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Use singular perturbation techniques to find the leading order uniform approximation to the solution to the boundary value problem

$$\epsilon y'' + \frac{2 \epsilon}{t} y'-y=0$$

$0<t<1$ and $y(0)=0 , y'(1)=1$

This has a boundary layer near $t=1$

I am having trouble figuring out how to compute the inner approximation. I am used to doing these problems with $t=0$ as the boundary layer.

Using these methods, I get the outer approximation to be $y_0=0$, which I know is correct, but I'm at a loss on how to proceed with the inner approximation. I would appreciate if someone could work out the inner layer so that I could see how it is done.

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    $\begingroup$ Let $x=1+\epsilon^\alpha X$ ($\alpha>0$) and $Y(X)=y(x)$, and re-write the equation in terms of $Y$ and $X$. Is that enough to get you going? $\endgroup$ – David Dec 2 '15 at 5:36
  • $\begingroup$ Yep. I started the ball rolling on that, now I'm stuck at trying to figure out my balancing. I want to say I need $$\epsilon ^ \frac{1}{2}$$, but I'm not sure. $\endgroup$ – MC989 Dec 2 '15 at 6:18
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    $\begingroup$ Yep $\alpha=1/2$ does give a dominant balance. $\endgroup$ – David Dec 2 '15 at 7:48
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First, something about the boundary conditions. The general solution to the ODE is \begin{equation} y(t) = c_1 \frac{1}{t}e^{\frac{t}{\sqrt{\epsilon}}}+ c_2 \frac{1}{t}e^{-\frac{t}{\sqrt{\epsilon}}}, \qquad (1) \end{equation} which has a singularity at $t=0$ for general $c_{1,2}$. Furthermore, you can show that, regardless of approximation techniques, the only solution obeying $y(0) = 0$ is the trivial solution. However, the solution for which $y'(1)=1$ and which doesn't blow up at $t=0$ is exponentially small in $\sqrt{\epsilon}$ near the origin. To summarise: you should take $y(0) = 0$ with a grain of salt, it should be $y(0) = \mathcal{O}(e^{-\frac{1}{\sqrt{\epsilon}}})$. The value of $y$ at $0$ is smaller than any (positive) power of $\epsilon$ though, so from the singular perturbation point of view, it's indistinguishable from $0$.

That being said, your approach is correct. The leading order equation gives $y = 0$, so you're correct to take that as an outer approximation. Then, following @David's advice, you can look at values of $t$ close to $1$, i.e. write \begin{equation} t = 1 + \epsilon^{\alpha} T, \end{equation} for some as yet undetermined $\alpha > 0$, where $T$ is now the free variable. You can invert this equation to get an expression for $T$ in terms of $t$, yielding \begin{equation} T(t) = \frac{1-t}{\epsilon^{\alpha}}. \end{equation} Now, again following @David, you would like to study how the solution depends on this 'zoomed in' variable $T$. So, you write \begin{equation} y(t) = Y(T(t)) = Y\left(\epsilon^{-\alpha}(1-t)\right), \end{equation} and substitute that in the ODE. The result is \begin{equation} \epsilon^{1-2\alpha} \frac{\text{d}^2 Y}{\text{d} T^2} + \frac{2 \epsilon^{1-\alpha}}{1+\epsilon^{\alpha} T} \frac{\text{d} Y}{\text{d} T} - Y = 0. \qquad (2) \end{equation} Since $\alpha > 0$ (and $0 < \epsilon \ll 1$), the term $\epsilon^{1-2\alpha}$ is larger than the term $\epsilon^{1-\alpha}$, so dominant balance implies that the term $\epsilon^{1-2\alpha} \frac{\text{d}^2 Y}{\text{d} T^2}$ should be of the same order as the third term, $-Y$. This implies that $\alpha = \frac{1}{2}$; note that this choice of time scaling is also present in the general solution to the ODE (1).

Substituting $\alpha = \frac{1}{2}$ in the rescaled ODE (2), we obtain \begin{equation} \frac{\text{d}^2 Y}{\text{d} T^2} - Y + \mathcal{O}(\sqrt{\epsilon}) = 0. \end{equation} This leading order equation is easy to solve; we obtain \begin{equation} Y(T) = c_1 e^{T} + c_2 e^{-T}. \end{equation} Now the only thing left is to incorporate the boundary conditions. That I leave up to you, but keep in mind that the boundary conditions are given in terms of $t$, instead of $T$. If you can reformulate the boundary conditions in terms of $T$, you should be able to solve the layer problem for $Y$ and then give a leading order expression for the solution $y$.

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