MATH 551 LECTURE NOTES FREDHOLM INTEGRAL EQUATIONS (A BRIEF INTRODUCTION) 微积分 551 讲义弗雷德霍姆积分方程(简介)
TOPICS COVERED 主要内容
Fredholm integral operators 弗雷德霍尔姆积分算子
Integral equations (Volterra vs. Fredholm) 积分方程(伏尔特拉与弗雷德霍尔姆)
Eigenfunctions for separable kernels 可分离核函数的本征函数
Adjoint operator, symmetric kernels 伴随算子，对称核
Solution procedure (separable) 分离解决程序
Solution via eigenfunctions (first and second kind) 基于本征函数（第一类和第二类）的解决方案
Shortcuts: undetermined coefficients 未定系数
An example (separable kernel, ) 一个示例(可分离核函数, )
Non-separable kernels (briefly) 不可分离核心(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,请阅读弗雷德霍尔姆替代说明。这在书中有涉及(9.4 节),但对积分方程的内容没有涉及。有关积分方程(以及本书中涉及的其他主题)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) 瑞利和霍布森,物理与工程的数学方法(这是一本广泛的参考资料,也适用于本课程的其他主题)
Guenther and Lee, Partial differential equations of mathematical physics and integral equations (more technical; not the best first reference) 古尔特和李,数学物理偏微分方程和积分方程(更技术性;不是最佳首选参考)
J.D. Logan, Applied mathematics (more generally about applied mathematics techniques, with a good section on integral equations) 洛根, 应用数学(更广泛地涉及应用数学技术,并有关于积分方程的良好部分)
1. FREDHOLM INTEGRAL EQUATIONS: INTRODUCTION 1. 弗雷德霍姆积分方程：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 是涉及线性算子 的更一般方程的一个子集。在此,我们简要介绍一下积分方程的推广。
To motivate this, every ODE IVP can be written as an 'integral equation' by integrating. For instance, consider the first order IVP 为了激励这个, 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 从 到 整合双方,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)；它们在这个意义上是"等价的"。然而，它更一般一些，因为 不需要是可微的。
Definition: A Volterra integral equation for has the form 伏尔特拉积分方程的形式为
The function is the kernel. Note the integral upper bound is (the ind. variable). 函数 是内核。注意积分上界为 (独立变量)。
Volterra integral equations are 'equivalent' to ODE initial value problems on for linear ODEs. They will not be studied here. 伏尔泰拉积分方程等同于在 上的线性常微分方程组的初值问题。这里不会研究它们。
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 边值问题可推广到弗雷德霍姆积分方程。这样的方程包含整个域上的积分(不仅限于 ):
Definition: A Fredholm integral equation (FIE) has two forms: 弗雷德霍姆积分方程(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 第二类"是第一类加上 的倍数。请注意运算符
is linear; this is called a Fredholm integral operator. 是线性的;这被称为弗雷德霍姆积分算子。
The simplest kernels are separable ('degenerate'), which have the form 最简单的核函数是可分离的('退化的'),它们的形式为
More complicated kernels are non-separable. Examples include: 更复杂的核函数是非可分离的。例如包括:
(Hilbert transform)
(Fourier transform) 傅里叶变换
(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. 我们稍后再研究最后两个问题。以形式(1.5)编写非分离核将需要无限级数,这使分析更加复杂。相反,我们将在稍后使用积分形式和一些复杂分析来正确处理非分离核。
2. SolVing SEParable FIEs 2. 求解分离型微分方程
Suppose is separable and is a Fredholm integral operator of the first kind: 假设 是可分离的,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: 对于可分离核函数,这总是可以做到。主要结果如下：
Eigenfunctions for FIE-1: For the FIE (2.1) with separable kernel, 对于可分离核的 FIE (2.1)，本征函数为:
there are non-zero eigenvalues with eigenfunctions . Each eigenfunction is a linear combination of the 's, i.e. 存在 个非零特征值 ,对应特征函数 。每个特征函数是 的线性组合,即 。
is an eigenvalue with infinite multiplicity: an infinite set of (orthogonal) eigenfunctions characterized by 是一个无限重数的特征值:一个由以下特征函数 构成的无限集合,其中 C 为任意常数
The eigenfunctions from (1) and (2) together are a basis for . 方程式(1)和(2)的本征函数共同构成了 的基底。
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:将此表达式代入算子 :
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
so the eigenvalue problem for is equivalent to a matrix eigenvalue problem for an matrix. Assuming is invertible, the result follows. 所以 的特征值问题等同于 矩阵的特征值问题。假设 可逆，结果即得。
Claim 2 (zero eigenvalue): We show that is always an eigenvalue with infinite multiplicity. Observe that 声明 2（零特征值）：我们证明 总是一个具有无穷多重性的特征值。观察到
This must be true for all , so it follows that 这对所有 必须是真实的,因此可以得出
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, 的特征函数集正好是正交于 跨度空间的函数空间。但是 是无穷维的,而 的跨度维度为 ,所以正交补充是无穷维的,这就证明了断言。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). 基于 是可数的事实(由 索引的序列)也源自 具有此性质的事实。 通过 Graham-Schmidt 构造特征函数 (见 2.1 小节中的例子)。
2.1. Example (eigenfunctions). Define the (separable) kernel 2.1. 示例(本征函数)。定义 (可分离的) 核函数
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 结果说 有 个非零特征值。寻找特征函数。
then plug in and integrate to get 然后插入并集成以获得
which gives (equating coefficients of on the LHS/RHS as in (2.3)) 根据(2.3)式中的等系数关系，
Plugging in the values from (2.4), the eigenvalue problem is 将(2.4)中的值代入，特征值问题为
Translating back to the integral equation with given by (2.5), the non-zero 's and 's are 将 (2.5) 式给出的积分方程转回得到的非零 和 。
along with the zero eigenvalue and eigenfunctions . 伴随着零本征值和本征函数 。
Zero eigenvalue (details): For , use the characterization (2.2): 零特征值(详细信息):对于 ，使用特征表示(2.2):
First, we need to orthogonalize. Define and 首先,我们需要进行正交化。定义 和
It follows that if is any function then a zero eigenfunction is 如果 是任何函数,那么零特征函数是
The whole set can be found using Gram-Schmidt, e.g. with . For instance, 使用格拉姆-施密特方法可以找到完整的集合,例如使用 。
is an eigenfunction for (orthogonal to and ). 是 的本征函数(正交于 和 )。
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 .
