Math 2.0

Integral equations Definition

Integral equation

Builds on FunctionRiemann integral

Definition

Let a<ba < b and let φ ⁣:[a,b]→ℂ\varphi \colon [a, b] \to \C be an unknown function. For each x∈[a,b]x \in [a, b], let Ix⊆[a,b]I_x \subseteq [a, b] be an interval of integration, either the fixed interval [a,b][a, b] or the interval [a,x][a, x]. An integral equation for φ\varphi is an equation of the form F(x,φ(x),∫IxH(x,s,φ(s))ds)=0for all x∈[a,b],F\left(x, \varphi(x), \int_{I_x} H\left(x, s, \varphi(s)\right) ds\right) = 0 \quad \text{for all } x \in [a, b], where FF and HH are given functions. A solution is a function φ\varphi for which the integrals exist and the equation holds for every x∈[a,b]x \in [a, b].

Remarks

Example: φ(x)=x+∫01xs φ(s) ds\varphi(x) = x + \int_0^1 x s \, \varphi(s) \, ds on [0,1][0, 1] is an integral equation with Ix=[0,1]I_x = [0, 1]. Try φ(x)=cx\varphi(x) = c x: the integral is cx/3c x / 3, so c=1+c/3c = 1 + c/3 and c=3/2c = 3/2. So φ(x)=3x/2\varphi(x) = 3x/2 is a solution.

Non-example: φ′(x)=φ(x)\varphi'(x) = \varphi(x) is a differential equation, not an integral equation, since φ\varphi does not appear under an integral. Its integral form φ(x)=φ(0)+∫0xφ(s) ds\varphi(x) = \varphi(0) + \int_0^x \varphi(s) \, ds is an integral equation.

The same definition is used with a region of ℝm\R^m in place of [a,b][a, b], and with vector-valued unknowns. When Ix=[a,b]I_x = [a, b] the equation is of Fredholm type; when Ix=[a,x]I_x = [a, x] it is of Volterra type.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Integral_equation
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DefinesIntegral equation
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