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MATH 551 LECTURE NOTES FREDHOLM INTEGRAL EQUATIONS (A BRIEF INTRODUCTION)
数学 551 讲义 FREDHOLM 积分方程(简介)

TOPICS COVERED 涵盖的主题

  • Fredholm integral operators
    Fredholm 积分算子
  • Integral equations (Volterra vs. Fredholm)
    积分方程(Volterra 与 Fredholm)
  • Eigenfunctions for separable kernels
    可分离核的特征函数
  • Adjoint operator, symmetric kernels
    伴随算子,对称核
  • Solution procedure (separable)
    求解过程(可分离)
  • Solution via eigenfunctions (first and second kind)
    通过特征函数(第一类和第二类)求解
  • Shortcuts: undetermined coefficients
    快捷方式:未确定的系数
  • An example (separable kernel, )
    示例(可分离内核,
  • Non-separable kernels (briefly)
    不可分离的内核(简要)
  • Hilbert-Schmidt theory 希尔伯特-施密特理论

PREFACE 前言

Read the Fredholm alternative notes before proceeding. This is covered in the book (Section 9.4), but the material on integral equations is not. For references on integral equations (and other topics covered in the book too!), see:
在继续之前请阅读 Fredholm 替代说明。本书(第 9.4 节)对此进行了介绍,但积分方程的材料并未介绍。有关积分方程的参考(以及本书中涵盖的其他主题!),请参阅:
  • Riley and Hobson, Mathematical methods for physics and engineering (this is an extensive reference, also for other topics in the course)
    Riley 和 Hobson,物理和工程的数学方法(这是一份广泛的参考资料,也适用于课程中的其他主题)
  • Guenther and Lee, Partial differential equations of mathematical physics and integral equations (more technical; not the best first reference)
    Guenther 和 Lee,数学物理的偏微分方程和积分方程(技术性更强;不是最好的第一参考)
  • J.D. Logan, Applied mathematics (more generally about applied mathematics techniques, with a good section on integral equations)
    J.D. Logan,应用数学(更一般地关于应用数学技术,其中有关于积分方程的很好的部分)

1. FREDHOLM INTEGRAL EQUATIONS: INTRODUCTION
1. FREDHOLM 积分方程:简介

Differential equations are a subset of more general equations involving linear operators . Here, we give a brief treatment of a generalization to integral equations.
微分方程 是涉及线性运算符 的更一般方程的子集。在这里,我们简要介绍积分方程的推广。
To motivate this, every ODE IVP can be written as an 'integral equation' by integrating. For instance, consider the first order IVP
为了激发这一点,每个 ODE IVP 都可以通过积分写成“积分方程”。例如,考虑一阶 IVP
Integrate both sides from to to get the integral equation
两边积分,得到积分方​​程
If solves (1.2) then it also solves (1.1); they are 'equivalent' in this sense. However, it is slightly more general as does not need to be differentiable.
如果 解决了(1.2),那么它也解决了(1.1);从这个意义上说,它们是“等价的”。然而,它稍微更通用,因为 不需要可微分。
Definition: A Volterra integral equation for has the form
定义: 的 Volterra 积分方程具有以下形式
The function is the kernel. Note the integral upper bound is (the ind. variable).
函数 是内核。请注意,积分上限为 (索引变量)。
Volterra integral equations are 'equivalent' to ODE initial value problems on for linear ODEs. They will not be studied here.
Volterra 积分方程与线性 ODE 的 上的 ODE 初值问题“等效”。这里不会对它们进行研究。
In contrast, ODE boundary value problems generalize to Fredholm integral equations. Such an equation involves an integral over the whole domain (not up to ):
相反,ODE 边值问题可推广到 Fredholm 积分方程。这样的方程涉及整个域的积分(不超过 ):
Definition: A Fredholm integral equation (FIE) has two forms:
定义:Fredholm 积分方程 (FIE) 有两种形式:

- First kind: (FIE-1)
- 第一类:(FIE-1)

  • Second kind: (FIE-2) 第二种:(FIE-2)
The 'second kind' is the first kind plus a multiple of . Note that the operator
“第二种”是第一种加上 的倍数。请注意,操作员
is linear; this is called a Fredholm integral operator.
是线性的;这称为 Fredholm 积分算子。
The simplest kernels are separable ('degenerate'), which have the form
最简单的内核是可分离的(“简并”),其形式为
More complicated kernels are non-separable. Examples include:
更复杂的内核是不可分离的。示例包括:
  • (Hilbert transform)
    (希尔伯特变换)
  • (Fourier transform)
    (傅里叶变换)
  • (Laplace transform)
    (拉普拉斯变换)
We will study the last two later. Writing a non-separable kernel in the form (1.5) would require an infinite series, which complicates the analysis. Instead, we will utilize an integral form and some complex analysis later to properly handle non-separable kernels.
我们稍后将研究后两个。以 (1.5) 的形式编写不可分离的核需要无限级数,这使分析变得复杂。相反,我们稍后将利用积分形式和一些复杂的分析来正确处理不可分离的内核。

