# Perturbation question

• Aug 24th 2009, 06:36 AM
bubblygirl90
Perturbation question

y" + (1+ey)y = 0. Subject to y(0)=1, y'(0)=0

I've used the expansion y(x)=y0(x) + ey1(x) + ...

when i differentiate this twice and substitute into my eqn and collect powers of e i get two equations:

y0" + y0 = 0
y1" + y1 = 0

I think this is wrong and not really sure what to do.

Thanks for any help (and sorry about the poor layout of the problem)
• Aug 24th 2009, 09:49 AM
chisigma
The DE has the form...

$y^{''} = f(y)$ (1)

... and its solution is...

$x= \int \frac{dy}{\sqrt {2 \phi(y) + c_{1}}} + c_{2}$ (2)

... where...

$\phi(y) = \int f(y)\cdot dy$ (3)

Here we have...

$f(y)= - y\cdot (1+e^{y})$ (4)

... so that is...

$x= \int \frac{dy}{\sqrt{- y^{2} - 2y e^{y} +2 e^{y} + c_{1}}} + c_{2}$ (5)

The integral in (5) has no elementary approach (Thinking) ...

Kind regards

$\chi$ $\sigma$
• Aug 24th 2009, 03:18 PM
Jester
Quote:

Originally Posted by bubblygirl90

y" + (1+ey)y = 0. Subject to y(0)=1, y'(0)=0

I've used the expansion y(x)=y0(x) + ey1(x) + ...

when i differentiate this twice and substitute into my eqn and collect powers of e i get two equations:

y0" + y0 = 0
y1" + y1 = 0

I think this is wrong and not really sure what to do.

Thanks for any help (and sorry about the poor layout of the problem)

If $y'' + y + \varepsilon y^2 = 0$, then substituting $y = y_0 + \varepsilon y_1 + O\left(\varepsilon^2\right)$ gives

$
y''_0 + \varepsilon y''_1 + y_0 + \varepsilon y_1 + \varepsilon \left(y_0 + \varepsilon y_1 \right)^2 + O\left(\varepsilon^2\right) = 0
$

or

$
O(1)\;\;\;y''_0 + y_0 = 0
$

$
O(\varepsilon)\;\;\;y''_1 + y_1 + y^2_0 = 0
$

with ICs $y_0(0) = 1, y'_0(0) = 0, y_1(0) = 0, y'_1(0) = 0,$.

The first one is easy - soln $y_0 = \cos x$. Use this in the second and solve for $y_1$.
• Aug 25th 2009, 03:36 AM
bubblygirl90
Thank you Danny, that really helped me out a lot!
I'm just having some trouble with this last bit now.

If i have: y1" + y1 = -(cosx)^2

then would the homogeneous solution be cos x? or A + B(e^x)?
and also I'm completely unsure how to find the particular solution (Worried)
• Aug 25th 2009, 04:11 AM
bubblygirl90
Quote:

Originally Posted by Danny
If $y'' + y + \varepsilon y^2 = 0$, then substituting $y = y_0 + \varepsilon y_1 + O\left(\varepsilon^2\right)$ gives

$
y''_0 + \varepsilon y''_1 + y_0 + \varepsilon y_1 + \varepsilon \left(y_0 + \varepsilon y_1 \right)^2 + O\left(\varepsilon^2\right) = 0
$

or

$
O(1)\;\;\;y''_0 + y_0 = 0
$

$
O(\varepsilon)\;\;\;y''_1 + y_1 + y^2_0 = 0
$

with ICs $y_0(0) = 1, y'_0(0) = 0, y_1(0) = 0, y'_1(0) = 0,$.

The first one is easy - soln $y_0 = \cos x$. Use this in the second and solve for $y_1$.

Thank you so much ive got an answer which when i check it it works :)
y = cos(x) + e( cos(x) - 1/2 - 1/6 cos(2x) )

The only thing is i havent used the initial conditions anywhere other than working out the solution of y0 = cos(x).. do i need them for working out the value of epsilon?
• Aug 25th 2009, 04:40 AM
Jester
Quote:

Originally Posted by bubblygirl90
Thank you so much ive got an answer which when i check it it works :)
y = cos(x) + e( cos(x) - 1/2 - 1/6 cos(2x) )

The only thing is i havent used the initial conditions anywhere other than working out the solution of y0 = cos(x).. do i need them for working out the value of epsilon?

$y_1 = c_1 \cos x + c_2 \ sin x - \frac{1}{2} - \frac{1}{6} \cos 2x$

Here you'd use the IC's.
• May 29th 2010, 02:23 PM
russel
Another perturbation question
I was going to open a new thread, but I'll post my question here! Does anybody know what singular perturbation methods we could apply and what will get for the bvp problem below?

$\varepsilon y''+(t-\frac{1}{2})y=0$
$0
$0<\varepsilon<<1$
with boundary values
$y(0)=1$
$y(1)=2$