# Thread: How do I solve this?

1. ## How do I solve this?

Ok sorry for posting a picture again, but LateX isn't very familiar. This is the problem I have:

It's confusing how I have to go about with this question?

Would really appreciate if someone could start me off.

2. Originally Posted by LooNiE
Ok sorry for posting a picture again, but LateX isn't very familiar. This is the problem I have:

It's confusing how I have to go about with this question?

Would really appreciate if someone could start me off.
Try a separation of variables.

3. Well yes it does say to do that, but I'm not certain how to do that. Please help if you are able to. Cheers

4. Originally Posted by LooNiE
Well yes it does say to do that, but I'm not certain how to do that. Please help if you are able to. Cheers
Your text should have a section on this. I'll walk you through the general solution for both initial displacement AND initial velocity.

$z(r,t)=F(r)T( \theta )$

Do your seperation of variables and equate it to -lamda. We will get

$T^{ \prime \prime } + \lambda T = 0$

$F^{ \prime \prime } + \frac{ F^ \prime}{r} + \frac{ \lambda F}{c^2} = 0$

If $\lambda \ge 0$ then let $\lambda = w^2$ the equation for F then becomes a zero order Bessel equation, with general solution

$F(r)=aJo( \frac{wr}{c} ) + bYo( \frac{wr}{c} )$

Select $b=0$

$F(r)=aJo( \frac{wr}{c} )$

$T^{ \prime \prime } + w^2 T = 0$

From regular methods, let T $=e^{rt}$ we find that

$T(t) = dcos(wt) + ksin(wt)$

Of course d,k will be taken care in our later summation so what we have is

$Z(r,t) = F(r)T( \theta ) = aJo ( \frac{wr}{c} ) cost(wt) + BwJo ( \frac{wr}{c} ) sin(wt)$

Note that Bw is not B. Can you solve from here?

5. sadly, none of this was taught to me in class. I still don't understand where you got that separation of variables from. It's my first time looking at a problem like this.

6. Originally Posted by LooNiE
sadly, none of this was taught to me in class. I still don't understand where you got that separation of variables from. It's my first time looking at a problem like this.
I can't believe that your prof would assign this question without first going threw the basic heat/wave equations. That just doesn't make sense.

Also, if you can pick up a book on advanced engineering mathematics, they pretty much have all the information you're looking for (in terms of DEs). I'm working out of "Advanced Engineering Mathematics" by O'Niel. If you want, I can scan some pages on this topic and send them your way. They will describe all you need to know. But for now,

I'll walk you through the seperation of variables in a basic wave equation, and then you must apply it to this situation.

Let us suppose that,

$\frac{d^2y}{dt^2} = c^2 \frac{d^2 y}{d^2x}$

$y(0,t)=y(L,t)=0$ for $t \ge 0$

$y(x,0)=f(x)$ and $\frac{dy}{dt} (x,0) = 0$ for $0 \le x \le L$

For our purposes, we will let

$y(x,t)=X(x)T(t)$

In words this means we are letting our function equal some function only dependent on X multiplied by some function only dependent on T.

Let us get this into the form of the equation above.

$\frac{d^2y}{dt^2} = XT^{\prime \prime}$

$\frac{d^2y}{dx^2} = X^{\prime \prime} T$

Subbing back into

$\frac{d^2y}{dt^2} = c^2 \frac{d^2 y}{d^2x}$

We get

$XT^{\prime \prime} = c^2 X^{\prime \prime} T$

Which we can manipulate to

$\frac {T^{\prime \prime} }{c^2 T} = \frac {X^{\prime \prime} }{X}$

Of course what this means is, we have one side only dependent on T and one side only dependent on X. This will only happen if these are both equal to a constant, so we will let that constant equal -lambda

$\frac {T^{\prime \prime} }{c^2 T} = \frac {X^{\prime \prime} }{X} = - \lambda$

We then get 2 equations from this

$T^{ \prime \prime} + c^2 \lambda T = 0$

$X^{ \prime \prime} + \lambda X = 0$

So let us solve our equation involving X. We will let

$X=e^{rx}$

Subbing this into our equation we get

$r^2 e^{rx} + \lambda e^{rx} = 0$

$r^2 + \lambda = 0$

$r = \sqrt {\lambda i}$

Of course this means,

$X(x) = e^{- \sqrt{ \lambda i } x } + e^{ \sqrt{ \lambda i } x} = acos( \sqrt {\lambda} x ) + bi sin( \sqrt {\lambda} x ) + ccos( \sqrt {\lambda} x ) -di sin ( \sqrt {\lambda} x )$

Lets simplify this by

$F= a + c$

$G = bi - di$

So then we get

$X(x) = Fcos( \sqrt {\lambda} x ) + G sin( \sqrt {\lambda} x )$

But we have initial conditions that state

$y(0,t) = 0 = X(0)T(t) = 0$

So,

$X(0) = 0 = F$

$X(x) = G sin( \sqrt {\lambda} x )$

Again from initial conditions,

$X(L)=0$

So,

$X(L) = 0= G sin( \sqrt {\lambda} L )$

This is only true when,

$\sqrt {\lambda} L = n \pi$

This means,

$\lambda = \frac {n^2 \pi^2}{L^2}$

And finally,

$X(x) = sin ( \frac{n \pi x}{L} )$

Do the same for T.

7. yes, if you are able to scan some pages by tonight that would be useful. None of this work that I am trying to understand was taught in the module. Our lecturer was away for about a month, so we missed about 10hrs of lectures, and he recently got replaced so we are moving on to a different topic. This all seems mind boggling. The new lecturer said that this Vibrating Membrane problem I am trying to solve is pretty straight forward, but the way you are explaining it is very confusing.

8. Originally Posted by LooNiE
yes, if you are able to scan some pages by tonight that would be useful. None of this work that I am trying to understand was taught in the module. Our lecturer was away for about a month, so we missed about 10hrs of lectures, and he recently got replaced so we are moving on to a different topic. This all seems mind boggling. The new lecturer said that this Vibrating Membrane problem I am trying to solve is pretty straight forward, but the way you are explaining it is very confusing.
Actually what I just explained is not a membrane problem. What I explained is how to solve the wave equation in 2D. Much easiar then the membrane question, but the membrane one follows a similar pattern.

What is it that you don't understand about the change of variables?

I'll have some stuff for you tonight.

9. can anyone else input into this problem? I have read the info on how to solve the general case but applying it to my question was hard!