# Fourth-order eigenvalue problem

1. Mar 3, 2005

### meteorologist1

I'm stuck on the following eigenvalue problem:
$$u^{iv} + \lambda u = 0, 0 < x < \pi$$
with the boundary conditions u = u'' = 0 at x = 0 and pi.
("iv" means fourth derivative)

I look at the characteristic polynomial for lambda > 0 and < 0 and I get fourth roots for each of them. In the case for lambda < 0, I get 2 real roots and 2 complex roots. In the case lambda > 0, I get 4 complex roots. But what I need to show is that all eigenvalues are real, and what the signs of the eigenvalues are, and what the eigenvalues are.

2. Mar 3, 2005

### dextercioby

Impose the boundary conditions on the the solution.Then U'll get a system of 4 eqns with 4 unknowns (the constants of integration:C_{1},C_{2},...) and one parameter,namely "\lambda".What should you be doing next...?

Daniel.

3. Mar 3, 2005

### meteorologist1

Well, ok for lambda < 0, I get the roots of the characteristic polynomial to be:
$$\sqrt[4]{-\lambda}, -\sqrt[4]{-\lambda}, i\sqrt[4]{-\lambda}, -i\sqrt[4]{-\lambda}$$
which means the basis of solutions is:
$$u(x) = ae^{\sqrt[4]{-\lambda}x} + be^{-\sqrt[4]{-\lambda}x} + ce^{i\sqrt[4]{-\lambda}x} + de^{-i\sqrt[4]{-\lambda}x}$$ or
$$u(x) = ae^{\sqrt[4]{-\lambda}x} + be^{-\sqrt[4]{-\lambda}x} + ccos(\sqrt[4]{-\lambda}x) + dsin(\sqrt[4]{-\lambda}x)$$

Then I looked at the boundary conditions and I got a = b = c = d = 0 if I'm not mistaking. So lambda < 0 does not provide any eigenvalues.

For lambda > 0, I get the roots of the characteristic polynomial to be:
$$\sqrt[4]{\lambda}e^{\frac{i\pi}{4}}, \sqrt[4]{\lambda}e^{\frac{3i\pi}{4}}, \sqrt[4]{\lambda}e^{\frac{5i\pi}{4}}, \sqrt[4]{\lambda}e^{\frac{7i\pi}{4}}$$
but now I don't know how to write the basis of solutions and apply the boundary conditions. It's going to be extremely complicated. I have a feeling that a = b = c = d = 0 again? If so, what are the eigenvalues then?

Thanks.

Last edited: Mar 3, 2005
4. Mar 3, 2005

### dextercioby

How did u get a=b=c=d=0 in the first case...?It looks like there should be a dependence on "lambda"...

Daniel.

P.S.Did u really solve the 4*4 system...?U know how u find the eigenvalues...?

5. Mar 3, 2005

### meteorologist1

For the first one I got relations: 1) a + b + c = 0. 2) a + b - c = 0. 3) a(y) + b(-y) + ccos(y) + dsin(y) = 0. 4) a(y) + b(-y) - ccos(y) - dsin(y) = 0
where y = exp(4sqrt(-lambda))pi)

From 1) and 2), I get c = 0. By 1) and 3) I get a = b = 0. By 3) and 4) I get d = 0.

6. Mar 3, 2005

### dextercioby

Shouldn't lambda be multiplied with the a,b,...It should come from diff.the solution wrt
"x"...

Daniel.

7. Mar 3, 2005

### meteorologist1

Yes, but I got the same thing for each of the four terms, so I factored it out and divided both sides by it.

8. Mar 3, 2005

### dextercioby

What are you saying...?It should be in only 2 of the equations,namely those which imply the derivative of the solution...

Daniel.

9. Mar 3, 2005

### meteorologist1

I'm sorry... which two equations? Could you write it out...

10. Mar 3, 2005

### meteorologist1

Oh I see, what I meant was each of the four terms in 2) and 4) produced the same lambda factors when I took the derivative twice, so since the right side of the equations are 0, I can divide both sides of each equation by the lambda factor and obtain what I got: a + b - c = 0, etc..

11. Mar 3, 2005

### dextercioby

You have the solution:y=y(x) as a sum of 4 terms.Compute the derivative:y':y'(x).
Impose Cauchy-type conditions.y(0)=0;y(\pi)=0;y'(0)=0;y'(\pi)=0.You'll get the system i mentioned a gazzilion posts ago.

