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Fourth order PDEby fluidistic
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#1
Nov2312, 07:26 PM

PF Gold
P: 3,188

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? 


#2
Nov2312, 10:04 PM

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#3
Nov2312, 10:20 PM

PF Gold
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Whoops, I made a mistake in the equation and boundary condition (mixed ^3 for ^2 and ^2 for ^3, I edited the post)



#4
Nov2312, 10:33 PM

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Fourth order PDE
$$a^4 X'' q X =0?$$ 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
Nov2312, 11:07 PM

PF Gold
P: 3,188

Ok thanks a lot guys!!!
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
Nov2412, 12:20 AM

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#7
Nov2412, 03:16 PM

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#8
Nov2412, 05:33 PM

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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
Nov2412, 10:07 PM

PF Gold
P: 3,188

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
Nov2612, 03:25 PM

PF Gold
P: 3,188

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
Nov2612, 08:34 PM

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


#12
Nov3012, 05:15 PM

PF Gold
P: 3,188

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
Nov3012, 06:32 PM

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#14
Nov3012, 10:32 PM

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