intsolve - Help

intsolve

solve a linear integral equation

 Calling Sequence intsolve(Inteqn, funcn ) intsolve(Inteqn, funcn, method = ..., order = ... )

Parameters

 Inteqn - linear integral equation in funcn funcn - unknown function to be solved for method = ... - optional, the right-hand side can be any of: Neumann, Laplace, differentialequation, eigenfunction order = ... - optional, the right-hand side can be a positive integer, specifying the order of the series approximation to use with the Neumann method

Description

 • The intsolve command is able to find exact closed-form solutions to various types of integral equations as explained below. The solution is returned as an explicit equation with the unknown function in the left-hand side and the solution in the right-hand side. In some cases arbitrary constants are returned in the solution; they are represented as $\mathrm{_C1},\mathrm{_C2},\dots ,\mathrm{_Cn}$, as in the output of dsolve. When the solution contains integrals, they are represented with the inert Int.
 • By default, intsolve converts the linear integral equation into an equivalent ODE problem with initial values (ODE-IVP), then attempts solving this problem; if it fails, it uses an eigenfunction approach. The method Laplace is optional and tackles the problem using a Laplace transform. The Neumann method is also optional and computes a series approximation to the solution.
 • intsolve can recognize Fredholm and Volterra integral equations of the following kinds:
 Fredholm equation of the first kind (F1):

${\int }_{a}^{b}k\left(x,y\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆy=g\left(x\right)$

 Fredholm equation of the second kind (F2):

$f\left(x\right)+{\int }_{a}^{b}k\left(x,y\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆy=g\left(x\right)$

 Fredholm equation of the third (homogeneous) kind (F3):

$f\left(x\right)+{\int }_{a}^{b}k\left(x,y\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆy=0$

 Volterra equation of the first kind (V1):

${\int }_{a}^{x}k\left(x,y\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆy=g\left(x\right)$

 Volterra equation of the second kind (V2):

$f\left(x\right)+{\int }_{a}^{x}k\left(x,y\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆy=g\left(x\right)$

 Volterra equation of the third (homogeneous) kind (V3):

$f\left(x\right)+{\int }_{a}^{x}k\left(x,y\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆy=0$

 • The Laplace method is applicable to equations of V1, V2 and V3 type. One advantage to using this option is that the integral equation may contain the Dirac or Heaviside functions. This method is not suitable for finding a solution when the equation does not possess a convolution type kernel.
 • The Neumann method is applicable to equations of V2, F2, and some of V1 type. The default order of the Neumann series in the solution is Order - this value can be overridden by passing the optional argument order = n, where n is a positive integer.
 • infolevel[intsolve] := 1,2,3 or 5 gives descriptive information on the progress of the computation. 5 may result in excessive printing.

Examples

 > $\mathrm{eq1}≔p\left(x\right)-\frac{1\left({{∫}}_{0}^{1}xyp\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}y\right)}{2}=\frac{5x}{6}$
 ${\mathrm{eq1}}{≔}{p}{}\left({x}\right){-}\frac{\left({{\int }}_{{0}}^{{1}}{x}{}{y}{}{p}{}\left({y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}\right)}{{2}}{=}\frac{{5}{}{x}}{{6}}$ (1)
 > $\mathrm{intsolve}\left(\mathrm{eq1},p\left(x\right)\right)$
 ${p}{}\left({x}\right){=}{x}$ (2)

If we instead use the 'Neumann' method, an approximation to this result is returned.

 > $\mathrm{intsolve}\left(\mathrm{eq1},p\left(x\right),\mathrm{method}=\mathrm{Neumann}\right)$
 ${p}{}\left({x}\right){=}\frac{{279935}{}{x}}{{279936}}$ (3)
 > $\mathrm{intsolve}\left(\mathrm{eq1},p\left(x\right),\mathrm{method}=\mathrm{Neumann},\mathrm{order}=20\right)$
 ${p}{}\left({x}\right){=}\frac{{21936950640377855}{}{x}}{{21936950640377856}}$ (4)

