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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