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] \\ &\approx (-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] \\ &\approx (-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] \\ &\approx (-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] \\ &\approx (-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? and if it is incorrect, what should I do to get the correct solution? I would appreciate your help.
2 Answers
Your result seems incorrect in view of
LaplaceTransform[-(x1*DiracDelta[x3]*DiracDelta[x2] -
1/6*x1^3 DiracDelta''[x3]*DiracDelta[x2] -
1/6 x1^3*DiracDelta''[x2]*DiracDelta[x3]), {x1, x2, x3}, {s1, s2, s3}]
-(1/s1^2) + s2^2/s1^4 + s3^2/s1^4
I have another way to solve the problem by using the Taylor series of three variables and I am wondering whether it is correct or not: \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] \\ &= L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ L^{-1}_{x_{1}} \left[ \frac{-1}{a^2 + b^2 + c^2} + \frac{2}{a^2 + b^2 + c^2} \left[ a(s_{1} - a) + b(s_{2} - b) + c(s_{3} - c) \right] + \ldots \right] \right] \right] \\ & \approx L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-1}{a^2 + b^2 + c^2} + \frac{2}{a^2 + b^2 + c^2} \left[ a(s_{1} - a) + b(s_{2} - b) + c(s_{3} - c) \right] \right] \right] \right] \\ &\approx \frac{-1}{a^2 + b^2 + c^2} \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \\ & \ \ \ + \frac{2}{a^2 + b^2 + c^2} \Big[ a \left(\delta^{\prime}(x_{1}) \delta(x_{2}) \delta(x_{3}) - a \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \right) \\ & \ \ \ + b \left(\delta(x_{1}) \delta^{\prime}(x_{2}) \delta(x_{3}) - b \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \right) \\ & \ \ \ + c \left(\delta(x_{1}) \delta(x_{2}) \delta^{\prime}(x_{3}) - c \delta(x_{1}) \delta(x_{2}) \delta(x_{3}) \right) \Big], \end{align*} for arbitrary $a, b, c \in [0, \infty)$.
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$\begingroup$ What do you mean by $\approx$? $\endgroup$ Commented Jul 12, 2022 at 15:29
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$\begingroup$ $\approx$ means "approximation" because I took only the first two terms of the Taylor series of three variables. $\endgroup$ Commented Jul 12, 2022 at 15:36
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2$\begingroup$ Are you using Mthematica software for your question and answer? If not, this question should be deleted and possibly posted instead in Mathematics. $\endgroup$ Commented Jul 12, 2022 at 15:38
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$\begingroup$ This was posted on Mathematics SE before being posted here. $\endgroup$ Commented Jul 14, 2022 at 17:14
InverseLaplaceTransform[-1/(s1^2 + s2^2 + s3^2), {s1, s2, s3}, {x1, x2, x3}]results in-InverseLaplaceTransform[ InverseLaplaceTransform[Sin[Sqrt[s2^2 + s3^2] x1]/Sqrt[s2^2 + s3^2], s2, x2], s3, x3]. In any case, the final result should be symmetric in $x_1,x_2,x_3$. $\endgroup$