2. SolVing SEParable FIEs
2. 解决可分离的外商投资企业

Suppose is separable and is a Fredholm integral operator of the first kind:
假设 是可分的,并且 是第一类 Fredholm 积分算子:
We wish to solve the eigenvalue problem
我们希望解决特征值问题
This can always be done for a separable kernel. The main result is the following:
对于可分离内核始终可以这样做。主要结果如下:
Eigenfunctions for FIE-1: For the FIE (2.1) with separable kernel,
FIE-1 的特征函数:对于具有可分离核的 FIE (2.1),
  1. there are non-zero eigenvalues with eigenfunctions . Each eigenfunction is a linear combination of the 's, i.e.
    个非零特征值 和特征函数 。每个特征函数都是 的线性组合,即
  2. is an eigenvalue with infinite multiplicity: an infinite set of (orthogonal) eigenfunctions characterized by
    是具有无限重数的特征值:(正交)特征函数 的无限集合,其特征在于
The eigenfunctions from (1) and (2) together are a basis for .
(1) 和 (2) 的特征函数一起构成 的基础。
Proof, part 1 (non-zero eigenvalues): Consider ; look for an eigenfunction
证明,第 1 部分(非零特征值):考虑 ;寻找特征函数
Derivation: Plug in this expression into the operator :
推导:将此表达式代入运算符
where . Now set this equal to
其中 。现在将其设置为等于
The coefficients on must match, leaving the linear system
上的系数必须匹配,离开线性系统
so the eigenvalue problem for is equivalent to a matrix eigenvalue problem for an matrix. Assuming is invertible, the result follows.
因此 的特征值问题相当于 矩阵的矩阵特征值问题。假设 可逆,结果如下。
Claim 2 (zero eigenvalue): We show that is always an eigenvalue with infinite multiplicity. Observe that
主张 2(零特征值):我们证明 始终是具有无限重数的特征值。观察一下
This must be true for all , so it follows that
这对于所有 都必须成立,因此可以得出结论
That is, the set of eigenfunctions for is precisely the space of functions orthogonal to the span of the 's. But is infinite dimensional and the span of has dimension , so the orthogonal complement is infinite dimensional - this proves the claim. That is,
也就是说, 的特征函数集恰好是与 的跨度正交的函数空间。但是 是无限维的,并且 的跨度具有维度 ,因此正交补是无限维的 - 这证明了这一说法。那是,
The fact that the basis for is countable (a sequence indexed by ) also follows from the fact that has this property. The eigenfunctions can be constructed using Gram-Schmidt (see example in subsection 2.1).
的基础是可数的(由 索引的序列)这一事实也源于 具有此属性的事实。 特征函数 可以使用 Gram-Schmidt 构造(参见第 2.1 小节中的示例)。
2.1. Example (eigenfunctions). Define the (separable) kernel
2.1.示例(本征函数)。定义(可分离)内核
and consider the FIE of the first kind
并考虑第一类FIE
The result says has non-zero eigenvalues. Look for an eigenfunction
结果表明 具有 非零特征值。寻找特征函数
then plug in and integrate to get
然后插上并集成即可得到
which gives (equating coefficients of on the LHS/RHS as in (2.3))
给出(使 LHS/RHS 上的 系数相等,如 (2.3) 中所示)
Plugging in the values from (2.4), the eigenvalue problem is
代入 (2.4) 中的值,特征值问题为
Translating back to the integral equation with given by (2.5), the non-zero 's and 's are
转换回由 (2.5) 给出的 的积分方程,非零
along with the zero eigenvalue and eigenfunctions .
以及零特征值和特征函数
Zero eigenvalue (details): For , use the characterization (2.2):
零特征值(详细信息):对于 ,使用表征(2.2):
First, we need to orthogonalize. Define and
首先,我们需要正交化。定义
It follows that if is any function then a zero eigenfunction is
由此可见,如果 是任意函数,则零特征函数是
The whole set can be found using Gram-Schmidt, e.g. with . For instance,
整个集合可以使用 Gram-Schmidt 找到,例如与 。例如,
is an eigenfunction for (orthogonal to and ).
的特征函数(与 正交)。
2.2. Adjoint operator. The operator is not self-adjoint in general. To solve FIEs with an eigenfunction expansion, we need the adjoint and its eigenfunctions . To obtain it, let denote the inner product .
2.2.伴随算子。运算符 通常不是自伴的。为了用特征函数展开求解 FIE,我们需要伴随 及其特征函数 。要获得它,让 表示 内积
We look for an integral operator such that
我们寻找一个积分运算符 使得
To find the adjoint, carefully exchange integration order (pay attention to which of or is the 'integration variable' here!). Compute
要找到伴随项,请仔细交换积分顺序(请注意 中的哪一个是此处的“积分变量”!)。计算
Note the inner integration variable above is now and not . From this result, we see that
请注意,上面的内部集成变量现在是 而不是 。从这个结果我们可以看出
after swapping symbols for and . Comparing to , we have the following characterization for the adjoint:
交换 符号后。与 相比,我们对伴随有以下特征:
FI operator adjoint: The FI operator of the first kind and its adjoint are
FI 算子伴随:第一类 FI 算子及其伴随为
i.e. the adjoint has kernel (the kernel of with the variables swapped). Moreover,
即伴随有内核 的内核与交换的变量)。而且,
That is, is self-adjoint if and only if the kernel is symmetric (e.g. ).
也就是说,当且仅当核对称时, 才是自伴的(例如 )。

