l******r
发帖数: 18699 | | l******r
发帖数: 18699 | 2 自己先顶一个
b
【在 l******r 的大作中提到】
| s*****e
发帖数: 115 | 3 的常数M能够控制住所有的这些特征向量函数。
I don't quite see why this is a question.
It seems that you require g(t) satisfies (1) at every t in [0,1], then g(t) must be continuous, and hence bounded on [0,1].
b | l******r
发帖数: 18699 | 4 非常赞同。其实我想表达的就是第二个意思。这里面考虑的函数类是2阶Sobolev space
，也就是特征向量g所在的空间。为了更清楚，一楼原问题已做修改，参见一楼附件。
谢谢。
的确可以把这个高阶ODE写成Hamilton system,也就是G'=QG.当 G是一维，Gronwall
inequality可以用。不过对于G是高维的也能用吗？
) must be continuous, and hence bounded on [0,1].
【在 s*****e 的大作中提到】
: 的常数M能够控制住所有的这些特征向量函数。
: I don't quite see why this is a question.
: It seems that you require g(t) satisfies (1) at every t in [0,1], then g(t) must be continuous, and hence bounded on [0,1].
: b
| l******r
发帖数: 18699 | 5 谢谢！
原来的叙述不清楚。我已经重新上传了这个问题在一楼。这里我们考虑的是所有的特征
向量构成的函数序列是否在(0,1)上一致有界。当然这个序列里的每一个函数由于连续
当然有界。可是这个序列整体是否一致有界不是很清楚。
) must be continuous, and hence bounded on [0,1].
【在 s*****e 的大作中提到】
: 的常数M能够控制住所有的这些特征向量函数。
: I don't quite see why this is a question.
: It seems that you require g(t) satisfies (1) at every t in [0,1], then g(t) must be continuous, and hence bounded on [0,1].
: b
| s*****e
发帖数: 115 | 6 Still not able to see the point of the question. For the boundary value problem you have, associated with each eigenvalue there are infinitely many eigenfunctions, one differing from another just by a scalar multiple. So for each $\lambda$ you can always pick $g_{\lambda}$ such that its sup-norm is exactly 1, unless you have to pick $g_{\lambda}$ in curtain special way that is not described here.
【在 l******r 的大作中提到】
: 谢谢！
: 原来的叙述不清楚。我已经重新上传了这个问题在一楼。这里我们考虑的是所有的特征
: 向量构成的函数序列是否在(0,1)上一致有界。当然这个序列里的每一个函数由于连续
: 当然有界。可是这个序列整体是否一致有界不是很清楚。
:
: ) must be continuous, and hence bounded on [0,1].
| l******r
发帖数: 18699 | 7 对了，这些eigenfunctions都是normalized,使得它们的L2-norm都是1.
在这种normalization之下，是否可以证明它们一致有界？请看一楼更正，谢谢。
当然不同的eigenvalue对应的eigenfunctions都是正交的，所以它们构成L2[0,1]空间
里的标准正交基。只不过eigenfunctions的Sobolev norm不是1，而是一个依赖于特征
值的常数，而且没有上界。
不好意思，这个问题一开始设计的不严谨。这貌似是微分方程里很重要的问题，可
是文献里却找不到。谢谢热心帮助！
problem you have, associated with each eigenvalue there are infinitely many
eigenfunctions, one differing from another just by a scalar multiple. So for
each $\lambda$ you can always pick $g_{\lambda}$ such that its sup-norm is
exactly 1, unless you have to pick $g_{\lambda}$ in curtain special way that
is not described here.
