Integro-differential equation

In mathematics, an integro-differential equation is an equation that involves both integrals and derivatives of a function.

General first order linear equations

The general first-order, linear (only with respect to the term involving derivative) integro-differential equation is of the form

{\frac {d}{dx}}u(x)+\int _{{x_{0}}}^{x}f(t,u(t))\,dt=g(x,u(x)),\qquad u(x_{0})=u_{0},\qquad x_{0}\geq 0.

As is typical with differential equations, obtaining a closed-form solution can often be difficult. In the relatively few cases where a solution can be found, it is often by some kind of integral transform, where the problem is first transformed into an algebraic setting. In such situations, the solution of the problem may be derived by applying the inverse transform to the solution of this algebraic equation.

Example

Consider the following first-order problem,

u'(x)+2u(x)+5\int _{{0}}^{{x}}u(t)\,dt=\left\{{\begin{array}{ll}1,\qquad x\geq 0\\0,\qquad x<0\end{array}}\right.\qquad {\text{with}}\qquad u(0)=0.

The Laplace transform is defined by,

U(s)={\mathcal {L}}\left\{u(x)\right\}=\int _{0}^{{\infty }}e^{{-sx}}u(x)\,dx.

Upon taking term-by-term Laplace transforms, and utilising the rules for derivatives and integrals, the integro-differential equation is converted into the following algebraic equation,

sU(s)-u(0)+2U(s)+{\frac {5}{s}}U(s)={\frac {1}{s}}.

Thus,

U(s)={\frac {1}{s^{2}+2s+5}}

.

Inverting the Laplace transform using contour integral methods then gives

u(x)={\frac {1}{2}}e^{{-x}}\sin(2x)

.

Alternatively, one can complete the square then use a table of Laplace transforms.

Applications

Integro-differential equations model many situations from science and engineering, such as in circuit analysis. By Kirchhoff's second law, the net voltage drop across a closed loop equals the voltage impressed $E(t)$ . (It is essentially an application of energy conservation.) An RLC circuit therefore obeys

$L{\frac {dI}{dt}}+RI(t)+{\frac {1}{C}}\int _{0}^{t}I(\tau )d\tau =E(t),$

where $I(t)$ is the current as a function of time, $R$ is the resistance, $L$ the inductance, and $C$ the capacitance.^[1]

The activity of interacting inhibitory and excitatory neurons can be described by a system of integro-differential equations, see for example the Wilson-Cowan model.

References

↑ Zill, Dennis G., and Warren S. Wright. “Section 7.4: Operational Properties II.” Differential Equations with Boundary-Value Problems, 8th ed., Brooks/Cole Cengage Learning, 2013, p. 305. Chapter 7 concerns the Laplace transform.

External links

Interactive Mathematics
Numerical solution of the example using Chebfun

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[1] Zill, Dennis G., and Warren S. Wright. “Section 7.4: Operational Properties II.” Differential Equations with Boundary-Value Problems, 8th ed., Brooks/Cole Cengage Learning, 2013, p. 305. Chapter 7 concerns the Laplace transform.