3. SolVing FIEs OF THE FIRST KIND (SEPARABLE)
3. 解决第一类 FIE(可分离)

We are now equipped to solve
我们现在有能力解决
where has a separable kernel (1.5). Note that from (2.8), it follows that also has a separable kernel.
其中 有一个可分离的内核 (1.5)。请注意,从 (2.8) 可知 也有一个可分离的内核。
Thus the adjoint also has non-zero eigenvalues with eigenfunctions (same structure). Moreover, the 's and 's are bi-orthogonal. That is, for and the same for the 's and 's and between the two sets (e.g. for all .
因此伴随 也有 非零特征值和特征函数 (相同的结构)。此外, 是双正交的。也就是说, 对于 ,对于 以及两个集合之间的情况相同(例如 < b9> 对于所有
Reminder: to get the coefficient of , take the inner product with , e.g.
提醒:要获得 的系数,请将 进行内积,例如
Solving the FIE: Project by taking to get
求解 FIE: 项目通过取 得到
We need to write and in terms of the eigenfunctions.
我们需要根据特征函数来写
  • For the RHS (f): This is direct:
    对于 RHS (f):这是直接的:
Let . Take the inner product with to get
。与 取内积得到
The other coefficients will only be needed later.
其他系数稍后才会需要。
  • For the (Lu): First write
    对于 (Lu):首先写
The second term is sent to zero by :
第二项由 发送为零:
Taking the inner product with , we get
进行内积,我们得到
  • Combining: Equating the two parts, we find that
    结合:将两部分等同,我们发现
Solvability: What about the 's? The adjoint has a zero eigenvalue, so we are in FAT case (B). A solvability condition decides between no solution or infinitely many solutions.
可解性: 怎么样?伴随 的特征值为零,因此我们处于 FAT 情况 (B)。可解性条件决定无解或无限多个解。
To find this condition, project onto one a eigenfunction for , i.e. take the inner product of with one of the 's:
要找到此条件,请将 的特征函数投影到一个特征函数上,即取 之一的内积:
Thus a solution exists only when has no component for any , i.e. if and only if
因此,仅当 没有任何 组件时才存在解决方案,即当且仅当
Summary of solution (separable FIE-1): If is of the form (3.4), the solution is
解决方案摘要(可分离 FIE-1):如果 的形式为 (3.4),则解决方案为
and the 's are arbitrary. If is not of the form (3.4) then there is no solution.
并且 是任意的。如果 不是 (3.4) 的形式,则无解。
3.1. The shortcut. There is a way to get around the full expansion. Recall that
3.1.捷径。有一种方法可以绕过全面扩张。回想起那个
i.e each eigenfunction (for non-zero ) is a linear combination of the 's. Then
即每个特征函数(对于非零 )是 的线性组合。然后
It follows that we could anticipate the solution and guess
由此可见,我们可以预测解并猜测
knowing that the result for will be in this form. We lose the bi-orthogonal structure (no eigenfunctions), but the 's are not needed to calculate the solution.
知道 的结果将采用这种形式。我们失去了双正交结构(无特征函数),但不需要 来计算解。
If is small, the shortcut is much easier (the linear system is small).
如果 很小,捷径就容易得多(线性系统很小)。
3.2. Example: Return to the previous example (subsection 2.1),
3.2.示例:返回前面的示例(第 2.1 小节),
with separable kernel 具有可分离内核
and non-zero eigenvalues/functions
和非零特征值/函数
along with the zero eigenvalue and eigenfunctions . Suppose the problem is
以及零特征值和特征函数 。假设问题是
Note that is always an eigenvalue, so we are in case (B) of the FAT. Here, the RHS is in the span of and (both linear combinations of and ), so there is a solution. It is unique up to an arbitrary sum of the form .
请注意, 始终是特征值,因此我们处于 FAT 的情况 (B) 中。在这里,RHS 位于 的范围内( 的线性组合),因此有解决方案。它在 形式的任意总和范围内是唯一的。
Direct method: Note that the kernel is symmetric here. Thus is self-adjoint and . Write the solution as
直接法:注意这里的核是对称的。因此 是自伴的并且 。将解写为
Take the inner product of the equation with :
计算方程的内积:
The coefficients can be computed from here; then
可以从这里计算系数;然后
for arbitrary 's (details not carried out here).
对于任意 的(此处不进行详细说明)。
Shortcut method: Instead look for the 'unique part' as a linear combination of 's:
捷径方法:而是寻找“独特部分”作为 的线性组合:
Plug into the equation and write as a linear system:
代入方程并写成线性系统:
so and . The solution is then (for arbitrary 's)
所以 。解决方案是(对于任意

