anonymous
  • anonymous
Problem with partial differential equation. Solution included:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
boundary conditions of u(x, t): u(0, t) = 0 u(pi, t) = 0 u(x, 0) = 0 \[\frac{ \delta u }{ \delta t } |_{t = 0} = sinx\] delta is partial derivative (because o.s. doesn't have the symbol yet) this is a wave equation problem: \[a^2\frac{ \delta^2 u }{ \delta x^2 } = \frac{ \delta^2 u }{ \delta t^2 }\] solution is here: http://img23.imageshack.us/img23/6848/l3tj.jpg i understand everything, and I have solved others. my only problem is the B1 term they introduced when solving for Bn (which turned out to = 0) using the fourier series method. I don't understand how it was obtained. Please explain or point me to the right resource :)
anonymous
  • anonymous
@Euler271 "openstudy" just uses MathJax which most definitely supports \(\partial\)
anonymous
  • anonymous
thanks for letting me know ^_^

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anonymous
  • anonymous
@Euler271 \(\partial=\text{\partial}\)
Jhannybean
  • Jhannybean
Oh that's what it is o_o
whpalmer4
  • whpalmer4
If you see something on OpenStudy and don't know how to produce it in LaTex, just select the offending math, right click and do Show Math As>TeX Commands
anonymous
  • anonymous
This solution manual text really sucks... :-p
experimentX
  • experimentX
forget the solution manual ... do it on your own. there are couple of inconsistencies with what you mention. for eg; you have u(x,0) = 0, solution assumes u_t(x,0) = 0
anonymous
  • anonymous
i don't understand what you see wrong with it. solution in textbook respects u_t(x, 0) = sinx I just don't understand what B1 is. Bn is solved using fourier sine series. How do you find B1?
experimentX
  • experimentX
You find it via comparison ... there is \(T'(0) = 0 \) . it's supposed to be \(T(0) = 0 \)
experimentX
  • experimentX
the solution is of the form X*T ... all those values of n consist of your solution. so you using superposition principle. Use boundary condition on X and T to simplify. |dw:1371323294729:dw||dw:1371323386702:dw|
experimentX
  • experimentX
|dw:1371323482696:dw||dw:1371323650255:dw|
experimentX
  • experimentX
you final solution is u(x,t) = 1/a sin(at) sin(x) which is a standing wave.
anonymous
  • anonymous
thank you very much :) I understand it now ^_^
experimentX
  • experimentX
don't know ... probably typo or error. there has to be a term there on time variable.
anonymous
  • anonymous
it's because of the a already in the derivative of the series (wrt t)
experimentX
  • experimentX
No i meant on 'u' of your solution manual *has to be term 'a' also note that you can compare sin(nx) on both sides because they are orthogonal.
anonymous
  • anonymous
since you've mentioned it, do you know what it means when two functions are "orthogonal" to each other? what it looks like graphically or what it represents
experimentX
  • experimentX
this comes from linear algebra ... sin(nx) and cos(nx) forms infinite dimensional vector space (of functions) usually you have dot product of two vectors, in case of two functions, inner product is defined as int_0^T f(x) g(x) dx. Fourier series is just projection of any function on this infinite dimensional vector space.
experimentX
  • experimentX
more on wikipeida http://en.wikipedia.org/wiki/Orthogonal_functions
anonymous
  • anonymous
awesome :) thanks
experimentX
  • experimentX
honestly i find it hard to interpret inner product geometrically for general case. it has been formulated quite broad. http://en.wikipedia.org/wiki/Inner_product good luck!
anonymous
  • anonymous
You can compare them because they're linearly independent, not orthogonal, surely? :-) It's just like when you equate coefficients for solving a system of polynomial equations -- the various powers of \(x\) are all linearly independent.
anonymous
  • anonymous
The \(\sin\cdot\) are indeed orthogonal, too, of course, but this implies linear independence.
experimentX
  • experimentX
yeah ... sorry on my part!! i was too narrow.
anonymous
  • anonymous
Good work on the PDE though... I kept looking at it and concluded it couldn't be \(B_1=1/a^2\). Why did they swap to using \(B_n\), though? They forgot the \(a\) in the trigonometric arguments, too! I've never seen a solution manual with so many errors.
anonymous
  • anonymous
Here we're interested in solving the one-dimensional case of the wave equation: $$\frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2}$$Since this is a boundary-value problem we find Dirichlet boundary conditions for \(x\) and Cauchy for \(t\):$$u(0,t)=0,u(\pi,t)=0\\u(x,0)=0, \partial u/\partial t(x,0)=\sin x$$Presume our solution \(u(x,t)\) is completely separable i.e. \(u(x,t)=F(x)G(t)\) and thus \(\partial^2u/\partial x^2=F''(x)G(t),\partial^2u/\partial t^2=F(x)G''(t)\):$$F(x)G''(t)=a^2 F''(x)G(t)\implies \frac{F''(x)}{F(x)}=\frac{G''(x)}{a^2G(x)}$$Since \(x,t\) are independent, we know these ratios are equal to a constant -- in particular, a *separation constant* \(-\lambda^2\):$$\frac{F''(x)}{F(x)}=-\lambda^2\implies F''(x)+\lambda^2 F(x)=0\\\frac{G''(t)}{a^2G(t)}=-\lambda^2\implies G''(t)+a^2\lambda^2G(t)=0$$These are both trivial and yield solutions \(F(x)=c_1\cos\lambda x+c_2\sin\lambda x,G(t)=c_3\cos a\lambda t+c_4\sin a\lambda t\). Now, let's satisfy our boundary conditions:$$u(0,t)=0\implies F(0)=0\implies c_1=0\\u(\pi, t)=0\implies F(\pi) = 0\implies \lambda \text{ is an integer}\\u(x,0)=0\implies G(0)=0\implies c_3=0$$.. so we have so far \(F(x)=c_2\sin\lambda x,G(t)=c_4\sin a\lambda t\). Since \(\lambda\) can be any integer, by super position we have:$$F(x)=\sum_{\lambda=1}^{\infty}c_{2,\lambda}\sin\lambda x\\G(t)=\sum_{\lambda=1}^\infty c_{4,\lambda}\sin a\lambda t\\\implies u(x,t)=\sum_{\lambda=1}^\infty c_\lambda\sin\lambda t\sin a\lambda t$$
anonymous
  • anonymous
Now we consult our last condition \(\partial u/\partial t(x,0)=\sin x\):$$\begin{align*}\frac{\partial u}{\partial t}(x,t)&=\frac{\partial}{\partial t}\sum_{\lambda=1}^\infty c_\lambda\sin\lambda x\sin a\lambda t\\&=\sum_{\lambda=1}^\infty c_\lambda\sin\lambda x\frac{\partial}{\partial t}\sin a\lambda t\\&=\sum_{\lambda=1}^\infty a\lambda c_\lambda\sin\lambda x\cos a\lambda t\\\frac{\partial u}{\partial t}(x,0)&=\sum_{\lambda=1}^\infty a\lambda c_\lambda \sin\lambda x\\\sin x&=\sum_{\lambda=1}^\infty a\lambda c_\lambda\sin \lambda x\end{align*}$$Since \(\sin\lambda x\) are linearly independent, it follows \(c_\lambda=0\) for \(\lambda>1\); discarding those terms we find$$\sin x=ac_1\sin x\\\implies c_1=\frac1a$$
anonymous
  • anonymous
Thus our solution is merely \(u(x,t)=\dfrac1a\sin t\sin at\) :-) sorry for wasting time I just wanted to try solving a PDE.

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