Consider the linear
partial differential equation :\boldsymbol{L}_{xy}f(x,y)=0, where \boldsymbol{L}_{xy} is a linear operator which contains derivatives with respect to and , subject to the mixed conditions on = 0, for some prescribed function , :f=g(x)\text{ for }x\leq 0, \quad f_y=0\text{ when }x>0 and decay at infinity i.e. → 0 as \boldsymbol{x}\rightarrow \infty. Taking a
Fourier transform with respect to results in the following
ordinary differential equation : \boldsymbol{L}_y \widehat{f}(k,y)-P(k,y)\widehat{f}(k,y)=0, where \boldsymbol{L}_{y} is a linear operator containing derivatives only, is a known function of and and : \widehat{f}(k,y)=\int_{-\infty}^\infty f(x,y)e^{-ikx} \, \textrm{d}x. If a particular solution of this ordinary differential equation which satisfies the necessary decay at infinity is denoted , a general solution can be written as : \widehat{f}(k,y)=C(k)F(k,y), where is an unknown function to be determined by the boundary conditions on =0. The key idea is to split \widehat{f} into two separate functions, \widehat{f}_{+} and \widehat{f}_{-} which are analytic in the lower- and upper-halves of the complex plane, respectively, : \widehat{f}_{+}(k,y)=\int_0^\infty f(x,y)e^{-ikx}\,\textrm{d}x, : \widehat{f}_{-}(k,y)=\int_{-\infty}^0 f(x,y)e^{-ikx}\,\textrm{d}x. The boundary conditions then give : \widehat{g\,}(k)+\widehat{f}_{+}(k,0) = \widehat{f}_{-}(k,0)+\widehat{f}_{+}(k,0) = \widehat{f}(k,0) = C(k)F(k,0) and, on taking derivatives with respect to y, : \widehat{f}'_{-}(k,0) = \widehat{f}'_{-}(k,0)+\widehat{f}'_{+}(k,0) = \widehat{f}'(k,0) = C(k)F'(k,0). Eliminating C(k) yields : \widehat{g\,}(k)+\widehat{f}_{+}(k,0) = \widehat{f}'_{-}(k,0)/K(k), where : K(k)=\frac{F'(k,0)}{F(k,0)}. Now K(k) can be decomposed into the product of functions K^{-} and K^{+} which are analytic in the upper and lower half-planes respectively. To be precise, K(k)=K^{+}(k)K^{-}(k), where : \log K^{-} = \frac{1}{2\pi i}\int_{-\infty}^\infty \frac{\log(K(z))}{z-k} \,\textrm{d}z, \quad \operatorname{Im}k>0, : \log K^{+} = -\frac{1}{2\pi i}\int_{-\infty}^\infty \frac{\log(K(z))}{z-k} \,\textrm{d}z, \quad \operatorname{Im}k (Note that this sometimes involves scaling K so that it tends to 1 as k\rightarrow\infty.) We also decompose K^{+}\widehat{g\,} into the sum of two functions G^{+} and G^{-} which are analytic in the lower and upper half-planes respectively, i.e., : K^{+}(k)\widehat{g\,}(k)=G^{+}(k)+G^{-}(k). This can be done in the same way that we factorised K(k). Consequently, : G^{+}(k) + K_{+}(k)\widehat{f}_{+}(k,0) = \widehat{f}'_{-}(k,0)/K_{-}(k) - G^{-}(k). Now, as the left-hand side of the above equation is analytic in the lower half-plane, whilst the right-hand side is analytic in the
upper half-plane, analytic continuation guarantees existence of an
entire function which coincides with the left- or right-hand sides in their respective half-planes. Furthermore, since it can be shown that the functions on either side of the above equation decay at large , an application of
Liouville's theorem shows that this entire function is identically zero, therefore : \widehat{f}_{+}(k,0) = -\frac{G^{+}(k)}{K^{+}(k)}, and so : C(k) = \frac{K^{+}(k)\widehat{g\,}(k)-G^{+}(k)}{K^{+}(k)F(k,0)}. == See also ==