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Green's function

 Definition/Summary Green's function $G\left(\mathbf{x},\mathbf{\xi}\right)$ can be defined thus $$\mathcal{L}G\left(\mathbf{x},\mathbf{\xi}\right) + \delta\left(\mathbf{x} - \mathbf{\xi}\right) = 0\;\;\; \mathbf{x},\mathbf{\xi} \in \mathbb{R}^n$$ Where $\mathcal{L}$ is a linear differential operator, $\mathbf{\xi}$ is an arbitrary point in $\mathbb{R}^n$ and $\delta$ is the Dirac Delta function.

 Equations

 Scientists George Green (1793-1841)

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 Breakdown Mathematics > Calculus/Analysis >> Differential Equations

 Extended explanation The Green's function for a particular linear differential operator $\mathcal{L}$, is an integral kernel that can be used to solve differential equations with appropriate boundary conditions. Given a linear differential equation $$\mathcal{L}u\left(\mathbf{x}\right) = \psi\left(\mathbf{x}\right)\ \ \ \ \ \ \left(*\right)$$ and the Green's function for the linear differential operator $\mathcal{L}$, one can derive an integral representation for the solution $u\left(\mathbf{x}\right)$ as follows: Consider the definition of Green's function above $$\mathcal{L}G\left(\mathbf{x},\mathbf{\xi}\right) + \delta\left(\mathbf{x} - \mathbf{\xi}\right) = 0 \Leftrightarrow \mathcal{L}G\left(\mathbf{x},\mathbf{\xi}\right) = -\delta\left(\mathbf{x} - \mathbf{\xi}\right)$$ Now multiplying through by $\psi\left(\mathbf{\xi}\right)$ and integrating over some bounded measure space $\Omega$ with respect to $\mathbf{\xi}$ $$-\int_\Omega \delta\left(\mathbf{x} - \mathbf{\xi}\right)\psi\left(\mathbf{\xi}\right) d\mathbf{\xi} = \int_\Omega \mathcal{L}G \left(\mathbf{x},\mathbf{\xi}\right) \psi\left(\mathbf{\xi}\right) d\mathbf{\xi}\hspace{2cm}\left(**\right)$$ Noting that by the properties of the Dirac Delta function $$\int_\Omega \delta\left(\mathbf{x} - \mathbf{\xi}\right)\psi\left(\mathbf{\xi}\right) d\mathbf{\xi} = \psi\left(\mathbf{x}\right)$$ And from (*) $$\int_\Omega \delta\left(\mathbf{x} - \mathbf{\xi}\right)\psi\left(\mathbf{\xi}\right) d\mathbf{\xi} = \mathcal{L}u\left(\mathbf{x}\right)$$ Hence (**) may be rewritten thus $$\mathcal{L}u\left(\mathbf{x}\right) = - \int_\Omega \mathcal{L}G \left(\mathbf{x},\mathbf{\xi}\right) \psi\left(\mathbf{\xi}\right) d\mathbf{\xi}$$ Since $\mathcal{L}$ is a linear differential operator, which does not act on the variable of integration we may write: $$u\left(\mathbf{x}\right) = - \int_\Omega G\left(\mathbf{x},\mathbf{\xi}\right) \psi\left(\mathbf{\xi}\right) d\mathbf{\xi}$$ Hence we have obtained an integral representation for $u\left(x\right)$. However, to evaluate this integral knowlagde of the explicit form of both $G\left(\mathbf{x},\mathbf{\xi}\right)$ and $\psi\left(\mathbf{\xi}\right)$ are required. Furthermore, even if both $G\left(\mathbf{x},\mathbf{\xi}\right)$ and $\psi\left(\mathbf{\xi}\right)$ are known, the associated integral may not be a trivial exercise. In addition, every linear differential operator does not admit a Green's Function. It would also be prudent to point out at this point that in general, Green's functions are distributions rather than classical functions.