0
$\begingroup$

$\epsilon \frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=0, y(0) = \alpha, y(1) = \beta$

Since we have the general form:

$\epsilon \frac{d^2y}{dx^2}+a(x)\frac{dy}{dx}+b(x)y=0$

we can see that $a(x) = 2 >0$ so our boundary layer is near $x=0$.

Outer Solution:

Setting $\epsilon = 0$ we get

$2y' + 2y = 0, y(1) = \beta$

$y(x) = \beta e^{1-x}$

Inner Solution:

Here I am unsure how to make the following transformation

let $y(x) = Y(X)$, $x = \delta X$, $\delta = \delta(\epsilon)$

Following my lecture notes they went from

$\epsilon y'' + x^2y' - y = 0$ -> $\frac{\epsilon}{\delta^2}Y'' + \delta X^2Y' - Y = 0$

$\endgroup$
2
$\begingroup$

What you are doing is going from $x$, which when changed a little bit in the boundary layer makes $y$ change a lot, to a variable $X$, which is like $x$, but stretched out so that $y$ doesn't change as quickly.

So, you have $x=\delta(\epsilon)X$, so you have to work out how the derivatives change. Use the chain rule $$ \frac{\mathrm d}{\mathrm dx}=\frac{\mathrm dX}{\mathrm dx}\frac{\mathrm d}{\mathrm dX}=\frac{1}{\delta}\frac{\mathrm d}{\mathrm dX}. $$ So the new equation is $$\frac{\epsilon}{\delta^2}Y''(X)+\frac{2}{\delta}Y'(X)+2Y=0 $$ and the boundary condition is $Y(0)=0$. You next have to use the method of dominant balance to work out what $\delta(\epsilon)$ should be.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.