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I am reffering to this paper on page 9.

Consider the equation $$ \partial_t F-\partial_v(\partial_v+v)F=0,\qquad F_{| t=0}=F^0\tag{2} $$ where $F=F(t,v)$.

If we write the ansatz $F=\mu+\mu f$, where $$ \mu(v)=\frac{1}{\sqrt{2\pi}}e^{-v^2/2} $$ it is said that the equation $(2)$ above reads $$ \partial_t f+(-\partial_v + v)\partial_v f=0,\qquad f_{| t=0}=f^0\tag{14} $$

I do not understand how they get $(14)$ from $(2)$.

If I subtitute $F=\mu+\mu f$ into $(2)$, what I get is $$ \mu\partial_t f-\mu_{vv}-\mu-v\mu_v-\mu_{vv}f-\mu_v f-\mu_v f_v -\mu f_{vv} - \mu f-v\mu_v f-v\mu f_v=0 $$

selector
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1 Answers1

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As $\mu(v)$ is explicitly defined, you can calculate $\mu_v$ and $\mu_{vv}$ to further simplify the equation you got.

For example, you will find that $\mu_{vv} + \mu + v\mu_v = 0$ .

Hetebrij
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  • Ah, I see. But why do we have $\partial_v^*=(-\partial_v +v)$? – selector May 07 '23 at 12:43
  • I think, this is just integration by parts: $$ \langle\partial_v g,h\rangle_{L^2(\mu dv)}=\int_{-\infty}^\infty (\partial_v g)h\mu, dv=-\int_{-\infty}^{\infty}g(h_v\mu + h\mu_v), dv=\langle g,-\partial_v h+vh\rangle_{L^2(\mu dv)} $$ – selector May 07 '23 at 13:03
  • I haven't looked at the paper, but you can also get it without integration by parts: $\mu_v f_v + \mu f_{vv} = -v \mu f_v + \mu f_{vv} = \mu (-v + \partial_v) f_v = \mu (- v + \partial_v) \partial_v f$. – Hetebrij May 07 '23 at 13:17
  • Also note that you have $\mu_v f$, which I suppose is a typo. – Hetebrij May 07 '23 at 13:17