Solving Asymptotic Matching for LaTeX Boundary Layer

  • Thread starter Thread starter rsq_a
  • Start date Start date
Click For Summary
The discussion revolves around finding the correct matching condition for a boundary-layer problem in the context of asymptotic analysis. The outer solution is given as \sim x^{1/4}, with the boundary layer thickness defined as x = O(\epsilon). The inner solution, when re-scaled as x = \epsilon X, is expected to behave like y \sim \epsilon^{1/4} X^{1/4} as X approaches infinity. However, concerns are raised about the inner solution potentially blowing up, indicating a need for clarification on the appropriate matching condition. The thread seeks guidance on resolving this apparent contradiction in the asymptotic matching process.
rsq_a
Messages
103
Reaction score
1
There seems to be something wrong with the LaTeX

This seems like an absolutely trite question, but I can't seem to figure it out.

Suppose you had a boundary-layer problem, say at x = 0. Suppose the first term of the outer solution, valid away from x = 0 was \sim x^{1/4}. Suppose that the boundary layer was of thickness x = O(\epsilon).

Suppose that you have solved for the inner solution near x = 0. What would be the required matching condition?

So in this example, we would re-scale x = \epsilon X. Then wouldn't the inner solution need to behave like, y \sim \epsilon^{1/4} X^{1/4} as X \to \infty? But this doesn't seem possible, since we can't allow the inner solution to blow up. What is the correct matching condition as X \to \infty?
 
Last edited:
Physics news on Phys.org
I've amended your equations, so they are viewable.

This seems like an absolutely trite question, but I can't seem to figure it out.

Suppose you had a boundary-layer problem, say at x = 0. Suppose the first term of the outer solution, valid away from x = 0 was \sim x^{1/4} . Suppose that the boundary layer was of thickness x = O(\epsilon) .

Suppose that you have solved for the inner solution near x = 0. What would be the required matching condition?

So in this example, we would re-scale x = \epsilon X . Then wouldn't the inner solution need to behave like, y \sim \epsilon^{1/4} X^{1/4} as X \rightarrow \infty ? But this doesn't seem possible, since we can't allow the inner solution to blow up. What is the correct matching condition X\rightarrow\infty ?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
5K