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 $$f''(x)$$?
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Hint: use integration by parts. See here for a full solution. |
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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. |
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