PlanetPhysics/Integral Equation

An integral equation involves an unknown function under the integral sign.\, Most common of them is a linear integral equation $$\begin{matrix} \alpha(t)\,y(t)+\!\int_a^bk(t,\,x)\,y(x)\,dx = f(t), \end{matrix}$$ where $$\alpha,\,k,\,f$$ are given functions.\, The function\, $$t \mapsto y(t)$$\, is to be solved.

Any linear integral equation is equivalent to a linear differential equation; e.g. the equation\, $$ y(t)\!+\!\int_0^t(2t-2x-3)\,y(x)\,dx = 1+t-4\sin{t}$$\, to the equation\, $$y''(t)-3y'(t)+2y(t) = 4\sin{t}$$\, with the initial conditions \,$$y(0) = 1$$\, and\, $$y'(0) = 0$$.\\

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 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 homogeneous .\\

Example. \, Solve the Volterra equation\, $$ y(t)\!+\!\int_0^t(t\!-\!x)\,y(x)\,dx = 1$$\, by using Laplace transform.

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

Solutions on some integral equations in EqWorld.