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In the PDE as below$$ \partial_t u - \frac{i}{\rho} ( - \partial_x^2 )^{\rho /2} u = 0 \;\;\;(t,x) \in \Bbb R^2 $$ How can I prove that $$ (- \partial_x ^2 )^{\rho/ 2} = \scr F ^{-1} | \xi |^\rho \scr F $$ where $\scr F$ is Fourier transform. Is this just definition?

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What is the meaning of $(-\partial_x^2)^{\frac{\rho}{2}}u$ ? –  doraemonpaul Oct 2 '12 at 9:42

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Using properties of the Fourier transform, we can write the Laplacian operator as a Fourier multiplier $$(-\Delta)f = [|\xi|^2\hat{f}(\xi)]^{\vee}$$ Generalizing this idea, define the operator $D_x^\alpha$ $$D_x^\alpha f = (-\Delta)^{\alpha/2}f = [|\xi|^\alpha\hat{f}(\xi)]^{\vee}$$ which is referred to as the homogeneous derivative or fractional Laplacian. The inverse is the Riesz potential.

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