Eigenfunctions from eigenvalues unsure

WtKemper
Messages
6
Reaction score
0

Homework Statement


using X''(x)+ lambda*X(x)=0 find the eigenvalues and eigenfunctions accordingly.
Use the case lambda=0, lambda=-k2, lambda=k2
where k>0

Homework Equations


X(0)=0, X'(1)+X(1)=0


The Attempt at a Solution


I know that for lambda=0
X(x)=C1x+C2
which applying the conditions gives no E.V.

also for lambda=-k2
X(x)=C1cosh(kx)+C2sinh(kx)
and applying the conditions gives no E.V.

for the final case lambda=k2
X(x)=C1cos(kx)+C2sin(kx)

using X(0)=C1=0

X'(x)=C2kcos(kx)

applying second condition then
C2(kcos(k)+sin(k))=0 so if we make the assumption that C2 is not 0 then kcos(k)+sin(k)=0

I've tried multiple things and finally came to dividing by cos(k) so that it becomes
k+tan(k)=0 or k=-tan(k)

but, this is where I get confused. My professor offered the hint that k becomes an approximation so I plotted x and -tan(x) and found where they intersect. This gives a few values but I don't understand how to get a value for k. Normally k=n*pi or some sort of thing. So, my question is how do I use this information to find lambda and the Eigenfunctions for this problem. Any help is appreciated.
 
Physics news on Phys.org
It doesn't "give" general values for k. As your professor told you, the best you can do is approximate them, perhaps by graphing as you did. Obviously k= 0 is one value but that is the only one that can be written in a simple way.
 
I figured out what was meant by approximation. As if you plot y=x and y=-tan(x) the points at which they cross close in on the values of pi*n/2 at odd n values. so k~(2n-1)*pi/2.
which makes the Eigenfunction sin(kx)=sin[((2n-1)*pi*x)/2]
 
NO, it does not! All you can say is that the correct eigenvalues are close to (2n-1)pi/2. And, by the way, that is only true for small eigenvalues. As the eigenvalues get larger, that approximation becomes very poor.
 
Sorry I didn't mean to say that it closes in but that it is close for (as you stated) small values. But, for some reason the professor didn't tell us that this was the case and I'm unsure why they are only concerned with the smaller values Thank You for the input and help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top