【在 s*****e 的大作中提到】
: Still not able to see the point of the question. For the boundary value problem you have, associated with each eigenvalue there are infinitely many eigenfunctions, one differing from another just by a scalar multiple. So for each $\lambda$ you can always pick $g_{\lambda}$ such that its sup-norm is exactly 1, unless you have to pick $g_{\lambda}$ in curtain special way that is not described here.
| E*****T
发帖数: 1193 | 8 y(4)=lambda q y是在什么空间？
2阶sobolev space求四阶导那边值还有什么用？ | l******r
发帖数: 18699 | 9 实际上eigenfunctions都是属于无穷次连续可导的空间的，
虽然我们考虑的是4阶微分方程，可是这个4阶方程加上boundary condition刚刚吻合2
阶sobolev空间。一般来说2m阶微分方程加上类似的boundary condition刚好吻合m阶
sobolev space，用的是integration by parts.我越来越认为这是一个research topic
，其答案不是一两句话能说的清楚的。要想给出完整解答，恐怕得写一个20页的paper
才行。下面是一个背景paper，其中讨论的是这里面q(t)=1的情形，发在journal of
approximation上面。
Boundary Effects on Convergence for Tikhonov Regularization (by FLORENCIO I.
UTRERAS). JOURNAL OF APPROXIMATION THEORY 54, 235-249 (1988)
【在 E*****T 的大作中提到】
: y(4)=lambda q y是在什么空间？
: 2阶sobolev space求四阶导那边值还有什么用？
| B********e
发帖数: 10014 | 10 nice problem.
Define operator L, Ly=y''''/q in W^{2,2}(H^2) with domain W^{4,2} (H4).
I think you can mimick the Sturm-Liouville theory(for 2nd order ode bvp)
see Michael Renardy's <>
I think the answer should be a Yes.
The uniform bound simply comes from the Sobolev Imbedding theorem:
H^2[a,b](actually H^1 is enough) is continuously imbedded in C[a, b], a ,b
finite.
【在 l******r 的大作中提到】
: 实际上eigenfunctions都是属于无穷次连续可导的空间的，
: 虽然我们考虑的是4阶微分方程，可是这个4阶方程加上boundary condition刚刚吻合2
: 阶sobolev空间。一般来说2m阶微分方程加上类似的boundary condition刚好吻合m阶
: sobolev space，用的是integration by parts.我越来越认为这是一个research topic
: ，其答案不是一两句话能说的清楚的。要想给出完整解答，恐怕得写一个20页的paper
: 才行。下面是一个背景paper，其中讨论的是这里面q(t)=1的情形，发在journal of
: approximation上面。
: Boundary Effects on Convergence for Tikhonov Regularization (by FLORENCIO I.
: UTRERAS). JOURNAL OF APPROXIMATION THEORY 54, 235-249 (1988)
| | | l******r
发帖数: 18699 | 11 Thanks for nice solution!
Just have a quick question. Is the Sobolev Imbedding theorem saying
||f||_sup<=constant||f||_soblev_norm ?
The sobolev norm of the eigenfunctions are usually depending on the
eigenvalues, which is an increasing sequence. So, it seems that directly
using Sobolev embedding thm cannot give us the desired result.
I agree with your first suggestion, mimick the SL theory to get the
conclusion. Thx!
【在 B********e 的大作中提到】
: nice problem.
: Define operator L, Ly=y''''/q in W^{2,2}(H^2) with domain W^{4,2} (H4).
: I think you can mimick the Sturm-Liouville theory(for 2nd order ode bvp)
: see Michael Renardy's <>
: I think the answer should be a Yes.
: The uniform bound simply comes from the Sobolev Imbedding theorem:
: H^2[a,b](actually H^1 is enough) is continuously imbedded in C[a, b], a ,b
: finite.
| B********e
发帖数: 10014 | 12 sobolev imbedding thm says, on whole space R^n, or on domain with nice
boundary, if we have a function with high enough sobolev order, say H^r, r>n
/2, then f is continuous and we have
||f|_sup <= C ||f||_{H^r}.
for 1-dim space, r=1 is good enough.
so you are right on that it's not enough to normalize eigenfunctions in L^2
norm.
see if your work needs to normalize the eigenfunctions in the configuration
functional space H^2 instead of L^2. Or at least in H^1.
if not so, i doubt about the conjecture.
【在 l******r 的大作中提到】
: Thanks for nice solution!
: Just have a quick question. Is the Sobolev Imbedding theorem saying
: ||f||_sup<=constant||f||_soblev_norm ?
