1. Eigenvalues in PDE

After some calculations in a PDE I have the following equations:

$\displaystyle X'' + \lambda X = 0$

$\displaystyle T' + \lambda k T = 0$

For the first equation, there are 3 possible cases:
(i) $\displaystyle \lambda$ is positive

(ii) $\displaystyle \lambda$ is negative

(iii) $\displaystyle \lambda$ equals 0.

I also have the boundary conditions $\displaystyle X'(0) = 0, X'(L) = 0$

I was hoping someone could answer some general questions I have about these cases.

First of all, by starting with case (iii) we have:

$\displaystyle X'' = 0 \ \Rightarrow \ X' = A \ \Rightarrow X = Ax + B$

Then with BC's I get

$\displaystyle X'(0) = 0 = A \ \Rightarrow \ \A = 0$

so $\displaystyle X = B$ is a solution.

Also, $\displaystyle T' = 0 \ \Rightarrow \ T = C$

so $\displaystyle u(x,t) = BC$......my question here is if $\displaystyle \lambda = 0$ leads to a solution as just shown, does this automatically mean that $\displaystyle a_0$ need to be included in my fourier series solution to the PDE?

Now for case (i):

$\displaystyle X'' + \lambda X = 0$ leads to solutions

$\displaystyle X = A \cos (\sqrt{\lambda} x) + B \sin (\sqrt{\lambda} x)$

$\displaystyle X' = - \sqrt{\lambda} A \sin (\sqrt{\lambda} x) + \sqrt{\lambda} B \cos (\sqrt{\lambda} x)$

so with BC's

$\displaystyle X'(0) = \sqrt{\lambda} B = 0 \ \Rightarrow B = 0$

$\displaystyle X'(L) = - \sqrt{\lambda} A \sin (\sqrt{\lambda} L) = 0$

Now $\displaystyle A = 0$ leads to a trivial solution, so

$\displaystyle \sqrt{\lambda} L = n \pi$ for $\displaystyle n = 0,1,...$

and $\displaystyle \lambda = \frac{n^2 \pi^2}{L^2}$

so $\displaystyle X_n = A \cos (\frac{n \pi x}{L})$

For the third case,

$\displaystyle X'' - \lambda X = 0$

$\displaystyle X = A \cosh (\sqrt{\lambda} x) + B \sinh (\sqrt{\lambda} x)$

$\displaystyle X' = \sqrt{\lambda} A \sinh (\sqrt{\lambda} x) + \sqrt{\lambda} B \cosh (\sqrt{\lambda} x)$

Now I get

$\displaystyle X'(0) = \sqrt{\lambda} B = 0 \ \Rightarrow \ B = 0$

$\displaystyle X'(L) = \sqrt{\lambda} A \sinh (\sqrt{\lambda} L) = 0$

Apparently this should be a trivial solution, but I don't understand how it is different from case (ii). I would get

$\displaystyle \lambda = \frac{n^2 \pi^2}{L^2}$

so $\displaystyle X_n = A \cosh (\frac{n \pi x}{L})$

What am I missing here? How can I have two values for $\displaystyle X_n$?

Thanks to anyone who can help me understand this.

2. From this equation
$\displaystyle X'(L) = \sqrt{\lambda} A \sinh (\sqrt{\lambda} L) = 0$

and as
$\displaystyle \sinh (\sqrt{\lambda} L) \ne 0$

if follows that
$\displaystyle A= 0.$

Pauls Online Notes : Differential Equations - Solving the Heat Equation

3. I have really confused myself now....

Originally Posted by zzzoak
From this equation
$\displaystyle X'(L) = \sqrt{\lambda} A \sinh (\sqrt{\lambda} L) = 0$

$\displaystyle X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0$

and as
$\displaystyle \sinh (\sqrt{\lambda} L) \ne 0$

if follows that $\displaystyle A= 0.$
So why can't $\displaystyle \sinh (\sqrt{- \lambda} L) = 0$?

The way I'm thinking is that for case 1, we have

$\displaystyle X'(L) = - \sqrt{\lambda} A \sin (\sqrt{\lambda} L) = 0$

and then we can have $\displaystyle \sin (\sqrt{\lambda} L) = 0$ for $\displaystyle \lambda = \frac{n^2 \pi^2}{L^2}$ because we are looking for non-trivial solutions.

And then in case 2, we have

$\displaystyle X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0$

Now $\displaystyle \sinh 0 = \sin 0 = 0$, and they only difference I can see is that $\displaystyle \lambda$ is positive in case 1 and negative in case 2...

so why can't you still end up with the possibility that $\displaystyle \sinh (\sqrt{- \lambda} L) = 0$? Is it because that $\displaystyle \lambda$ is negative?

Is it because perhaps $\displaystyle \sqrt{- \lambda } L \ne 0$?

4. Originally Posted by OliviaB
I have really confused myself now....

$\displaystyle X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0$

So why can't $\displaystyle \sinh (\sqrt{- \lambda} L) = 0$?

The way I'm thinking is that for case 1, we have

$\displaystyle X'(L) = - \sqrt{\lambda} A \sin (\sqrt{\lambda} L) = 0$

and then we can have $\displaystyle \sin (\sqrt{\lambda} L) = 0$ for $\displaystyle \lambda = \frac{n^2 \pi^2}{L^2}$ because we are looking for non-trivial solutions.

And then in case 2, we have

$\displaystyle X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0$

Now $\displaystyle \sinh 0 = \sin 0 = 0$, and they only difference I can see is that $\displaystyle \lambda$ is positive in case 1 and negative in case 2...

so why can't you still end up with the possibility that $\displaystyle \sinh (\sqrt{- \lambda} L) = 0$? Is it because that $\displaystyle \lambda$ is negative?

Is it because perhaps $\displaystyle \sqrt{- \lambda } L \ne 0$?
I think you are forgetting what the $\displaystyle \sinh(x)$ function is. First it only has ONE zero and it is at zero. It is not periodic like the sine function.

Remember $\displaystyle \displaystle \sinh(x)=\frac{e^{x}-e^{-x}}{2}$.

Remember $\displaystyle \displaystle \sinh(x)=\frac{e^{x}-e^{-x}}{2}$.