Math 2.0

Integral equations Definition

Volterra equation of the first kind

Builds on Volterra integral equationKernel of an integral equationRiemann integral

Definition

Let a<ba < b, let Δ={(t,s):a≤s≤t≤b}\Delta = \set{(t, s) : a \le s \le t \le b}, let K ⁣:Δ→ℂK \colon \Delta \to \C be a kernel, and let f ⁣:[a,b]→ℂf \colon [a, b] \to \C be a given function. A linear Volterra equation of the first kind is the equation ∫atK(t,s) x(s) ds=f(t),t∈[a,b],\int_a^t K(t, s) \, x(s) \, ds = f(t), \quad t \in [a, b], for an unknown function xx on [a,b][a, b]. The unknown appears only inside the integral, and the upper limit of integration is the variable tt.

Remarks

Example: ∫0t(t−s) x(s) ds=t3\int_0^t (t - s) \, x(s) \, ds = t^3 on [0,1][0, 1] is of the first kind with K(t,s)=t−sK(t, s) = t - s. Differentiating twice gives x(t)=6tx(t) = 6t, and indeed ∫0t(t−s) 6s ds=3t3−2t3=t3\int_0^t (t - s) \, 6s \, ds = 3t^3 - 2t^3 = t^3. Non-example: x(t)−∫0tx(s) ds=1x(t) - \int_0^t x(s) \, ds = 1 is not of the first kind, since xx also appears outside the integral; it is a Volterra equation of the second kind. A necessary condition for a continuous solution when KK is continuous is f(a)=0f(a) = 0, since the integral vanishes at t=at = a. Often KK is required to be continuous on Δ\Delta, but weakly singular kernels such as (t−s)−1/2(t - s)^{-1/2} are also allowed, as in Abel's integral equation.

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  1. Kernel of an integral equation concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Volterra_integral_equation
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MSC45D05
DefinesVolterra equation of the first kind
UsesVolterra integral equation, Kernel of an integral equation, Riemann integral

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