Abdulhameed Qahtan Abbood Altai

2022-07-11

I want to find the inverse triple Laplace transform of $L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right]$. I did
\begin{align*}
L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] &= L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[\frac{1}{a} L^{-1}_{x_{1}} \left[ \frac{a}{s^2_{1} + a^2} \right] \right] \right], \ \  a^2 = s^2_{2} + s^2_{3}
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[\frac{ \sin \left( x_{1} \sqrt{ \left( s^2_{2} + s^2_{3}\right)} \right) }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}}  \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \frac{ \displaystyle\sum_{k = 0}^{\infty} \frac{(-1)^k \left(x_{1} \sqrt{s^2_{2} + s^2_{3}} \right)^{2k+1}}{(2k+1)!} }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&\approx (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \frac{ x_{1} \sqrt{s^2_{2} + s^2_{3}} - \frac{1}{6} \left(x_{1} \sqrt{s^2_{2} + s^2_{3}} \right)^3 }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ x_{1} - \frac{1}{6} x_{1}^3 \left( s^2_{2} + s^2_{3} \right) \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \left( x_{1} - \frac{1}{6} x_{1}^3 s^2_{3} \right) - \frac{1}{6} x_{1}^3 s^2_{3} \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ \left( x_{1} - \frac{1}{6} x_{1}^3 s^2_{3} \right) \delta(x_{2}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{2}) \right]
\\
&= (-1) \left( \left( x_{1} \delta(x_{3}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{3}) \right) \delta(x_{2}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{2}) \delta(x_{3}) \right)
\end{align*}
I am wondering if this solution is correct or not? I would appreciate your help.

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