integral equation

An integral equation involves an unknown function under the .  Most common of them is a linear integral equation

$\alpha (t)y(t)+{\displaystyle {\int }_a^b}k(t,x)y(x)𝑑x=f(t),$(1)

where $\alpha ,k,f$ are given functions.  The function  $t\mapsto y(t)$  is to be solved.

Any linear integral equation is equivalent (http://planetmath.org/Equivalent3) to a linear differential equation; e.g. the equation  $y(t)+{\displaystyle {\int }_0^t}(2t-2x-3)y(x)𝑑x=1+t-4\sin t$  to the equation  $y^{\prime \prime }(t)-3y^{\prime }(t)+2y(t)=4\sin t$  with the initial conditions  $y(0)=1$  and  $y^{\prime }(0)=0$.

The equation (1) is of

• 1st kind if  $\alpha (t)\equiv 0$,

• 2nd kind if $\alpha (t)$ is a nonzero constant,

• 3rd kind else.

If both limits (http://planetmath.org/UpperLimit) of integration in (1) are constant, (1) is a Fredholm equation, if one limit is variable, one has a Volterra equation.  In the case that  $f(t)\equiv 0$,  the linear integral equation is .

Example.  Solve the Volterra equation  $y(t)+{\displaystyle {\int }_0^t}(t-x)y(x)𝑑x=1$  by using Laplace transform.

Using the convolution (http://planetmath.org/LaplaceTransformOfConvolution), the equation may be written  $y(t)+t*y(t)=1$.  Applying to this the Laplace transform, one obtains  $Y(s)+{\displaystyle \frac{1}{s^2}}Y(s)={\displaystyle \frac{1}{s}}$,  whence  $Y(s)={\displaystyle \frac{s}{s^2+1}}$.  This corresponds the function  $y(t)=\cos t$,  which is the solution.

http://eqworld.ipmnet.ru/en/solutions/ie.htmSolutions on some integral equations in EqWorld.