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PDEs

Post Icon Posted: Submitted by sooz on 24 May 2008 - 4:52pm.

Joined: 2008-05-09
Posts: 7

Hi, I'm really stuck on Q2 a) of the 2007 PDE exam paper.
I would post the question but I'm awful when it comes to typing maths!
Any ideas? I'm trying to work through it like we did in class but l is pi
and theres a's coming into it!
Thanks,
Sooz

‹ Revision Guides Why does addition seem to fail in general? ›
Post Icon Posted: 25 May 2008 - 9:21pm

Joined: 2007-10-01
Posts: 170

I don't have a solution, but I'll mathtype the question so you don't have to go hunting for it:

$ \phi, \psi \in C^4(\mathbb{R}) $ are both $ \pi\, $ periodic (i.e. $ \phi(0) = \phi(\pi) \, $ etc) . Construct a sine series which is a solution to the initial boundary problem:

$ p u_{tt} = T u_{xx}, \, x \in (0,\pi), t \in \mathbb{R} $

$ u(0,t) = 0, \, u(\pi,t) = a $

$ u(x,0) = \phi(x) + ax, \, u_t(x,0) = \psi(x) $

Hint: Start with a=0
Post Icon Posted: 25 May 2008 - 11:14pm

Joined: 2007-10-01
Posts: 170

just had a flick through my notes and I have a feeling you have to use the Dirichlet conditions so you end up with something like $ u(x,t) = \Sigma sin(nx)(A_n cos(nct) + b_n sin(nct) )\, $. This looks slightly different to the one in the notes however and I'm not entirely sure what to do.

EDIT:

Aha!! I think I see what to do. This is the Wave Equation under the Dirichlet initial conditions. This is basically $ x \in [0,L] $, $ u(x,0) = \pi(x), u_t(x,0) = \psi(x)\, $ and the Dirichlet condition $ u(0,t) = u(L,t) = 0\, $ where $ L=\pi $. This is the case when $ a=0 $ and is identical to the one in the notes. We are working under the assumption that some solution to the wave equation of the form $ u(x,t) = X(x)T(t) $.

Using this we can deduce that T is the solution to $ -T'' = k^2T \, $ and X solves $ -X'' = k^2 X $ for some constant k (can't remember exactly what it is. These are just ODE's with solution:
$ X(x) = Ccos(kx) + Dsin(kx) $ and $ T(t) = A cos(kt) + B sin(kt) $. I have a feeling the constants are very important so it's probably best to check up on them sometime.

so we know that $ X(0) = X(L) = u(0,t) = u(L,t) = 0 \, $ because of the Dirichlet condition and the fact that u=XT by our assumption. This means that $ C=0 $ and $ Dsin(kL)=0 $ so $ kL=n\pi $ giving us $ k=\frac{n\pi}{L} = \frac{n\pi}{\pi} = n \in \mathbb{Z} $. Also $ X'(0) = Dk = 0 $ so $ D=0 $ and $ X'(L) = Ck(-sin(kL)) = 0 $ so $ k=n $ for the same reason.

so remembering that $ u=XT $ we get $ u(x,t) = sin(nx)(A cos(n \sqrt{ \frac{T}{p} }t) + B sin(n\sqrt{ \frac{T}{p}t) )\, $. Then the most generalised form is to take this as a series so the general solution is:

$ u(x,t) = \Sigma sin(nx)(A_n cos(n \sqrt{ \frac{T}{p} }t) + B_n sin(n\sqrt{ \frac{T}{p}}t) )\, $

Given this, we can now work out the initial conditions, $ \phi\, $ and $ \psi\, $ are both $ \pi $ periodic so we know they have a representation $ \phi(x) = \Sigma \alpha_n sin(nx) \, $ and $ \psi(x) = \Sigma \gamma_n sin(nx) \, $ we also know $ \phi(x) = u(x,0) = \Sigma A_n sin(nx) \, $ and $ \psi(x) = u_t(x,0) = \Sigma B_n n \sqrt{ \frac{T}{p}} sin(nx) \, $ then we equate coefficients to deduce that $ A_n =\alpha_n\, $ and $ B_n = \frac{\gamma_n}{ n\sqrt{ \frac{T}{p}} } $.

I THINK that is the case that a=0, but I could be very very wrong indeed. When we generalise this to $ a \in \mathbb{R}\, $ we get something like $ \phi(x) + ax = u(x,0) = \Sigma A_n sin(nx) \, $ and $ \psi(x) = u(x,\pi) = \Sigma B_n n \sqrt{ \frac{T}{p}} sin(nx) \, $ so nothing changes with $ \psi \, $ however we now have to solve $ \phi(x) + ax = ax + \Sigma \alpha_n sin(nx) = \Sigma A_n sin(nx) = u(x,0) \, $ i.e. how do we work out $ A_n\, $ in terms of $ \alpha_n $?

Don't know how to solve that but I'm hoping someone else can come to the rescue on this. Has any of this helped or have I just said stuff you already knew (I did basically copy it al out of the notes).
Post Icon Posted: 25 May 2008 - 11:41pm

Joined: 2006-10-01
Posts: 427

Finally my internet is behaving itself!

The key to getting started on this is D'Alembert's formula , It's a super hot way of solving the wave equation in 1-d and it pretty much sets you up for the answer, though I'm not sure what it means by "by matching coefficients": presumably if you work the thing through, all will be well...

Edit: Beaten to it! And Richard seems to know what he's talking about far more than I do... I didn't even realise T was sposed to be a function... Mind you I haven't solved a PDE in donkey's so it's not surprising!
Post Icon Posted: 26 May 2008 - 10:00am

Joined: 2007-10-01
Posts: 170

I think you're right tom, that is a way of solving WE but then we did it with fourier series as well which is why it talk about sine. In this case I don't think T is a function, but when we solved this problem there was a function called T.
Post Icon Posted: 26 May 2008 - 12:53pm

Joined: 2006-10-01
Posts: 427

Oh yeah... reread you're working and all made much more sense, the point still stands that I suck at this...
Post Icon Posted: 26 May 2008 - 6:15pm

Joined: 2008-05-09
Posts: 7

thanks for your help! Reeeally wish there was a revision lecture for this!
Post Icon Posted: 26 May 2008 - 10:15pm

Joined: 2006-10-29
Posts: 9

I have only skim read this, and I may be well off with this, but to calculate the $ A_{n} $ in terms of the $ \alpha_{n} $, do you not just calculate the Fourier series for $ f(x) = ax $ and combine that with the solution you got for $ a=0 $? Linearity should follow from the definition of the coefficients of a Fourier series (obtained from an integral, which we know from Analysis III is linear). Since this function is odd, it will have a Fourier series containing only sine terms, so the solution you obtain should only contain sine terms, $ W^{5} $ (to quote someone else).

Assuming the change in boundary values does not affect the calculations made above, I think this will work.

It is even longer since I solved a PDE. In fact, I'm not sure I ever have solved one. I probably still haven't.
Post Icon Posted: 27 May 2008 - 9:51am

Joined: 2007-10-01
Posts: 170

that would seem to work, but then you still get a cosine in the final answer which confuses me a little

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