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I was reading an article about pertubation in advection-transport equations, nad so they have defined the following equation with the perturbation ($\epsilon). $

$$y_t+-\epsilon .y_{xx}+ M.y_x=0\, ;(x,t) \in (0,1)\times(0,T)$$ $$y(0,t)=v(t), \text{if} \, M>0 \, ;t\in (0,T)$$

$$y(1,t)=0 \,\text{if}\,M<0 \, ;t\in (0,T)$$ $$y(x,0)=y_0(x) \, ; x\in (0,1)$$ And they took the case of$\epsilon=0 $, we will get the transport equation. $$y_t+M.y_x=0\, ;(x,t) \in (0,1)\times(0,T)$$ $$y(0,t)=v(t), \text{if} \, M>0 \, ;t\in (0,T)$$

$$y(1,t)=0 \,\text{if}\,M<0 \, ;t\in (0,T)$$ $$y(x,0)=y_0(x) \, ; x\in (0,1)$$

and said that we have two boundary layers :

1/ In $x=1$ of size $O(\epsilon)$

2/ In the characterestic $\{(x,t)\in(0,1)\times (0,T): x-M.t=0\}$ of size $O(\sqrt{\epsilon})$

And I did not get how they concluded these results.

BrianTag
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  • The system needs another boundary condition. (Or two more if $M=0$). – David Oct 09 '19 at 22:41
  • @David Well it's what I ve found in the article ; https://arxiv.org/pdf/1904.12669.pdf ; And if you can tell me how in general do we find those boudary layers. – BrianTag Oct 10 '19 at 08:32
  • In that article, the general equation (equation (1)) with nonzero $\epsilon$ has both boundary conditions, there is no dependence on $M$. It is only the weak formulation (equation (2)) with $\epsilon=0$ where there is only one boundary condition (and is chosen based on the sign of $M$). – David Oct 10 '19 at 22:03
  • @David yes, I understand, but how did they find the boundary layers ? because we need to solve equation and see where we might have a problem. and in this case why the problem is in the characterestion ${(x,t) ; x-Mt=0}$ and $x=1$ – BrianTag Oct 12 '19 at 16:49

1 Answers1

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This is a partial answer, the details are in the linked paper, and references therein, I'm not going to attempt to rewrite it.

With $\epsilon=0$ the equation is order 1 in space, so can't satisfy 2 general boundary conditions. To find the location/size of the boundary layer, you can rescale $\eta=(x-x_0)/\epsilon^\alpha$ to give $$ y_t-\epsilon^{1-2\alpha}y_{\eta\eta}+\epsilon^{-\alpha}My_\eta=0 $$ and so if $\alpha=1$ you have a dominant balance, so there is a boundary layer width $O(\epsilon)$.

If $M>1$ the boundary layer must be at $x_0=1$, and if $M<0$ it will be at $x_0=0$. The paper you linked to does go through these calculations in limited detail for the $M>0$ case.

Section 2 of the paper gives some more details about the different substitutions.

Also see this answer for general information about finding the location/width of boundary layers.

David
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  • 2 general boundary conditions ? which ones ?, because I think it doesn't satify only to $y(1,t)=0$ because transport equation, we only have initial condition, and one boundary condition for $x=0$. Correct me if I'm wrong. – BrianTag Oct 15 '19 at 05:50
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    Exactly, when $\epsilon=0$ you can only satisfy one boundary condition (unless you get lucky and the other one is satisfied by chance, that's what I mean by general). The actual boundary layer you will satisfy depends on $M$, at the other end of the domain will be a boundary layer. – David Oct 15 '19 at 05:55
  • Yes, well I'm having difficulties in determining boundary layers (exactly) and their length. in this case the article (page 3), said that we have boundary layers at $x=1$ (I agree because the boundary coundition is not satisfied) but how the length it's $\epsilon$ and why the characterestic ${x-Mt=0}$ is a layer of size $\sqrt{\epsilon}$ – BrianTag Oct 15 '19 at 06:10
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    My second paragraph gives the rescaled equations for a boundary layer of width $\epsilon^\alpha$, then dominant balance tells us that $\alpha=1$. If these concepts are unfamiliar to you, I suggest studying boundary layer theory from a book before coming back to this particular equation. – David Oct 15 '19 at 21:24