: The sobolev norm of the eigenfunctions are usually depending on the
: eigenvalues, which is an increasing sequence. So, it seems that directly
: using Sobolev embedding thm cannot give us the desired result.
: I agree with your first suggestion, mimick the SL theory to get the
: conclusion. Thx!
| B********e
发帖数: 10014 | 13 sorry I should have emphasized that C does not depend on f.
so if f is bounded under H^r, r>n/2, so is the sup norm of f.
>n
2
configuration
【在 B********e 的大作中提到】
: sobolev imbedding thm says, on whole space R^n, or on domain with nice
: boundary, if we have a function with high enough sobolev order, say H^r, r>n
: /2, then f is continuous and we have
: ||f|_sup <= C ||f||_{H^r}.
: for 1-dim space, r=1 is good enough.
: so you are right on that it's not enough to normalize eigenfunctions in L^2
: norm.
: see if your work needs to normalize the eigenfunctions in the configuration
: functional space H^2 instead of L^2. Or at least in H^1.
: if not so, i doubt about the conjecture.
| s*****e
发帖数: 115 | 14 Are there any additional information on q(t) besides it is bounded between a
and b? Is it continuous/smooth?
【在 l******r 的大作中提到】
: Thanks for nice solution!
: Just have a quick question. Is the Sobolev Imbedding theorem saying
: ||f||_sup<=constant||f||_soblev_norm ?
: The sobolev norm of the eigenfunctions are usually depending on the
: eigenvalues, which is an increasing sequence. So, it seems that directly
: using Sobolev embedding thm cannot give us the desired result.
: I agree with your first suggestion, mimick the SL theory to get the
: conclusion. Thx!
| l******r
发帖数: 18699 | 15 Yes, it can be arbitrarily smooth
a
【在 s*****e 的大作中提到】
: Are there any additional information on q(t) besides it is bounded between a
: and b? Is it continuous/smooth?
| l******r
发帖数: 18699 | 16 The normaization is only under L^2 norm, i.e., all the eigenfunctions have L
^2 norm equal to one. This will make their H^r norms inflating with the
corresponding eigenvalues since the eigenvalues must approach +infty.
I agree with you about mimicking SL's theory, probably that is the right
track. So little about this in literature:-)
>n
2
configuration
【在 B********e 的大作中提到】
: sobolev imbedding thm says, on whole space R^n, or on domain with nice
: boundary, if we have a function with high enough sobolev order, say H^r, r>n
: /2, then f is continuous and we have
: ||f|_sup <= C ||f||_{H^r}.
: for 1-dim space, r=1 is good enough.
: so you are right on that it's not enough to normalize eigenfunctions in L^2
: norm.
: see if your work needs to normalize the eigenfunctions in the configuration
: functional space H^2 instead of L^2. Or at least in H^1.
: if not so, i doubt about the conjecture.
| E*****T
发帖数: 1193 | 17 你确定边值条件是2，3阶导而不是原函数以及1阶导的？ | l******r
发帖数: 18699 | 18 边值条件确实是2，3阶可导,不过可以考虑特征向量的2阶导数满足的ODE,其边值条件就
是0阶和一阶的了。
这个问题的证明可能要用到Green's function的性质。一个well known的结果是green
function 的Laurent series的pole都是simple的，它的留数就是原边值问题的特征向
量除以特征根。如果能证明Green functions的留数一致有界则问题就大部分解决了。
我查到MH Stone 1925年一个论文讲的是常系数ODE的green function留数一致有界，不
知道对于系数是函数的时候是否也有类似结果。这个论文67页，真不打算把它的结果注
意推广，太复杂了。。。
基本上我们要考虑 L=（1/q）*D^(4),我们要证明L的特征向量一致有界。为此需要考虑
复合算子 L(D^(2))的特征向量的性质。可以通过研究L(D^(2))的Green function去实
现。我也正在搜文献想怎么证，等想到了给大家汇报一下，同时也感谢各位大牛的建议
呵呵。
【在 E*****T 的大作中提到】
: 你确定边值条件是2，3阶导而不是原函数以及1阶导的？
| E*****T
发帖数: 1193 | 19 边值条件是二三阶导确定能做regularity? 我是很怀疑。
如果能得到y smooth, 你把微分方程两边乘y 然后分部积分，左边是0右边lambda如何
解释？ | l******r
发帖数: 18699 | 20 用分部积分在y^(4)*y上(over [0,1])可以得到|y(2)|^2在(0,1)上积分，
如果继续使用分部积分则又得回y*y(4)的积分了:-)
【在 E*****T 的大作中提到】
: 边值条件是二三阶导确定能做regularity? 我是很怀疑。
: 如果能得到y smooth, 你把微分方程两边乘y 然后分部积分，左边是0右边lambda如何
: 解释？
| s*****e
发帖数: 115 | 21 Not so familiar with the Green's function approach, so I cannot comment on that.
