Eigenvalue problem

[sigmund]

I have this eigenvalue problem:
\frac{\mbox{d}^2y}{\mbox{d}x^2}+\left(1-\lambda\right)\frac{\mbox{d}y}{\mbox{d}x}-\lambda y = 0 \ , \ x\in[0,1], \ \lambda\in\mathbb{R}

y(0)=0
\frac{\mbox{d}y}{\mbox{d}x}(1)=0
Then, I have to show that there exists only one eigenvalue \lambda , and find this eigenvalue and write the corresponding eigenfunctions.

Thus far, I have solved the ODE's characteristic equation
r^2+(1-\lambda)r-\lambda=0 .
This gives me two solutions
r=-1 and r=\lambda .
Thus the solution to the ODE is

y=c_1\exp(-x)+c_2\exp(\lambda x) \ , \ x\in\mathbb{R} \ , \ c_1, \ c_2\in\mathbb{R}
.

Can I now conclude that because we only have real solutions to the characteristic equation, only one eigenvalue exists?

Secondly, I am not completely sure how to find the sought eigenvalue. I know how to find the eigenvalue when I have a solution that involves \sin and \cos , but here I am not sure how to do it. Could anyone give me a hint?

Thirdly, when I have the eigenvalue there should not be any problems in writing the corresponding eigenfunctions, or is it?

I would appreciate any help. I am not looking for a solution to my homework problem, but any hints to the problem mentioned above are welcome.

[dextercioby]

You miss text from the problem.In order for that to be a genuine Cauchy problem,you need to specify 2 initial conditions.Besides,you didn't use the one which u have posted...:wink:

Daniel.

[HallsofIvy]

An "eigenvalue" is a value of &lamba; such that the problem has a non-trivial solution. Have you determined what functions satisfy your other conditions?
(You have y(0)= 0 but then is see only \frac{dy}{dx}. Did you mean \frac{dy}{dx}(0)= 0 or \frac{dy}{dx}(1)= 0?)

The problem is a lot easier if it is \frac{dy}{dx}(0)= 0!

Put x= 0 into the formulas for y and dy/dx and you get two equations to solve for c1 and c2. For what values of &lamda; can you NOT solve for specific value of c1 and c2?

[dextercioby]

He corrected his typo and now he can follow your advice for finding "lambda"...:approve:

Daniel.

[sigmund]

Has anyone tried to calculate the eigenvalue? My suggestion is \lambda=-1 . Can anyone either confirm or refute this?

[dextercioby]

What is the system of equations for C_{1} & C_{2}...?

Daniel.

[sigmund]

When I use the initial conditions, I get this system of equations:

c_1+c_2=0
-c_1\exp(-1)+c_2\lambda\exp(\lambda)=0

I then guess that \lambda=-1 , and find out that with this eigenvalue you cannot solve the system for any particular value of c_1 and c_2 .

[dextercioby]

Yes,indeed,lambda=-1 makes the 2 equations identical.

Now find the eigenfunctions corresponding to \lambda=-1

Daniel.

[HallsofIvy]

Deleted- DexterCioby beat me to it!

[dextercioby]

Halls,it has lambda =-1...:wink:

I "guess" -1 can be called elementary,huh...?

Daniel.

[sigmund]

Well, the eigenfunctions corresponding to \lambda=-1 must then be

y=c_1\exp(-x)+c_2\exp(-x)=(c_1+c_2)\exp(-x)=c\exp(-x),~~c_1,\,c_2,\,c\in\mathbb{R}

[dextercioby]

Yes. y(x)=C\exp(-x),C\in\mathbb{R} is the awaited sollution.

Daniel.

[sigmund]

After I have talked to the teacher, I have realised that the eigenfunction written in post #11 is wrong. The reason is that when \lambda=-1, the characteristic equation has a double root r=-1. Thus, the solution is y=c_1\exp(-x)+c_2x\exp(-x), and NOT y=c\exp(-x) as written earlier. Do you agree dextercioby?

[saltydog]

After I have talked to the teacher, I have realised that the eigenfunction written in post #11 is wrong. The reason is that when \lambda=-1, the characteristic equation has a double root r=-1. Thus, the solution is y=c_1\exp(-x)+c_2x\exp(-x), and NOT y=c\exp(-x) as written earlier. Do you agree dextercioby?

I'm confused: To meet the initial condition y(0)=0, c_1 has to be zero. But if that's the case, then any value of c_2 meets the derivative at the boundary condition specified above and thus we loose uniqueness.

[HallsofIvy]

Yes, that's the whole point of "eigenvalue". If λ is not an eigenvalue, then the equation Ax= &lamda;x has a unique solution: x= 0. If λ is an eigenvalue, then there exist an infinite number of solutions: the set of eigenvectors corresponding to a given eigenvalue is is a subspace of the original vectdor space.