PlanetPhysics/Integral Equation

An integral equation involves an unknown [[../Bijective/|function]] under the integral sign.\, Most common of them is a linear integral equation

$\begin{array}{c}\alpha (t)y(t)+\underset{a}{\overset{b}{\int }}k(t,x)y(x)dx=f(t),\end{array}$

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 [[../DifferentialEquations/|differential equation]]; e.g. the equation\, $y(t)+\underset{0}{\overset{t}{\int }}(2t-2x-3)y(x)dx=1+t-4\sin t$\, to the equation\, $y^{″}(t)-3y^{\prime }(t)+2y(t)=4\sin t$\, 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 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\, $y(t)+\underset{0}{\overset{t}{\int }}(t-x)y(x)dx=1$\, by using [[../2DLT/|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.