Math 2.0

Integral equations Definition

Integro-differential equation

Builds on Ordinary differential equationRiemann integralVolterra integral equation

Definition

Let I=[t0,t0+T]I = [t_0, t_0 + T] with T>0T > 0, let n≥1n \ge 1, and let f ⁣:I×ℝn→ℝnf \colon I \times \R^n \to \R^n and K ⁣:I×I×ℝn→ℝnK \colon I \times I \times \R^n \to \R^n be continuous. A (Volterra type) integro-differential equation is an equation x′(t)=f(t,x(t))+∫t0tK(t,s,x(s)) ds,t∈I,x'(t) = f(t, x(t)) + \int_{t_0}^{t} K(t, s, x(s)) \, ds, \quad t \in I, for an unknown continuously differentiable x ⁣:I→ℝnx \colon I \to \R^n. More generally, an integro-differential equation is any equation with the unknown both under a derivative and under an integral. When the integral is over a fixed interval, the equation is of Fredholm type.

Remarks

Example: x′(t)=−∫0tx(s) dsx'(t) = -\int_0^t x(s) \, ds, x(0)=1x(0) = 1. Differentiating gives x′′=−xx'' = -x with x(0)=1x(0) = 1 and x′(0)=0x'(0) = 0, so x=cos⁡tx = \cos t. Check: −∫0tcos⁡s ds=−sin⁡t=x′(t)-\int_0^t \cos s \, ds = -\sin t = x'(t).

Non-example: x′(t)=−x(t)x'(t) = -x(t) contains no integral of the unknown, so it is an ordinary differential equation. Also x(t)=1+∫0tx(s) dsx(t) = 1 + \int_0^t x(s) \, ds contains no derivative; it is a Volterra integral equation.

Integrating the Volterra form from t0t_0 to tt turns the problem with x(t0)=x0x(t_0) = x_0 into a Volterra integral equation for xx. For a convolution kernel K(t,s,x)=k(t−s)xK(t, s, x) = k(t - s) x, the Laplace transform turns the linear equation into an algebraic one.

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  1. Ordinary differential equation concept
  2. Riemann integral concept
  3. Volterra integral equation concept
  4. Integro-differential equation definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Integro-differential_equation
About this page
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Kinddefinition
WritingW0 stub
MSC45J05
DefinesIntegro-differential equation
UsesOrdinary differential equation, Riemann integral, Volterra integral equation
Changed2 November 2026

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