# Fourth order PDE

1. Nov 23, 2012

### fluidistic

Hi guys!
This is not a homework question, it was a question on my test a few days ago. I could not solve it.
Out of memory, the problem was a rod of length L with an end fixed in a wall and the other end free. Its motion satisfies the PDE $a^4\frac{\partial ^4 u }{\partial x^4} + \frac{\partial ^2 u}{\partial t^2}=0$ where $a>0$.
The boundary conditions are $u(0,t)=\frac{\partial u}{\partial x}(0,t)=0$ and $\frac{\partial ^2 u}{\partial x^2} (L,t)=\frac{\partial ^3 u}{\partial x^3 }(L,t)=0$.
I had to show that its eigenfrequencies satisfy the following relation: $\cosh \left ( \frac{\sqrt \omega L }{a} \right ) = \sec \left ( \frac{\sqrt \omega L }{a} \right )$.
Attempt at solution: First off the problem struck me like a hammer since there was no similar problem in my assignments.
I tried to solve the PDE usng separation of variables, until the professor told us "there's no need to solve the PDE. Of course you can do it but it's extra work" which basically means to me that there's some trick I totally missed.
Anyway what I had reached after separation of variable $u(x,t)=X(x)T(t)$ is $T(t)=A\cos (\lambda t )+B\sin (\lambda t)$ where lambda is a constant of separation.
The remaining ODE I had to solve was $a^4 \frac{X''''}{X}-\lambda ^2 =0$. I assumed the solution was of the form $X(x)=Ae^{kx}+B^{-kx}$ (now I realize it's wrong since it's of order 4 so there should be 4 linearly independent terms, not 2) and I reached after plugging it back into the ODE that $k=\pm \frac{\lambda }{a}$ which gave me $X(x)$. I tried a few other things like trying to find out the constants of the general solution to the PDE (but I've got X(x) wrong since it's of order 4, not 2) but I reached nothing. Obviously I missed a trick.
So I don't really know how to proceed to solve the problem, even at home. Has anyone an idea?

Last edited: Nov 23, 2012
2. Nov 23, 2012

### haruspex

cos(kx) would also be a solution, yes?

3. Nov 23, 2012

### fluidistic

Whoops, I made a mistake in the equation and boundary condition (mixed ^3 for ^2 and ^2 for ^3, I edited the post)
Yeah I'll have to check this out, thanks for the tip. I'll come back to this probably tomorrow.

4. Nov 23, 2012

### Mute

I'll give you a slightly different angle than haruspex. What if you had just been given the ODE

$$a^4 X''- q X =0?$$

I've replaced $\lambda^2$ with $q$. Say you don't know the sign of q. Do you remember how you first went about solving that equation to get the general solution?

If you can recall that, then suppose you were given the ODE you found,

$$a^4 X''''- q X =0.$$

Use the same trick to find the four terms that should be present in your general solution.

By the way, when you separated variables, how are you sure that the separation constant should be positive? Why can't it be negative?

5. Nov 23, 2012

### fluidistic

Ok thanks a lot guys!!!
Yes, absolutely. If q is negative you get oscillatory motion (Acos $\sqrt q$ x + B sin $\sqrt q$ x). If q=0 you get the equation of a straight line, namely Ax+b and if q is postive you get growing and decaying exponential whose arguments are plus and minus the square root of q times x.

Hmm ok I will try to see if it can help me for the fourth order ODE.
About the constant of separation, I reached $a^4\frac{X''''}{X}+\frac{T''}{T}=0$. So that T''/T is worth a constant. In my mind it was clear that the rod would have a periodic motion in time because of the physical problem. So I assumed the constant of separation worth to be $-\lambda ^2$ so that I could express the solution $T(t)$ as oscillatory and without square root. If the sign of q, or lambda would have been different, then the motion of the rod would have been different.

6. Nov 24, 2012

### Mute

Yes, that is correct, but do you remember how you derive those solutions for $q \neq 0$ in the first place? It's similar to what you tried to do to solve the fourth order ODE. Say that instead of plugging in a trial solution $A\exp(kx) + B\exp(-kx)$ you had only plugged in $exp(kx)$... then what?

Good. Why does the beam only oscillate in time, though? Why can't the oscillations decay? What kind of term in the PDE do you think would give you decaying oscillations? (Or possibly no oscillations at all, depending on parameter values)

7. Nov 24, 2012

### fluidistic

Ah yes, with the characteristic equation. So here if my equation is $X''''\underbrace{-\frac{\lambda^2}{a^4}}_{=q}X=0$ the char. eq. is $r^4+qr=0$ which yields $r=0 \text{ or } r=(-q)^{3/2}$. Apparently I'll have to use variation of parameters to get the solution because of the triply repeated root and $r=0$.

