Can anyone help me get this Laplace transform, L[(f''(x))^n] where f'(0)=0 and f''(0)=0 and $n$ is power of f''(x)?
Laplace transform of $[f''[x]]^n$
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1Whether or not relevant, it is interesting to note the following (not too well-known) result: the Laplace transform of the product $f_1 (t) f_2 (t)$ is equal to the convolution of $F_1 (s)$ and $F_2 (s)$. See http://www.atp.ruhr-uni-bochum.de/rt1/syscontrol/node145.html#sec:AConvolutionInFrequencyDomain – 2011-04-13
2 Answers
Hint: use integration by parts.
See here for a full solution.
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0He means the n-fold convolution (on a vertical line in the complex plane). – 2011-04-13
I don't think there is a simple enough closed form (i.e. the case f''(x)^{10} would be a sucession of integrals involving products and sums of the powers of all derivatives from $10$ to $0$ (the original function).
Integrating by parts gives:
\mathcal{L}\left\{ {f{{\left( t \right)}^n}} \right\} = \frac{{f{{\left( 0 \right)}^n}}}{s} + \frac{n}{s}\int\limits_0^\infty {{e^{ - st}}f{{\left( t \right)}^{n - 1}}f'\left( t \right)dt}
But then you'd need a new formula for
\mathcal{L}\left\{ {f{{\left( t \right)}^{n - 1}}f'\left( t \right)} \right\}\left( s \right)
Which would be, if I'm not mistaken
L\left\{ {f{{\left( t \right)}^{n - 1}}f'\left( t \right)} \right\}\left( s \right) = \frac{1}{s}f{\left( 0 \right)^{n - 1}}f'\left( 0 \right) + \frac{{n - 1}}{s}L\left\{ {f{{\left( t \right)}^{n - 2}}f'{{\left( t \right)}^2}} \right\}\left( s \right) + \frac{1}{s}L\left\{ {f{{\left( t \right)}^{n - 1}}f''\left( t \right)} \right\}\left( s \right)
And now you'd need one for the two new arguments. The recursion would be rather chaotic.
EDITED: Didn't read question properly.