复杂的拉普拉斯逆变换
Laplace Inverse Transformation (LIT)
L ( 1 π t e − a 2 / 4 t ) = e − a s s , a > 0 \mathcal{L}(\frac{1}{\sqrt{\pi t}}\text{e}^{-a^2/4t})=\frac{\text{e}^{-a\sqrt{s}}}{\sqrt{s}},~~ a>0 L(πt1e−a2/4t)=se−as, a>0
k ⋅ s − 1 / 2 e − 2 s k\cdot s^{-1/2} e^{-2\sqrt{s}} k⋅s−1/2e−2s
where
The difficulty with this ILT is that there is a branch point singularity at s=0 that a Bromwich contour must avoid. That is, we are tasked with evaluating
∮ C d s s − 1 / 2 e − 2 s e s t \oint_C ds \: s^{-1/2} e^{-2 \sqrt{s}} e^{s t} ∮Cdss−1/2e−2sest
where C is the following contour
We will define Argz∈(−π,π], so the branch is the negative real axis. There are 6 pieces to this contour, Ck, k∈{1,2,3,4,5,6}, as follows.
C1 is the contour along the line z∈[c−iR,c+iR] for some large value of R.
C2 is the contour along a circular arc of radius R from the top of C1 to just above the negative real axis.
C3 is the contour along a line just above the negative real axis between [−R,−ϵ] for some small ϵ.
C4 is the contour along a circular arc of radius ϵ about the origin.
C5 is the contour along a line just below the negative real axis between [−ϵ,−R].
C6 is the contour along the circular arc of radius R from just below the negative real axis to the bottom of C1.
We will show that the integral along C2,C4, and C6 vanish in the limits of R→∞ and ϵ→0.
On C2, the real part of the argument of the exponential is
R t cos θ − 2 R cos θ 2 R t \cos{\theta} - 2 \sqrt{R} \cos{\frac{\theta}{2}} Rtcosθ−2Rcos2θ
where θ∈[π/2,π). Clearly, cosθ<0 and cosθ2>0, so that the integrand exponentially decays as R→∞ and therefore the integral vanishes along C2.
On C6, we have the same thing, but now θ∈(−π,−π/2]. This means that, due to the evenness of cosine, the integrand exponentially decays again as R→∞ and therefore the integral also vanishes along C6.
On C4, the integral vanishes in the limit as ϵ→0. Thus, we are left with the following by Cauchy’s integral theorem (i.e., no poles inside C):
[ ∫ C 1 + ∫ C 3 + ∫ C 5 ] d s s − 1 / 2 e − s e s t = 0 \left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] ds \: s^{-1/2} e^{-\sqrt{s}} e^{s t} = 0 [∫C1+∫C3+∫C5]dss−1/2e−sest=0
On C3, we parametrize by s = e i π x s=e^{iπx} s=eiπx and the integral along C3 becomes
∫ C 3 d s s − 1 / 2 e − 2 s e s t = − i e i π ∫ ∞ 0 d x x − 1 / 2 e − i 2 x e − x t \int_{C_3} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = -i e^{i \pi} \int_{\infty}^0 dx \:x^{-1/2} e^{-i 2\sqrt{x}} e^{-x t} ∫C3dss−1/2e−2sest=−ieiπ∫∞0dxx−1/2e−i2xe−xt
On C5, however, we parametrize by s = e − i π x s=e^{−iπx} s=e−iπx and the integral along C5 becomes
∫ C 5 d s s − 1 / 2 e − 2 s e s t = i e − i π ∫ 0 ∞ d x x − 1 / 2 e i 2 x e − x t \int_{C_5} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = i e^{-i \pi} \int_0^{\infty} dx \: x ^{-1/2} e^{i 2 \sqrt{x}} e^{-x t} ∫C5dss−1/2e−2sest=ie−iπ∫0∞dxx−1/2ei2xe−xt
We may now write
1 i 2 π ∫ c − i ∞ c + i ∞ d s s − 1 / 2 e − 2 s e s t = 1 π ∫ 0 ∞ d x x − 1 / 2 cos ( 2 x ) e − x t \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = \frac{1}{\pi} \int_0^{\infty} dx \: x^{-1/2} \cos{(2 \sqrt{x})}\, e^{-x t}\\ i2π1∫c−i∞c+i∞dss−1/2e−2sest=π1∫0∞dxx−1/2cos(2x)e−xt
= 1 π ∫ − ∞ ∞ d u cos ( 2 u ) e − t u 2 ⎵ x = u 2 = \frac{1}{\pi} \underbrace{\int_{-\infty}^{\infty} du \: \cos{(2 u)} \, e^{-t u^2}}_{x=u^2} \\ =π1x=u2 ∫−∞∞ducos(2u)e−tu2
= 1 π ℜ [ ∫ − ∞ ∞ d u e − t u 2 e i 2 u ] = \frac{1}{\pi} \Re{\left [\int_{-\infty}^{\infty} du \: e^{-t u^2} e^{i 2 u} \right ]} \\ =π1ℜ[∫−∞∞due−tu2ei2u]
= e − 1 / t π ℜ [ ∫ − ∞ ∞ d u e − t ( u − i / t ) 2 ] = \frac{e^{-1/t}}{\pi} \Re{\left [\int_{-\infty}^{\infty} du \: e^{-t (u-i/t)^2} \right ]}\\ =πe−1/tℜ[∫−∞∞due−t(u−i/t)2]
= e − 1 / t π π t = \frac{e^{-1/t}}{\pi} \sqrt{\frac{\pi}{t}} =πe−1/ttπ
Therefore the ILT is
1 i 2 π ∫ c − i ∞ c + i ∞ d s s − 1 / 2 e − 2 s e s t = ( π t ) − 1 / 2 e − 1 / t \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = (\pi t)^{-1/2} e^{-1/t} i2π1∫c−i∞c+i∞dss−1/2e−2sest=(πt)−1/2e−1/t