Hmm yeah of course the motion could be different from oscillating. If your q was positive then $T(t)=Ae^{-\sqrt q t}$ and the characteristic equation for $X(x)$ yields $r=0$ or $r =\left ( \frac{q}{a^4} \right ) ^{3/2}$ with again a triple root.

8. Nov 24, 2012

### Mute

Careful! Only derivatives will bring down powers of r. The characteristic equation is actually just $\alpha^4 r^4 +q = 0$. Also be careful when taking roots here. In your first attempt with the accidental extra factor of r, $r(\alpha^4 r^3 + q) = 0$, the non-zero roots are $r = (q/\alpha^4)^{1/3}$. This is not a triple root because the cube root actually produces three distinct roots in the complex plane, not a single triple root. Similarly, with the correct characteristic equation, you will have four distinct roots in the complex plane.

Ok, what I was actually getting at with that question was that there was no damping in the original PDE, so you wouldn't expect any decaying oscillatory behavior. I was just posing the question to give you some more practice thinking about the form of the PDE/ODEs in case a similar question ever came up on a future exam.

That said, while we're on the topic of what we expect the behavior of the solutions to be based on physical reasoning, are you sure the eigenfrequency relation you were asked to show was $\cosh(\sqrt{\omega}L/\alpha) = \mbox{sec}(\sqrt{\omega}L/\alpha)$? That doesn't have any real solutions other than $\omega = 0$.

9. Nov 24, 2012

### fluidistic

Whoops, you are right. I made a stupid mistake here.

Ah nice, thank you. I'm preparing for an upcoming final exam worth 100% of my grade for the course. Here I sought oscillatory solutions for no particular reason. I did not notice that the PDE had no damping term so next time I'll be more careful.

I'm 99% sure that yes. I can't access the test to confirm though but I'm sure that yes.
I just "solved" $a^4X''''-\lambda ^2 X=0$. The characteristic equation gave me $r= \pm \frac{\sqrt \lambda}{a}$ or $r=\pm \frac{\sqrt {-\lambda}}{a}$. Hmm so what does this tell me.
That $X(x)=Ae^{\frac{\sqrt \lambda x}{a}}+Be^{\frac{-\sqrt \lambda x}{a}}+Ce^{\frac{i\sqrt \lambda x }{a}}+De^{\frac{-i\sqrt \lambda x}{a}}$. Now I guess I must toss out some coefficients.

10. Nov 26, 2012

### fluidistic

I just found out the problem in a book by chance! Riley's mathematical methods for physics and engineering.
The problem is as I wrote but the relation is $\cosh \left ( \frac{\sqrt \omega L }{a} \right ) = - \sec \left ( \frac{\sqrt \omega L }{a} \right )$ (notice the - sign) where omega stands for the angular frequency of vibration. (So that there does not seem to have more than 1 omega).

11. Nov 26, 2012

### Mute

Ok, that's comforting, because before the only real solution was $\omega = 0$, which would mean the displacement was either constant in time or linearly increasing in time! That didn't seem right, hence why I suspected there must be something missing. =)

Are you now able to derive that condition from applying the boundary conditions to your general solution?

12. Nov 30, 2012

### fluidistic

I've changed my lambda for omega in order to get a temporal solution under the form $T(t)=Ae^{i\omega t}+Be^{-i\omega t}$.
So that $X(x)=Ce^{\frac{\sqrt \omega x}{a}}+De^{\frac{-\sqrt \omega x}{a}}+Ee^{\frac{i\sqrt \omega x }{a}}+Fe^{\frac{-i\sqrt \omega x}{a}}$.
The condition $u(0,t)=0$ gives $C+D+E+F=0$. The other 3 boundary conditions gives slightly more complicated conditions over C, D, E and F. So I must solve a system of 4 homogeneous equations. I have basically $\mathbf A \begin{bmatrix} C \\ D \\E \\F \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\0 \\0 \end{bmatrix}$ where A is a 4x4 (not beautiful) matrix. Now if A is singular then the only solution is the trivial one (C=D=E=F=0) and I get no condition over omega. However if A is non singular, then I should obtain the desired condition over omega. The problem is that the A matrix is pretty "ugly" to me and it involves calculating its determinant and equate it to 0 and solve for omega. I've done a few terms and it's not like there's something that cancels out... sigh.

13. Nov 30, 2012

### Mute

What if you just try eliminating variables step by step from the four equations? I think I tried it a while back and it boiled down to C(something) = C(something else), though I wasn't being particularly careful.

14. Nov 30, 2012

### haruspex

You meant that the other way about, right?

15. Nov 30, 2012

### fluidistic

Oops, yes. Change singular for non singular and non singular for singular. Or simply singular for inversible.