Finding eigenvalue and eigenfunctions with exponential solutions

[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...

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"...

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...

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 realized 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 realized 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.