Perturbation Methods: Asymptotic Matching Question

[Master1022]

Homework Statement

Consider the problem (shown below) where $y(0) = 0$ and $y(1) = 3$.
(a) Find a composite expansion of the problem
(b) Find a two-term composite expansion

Relevant Equations

Asymptotic matching

Hi,

I was working on a problem and I can't seem to make much progress with it. From a high level my steps are:
1. Use the naive method to find the outer solution which can satisfy one of the two boundary conditions
2. Introduce a boundary layer (via stretched coordinates) to find inner solution to satisfy other boundary condition
3. Match boundary layers to find parameters
4. Combine solutions and subtract overlap to find overall solution.

Question:
Consider the problem
\epsilon y'' + 6 \sqrt{x} y' - 3y = -3, for 0 < x < 1
where $y(0) = 0$ and $y(1) = 3$.
(a) Find a composite expansion of the problem
(b) Find a two-term composite expansion

Attempt:
I am not sure exactly what the question is asking for, but I am assuming it is asking for a perturbation series solution (I haven't come across those terms before).

At a high level, the pre-multiplication by $\epsilon$ suggests that there may be a boundary layer present in the solution.

1. Let us first start by using the naive method and define a series:
y(x) = y_{0} (x) + \epsilon y_{1} (x) + \epsilon ^2 y_{2} (x) + ... + \epsilon ^n y_{n} (x)

2. Substitute into the ODE:
\epsilon (y_{0} ''(x) + \epsilon y_{1} ''(x) + ...) + 6 \sqrt{x} (y_{0} '(x) + \epsilon y_{1} '(x) + ...) - 3 (y_{0} (x) + \epsilon y_{1} (x) + ...) = -3

3. Substitute into the initial conditions
y(0) = 0 \rightarrow 0 = y_{0} (0) + \epsilon y_{1} (0) + ...
y(1) = 3 \rightarrow 3 = y_{0} (1) + \epsilon y_{1} (1) + ...

4. Re-group and solve the sub-problems
I will take a look at the smallest power of $\epsilon$ problem ($\epsilon ^0$):
6 \sqrt{x} y_{0} ' - 3 y_{0} = -3
y_{0} (0) = 0
y_{0} (1) = 3
which I then solved (I hope) to give: $y_{0} (x) = y_{H} + y_{P} = A e^{\sqrt{x}} + 1$

5. Now apply one of the two boundary conditions:
y(0) = 0 \rightarrow 0 = A + 1 \rightarrow A = -1

We cannot also satisfy the other boundary condition at $x = 1$ and thus we should introduce a boundary layer (at $x = 1$)

6. Let us introduce stretched co-ordinates: $\bar x = \frac{1 - x}{\epsilon^{\alpha}}$
\frac{dy}{dx} = ( - \epsilon ^{- \alpha}) \cdot \left( \frac{dy}{d\bar{x}} \right)
\frac{d^2 y}{dx^2} = (\epsilon^{-2 \alpha}) \cdot \frac{d^2 y}{d \bar{x}^2}

7. Substitute into the ODE:
\epsilon (\epsilon^{-2 \alpha} (y_{0} ''(\bar{x}) + \epsilon y_{1} ''(\bar{x}) + ...) + 6 \sqrt{1 - \bar{x} \epsilon^{\alpha}} \cdot ( - \epsilon ^{-\alpha})(y_{0} '(\bar{x}) + \epsilon y_{1} '(\bar{x}) + ...) - 3 (y_{0} (\bar{x}) + \epsilon y_{1} (\bar{x}) + ...) = -3
which can be re-written as:
\epsilon ^{1 - 2\alpha} (y_{0} ''(x) + \epsilon y_{1} ''(x) + ...) - 6 \epsilon ^{-\alpha} (1 - \frac{1}{2} \bar{x} \epsilon ^{\alpha} + ...) (y_{0} '(\bar{x}) + \epsilon y_{1} '(\bar{x}) + ...) - 3 (y_{0} (\bar{x}) + \epsilon y_{1}(\bar{x}) + ...) = -3

