Question

Separable PDEs...

Answer 1

\[k u_{xx} - u = u_{t}, k >0\]

Answer 2

If we assume that it's separable, we get\[kX''T-XT=XT'\]Divide by \(X,T\), we get\[\frac{kX''}{X}=\frac{T+T'}{T}\]

Answer 3

using u = XY

the pde is supposed to yield, kX''Y - XY = XY'

but it seems to me that it should be kX''Y - XY = 0

Answer 4

Why does the notation change? and why do we get T'?

Answer 5

I was using \(T\) since we're taking derivatives with respect to \(x\) and \(t\). And we have a \(T'\) since we take the derivative with respect to \(t\) in the right side.

Answer 6

lol ok. I overlooked that.

Answer 7

shouldn't it be:
\[\frac{ T' }{ T } = \frac{kX'' - X}{X} = -\lambda\]

Answer 8

I think that's equivalent to what I had, but you could do it that way if you're more comfortable with it.

Answer 9

ok so for T it would be the same method, resulting in:
\[T = c_{1}e^{-\lambda t}\]

what about X?

Answer 10

Well we get \[kX''=X-\lambda X\implies X=\frac{k}{1-\lambda}X''=\mu X''\]

Answer 11

If I remember correctly, the general solution for this is either \[\large e^{\sqrt\mu x}\]or\[\cos(\sqrt\mu x)+\sin(\sqrt\mu x)\]with the appropriate constants.

Answer 12

Sorry, the cos/sin is not an actual solution, and the first equation should be\[\large c_1e^{\sqrt\mu x}+c_2e^{-\sqrt\mu x}.\]

Answer 13

from:
\[X = \frac{k}{1-\lambda}X''\]
\[\frac{k}{1 - \lambda}X'' - X = 0\]

characteristic equation:
\[\frac{k}{1 - \lambda}m^2 - m = 0\]

Answer 14

Shouldn't that be \[\frac{k}{1-\lambda}m^2-1=0?\]

Answer 15

yes lol

Answer 16

But I think I'm supposed to find solutions for each case of:
(lambda - 1)/k = 0
(lambda - 1)/k = - alpha^2 < 0
and (lambda - 1)/k = -alpha^2 > 0

Answer 17

Probably a good idea.

Answer 18

but why am I using these three cases with the X'' coeficient and not something else?

Answer 19

What else would you be using?

Answer 20

because it's the term with lambda in it?

Answer 21

I guess that's what you could say. Since \(\lambda\) can vary in such a way that you get all those cases, you technically need to look at them. I would be most careful with the case when \(\lambda=1\) since then you start dividing by 0.

Answer 22

that helps!

but dividing by zero does not seem to be an issue because in the solutions it's rearranged that k is in the denominator and the coefficient, (lambda - 1)/k is on X rather than X''

Answer 23

That would certainly make things easier.

Answer 24

So the first case is fine.

2nd case:

(lambda - 1)/k = -alpha^2 < 0

so lambda = 1 - k*alpha^2
and therefore X'' - (alpha^2)X = 0

Answer 25

this should lead to
\[X = c_{4}\cosh(\alpha x) + c_{5}\sinh(\alpha x)\]

but I'm not sure how I would find this.

Answer 26

Sorry, I've had alot of questions. But you've been a great help.

Answer 27

That would come from the solution \[\large c_4e^{\sqrt{-\alpha^2}x}+c_5e^{-\sqrt{-\alpha^2}x}\]Since you have negatives in the square root you get some \(i\)'s, and after some manipulation, you get the \(\sinh\) and \(\cosh\).

Answer 28

Yes! It all seems so obvious now :)

You sir, are a god among men.

Answer 29

I think this is all the help I need for now. Thanks again!

Answer 30

You're welcome.