Then see under which conditions this system admits solution (for {a,b,c,d}) and find lambda...

Daniel.

12. Mar 5, 2005

### saltydog

Meterologist,

I think I have this one. Can you post the four algebraic (homogeneous) equations for the determination of the various constants. Can you use subscripts ($c_1,c_2,c_3,c_4$) instead of a,b,c,d (is that too picky?). As Daniel said, you'll need to determine what values of lambda allow this system to have a non-zero solution. A homogeneous system has a solution if the determinant is zero so, well, equate the determinant to zero and figure out what values of lambda would make it zero without all the constants being zero. My solution has several being zero. Anyway if no one responds, I'll write it up in a few days, plot to boot.

13. Mar 6, 2005

### meteorologist1

Hi saltydog, here is what I have. Sorry Daniel, I didn't have time last week to post everything step by step. Please correct me where I'm making a mistake:

Since the boundary conditions are:
$$u(0) = u''(0) = u(\pi) = u''(\pi) = 0$$

For lambda < 0, my solution is:
$$u(x) = c_1e^{\sqrt[4]{-\lambda}x} + c_2e^{-\sqrt[4]{-\lambda}x} + c_3cos(\sqrt[4]{-\lambda}x) + c_4sin(\sqrt[4]{-\lambda}x)$$
so therefore,
$$u'(x) = c_1\sqrt[4]{-\lambda}e^{\sqrt[4]{-\lambda}x} + c_2\sqrt[4]{-\lambda}e^{-\sqrt[4]{-\lambda}x} - c_3\sqrt[4]{-\lambda}sin(\sqrt[4]{-\lambda}x) + c_4\sqrt[4]{-\lambda}cos(\sqrt[4]{-\lambda}x)$$
$$u''(x) = c_1\sqrt{-\lambda}e^{\sqrt[4]{-\lambda}x} + c_2\sqrt{-\lambda}e^{-\sqrt[4]{-\lambda}x} - c_3\sqrt{-\lambda}cos(\sqrt[4]{-\lambda}x) - c_4\sqrt{-\lambda}sin(\sqrt[4]{-\lambda}x)$$

Plugging in the boundary conditions, I have:
$$u(0) = c_1 + c_2 + c_3 = 0$$ (1)
$$u''(0) = c_1\sqrt{-\lambda} + c_2\sqrt{-\lambda} - c_3\sqrt{-\lambda} = 0$$
$$\Rightarrow u''(0) = c_1 + c_2 - c_3 = 0$$ (2)

$$u(\pi) = c_1e^{\sqrt[4]{-\lambda}\pi} + c_2e^{-\sqrt[4]{-\lambda}\pi} + c_3cos(\sqrt[4]{-\lambda}\pi) + c_4sin(\sqrt[4]{-\lambda}\pi) = 0$$ (3)
$$u''(\pi) = c_1\sqrt{-\lambda}e^{\sqrt[4]{-\lambda}\pi} + c_2\sqrt{-\lambda}e^{-\sqrt[4]{-\lambda}\pi} - c_3\sqrt{-\lambda}cos(\sqrt[4]{-\lambda}\pi) - c_4\sqrt{-\lambda}sin(\sqrt[4]{-\lambda}\pi) = 0$$
$$\Rightarrow u''(\pi) = c_1e^{\sqrt[4]{-\lambda}\pi} + c_2e^{-\sqrt[4]{-\lambda}\pi} - c_3cos(\sqrt[4]{-\lambda}\pi) - c_4sin(\sqrt[4]{-\lambda}\pi) = 0$$ (4)

From (1) and (2), I conclude that c3 = 0. From (3) and (4), I conclude that c4 = 0. From (1) and (3), the only way I think both (1) and (3) can be satisfied is that c1 = c2 = 0. Therefore, there are no eigenvalues less than 0 that provide nontrivial solutions.

For lambda > 0, I'm going to post it a little while...

