Separation of variables; PDEs

Hey guys, I'm having a bit of trouble understanding the working in my notes below (attatched picture just for context, but first line regarding X'' it corrcted below. I understand it until the point where it says that

when $\begin{array}{l} X'' - \alpha X = 0\\ Y'' + \alpha Y = 0 \end{array}$ and $\alpha \ne 0$ ,

the 2 equations have the general solution:

$\begin{array}{l} X(x) = A{e^{\sqrt \alpha x}} + B{e^{ - \sqrt \alpha x}}\\ Y(y) = C{e^{i\sqrt \alpha y}} + D{e^{ - i\sqrt \alpha y}} \end{array}$

Now I've started by putting $\alpha = {s^2}$ or $\alpha = - {s^2}$ to account for $\alpha$ being both positive or negative. From that I get 2 auxiliary equations: ${m^2} + {n^2} = 0$ and ${m^2} - {n^2} = 0$ each for both the X and Y equations.

However I can only achive
$X(x) = A{e^{\sqrt \alpha x}} + B{e^{ - \sqrt \alpha x}}$ when $\alpha = {s^2}$

and
$Y(y) = C{e^{i\sqrt \alpha y}} + D{e^{ - i\sqrt \alpha y}}$ when $\alpha = - {s^2}$

Can anyone help me understand where the other cases go?

ie.
$X(x)$ when $\alpha = - {s^2}$
and
$Y(y)$ when $\alpha = {s^2}$

Reply 1

ghostwalker

Original post by chris_d

However I can only achive
$X(x) = A{e^{\sqrt \alpha x}} + B{e^{ - \sqrt \alpha x}}$ when $\alpha = {s^2}$

and
$Y(y) = C{e^{i\sqrt \alpha y}} + D{e^{ - i\sqrt \alpha y}}$ when $\alpha = - {s^2}$

Perhaps I'm being thick, but don't both those equations arise when you have

\alpha = s^2

, not from one postive and one minus.

Reply 2

joostan

Original post by ghostwalker

Perhaps I'm being thick, but don't both those equations arise when you have

\alpha = s^2

, not from one postive and one minus.

I agree with this.

Reply 3

ghostwalker

Original post by joostan

I agree with this.

Thanks for the confirmation (PRSOM).

Reply 4

chris_d

Original post by ghostwalker

Perhaps I'm being thick, but don't both those equations arise when you have

\alpha = s^2

, not from one postive and one minus.

I;m not really sure :frown:

This is how much I understand this whole thing haha! These were my workings for the $Y(y) = C{e^{i\sqrt \alpha y}} + D{e^{ - i\sqrt \alpha y}}$

$Y'' + \alpha Y = 0$

let $\alpha = - {s^2}$ , therefore $s = i\sqrt \alpha$

$Y'' - {s^2}Y = 0$

$Y(y) = C{e^{sy}} + D{e^{ - sy}}$

then substituting $s = i\sqrt \alpha$

$Y(y) = C{e^{i\sqrt \alpha y}} + D{e^{ - i\sqrt \alpha y}}$

I'm not very good at all this...

Reply 5

ghostwalker

Original post by chris_d

...

Let me just check something first:

I was assuming your alpha was real and the same in both equations. Is that the case?

Reply 6

chris_d

Original post by ghostwalker

Let me just check something first:

I was assuming your alpha was real and the same in both equations. Is that the case?

In the case of the 2 equations, $\alpha$ is equal in both of them, though it can be zero, real, imaginary or complex. I understand the case where $\alpha = 0$ , it's the case where $\alpha \ne 0$ I don't really understand. Sorry if I'm not making any sense

Reply 7

ghostwalker

Original post by chris_d

I'm not an expert on pds's, but I think it's highly unlikely that alpha is anything other than real in this context - but I could be wrong.

I am getting confused myself looking at your working. I can't see the error, but "know" there is one. Sorry.

Reply 8

TeeEm

Original post by chris_d

please post a photo of the original PDE

separation of variables create F(x) = G(y)
therefore both are at most a constant positive, s²,
negative -s²
or zero

I need to see the PDE however

Reply 9

chris_d

Original post by ghostwalker

Aha, just like me then! I think I've reached a solution, though it's not one I'm very confident of. Here's the workings for X(x) (sorry it's so long) and I combined constants $A = {A_0} + {A_2}$ etc if thats ok?

