Eigenvalues in PDE

**OliviaB** · December 7th, 2010, 11:43 PM

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

$X'' + \lambda X = 0$

$T' + \lambda k T = 0$

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

(ii) $\lambda$ is negative

(iii) $\lambda$ equals 0.

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

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

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

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

Then with BC's I get

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

so $X = B$ is a solution.

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

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

Now for case (i):

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

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

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

so with BC's

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

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

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

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

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

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

For the third case,

$X'' - \lambda X = 0$

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

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

Now I get

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

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

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

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

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

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

Thanks to anyone who can help me understand this.

**zzzoak** · December 9th, 2010, 04:07 PM

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

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

if follows that
$ A= 0. $

Please read here (example 4):
Pauls Online Notes : Differential Equations - Solving the Heat Equation

**OliviaB** · December 9th, 2010, 08:07 PM

I have really confused myself now....

Originally Posted by zzzoak

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

First of all, I made a mistake. I should have had:

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

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

if follows that $A= 0.$

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

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

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

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

And then in case 2, we have

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

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

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

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

**TheEmptySet** · December 9th, 2010, 08:22 PM

Originally Posted by OliviaB

I have really confused myself now....

First of all, I made a mistake. I should have had:

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

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

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

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

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

And then in case 2, we have

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

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

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

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

I think you are forgetting what the $\sinh(x)$ function is. First it only has ONE zero and it is at zero. It is not periodic like the sine function.

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

Does this answer your question?

**OliviaB** · December 10th, 2010, 01:04 AM

Originally Posted by TheEmptySet

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

Does this answer your question?

Bingo. Thank-you... I wasn't thinking.

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