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Problem of the Week #202 - April 12, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
Here's this week's problem!

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Let $u$ be an $H^1(\Bbb R^d)$-solution of the semi-linear PDE

$$-\Delta u + au = b|u|^{\alpha}u\quad (a > 0,\, \alpha > 0,\, b\in \Bbb R)$$

Derive the Pohozaev identity

$$(d - 2)\int_{\Bbb R^d} \lvert \nabla_xu\rvert^2\, dx + da\int_{\Bbb R^d} \lvert u\rvert^2\, dx = \frac{2bd}{\alpha + 2}\int_{\Bbb R^d} \lvert u\rvert^{\alpha + 2}$$
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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
No one answered this week's problem. You can find my solution below.



Multiplying both sides of the PDE by $x\cdot \nabla u$ and integrating over $\Bbb R^d$ yields

$$\int_{\Bbb R^d} -\Delta u(x\cdot \nabla u)\, dx + a\int_{\Bbb R^d} \lvert u\rvert^2(x\cdot \nabla u)\, dx = b\int_{\Bbb R^d} \lvert u\rvert^\alpha u(x\cdot \nabla u)\, dx.$$

Now

$$\int_{\Bbb R^d} -\Delta u (x\cdot \nabla u)\, dx = -\int_{\Bbb R^d} u_{x_ix_j}x_ju_{x_j}\, dx $$
$$= \int_{\Bbb R^d} u_{x_i}(x_ju_{x_j})_{x_i}\, dx = \int_{\Bbb R^d} (u_{x_i}\delta_{ij}u_{x_i} + u_{x_i}x_ju_{x_jx_i})\, dx$$
$$=\int_{\Bbb R^d} \left(\lvert \nabla u\rvert^2 + \left(\frac{\lvert \nabla u\rvert^2}{2}\right)_{x_j}x_j\right)\, dx$$
$$= \left(1 - \frac{d}{2}\right)\int_{\Bbb R^d} \lvert \nabla u\rvert^2\, dx,$$

$$a\int_{\Bbb R^d}\lvert u\rvert^2(x\cdot \nabla u)\, dx = a\int_{\Bbb R^d} ux_ju_{x_j}\, dx = \frac{a}{2}\int_{\Bbb R^d} (\lvert u\rvert^2)_{x_j}x_j\, dx = \frac{-da}{2}\int_{\Bbb R^d} \lvert u\rvert^2\, dx,$$

$$b\int_{\Bbb R^d}\lvert u\rvert^\alpha u(x\cdot \nabla u)\, dx = b\int_{\Bbb R^d} \lvert u\rvert^\alpha ux_ju_{x_j}\, dx = b\int_{\Bbb R^d} \left(\frac{\lvert u\rvert^{\alpha+2}}{\alpha+2}\right)_{x_j}x_j\, dx = -\frac{db}{\alpha+2}\int_{\Bbb R^d}\lvert u\rvert^{\alpha+2}\, dx$$

Thus

$$\left(1 - \frac{d}{2}\right)\int_{\Bbb R^d} \lvert \nabla u\rvert^2\, dx - \frac{d}{2}\int_{\Bbb R^d} \lvert u\rvert^2\, dx = -\frac{db}{\alpha + 2}\int_{\Bbb R^d} \lvert u\rvert^{\alpha+2}\, dx$$

The result follows by multiplying both sides of this equation by $-2$.
 
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