Math 2.0

Integral equations Definition

Integral equation of the first kind

Builds on Linear integral equationKernel of an integral equation

Definition

Let a<ba < b, let K ⁣:[a,b]×[a,b]→ℂK \colon [a, b] \times [a, b] \to \C be a kernel, and let f ⁣:[a,b]→ℂf \colon [a, b] \to \C be given. For each x∈[a,b]x \in [a, b], let IxI_x be either [a,b][a, b] or [a,x][a, x]. An integral equation of the first kind is a linear integral equation of the form ∫IxK(x,s)φ(s) ds=f(x)for all x∈[a,b],\int_{I_x} K(x, s) \varphi(s) \, ds = f(x) \quad \text{for all } x \in [a, b], for the unknown function φ ⁣:[a,b]→ℂ\varphi \colon [a, b] \to \C.

Remarks

Example: ∫01(x+s)φ(s) ds=x\int_0^1 (x + s) \varphi(s) \, ds = x on [0,1][0, 1] is of the first kind. With α=∫01φ\alpha = \int_0^1 \varphi and β=∫01sφ(s) ds\beta = \int_0^1 s \varphi(s) \, ds it says αx+β=x\alpha x + \beta = x, so α=1\alpha = 1 and β=0\beta = 0. The function φ(s)=4−6s\varphi(s) = 4 - 6s satisfies both, since ∫01(4−6s) ds=1\int_0^1 (4 - 6s) \, ds = 1 and ∫01(4s−6s2) ds=0\int_0^1 (4s - 6s^2) \, ds = 0; so do infinitely many others.

Non-example: φ(x)=x+∫01(x+s)φ(s) ds\varphi(x) = x + \int_0^1 (x + s) \varphi(s) \, ds is not of the first kind, because φ\varphi also appears outside the integral.

With Ix=[a,b]I_x = [a, b] it is a Fredholm equation of the first kind; with Ix=[a,x]I_x = [a, x] it is a Volterra equation of the first kind. The example shows that the solution need not be unique. The right side f(x)=x2f(x) = x^2 shows that it need not exist, since the left side is always affine in xx.

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  1. Kernel of an integral equation concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Integral_equation
About this page
TagFWPYXMK52W
URIm2://FWPYXMK52W
Kinddefinition
WritingW0 stub
MSC45A05, 45B05
DefinesIntegral equation of the first kind
UsesLinear integral equation, Kernel of an integral equation

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