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复杂的拉普拉斯逆变换

發布時間：2023/12/10 编程问答 23 豆豆

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Laplace Inverse Transformation (LIT)
$L(1πte?a2/4t)=e?ass,a>0\mathcal{L}(\frac{1}{\sqrt{\pi t}}\text{e}^{-a^2/4t})=\frac{\text{e}^{-a\sqrt{s}}}{\sqrt{s}},~~ a>0$
$k?s?1/2e?2sk\cdot s^{-1/2} e^{-2\sqrt{s}}$
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
$∮Cds s?1/2e?2sest\oint_C ds \: s^{-1/2} e^{-2 \sqrt{s}} e^{s t}$
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,??] for some small ?.

C4 is the contour along a circular arc of radius ? about the origin.

C5 is the contour along a line just below the negative real axis between [??,?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 ?→0.

On C2, the real part of the argument of the exponential is
$\cos{\theta} - 2 \sqrt{R} \cos{\frac{\theta}{2}}$
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 ?→0. Thus, we are left with the following by Cauchy’s integral theorem (i.e., no poles inside C):
$[∫C1+∫C3+∫C5]ds s?1/2e?sest=0\left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] ds \: s^{-1/2} e^{-\sqrt{s}} e^{s t} = 0$

On C3, we parametrize by $s=e^{iπx}$ and the integral along C3 becomes
$∫C3ds s?1/2e?2sest=?ieiπ∫∞0dx x?1/2e?i2xe?xt\int_{C_3} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = -i e^{i \pi} \int_{\infty}^0 dx \:x^{-1/2} e^{-i 2\sqrt{x}} e^{-x t}$
On C5, however, we parametrize by $s=e^{?iπx}$ and the integral along C5 becomes

$∫C5ds s?1/2e?2sest=ie?iπ∫0∞dx x?1/2ei2xe?xt\int_{C_5} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = i e^{-i \pi} \int_0^{\infty} dx \: x ^{-1/2} e^{i 2 \sqrt{x}} e^{-x t}$
We may now write
$1i2π∫c?i∞c+i∞ds s?1/2e?2sest=1π∫0∞dx x?1/2cos?(2x) e?xt\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = \frac{1}{\pi} \int_0^{\infty} dx \: x^{-1/2} \cos{(2 \sqrt{x})}\, e^{-x t}\\$
$\frac{1}{\pi} \underbrace{\int_{-\infty}^{\infty} du \: \cos{(2 u)} \, e^{-t u^2}}_{x=u^2} \\$
$\frac{1}{\pi} \Re{\left [\int_{-\infty}^{\infty} du \: e^{-t u^2} e^{i 2 u} \right ]} \\$
$\frac{e^{-1/t}}{\pi} \Re{\left [\int_{-\infty}^{\infty} du \: e^{-t (u-i/t)^2} \right ]}\\$
$\frac{e^{-1/t}}{\pi} \sqrt{\frac{\pi}{t}}$

Therefore the ILT is
$1i2π∫c?i∞c+i∞ds s?1/2e?2sest=(πt)?1/2e?1/t\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: s^{-1/2} e^{-2\sqrt{s}} e^{s t} = (\pi t)^{-1/2} e^{-1/t}$

總結

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