In this topic we consider the analysis of solutions of the following system entitled Toda system
Following is our main result
Lemma 1. The following identities
hold.
In this topic we consider the analysis of solutions of the following system entitled Toda system
Following is our main result
Lemma 1. The following identities
hold.
Let us consider the following integral in
when varies.
Obviously, the sphere can be parametrized as the following
so
If we wish to work on the average, the formula is much simpler than that, precisely
that means
where the bar means the average.
More general, we get
.
Today, we try to evaluate the following surface integral in . 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 be a continuous function. Then we have
here we denote .
Proof. Remark that the integral only depends on , so that we may assume and introduce the following spherical coordinate
where
.
This system of coordinates can be seen via the picture below
Let me provide another proof of Example 2 in this topic.
Example 2. The conformal Laplacian operator acting on a smooth function as the following
is conformal invariant.
The proof relies on the following fact
Proposition. Under a conformal change , we have
where and are the volume elements with respect to metrics and respectively.
We are now in a position to prove example 2. For simplicity, let us consider the following conformal change . The proof is divided into three steps.
Step 1. Showing
.
Proof. Indeed, for any test function we have
.
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,
.
Therefore
Step 2. Showing the scalar curvature equation
.
Equivalently,
.
Step 3. Showing
.
Proof. Clearly
Let us consider a very simple class of schemes solving the convection equations entitled the upwind schemes. Precisely, let us consider
together with the following initial data
.
It is well-known that the above problem has a unique solution given by , which is a right traveling wave of speed .
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 defined by
.
for some spatial grid size and time step .
Let be an -dimensional differentiable connected Riemannian manifold with metric tensor . In order to distinguish from other metrics on , we shall denote the manifold with by .
We start with the following terminology called “a conformal change“.
Definition. If is another metric on , and there is a function on such that , then and are said to be conformally related or conformal to each other, and such a change of metric 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 defined operator conformally covariant of bi-degree if under the conformal change of metric , the pair of corresponding operators and are related by
.
Example 1. The Laplace-Beltrami operator is conformal invariant.
Proof. Let us recall from this topic that on , the Laplace-Beltrami operator is defined to be
and in local coordinates it is given as follows
.
By a change of metric we need to calculate and then compare with .
Indeed, in local coordinates
which implies
.
Besides,
.
Let us start with a given matrix
.
The aim of this entry is to compare nullspace, column space and row space between , and . Obviously, and are and matrices respectively.
Nullspaces. We start with the following result
Proposition 1. The following
holds.
Proof. Pick an arbitrary element , i.e. , we can see that
so
.
Conversely, assume is such that , as a consequence, . This gives us the fact
.
Consequently, which proves
.
In other words, .
Remark. is no longer true since these two matrices have different dimension.
Regarding to matrix , one has
.
Similarly, involving matrix , one gets
.
It now follows from Proposition 1 that
.
In other words, column and row spaces associated to and have the same dimension respectively.
Row spaces. We prove the following
Proposition 2. The following
holds.
Proof. The way to compare column spaces is to use the following facts
and
.
Equivalently, from the first fact we need to show that
.
In term of the second fact, once you have a suitable matrix the column space of is indeed contained in the column space of . Therefore
which turns out to be
since they have the same dimension, equality occurs.
Remark. This comes from the proof above. If you have a good matrix , the following is true
.
Column spaces. We prove the following
Proposition 3. The following
holds.
Proof. This is trivial by using Proposition 2.
Remark. If you have a good matrix , the following is true
.
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 by a polynomial of order at most in terms of derivatives of of order . Both the error of the approximation and the derivatives of are measured by norms on a bounded domain in .
Theorem. Over a sufficiently domain , there exists a constant such that
for every where and denote the norm and semi-norm of the Sobolev space .
This is similar to classical numerical analysis, where, for example, the error of linear interpolation can be bounded using the second derivative of . However, the Bramble-Hilbert lemma applies in any number of dimensions, not just one dimension, and the approximation error and the derivatives of 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 boundary.
The main use of the Bramble-Hilbert lemma is to prove bounds on the error of interpolation of function by an operator that preserves polynomials of order up to , in terms of the derivatives of of order . 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.
This entry deals with the conservation law
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
in the limit as decreases to zero.
We consider a jump discontinuity in the solution with left state and right state and shock speed satisfying the Rankine-Hugoniot condition
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
.
Substituting this into the equation gives
.
We can integrate this once to get
.
Now we get
and therefore we shall insist that
.
In particular, the right hand side of the above ODE has a limit as and so has a limit as well. If this limit is nonzero then of course itself cannot have a limit so in fact the limit must be zero. Thus we get
.
Clearly,
.
Besides, the above identity states that is the equation of the straight line joining the two points and on the graph of , that is the secant between those two points. So we have discovered that the graph must lie to one side of that secant for between and .
Theorem. The given equation admits a traveling wave solution of the form above satisfying
if and only if
for all between and where and are chosen so that the above expression is zero when or .
O. Oleinik entropy condition. The entropy condition given by the above theorem is equivalent to the inequalities
.
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
in the case when is either strictly convex or strictly concave. Of course the Lax condition has the advantage that it is very easy to check.
We now consider another kind of problem involving biharmonic operator. Let us assume a solution of the equation
in . We shall prove the following result
Theorem. The following identity
holds.