Editing Green's function (section)

==Motivation==
{{See also |Spectral theory|Volterra integral equation}}
Loosely speaking, if such a function {{mvar|G}} can be found for the operator {{math|''L''}}, then, if we multiply {{EquationNote|1|equation&nbsp;1}} for the Green's function by {{math|''f''(''s'')}}, and then integrate with respect to {{mvar|s}}, we obtain,
<math display="block">\int LG(x,s)\,f(s) \, ds = \int \delta(x-s) \, f(s) \, ds = f(x)\,.</math>
Because the operator <math>L = L(x)</math> is linear and acts only on the variable {{mvar|x}} (and ''not'' on the variable of integration {{mvar|s}}), one may take the operator <math>L</math> outside of the integration, yielding
<math display="block">L\left(\int G(x,s)\,f(s) \,ds \right) = f(x)\,.</math>
This means that
{{NumBlk|1=|2=<math display="block">u(x) = \int G(x,s)\,f(s) \,ds</math>|3={{EquationRef|3}}}}
is a solution to the equation <math>L u(x) = f(x)\,.</math>

Thus, one may obtain the function {{math|''u''(''x'')}} through knowledge of the Green's function in {{EquationNote|1|equation&nbsp;1}} and the source term on the right-hand side in {{EquationNote|2|equation&nbsp;2}}. This process relies upon the linearity of the operator {{math|''L''}}.

In other words, the solution of {{EquationNote|2|equation&nbsp;2}}, {{math| ''u''(''x'')}}, can be determined by the integration given in {{EquationNote|3|equation&nbsp;3}}. Although {{math|''f''(''x'')}} is known, this integration cannot be performed unless {{mvar|G}} is also known. The problem now lies in finding the Green's function {{mvar|G}} that satisfies {{EquationNote|1|equation&nbsp;1}}. For this reason, the Green's function is also sometimes called the [[fundamental solution]] associated to the operator {{math|''L''}}.

Not every operator <math>L</math> admits a Green's function. A Green's function can also be thought of as a [[Inverse function#Left and right inverses|right inverse]] of {{math|''L''}}. Aside from the difficulties of finding a Green's function for a particular operator, the integral in {{EquationNote|3|equation&nbsp;3}} may be quite difficult to evaluate. However the method gives a theoretically exact result.

This can be thought of as an expansion of {{mvar|f}} according to a [[Dirac delta function]] basis (projecting {{mvar|f}} over {{nowrap|<math>\delta(x - s)</math>;}} and a superposition of the solution on each [[Projection (mathematics)|projection]]. Such an integral equation is known as a [[Fredholm integral equation]], the study of which constitutes [[Fredholm theory]].