$X'' - \alpha X = 0$
for $\alpha = {s^2}$ and so $s = \sqrt \alpha$

$\begin{array}{l} X'' - {s^2}X = 0\\ X(x) = {A_0}{e^{sx}} + {B_0}{e^{ - sx}}\\ X(x) = {A_0}{e^{\sqrt \alpha x}} + {B_0}{e^{ - \sqrt \alpha x}} \end{array}$

$X(x) = {A_0}{e^{sx}} + {B_0}{e^{ - sx}}$

$X(x) = {A_0}{e^{\sqrt \alpha x}} + {B_0}{e^{ - \sqrt \alpha x}}$

and for $\alpha = - {s^2}$ where $s = i\sqrt \alpha$

$X'' + {s^2}X = 0$

$X(x) = {A_1}\cos sx + {B_1}\sin sx$

$X(x) = \frac{{{A_1}}}{2}({e^{isx}} + {e^{ - isx}}) + \frac{{{B_1}}}{{2i}}({e^{isx}} - {e^{ - isx}})$

$X(x) = (\frac{{{A_1}}}{2} + \frac{{{B_1}}}{{2i}}){e^{isx}} + (\frac{{{A_1}}}{2} - \frac{{{B_1}}}{{2i}}){e^{ - isx}}$

$X(x) = {B_2}{e^{ - \sqrt \alpha x}} + {A_2}{e^{\sqrt \alpha x}}$

and combining the 2 cases:

$X(x) = {A_0}{e^{\sqrt \alpha x}} + {B_0}{e^{ - \sqrt \alpha x}} + {B_2}{e^{ - \sqrt \alpha x}} + {A_2}{e^{\sqrt \alpha x}}$

$X(x) = A{e^{\sqrt \alpha x}} + B{e^{ - \sqrt \alpha x}}$

Reply 10

joostan

Original post by chris_d

I could well be being thick, and I've not done much on PDEs, but surely if

\alpha

is permitted to be complex then there is no necessity to consider the cases of

\alpha=\pm s^2

?
If you introduce this I feel that you are trying to end up with 4 linearly independent solutions for the CF, but there can only be two for each.

Reply 11

chris_d

Original post by TeeEm

please post a photo of the original PDE

separation of variables create F(x) = G(y)
therefore both are at most a constant positive, s²,
negative -s²
or zero

I need to see the PDE however

Here's the pages in my notes regarding this :smile:

The first equation is the 2D form of Laplace, and I'm looking at Example 1

MAS252 Lecture Notes.jpg

Reply 12

TeeEm

Original post by chris_d

Here's the pages in my notes regarding this :smile:

The first equation is the 2D form of Laplace, and I'm looking at Example 1

MAS252 Lecture Notes.jpg

this is standard Laplace in solution in cartesian

what is the problem exactly?

(edited 9 years ago)

Reply 13

DFranklin

Original post by chris_d

Here's the pages in my notes regarding this :smile:

I think tis isn't terribly well explained, and one thing is basically "wrong". He says

\alpha

can be imaginary but then talks about

\alpha

being positive or negative, which is essentially meaningless.

I've certainly never seen the sep. of vars. done where you'd allow the constant to be imaginary. (This isn't to say it can't work, but it's not what you'd normally do, and the 2nd page of your notes do actually seem to assume it is real);.

So, if it's real and non-zero, note that what you have is:

\dfrac{X''}{X} = \alpha = -\dfrac{Y''}{Y}

I think a lot of confusion has come up because of trying to distinguish between the alpha > 0 and alpha < 0 cases, because the way we'd naturally write the solution to

X''=\alpha X

has a different form for alpha > 0 and alpha < 0.

But it is in fact easy to verify that if alpha is non-zero, then

X =A e^{\sqrt{\alpha} x} + Be^{-\sqrt{\alpha}x}

is the general solution to

X''=\alpha X

(and similarly

Y =C e^{\sqrt{-\alpha}y} + Be^{-\sqrt{-\alpha}y}

is the general solution to

Y''=-\alpha X

).

Note that finding solutions in terms sin and cos and converting back into exponentials is basically a waste of time here. The sin/cos solutions are generally derived by finding the exponential solution and then combining to get real solutions. You're just undoing that again.

There's a symmetry here between X and Y;

Reply 14

TeeEm

Original post by chris_d

Here's the pages in my notes regarding this :smile:

The first equation is the 2D form of Laplace, and I'm looking at Example 1

MAS252 Lecture Notes.jpg

I hope you solved the issue wit the PDE
(DFranklin explained some things already)

If this is not clear look at some more questions in the Laplace section of this link
http://madasmaths.com/archive/maths_booklets/advanced_topics/partial_differential_equations.pdf

you may want to also look at separation of variables for the wave equation or the diffussion/heat equation