We look for an integral operator such that 我们寻找一个积分算子 ，使得
To find the adjoint, carefully exchange integration order (pay attention to which of or is the 'integration variable' here!). Compute 要找到伴随矩阵,请仔细交换积分次序(注意这里哪一个 或 是'积分变量'!).计算
Note the inner integration variable above is now and not . From this result, we see that 注意上方的内部积分变量现在是 而不是 。从这个结果来看,
after swapping symbols for and . Comparing to , we have the following characterization for the adjoint: 在替换 和 的符号后。与 相比,我们有以下关于伴随算子的表述:
FI operator adjoint: The FI operator of the first kind and its adjoint are
i.e. the adjoint has kernel (the kernel of with the variables swapped). Moreover, 即其伴随具有 核(即 变量对调后的核)。此外,
That is, is self-adjoint if and only if the kernel is symmetric (e.g. ). 也就是说, 当且仅当核函数是对称的(例如 )时,它才是自伴随的。
3. SolVing FIEs OF THE FIRST KIND (SEPARABLE) 解决第一类可分离常微分方程
We are now equipped to solve 我们现在已经有能力解决
where has a separable kernel (1.5). Note that from (2.8), it follows that also has a separable kernel.
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 .
Reminder: to get the coefficient of , take the inner product with , e.g.
Solving the FIE: Project by taking to get
We need to write and in terms of the eigenfunctions.
For the RHS (f): This is direct:
Let . Take the inner product with to get
The other coefficients will only be needed later.
For the (Lu): First write
The second term is sent to zero by :
Taking the inner product with , we get
Combining: Equating the two parts, we find that
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.
To find this condition, project onto one a eigenfunction for , i.e. take the inner product of with one of the 's:
Thus a solution exists only when has no component for any , i.e. if and only if
Summary of solution (separable FIE-1): If is of the form (3.4), the solution is
and the 's are arbitrary. If is not of the form (3.4) then there is no solution.
3.1. The shortcut. There is a way to get around the full expansion. Recall that
i.e each eigenfunction (for non-zero ) is a linear combination of the 's. Then
It follows that we could anticipate the solution and guess
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.
If is small, the shortcut is much easier (the linear system is small).
3.2. Example: Return to the previous example (subsection 2.1),
with separable kernel
and non-zero eigenvalues/functions
along with the zero eigenvalue and eigenfunctions . Suppose the problem is
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 .
Direct method: Note that the kernel is symmetric here. Thus is self-adjoint and . Write the solution as
Take the inner product of the equation with :
The coefficients can be computed from here; then
for arbitrary 's (details not carried out here).
Shortcut method: Instead look for the 'unique part' as a linear combination of 's:
Plug into the equation and write as a linear system:
so and . The solution is then (for arbitrary 's)
4. FIEs OF THE SECOND KIND (STILL SEPARABLE)
Now consider the FI operator of the second kind,
where is an FIE of the first kind with a separable kernel:
Note that the term added to only shifted the eigenvalues. The result:
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
with the same eigenfunctions and .
It follows that has all non-zero eigenvalues unless is an eigenvalue of .
Now we solve the FIE
under the assumption that
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.
Direct method (not ideal): Expand and in terms of the eigenfunctions:
Now plug into the FIE (4.2) to get
Equate this to (note that now ) to get the coefficients:
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.
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!
4.1. Shortcut (undetermined coefficients): Assume a solution of the form
Since the kernel is separable, for any function is a linear combination of 's:
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,
Now plug in and collect all the terms and the leftover part:
That is, the part must cancel, leaving just , so
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).
4.2. Example: again, return to the example. Now let
The eigenvalues of are and , with the same eigenfunctions as computer earlier. Consider
Look for a solution
After following the process (plug everything in), and
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.
5. WHAT IF THE KERNEL IS NON-SEPARABLE?
For a non-separable kernel such as
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
self-adjoint is 'nice' good properties for eigenfunctions/values.
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
which guarantees that the operator is 'nice' (precisely, a compact operator on . Hilbert-Schmidt theory then says that if
satisfies the conditions
is self-adjoint (in )
is not separable (although the separable case is similar)
The Hilbert-Schmidt condition (5.2) holds ( is 'compact') then the eigenvalue problem has nice properties:
The eigenvalues are real
The eigenvalues are a discrete sequence, each with finite multiplicity:
The eigenfunctions are an orthogonal basis for if
The Fredholm Alternative and eigenfunction expansions for apply (in the same way that we used them for BVPs with Sturm-Liouville theory)
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.
In functional analysis terms, it is a 'separable Hilbert space'.
The computations are messy and not included here.
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').