Computing the solution and verifying the result

 > $\mathrm{eq2}≔f\left(x\right)=x+1+{{∫}}_{0}^{x}\left(1+2\left(x-y\right)\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}y$
 ${\mathrm{eq2}}{≔}{f}{}\left({x}\right){=}{x}{+}{1}{+}{{\int }}_{{0}}^{{x}}\left({1}{+}{2}{}{x}{-}{2}{}{y}\right){}{f}{}\left({y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}$ (5)
 > $\mathrm{intsolve}\left(\mathrm{eq2},f\left(x\right)\right)$
 ${f}{}\left({x}\right){=}{{ⅇ}}^{{2}{}{x}}$ (6)

To verify this solution, transform it first into a procedure

 > $f=\mathrm{unapply}\left(\mathrm{rhs}\left(\right),x\right)$
 ${f}{=}\left({x}{↦}{{ⅇ}}^{{2}{}{x}}\right)$ (7)
 > $\mathrm{eval}\left(\mathrm{eq2},\right)$
 ${{ⅇ}}^{{2}{}{x}}{=}{x}{+}{1}{+}{{\int }}_{{0}}^{{x}}\left({1}{+}{2}{}{x}{-}{2}{}{y}\right){}{{ⅇ}}^{{2}{}{y}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}$ (8)
 > $\mathrm{value}\left(\right)$
 ${{ⅇ}}^{{2}{}{x}}{=}{{ⅇ}}^{{2}{}{x}}$ (9)

For some type of problems the solution obtained using the Laplace method contains Dirac functions

 > $\mathrm{eq3}≔{{∫}}_{0}^{x}{ⅇ}^{a\left(x-y\right)}f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}y=x$
 ${\mathrm{eq3}}{≔}{{\int }}_{{0}}^{{x}}{{ⅇ}}^{{a}{}\left({x}{-}{y}\right)}{}{f}{}\left({y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}{=}{x}$ (10)
 > $\mathrm{intsolve}\left(\mathrm{eq3},f\left(x\right),\mathrm{method}=\mathrm{Laplace}\right)$
 ${f}{}\left({x}\right){=}{-}{a}{}{x}{+}{1}$ (11)

Another type of integral equation automatically solved using the eigenfunction approach

 > $\mathrm{eq4}≔f\left(x\right)+b\left({{∫}}_{-1}^{1}\left(xy+{x}^{2}{y}^{2}\right)f\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}y\right)=d\left(x\right)$
 ${\mathrm{eq4}}{≔}{f}{}\left({x}\right){+}{b}{}\left({{\int }}_{{-1}}^{{1}}\left({{x}}^{{2}}{}{{y}}^{{2}}{+}{x}{}{y}\right){}{f}{}\left({y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}\right){=}{d}{}\left({x}\right)$ (12)
 > $\mathrm{intsolve}\left(\mathrm{eq4},f\left(x\right)\right)$
 ${f}{}\left({x}\right){=}{{\int }}_{{-1}}^{{1}}\left({-}\frac{{5}{}{{x}}^{{2}}{}{b}{}{{y}}^{{2}}}{{2}{}{b}{+}{5}}{-}\frac{{3}{}{x}{}{b}{}{y}}{{2}{}{b}{+}{3}}\right){}{d}{}\left({y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}{+}{d}{}\left({x}\right)$ (13)

An example where an arbitrary constant $\mathrm{_C1}$ is introduced with the solution

 > $\mathrm{eq5}≔3{x}^{2}+4x={{∫}}_{-1}^{1}\left(6{x}^{2}y+4x{y}^{2}\right)p\left(y\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}y$
 ${\mathrm{eq5}}{≔}{3}{}{{x}}^{{2}}{+}{4}{}{x}{=}{{\int }}_{{-1}}^{{1}}\left({6}{}{{x}}^{{2}}{}{y}{+}{4}{}{x}{}{{y}}^{{2}}\right){}{p}{}\left({y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{y}$ (14)
 > $\mathrm{intsolve}\left(\mathrm{eq5},p\left(x\right)\right)$
 ${p}{}\left({x}\right){=}{\mathrm{_C1}}{}{{x}}^{{2}}{+}\frac{{3}}{{4}}{}{x}{+}\frac{{3}}{{2}}{-}\frac{{3}}{{5}}{}{\mathrm{_C1}}$ (15)

References

 Chambers, LL. G. Integral Equations: A Short Course. London: International Textbook Company Ltd., 1976.
 Tricomi, F. G. Integral Equations. New York: Interscience, 1957.