14. Mar 6, 2005

### meteorologist1

I haven't tried the determinant method. Did you get something different? Anyway, for lambda > 0,

The roots of the characteristic polynomial were:
$$\sqrt[4]{\lambda}e^{\frac{i\pi}{4}}, \sqrt[4]{\lambda}e^{\frac{3i\pi}{4}}, \sqrt[4]{\lambda}e^{\frac{5i\pi}{4}}, \sqrt[4]{\lambda}e^{\frac{7i\pi}{4}}$$ or
$$\sqrt[4]{\lambda}(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}), \sqrt[4]{\lambda}(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}), \sqrt[4]{\lambda}(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}), \sqrt[4]{\lambda}(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2})$$

so the solution is:
$$u(x) = c_1e^{\sqrt[4]{\lambda}(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2})x} + c_2e^{\sqrt[4]{\lambda}(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2})x} + c_3e^{\sqrt[4]{\lambda}(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2})x} + c_4e^{\sqrt[4]{\lambda}(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2})x}$$
$$\Rightarrow u(x) = c_1e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x + c_2e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x + c_3e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x + c_4e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x$$
$$u'(x) = -c_1e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x - c_2e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x$$
$$+ c_3e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x + c_4e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x$$
$$u''(x) = -c_1e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt{\lambda}\frac{1}{2}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x - c_2e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt{\lambda}\frac{1}{2}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x$$
$$- c_3e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt{\lambda}\frac{1}{2}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x - c_4e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\sqrt{\lambda}\frac{1}{2}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}x$$

Applying the boundary conditions,
$$u(0) = c_1e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}} + c_2e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}} = 0$$ (1)
$$u''(0) = -c_1e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\frac{\sqrt{\lambda}}{2} - c_2e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}\frac{\sqrt{\lambda}}{2} = 0$$
$$\Rightarrow u''(0) = c_1e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}} + c_2e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}} = 0$$ (same as 1)
$$u(\pi) = c_1e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}\pi + c_2e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}cos\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}\pi + c_3e^{\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}\pi + c_4e^{-\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}}sin\sqrt[4]{\lambda}\frac{\sqrt{2}}{2}\pi = 0$$ (2)
too lazy to write, but u''(pi) will also give you the same thing as in (2).

From (1) and (2), I conclude that c1 = c2 = 0.
Now in order for (2) to be satisfied, we need
$$\sqrt[4]{\lambda}\frac{\sqrt{2}}{2} = n$$
where n = 1, 2, 3, ...
$$\Rightarrow \frac{1}{4}\lambda = n^4 \Rightarrow \lambda = 4n^4$$

These are the eigenvalues. They are clearly real and they are positive.

15. Mar 6, 2005

### saltydog

Dude, I'm sorry you had to type in all that. Allow me to make some suggestions and please don't take offense: it's too messy. You've got to "condense" things to make it easier to read. Here goes:

Let:

$$v=\sqrt[4]{-\lambda}$$

Now:

$$u(x)=c_1e^{vx}+c_2e^{-vx}+c_3\sin(vx)+c_4\cos(vx)$$

Now, when I differentiate twice, I'll get:

$$u^{''}(x)=c_1v^2e^{vx}+c_2v^2e^{-vx}-c_3v^2\sin(vx)-c_4v^2\cos(vx)$$

So that the four equations are:

$$c_1+c_2+ c_4=0$$
$$c_1+c_2 -c_4=0$$

This tells me that $c_4$ has to be zero which means $c_1=-c_2$

Thus yielding the remaining two equations to be:

$$-c_2e^{v\pi}+c_2e^{-v\pi}+c_3\sin(v\pi)=0$$

$$-c_2e^{v\pi}+c_2e^{-v\pi}-c_3\sin(v\pi)=0$$

So this is two equations in two unknowns:

$$c_2(e^{-v\pi}-e^{v\pi})+c_3\sin(v\pi)=0$$
$$c_2(e^{-v\pi}-e^{v\pi})-c_3\sin(v\pi)=0$$

In order for them to have a solution, the determinant has to be equal to zero. Can you set up the determinant, set it to zero, and then determine what values of v make it zero?

16. Mar 6, 2005

### meteorologist1

All right thanks no problem. Could you look at the second part for lambda > 0? I tested the solution and it does not work somehow. :( I'm thinking now that lambda > 0 does not yield eigenvalues?

17. Mar 6, 2005

### meteorologist1

Ok I got it now using the equations you got.

$$\lambda = -n^4$$
where n = 1, 2, 3, ...

This answer looks much more reasonable.

And I think the lambda > 0 case shouldn't even be considered since lambda is a real number, and you can't take the fourth root of negative real number.

Thanks a lot for your help, both of you.

18. Mar 7, 2005

### saltydog

I think n can be any integer and I'm not sure about $\lambda>0$. You know that you can take the fourth root of a negative number (DeMoivre's formula) and since working that ODE with variable coefficients with StatusX some time ago here, I've realized it's Ok to get lots of imaginary numbers when working with ODEs.

Perhaps Daniel can answer that for us.