)$.

Column spaces. We prove the following

Proposition 3. The following

$\displaystyle CS(A)=CS(AA^T)$

holds.

Proof. This is trivial by using Proposition 2.

Remark. If you have a good matrix $P$, the following is true

$\displaystyle RS(PA) \subset RS(A)$.

## April 22, 2010

### The Bramble-Hilbert lemma

Filed under: Giải tích 9 (MA5265) — Ngô Quốc Anh @ 14:27

In numerical analysis, the Bramble-Hilbert lemma, named after James H. Bramble and Stephen R. Hilbert, bounds the error of an approximation of a function $u$ by a polynomial of order at most $k$ in terms of derivatives of $u$ of order $k+1$. Both the error of the approximation and the derivatives of $u$ are measured by $L^p$ norms on a bounded domain in $\mathbb R^n$.

Theorem. Over a sufficiently domain $\Omega$, there exists a constant $C(\Omega)$ such that

$\displaystyle\mathop {\inf }\limits_{v \in {P_k}(\Omega )} {\left\| {u - v} \right\|_{{W^{k + 1,p}}(\Omega )}} \leqslant C(\Omega ){\left| u \right|_{{W^{k + 1,p}}(\Omega )}}$

for every $u \in W^{k+1,p}(\Omega)$ where $\| \cdot \|$ and $| \cdot|$ denote the norm and semi-norm of the Sobolev space $W^{k+1,p}(\Omega)$.

This is similar to classical numerical analysis, where, for example, the error of linear interpolation $u$ can be bounded using the second derivative of $u$. However, the Bramble-Hilbert lemma applies in any number of dimensions, not just one dimension, and the approximation error and the derivatives of $u$ are measured by more general norms involving averages, not just the maximum norm.

Additional assumptions on the domain are needed for the Bramble-Hilbert lemma to hold. Essentially, the boundary of the domain must be “reasonable”. For example, domains that have a spike or a slit with zero angle at the tip are excluded. Lipschitz domains are reasonable enough, which includes convex domains and domains with $C^1$ boundary.

The main use of the Bramble-Hilbert lemma is to prove bounds on the error of interpolation of function $u$ by an operator that preserves polynomials of order up to $k$, in terms of the derivatives of $u$ of order $k+1$. This is an essential step in error estimates for the finite element method. The Bramble-Hilbert lemma is applied there on the domain consisting of one element.

## April 20, 2010

### Entropy conditions for conservation laws

Filed under: PDEs — Ngô Quốc Anh @ 0:46

This entry deals with the conservation law

$u_t+F(u)_x=0$

and the problem of determining which weak solutions of the above equation are to be considered acceptable. We shall use the idea of vanishing viscosity whereby one considers instead the equation

$u_t^\varepsilon+F(u^\varepsilon)_x=\varepsilon u_{xx}^\varepsilon$

in the limit as $\varepsilon$ decreases to zero.

We consider a jump discontinuity in the solution $u$ with left state $u_l$ and right state $u_r$ and shock speed $s'$ satisfying the Rankine-Hugoniot condition

$\displaystyle s'=\frac{F(u_l)-F(u_r)}{u_l-u_r}$

and the problem is to determine further conditions on the shock that will hopefully lead to uniqueness as well of existence of weak solutions to the initial value problem for the problem.

Traveling waves. One way to get an answer to our problem is to look for traveling wave solutions to our problem, i.e., solutions on the form

$\displaystyle u^\varepsilon(x,t)=w\left( \frac{x-s' t}{\varepsilon}\right)$.

Substituting this into the equation gives

$\displaystyle -s' w'+F(w)'=w''$.

We can integrate this once to get

$\displaystyle w'(\xi) = F(w(\xi))-s' w(\xi)-C$.

Now we get

$\displaystyle\mathop {\lim }\limits_{\varepsilon \searrow 0} {u^\varepsilon }(x,t) = \left\{ \begin{gathered} \mathop {\lim }\limits_{\xi \to - \infty } w(\xi ), \quad x < s't, \hfill \\ \mathop {\lim }\limits_{\xi \to + \infty } w(\xi ), \quad x > s't, \hfill \\ \end{gathered} \right.$

and therefore we shall insist that

$\displaystyle\mathop {\lim }\limits_{\xi \to - \infty } w(\xi ) = {u_l}, \quad \mathop {\lim }\limits_{\xi \to + \infty } w(\xi ) = u_r$.

In particular, the right hand side of the above ODE has a limit as $\xi \to \pm\infty$ and so $w'(\xi)$ has a limit as well. If this limit is nonzero then of course $w$ itself cannot have a limit so in fact the limit must be zero. Thus we get

$\displaystyle F({u_l}) - s'{u_l} - C = 0 = F({u_r}) - s'{u_r} - C$.

Clearly,

$\displaystyle\int {\frac{{w'(\xi )}}{{F(w(\xi )) - s'w(\xi ) - C}}d\xi } = \int {d\xi } = \xi$.

Besides, the above identity states that $z= s'w +C$ is the equation of the straight line joining the two points $(u_l, F(u_l))$ and $(u_r,F(u_r))$ on the graph of $F$, that is the secant between those two points. So we have discovered that the graph $z=F(w)$ must lie to one side of that secant for $w$ between $u_l$ and $u_r$.

Theorem. The given equation admits a traveling wave solution of the form above satisfying

$\displaystyle\mathop {\lim }\limits_{\xi \to - \infty } w(\xi ) = {u_l}, \quad \mathop {\lim }\limits_{\xi \to + \infty } w(\xi ) = u_r$

if and only if

$\displaystyle (u_r-u_l)(F(w)-s'w-C)>0$

for all $w$ between $u_l$ and $u_r$ where $s'$ and $C$ are chosen so that the above expression is zero when $w=u_l$ or $w=u_r$.

O. Oleinik entropy condition. The entropy condition given by the above theorem is equivalent to the inequalities

$\displaystyle\frac{{F(w) - F({u_l})}}{{w - {u_l}}} > s' > \frac{{F(w) - F({u_r})}}{{w - {u_r}}}$.

This entropy condition was suggested by O. Oleinik but since she gave another entropy condition more commonly associated with her name this condition is often referred to as the viscous profile entropy condition.

Lax entropy condition. The Lax entropy condition is the pair of inequalities

$\displaystyle F'(u_l) \geqslant s' \geqslant F'(u_r)$

in the case when $F$ is either strictly convex or strictly concave. Of course the Lax condition has the advantage that it is very easy to check.

## April 18, 2010

### The Pohozaev identity: Elliptic problem with biharmonic operator

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 14:19

We now consider another kind of problem involving biharmonic operator. Let us assume $u>0$ a solution of the equation

$\displaystyle (\Delta^2u)(x)+Q(x)f(u(x))=0$

in $\mathbb R^n$. We shall prove the following result

Theorem. The following identity

$\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] + \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad= \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)} \right)dx}\hfill \\ \end{gathered}$

holds.

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