4. FIEs OF THE SECOND KIND (STILL SEPARABLE)
4. 第二类外商投资企业(仍可分离)

Now consider the FI operator of the second kind,
现在考虑第二类 FI 算子,
where is an FIE of the first kind with a separable kernel:
其中 是具有可分离内核的第一类 FIE:
Note that the term added to only shifted the eigenvalues. The result:
请注意,添加到 中的 项仅改变了特征值。结果:
For the second kind: For the FIE of the second kind (4.1), the eigenfunctions are the same as and the eigenvalues are shifted by . That is, the eigenvalues of are
对于第二类:对于第二类FIE(4.1),特征函数与 相同,特征值移位 。即 的特征值为
with the same eigenfunctions and .
具有相同的特征函数
It follows that has all non-zero eigenvalues unless is an eigenvalue of .
由此可见, 具有所有非零特征值,除非 的特征值。
Now we solve the FIE
现在我们解决FIE
under the assumption that
假设
If (4.3) holds then the FAT case (A) applies (since is non-zero) and there is a unique solution, unlike an FIE of the first kind.
如果 (4.3) 成立,则 FAT 情况 (A) 适用(因为 不为零)并且存在唯一的解决方案,与第一类 FIE 不同。
Direct method (not ideal): Expand and in terms of the eigenfunctions:
直接法(不理想):根据特征函数展开
Now plug into the FIE (4.2) to get
现在插入 FIE (4.2) 即可得到
Equate this to (note that now ) to get the coefficients:
将其等同于 (请注意,现在是 )以获得系数:
This can be continued, but there is a better shortcut, inspired by the FIE-1 case. The trick is that the 'infinite part' (with ) of the solution is really a multiple of , leaving only the 'finite part', which is a sum only over terms.
这可以继续,但有一个更好的捷径,受到 FIE-1 案例的启发。诀窍在于,解决方案的“无限部分”(带有 )实际上是 的倍数,只留下“有限部分”,它的总和仅超过 < b2> 条款。
Practical note: The direct method leads to trivial equations (from orthogonality), but there are an infinite number of them. Moreover, the part is best avoided if we can get away with a solution containing only a finite sum!
实用注意事项:直接方法会产生平凡的方程(根据正交性),但方程的数量是无限的。此外,如果我们能够得到仅包含有限和的解,那么最好避免 部分!