However, I believe your conjecture is true for the following reasons:
What you want to prove is equivalent to that if the eigenfunctions are normalized with respect to sup-norm then their L2 norms are always bounded away from zero.
Clearly, this not true for general functions. For example, both exp(-kt) and exp(-k)*exp(kt) (k>0) on [0,1] have L2 norms approaching to zero as k-->+ infinity. On the other hand, one can achieve this using some combination of cos(kt) and sin(kt) if k is chosen according to certain boundary conditions. This is exactly why your conjecture is true when q(t)=constant>0. In this case, for any eigenvalue lambda>0, the eigenfunctions are in the form of
f=c1*exp(-kt)+c2*exp(kt)+c3*sin(kt)+c4*cos(kt).
For very large k, if the sup-norm of f is normalized to 1, then on [0+delta,1-delta] the trig part dominates while the exponential part is almost zero.
Having seen this, the remaining technical issue is that q(t) is not constant. The standard way to handle this is the Lyapunov transformation, which, however, seems to be used more often in initial value problems and stability analysis than in BVPs.
My sketch for the rest of the proof is the following:
1. Let p(t) be the even periodic extension of q(t) with period T=2. Now consider y(4)=lambda*p(t)*y. Rewrite this as a system Y'=P(t)Y for Y in R^4.
2. By the Floquet theory, there is a periodic Lyapunov transformation L(t): R --> R^{4*4} with period 2T=4 such that Y'=P(t)Y becomes X'=AX, where A is a real constant matrix.
3. We can assume that L(t) is chosen in the way such that A is already in the Jordan canonical form. Then prove that A has a 2D center subspace, a 1D stable subspace, and a 1D unstable subspace. In addition, show that solutions in the center subspace are in the form of sin and cos and that solutions in the stable/unstable subspace are in the form of exp(-/+kt), where, I believe,
k=(the average of lambda*q(t) on [0,1])^(1/4).
There may be exact results like this already in the literature. I don't remember, but I believe this should be standard.
4. Rewrite your BCs for X. The transformations involved are L(0) and L(1), both of which are nonsingular. Then you just need to show that under the new boundary conditions, the sin and cos part never diminishes as lambda-->+infinity.
OR
Instead of doing 3 & 4 above, you just directly rewrite your BCs for X using the transformations L(0) and L(1). Then consider if MH Stone's result you mentioned can be applied to X'=AX with the BCs.
green
【在 l******r 的大作中提到】
: 边值条件确实是2，3阶可导,不过可以考虑特征向量的2阶导数满足的ODE,其边值条件就
: 是0阶和一阶的了。
: 这个问题的证明可能要用到Green's function的性质。一个well known的结果是green
: function 的Laurent series的pole都是simple的，它的留数就是原边值问题的特征向
: 量除以特征根。如果能证明Green functions的留数一致有界则问题就大部分解决了。
: 我查到MH Stone 1925年一个论文讲的是常系数ODE的green function留数一致有界，不
: 知道对于系数是函数的时候是否也有类似结果。这个论文67页，真不打算把它的结果注
: 意推广，太复杂了。。。
: 基本上我们要考虑 L=（1/q）*D^(4),我们要证明L的特征向量一致有界。为此需要考虑
: 复合算子 L(D^(2))的特征向量的性质。可以通过研究L(D^(2))的Green function去实
|
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