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We consider a diffusion equation of the form $$ \frac{\partial F}{\partial t} = \frac{\partial }{\partial x} \!\cdot\! \left[ - \mathcal{F} (t , x) \right] \, $$ where ${ \mathcal{F} (t ,x) }$ is the flux vector of the diffusion. The initial conditions of the diffusion is given by $$ F (x , t \!=\! 0) = F_{0} (x) \, . $$ I am interested in the evolution of the rms of the function $F$ during diffusion. As a consequence, I naturally introduce the well-defined function $g$ as $$ g (t) = \sqrt{\int \!\! \mathrm{d} x \, \left[ F (x , t) \!-\! F_{0} (x) \right]^{2}} \, . $$ One can note that ${ g (t \!=\! 0) \!=\! 0 }$. I am interested in the behavior of $g$ for small times, so that I introduce the limited development $$ g(t) \simeq g_{0} \!+\! g_{1} t \!+\! g_{2} \frac{t^{2}}{2} \!+\! ... \, , $$ where we have $$ \begin{cases} \displaystyle g_{0} = g (t \!=\! 0) \, , \\ \displaystyle g_{1} = \frac{\mathrm{d} g}{\mathrm{d} t} \bigg|_{t = 0} \, , \\ \displaystyle g_{2} = \frac{\mathrm{d}^{2} g}{\mathrm{d} t^{2}} \bigg|_{t = 0} \, . \end{cases} $$ To compute the coefficient $g_{1}$, I naively applied the chain rule and the derivation under the integral sign to write $$ \frac{\mathrm{d} g}{\mathrm{d} t} = \frac{\displaystyle \int \!\! \mathrm{d} x \, \left[ F (x , t) \!-\! F_{0} (x) \right] \frac{\partial F}{\partial t}}{\displaystyle \sqrt{\int \!\! \mathrm{d} x \, \left[ F (x , t) \!-\! F_{0} (x) \right]^{2}}} \, . $$ However, one should note that this expression when evaluated in ${ t \!=\! 0 }$ is ill-defined, since it takes the form ${ \sim 0 / 0 }$. However, the coefficient $g_{1}$ should be a perfectly well-defined function, since the function $g$ is a perfectly regular function.

How should one proceed to write a correct expression for $g_{1}$ ?

(Additional question : I have the same issue with $g_{2}$ that I also need to evaluate...)

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  • $\begingroup$ Maybe you could just consider $\bar{g}:=g^2$ obtain the series and then return back to $g$. $\endgroup$ – RTJ Apr 13 '15 at 9:44
  • $\begingroup$ @CTNT Yes indeed ! Your suggestion is exactly the simple idea I had missed. With your notations, one obtains : $\bar{g}_{0} \!=\! 0$, $\bar{g}_{1} \!=\! 0$, $\bar{g}_{2} \!>\! 0$, so that $g_{1} \!=\! \sqrt{\bar{g}_{2} / 2} $. $\endgroup$ – jibe Apr 13 '15 at 11:23

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