 #1
Master1022
 611
 116
 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 twoterm 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
[tex] \epsilon y'' + 6 \sqrt{x} y'  3y = 3, for 0 < x < 1 [/tex]
where ##y(0) = 0## and ##y(1) = 3##.
(a) Find a composite expansion of the problem
(b) Find a twoterm 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 premultiplication 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:
[tex] y(x) = y_{0} (x) + \epsilon y_{1} (x) + \epsilon ^2 y_{2} (x) + ... + \epsilon ^n y_{n} (x) [/tex]
2. Substitute into the ODE:
[tex] \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 [/tex]
3. Substitute into the initial conditions
[tex] y(0) = 0 \rightarrow 0 = y_{0} (0) + \epsilon y_{1} (0) + ... [/tex]
[tex] y(1) = 3 \rightarrow 3 = y_{0} (1) + \epsilon y_{1} (1) + ... [/tex]
4. Regroup and solve the subproblems
I will take a look at the smallest power of ##\epsilon## problem (##\epsilon ^0 ##):
[tex] 6 \sqrt{x} y_{0} '  3 y_{0} = 3 [/tex]
[tex] y_{0} (0) = 0 [/tex]
[tex] y_{0} (1) = 3 [/tex]
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:
[tex] y(0) = 0 \rightarrow 0 = A + 1 \rightarrow A = 1 [/tex]
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 coordinates: ## \bar x = \frac{1  x}{\epsilon^{\alpha}} ##
[tex] \frac{dy}{dx} = (  \epsilon ^{ \alpha}) \cdot \left( \frac{dy}{d\bar{x}} \right) [/tex]
[tex] \frac{d^2 y}{dx^2} = (\epsilon^{2 \alpha}) \cdot \frac{d^2 y}{d \bar{x}^2} [/tex]
7. Substitute into the ODE:
[tex] \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 [/tex]
which can be rewritten as:
[tex] \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 [/tex]
8. Now balance the values of ## \alpha ##:
[tex] 1  2 \alpha = \alpha \rightarrow \alpha = 1 [/tex]
[tex] 1  2 \alpha = 0 \rightarrow \alpha = \frac{1}{2} [/tex]
[tex]  \alpha = 0 \rightarrow \alpha = 0 [/tex]
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 coordinates into the boundary conditions
[tex] y(0) = 0 \rightarrow 0 = y_{0} ( \epsilon ^{\alpha}) + \epsilon y_{1} ( \epsilon ^{\alpha}) + ... [/tex]
[tex] y(1) = 3 \rightarrow 3 = y_{0} (0) + \epsilon y_{1} (0) + ... [/tex]
10. Now regroup and solve the subproblems
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} ##:
[tex] y_{0} ''  6 y_{0} '' = 0 [/tex]
[tex] y_{0} (0) = 3 [/tex]
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:
[tex] y_{0}(0) = 3 \rightarrow 3 = C + D \rightarrow C = 3  D [/tex]
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.
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
[tex] \epsilon y'' + 6 \sqrt{x} y'  3y = 3, for 0 < x < 1 [/tex]
where ##y(0) = 0## and ##y(1) = 3##.
(a) Find a composite expansion of the problem
(b) Find a twoterm 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 premultiplication 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:
[tex] y(x) = y_{0} (x) + \epsilon y_{1} (x) + \epsilon ^2 y_{2} (x) + ... + \epsilon ^n y_{n} (x) [/tex]
2. Substitute into the ODE:
[tex] \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 [/tex]
3. Substitute into the initial conditions
[tex] y(0) = 0 \rightarrow 0 = y_{0} (0) + \epsilon y_{1} (0) + ... [/tex]
[tex] y(1) = 3 \rightarrow 3 = y_{0} (1) + \epsilon y_{1} (1) + ... [/tex]
4. Regroup and solve the subproblems
I will take a look at the smallest power of ##\epsilon## problem (##\epsilon ^0 ##):
[tex] 6 \sqrt{x} y_{0} '  3 y_{0} = 3 [/tex]
[tex] y_{0} (0) = 0 [/tex]
[tex] y_{0} (1) = 3 [/tex]
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:
[tex] y(0) = 0 \rightarrow 0 = A + 1 \rightarrow A = 1 [/tex]
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 coordinates: ## \bar x = \frac{1  x}{\epsilon^{\alpha}} ##
[tex] \frac{dy}{dx} = (  \epsilon ^{ \alpha}) \cdot \left( \frac{dy}{d\bar{x}} \right) [/tex]
[tex] \frac{d^2 y}{dx^2} = (\epsilon^{2 \alpha}) \cdot \frac{d^2 y}{d \bar{x}^2} [/tex]
7. Substitute into the ODE:
[tex] \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 [/tex]
which can be rewritten as:
[tex] \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 [/tex]
8. Now balance the values of ## \alpha ##:
[tex] 1  2 \alpha = \alpha \rightarrow \alpha = 1 [/tex]
[tex] 1  2 \alpha = 0 \rightarrow \alpha = \frac{1}{2} [/tex]
[tex]  \alpha = 0 \rightarrow \alpha = 0 [/tex]
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 coordinates into the boundary conditions
[tex] y(0) = 0 \rightarrow 0 = y_{0} ( \epsilon ^{\alpha}) + \epsilon y_{1} ( \epsilon ^{\alpha}) + ... [/tex]
[tex] y(1) = 3 \rightarrow 3 = y_{0} (0) + \epsilon y_{1} (0) + ... [/tex]
10. Now regroup and solve the subproblems
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} ##:
[tex] y_{0} ''  6 y_{0} '' = 0 [/tex]
[tex] y_{0} (0) = 3 [/tex]
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:
[tex] y_{0}(0) = 3 \rightarrow 3 = C + D \rightarrow C = 3  D [/tex]
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.