Integro-differential equation

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

${\displaystyle \frac{d}{dx}u(x)+{\int }_{x_0}^{x}f(t,u(t))dt=g(x,u(x)),u(x_0)=u_0,x_0\ge 0.}$

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 second-order problem,

${\displaystyle u^{\prime }(x)+2u(x)+5{\int }_0^{x}u(t)dt=\theta (x)\text{with}u(0)=0,}$

where

${\displaystyle \theta (x)=\{\begin{array}{l}1,x\ge 0\\ 0,x<0\end{array}}$

is the Heaviside step function. The Laplace transform is defined by,

${\displaystyle U(s)=\mathcal{L}\left\{u(x)\right\}={\int }_0^{\mathrm{\infty }}{e}^{-sx}u(x)dx.}$

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,

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

Thus,

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

Inverting the Laplace transform using contour integral methods then gives

${\displaystyle u(x)=\frac{1}{2}{e}^{-x}\sin (2x)\theta (x)}$.

Alternatively, one can complete the square and use a table of Laplace transforms ("exponentially decaying sine wave") or recall from memory to proceed:

${\displaystyle U(s)=\frac{1}{s^2+2s+5}=\frac{1}{2}\frac{2}{(s+1{)}^2+4}\Rightarrow u(x)={\mathcal{L}}^{-1}\left\{U(s)\right\}=\frac{1}{2}{e}^{-x}\sin (2x)\theta (x)}$.