8. Now balance the values of $\alpha$:
1 - 2 \alpha = \alpha \rightarrow \alpha = 1
1 - 2 \alpha = 0 \rightarrow \alpha = \frac{1}{2}
- \alpha = 0 \rightarrow \alpha = 0

We will go with $\alpha = 1$ to keep the terms 'balanced' (still not fully sure if this is correct

9. Now we can substitute our stretched co-ordinates into the boundary conditions
y(0) = 0 \rightarrow 0 = y_{0} ( \epsilon ^{-\alpha}) + \epsilon y_{1} ( \epsilon ^{-\alpha}) + ...
y(1) = 3 \rightarrow 3 = y_{0} (0) + \epsilon y_{1} (0) + ...

10. Now re-group and solve the sub-problems
Note that we mainly care about the boundary condition at x = 1, because that was the one that wasn't satisfied by our original (outer) solution

Let us look at order $\epsilon ^{-1}$:
y_{0} '' - 6 y_{0} '' = 0
y_{0} (0) = 3

which I then solved to get: $y_{0} (\bar{x}) = C + D e^{+6 \bar{x}}$

11. Applying stretched coordinate versions of boundary conditions to our boundary layer equation:
y_{0}(0) = 3 \rightarrow 3 = C + D \rightarrow C = 3 - D

and thus: $y_{0} (\bar{x}) = 3 + D (e^{+6 \bar{x}} - 1)$

12. Find $D$ by matching the solutions such that:
$\lim_{x \rightarrow 1} y_{0} (x) = \lim_{\bar{x} \rightarrow \infty} y_{0} (\bar{x})$

This leads to: $1 - e = 3 + D \cdot ( \infty - 1)$ which has no solution for $D$

I am not sure where I went wrong, so any help would be appreciated.

[Note:] I can provide more explanations if necessary, but this problem was already quite long in its current form, so I decided to note put every tiny bit of calculation.

 

[pasmith]

If you define y_i as the outer solutions, then you need to use a different letter for the inner solution; Y_i for example. Otherwise your matching section is difficult to follow.

The fact that you get a De^{6\tilde x} term at zeroth order means that at first order your inhomogenous left hand side will require a resonant term K\tilde x e^{6\tilde x}. So unless D = 0 this first-order correction will swamp the leading order term as \tilde{x} \to \infty. This is a problem.

To fully match the expansions, you must compute them to higher order and introduce an intermediate variable \eta = x\epsilon^{-\beta} = \epsilon^{-\beta}(1 - \epsilon\tilde {x}) for 0 < \beta < 1 and match orders of \epsilon at fixed \eta.

 

[Master1022]

Thanks @pasmith !

If you define y_i as the outer solutions, then you need to use a different letter for the inner solution; Y_i for example. Otherwise your matching section is difficult to follow.

Apologies, so if I use $Y_i$ for the inner solution (i.e. the boundary layer), then those limits should read:
\lim_{x \rightarrow 1} y(x) = \lim_{\bar{x} \rightarrow \infty} Y_{i} (x)

Is it clear where those limits come from? Our boundary layer is at $x = 1$ so we need our outer and inner solutions need to match at that location. $x \rightarrow 1$ means that $\bar{x} = \frac{1 - x}{\epsilon} \rightarrow \infty$ as $\epsilon \rightarrow 0$.

The fact that you get a De^{6\tilde x} term at zeroth order means that at first order your inhomogenous left hand side will require a resonant term K\tilde x e^{6\tilde x}. So unless D = 0 this first-order correction will swamp the leading order term as \tilde{x} \to \infty. This is a problem.

To fully match the expansions, you must compute them to higher order and introduce an intermediate variable \eta = x\epsilon^{-\beta} = \epsilon^{-\beta}(1 - \epsilon\tilde {x}) for 0 < \beta < 1 and match orders of \epsilon at fixed \eta.

Oh okay, so this problem cannot be solved at just the leading order?