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MATH 551 讲座笔记弗雷德霍姆积分方程(简要介绍)


  • Fredholm integral operators
  • Integral equations (Volterra vs. Fredholm)
    积分方程(Volterra vs. Fredholm)
  • Eigenfunctions for separable kernels
  • Adjoint operator, symmetric kernels
  • Solution procedure (separable)
  • Solution via eigenfunctions (first and second kind)
  • Shortcuts: undetermined coefficients
  • An example (separable kernel, )
  • Non-separable kernels (briefly)
  • Hilbert-Schmidt theory 希尔伯特-施密特理论


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,请阅读弗雷德霍姆交替笔记。 这在书中有涵盖(第 9.4 节),但积分方程的材料没有。 有关积分方程(以及书中涵盖的其他主题!)的参考资料,请参见:
  • Riley and Hobson, Mathematical methods for physics and engineering (this is an extensive reference, also for other topics in the course)
  • 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(更技术性;不是最佳的第一参考资料)
  • 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(更普遍地涉及应用数学技术,有一个很好的部分关于积分方程)

1. 弗雷德霍姆积分方程:介绍

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 是涉及线性算子 的更一般方程的子集。在这里,我们对积分方程的一般化进行简要处理。
To motivate this, every ODE IVP can be written as an 'integral equation' by integrating. For instance, consider the first order IVP
为了激励这一点,每个 ODE IVP 都可以通过积分写成一个“积分方程”。例如,考虑一阶 IVP
Integrate both sides from to to get the integral equation
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(1.2),那么它也解决(1.1);在这个意义上它们是“等价的”。然而,它稍微更一般,因为 不需要可微。
Definition: A Volterra integral equation for has the form
定义: 的 Volterra 积分方程的形式
The function is the kernel. Note the integral upper bound is (the ind. variable).
函数 是核心。请注意积分上限是 (自变量)。
Volterra integral equations are 'equivalent' to ODE initial value problems on for linear ODEs. They will not be studied here.
Volterra 积分方程在 上与线性 ODE 的初值问题“等价”。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 ):
相比之下,ODE 边界值问题推广到 Fredholm 积分方程。这样的方程涉及整个域上的积分(不仅仅是 )。
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)
- 第一种:(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
“第二类”是第一类加上 的倍数。请注意操作符
is linear; this is called a Fredholm integral operator.
是线性的;这被称为 Fredholm 积分算子。
The simplest kernels are separable ('degenerate'), which have the form
More complicated kernels are non-separable. Examples include:
  • (Hilbert transform)
    (Hilbert 变换)
  • (Fourier transform)
  • (Laplace transform)
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.

2. SolVing SEParable FIEs
2. 解决可分离的有限元方程

Suppose is separable and is a Fredholm integral operator of the first kind:
假设 是可分离的, 是第一类 Fredholm 积分算子:
We wish to solve the eigenvalue problem
This can always be done for a separable kernel. The main result is the following:
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 ,具有特征函数 。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
    是具有无限重复性的特征值:由 特征化的无限集合(正交)特征函数
The eigenfunctions from (1) and (2) together are a basis for .
(1)和(2)中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 ;寻找一个特征函数
Derivation: Plug in this expression into the operator :
导出:将此表达式插入运算符 中:
where . Now set this equal to
The coefficients on must match, leaving the linear system
so the eigenvalue problem for is equivalent to a matrix eigenvalue problem for an matrix. Assuming is invertible, the result follows.
因此, 的特征值问题等价于一个 矩阵的特征值问题。假设 是可逆的,结果就会得出。
Claim 2 (zero eigenvalue): We show that is always an eigenvalue with infinite multiplicity. Observe that
声明 2(零特征值):我们展示 始终是一个具有无限重复性的特征值。观察到
This must be true for all , so it follows that
这对所有 都必须是真实的,因此它遵循
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,对于 的特征函数集合恰好是与 的张成正交的函数空间。但 是无限维的,而 的张成维度为 ,因此正交补空间也是无限维的 - 这证明了这个断言。也就是说,
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).
事实上, 的基础是可数的(由 索引的序列)也是因为 具有这个特性。 特征函数 可以使用 Gram-Schmidt 方法构建(请参见第 2.1 小节中的示例)。
2.1. Example (eigenfunctions). Define the (separable) kernel
2.1. 例子(特征函数)。定义(可分离)核
and consider the FIE of the first kind
考虑第一类的 FIE
The result says has non-zero eigenvalues. Look for an eigenfunction
结果显示 个非零特征值。寻找一个特征函数
then plug in and integrate to get
which gives (equating coefficients of on the LHS/RHS as in (2.3))
这给出了(将 LHS/RHS 上的 的系数等同于(2.3)中的系数)
Plugging in the values from (2.4), the eigenvalue problem is
Translating back to the integral equation with given by (2.5), the non-zero 's and 's are
along with the zero eigenvalue and eigenfunctions .
Zero eigenvalue (details): For , use the characterization (2.2):
零特征值(详细信息):对于 ,使用特征描述(2.2):
First, we need to orthogonalize. Define and
It follows that if is any function then a zero eigenfunction is
根据这个结论,如果 是任何函数,那么零特征函数是