4.1. Shortcut (undetermined coefficients): Assume a solution of the form
4.1.快捷方式(未确定的系数):假设以下形式的解

Since the kernel is separable, for any function is a linear combination of 's:
由于内核是可分离的,因此任何函数 都是 的线性组合:
Now plug the guess (4.4) into and keep the part of separate from the rest. All the other terms, via rule (4.5), will give various linear combinations of 's. To be precise,
现在将猜测 (4.4) 代入 中,并将 部分与其余部分分开。所有其他项,通过规则(4.5),将给出 的各种线性组合。准确地说,
Now plug in and collect all the terms and the leftover part:
现在插入并收集所有 项和剩余的 部分:
That is, the part must cancel, leaving just , so
也就是说, 部分必须取消,只留下 ,所以
which is linear system for of the form . Now recall that under the assumptions made at the start, the FAT guarantees a unique solution, so we are done (the guess is a solution, so it is the unique solution).
这是 形式 的线性系统。现在回想一下,在开始时所做的假设下,FAT 保证了唯一的解决方案,因此我们完成了(猜测是一个解决方案,因此它是唯一的解决方案)。
4.2. Example: again, return to the example. Now let
4.2.示例:再次回到示例。现在让
The eigenvalues of are and , with the same eigenfunctions as computer earlier. Consider
的特征值是 ,与早期计算机的 具有相同的特征函数。考虑
Look for a solution
寻找解决方案
After following the process (plug everything in), and
遵循流程 (插入所有内容)后,
This is the unique solution! Note the deals neatly with the RHS; no infinite sum required. The downside is that the algebra for and is messy, but it is a small linear system, and a computer algebra package has no trouble with it.
这是独特的解决方案!请注意, 与 RHS 完全相关;不需要无限总和。缺点是 的代数很混乱,但它是一个小型线性系统,计算机代数包可以轻松处理它。

5. WHAT IF THE KERNEL IS NON-SEPARABLE?
5. 如果内核是不可分离的怎么办?

For a non-separable kernel such as
对于不可分离的内核,例如
the theory is different. More conditions are required to ensure that the eigenvalue problem behaves well. If the kernel is 'nice', there is an infinite set of eigenvalues with structure similar to Sturm-Liouville theory. As for Sturm-Liouville operators the general idea is that
理论是不同的。需要更多的条件来确保特征值问题表现良好。如果核是“好的”,则存在无限组特征值,其结构类似于 Sturm-Liouville 理论。对于 Sturm-Liouville 算子,总体思路
self-adjoint is 'nice' good properties for eigenfunctions/values.
自伴 是“很好” 特征函数/值的良好属性。
For integral equations, Hilbert-Schmidt theory provides the answer for what is required to be 'nice' and what 'good properties' result. The Hilbert-Schmidt condition is
对于积分方程,希尔伯特-施密特理论提供了什么是“好的”以及什么是“好的性质”结果的答案。希尔伯特-施密特条件是
which guarantees that the operator is 'nice' (precisely, a compact operator on . Hilbert-Schmidt theory then says that if
这保证了运算符 是“好的”(准确地说,是 上的紧凑运算符。希尔伯特-施密特理论说,如果
satisfies the conditions
满足条件
  1. is self-adjoint (in )
    是自伴的(在 中)
  2. is not separable (although the separable case is similar)
    是不可分离的(尽管可分离的情况类似)
  3. The Hilbert-Schmidt condition (5.2) holds ( is 'compact') then the eigenvalue problem has nice properties:
    希尔伯特-施密特条件 (5.2) 成立( 为“紧”),则特征值问题具有良好的属性:
  • The eigenvalues are real
    特征值 是实数
  • The eigenvalues are a discrete sequence, each with finite multiplicity:
    特征值是一个离散序列,每个特征值都具有有限重数:
  • The eigenfunctions are an orthogonal basis for if
    特征函数 的正交基,如果
  • The Fredholm Alternative and eigenfunction expansions for apply (in the same way that we used them for BVPs with Sturm-Liouville theory)
    Fredholm 替代方案和 的特征函数展开适用(与我们将它们用于具有 Sturm-Liouville 理论的 BVP 的方式相同)
For further reading, see the references at the start of the notes. We will study non-separable kernels in a different way when addressing the Fourier and Laplace transforms.
如需进一步阅读,请参阅注释开头的参考文献。在解决傅里叶变换和拉普拉斯变换时,我们将以不同的方式研究不可分离核。

  1. In functional analysis terms, it is a 'separable Hilbert space'.
    用泛函分析术语来说,它是一个“可分离的希尔伯特空间”。
  2. The computations are messy and not included here.
    计算很混乱,这里不包括在内。
    Both the theory for integral operators and Sturm-Liouville theory are part of a more general theory; the structure comes from the theory of compact operators on a Hilbert space (the 'spectral theorem for compact operators').
    积分算子理论和 Sturm-Liouville 理论都是更一般理论的一部分;该结构来自希尔伯特空间上的紧算子理论(“紧算子谱定理”)。