Damped Wave equation, PDE.

[Spoony]

Homework Statement

I have the damped wave equation;

$$u_{tt} = 4 u_{xx} -2 u_{t}$$

which is to be solved on region 0 < x < 2
with boundary conditions;

$$u(0,t) = 2, u(2,t) = 1.$$

i must;
1) find steady state solution $$u_{steady}(x)$$ and apply boundary conditions.
2) find $$\theta(x,t) = u(x,t) - u_{steady}$$ and find PDE and boundary conditions obeyed by theta.
3) use theta = f(t)g(x) to obtain 2 separate ODE's with separation constant
4) then i must solve the homogeneous problem for theta and then recover the general solution for u
5) finally i must find the solution for u using intial conditions
$$u(x,0) = 2 - x/2 , u_{t} (x,0) = x^{2}(2-x)$$
and I am given
(this is between 2 and 0) $$\int x^{2} (2-x) sin(\frac{n\pi}{2} x) dx = \frac{32(1+2(-1)^{n}}{n^{3}\pi^{3}}$$

The Attempt at a Solution

1) steady state solution $$u_{steady}$$ implies $$u_{t} = 0$$ which reduces the PDE to $$4 u_{xx} = 0 \rightarrow u_{xx} = 0$$ after intergration, $$u_{steady} = ax + b$$ applying boundary conditions leads to $$u_{steady} = - \frac{1}{2}x + 2$$

2) As
$$\theta (x,t) = u(x,t) - u_{st}$$
and $$u_{st}$$ is a function of x only then $$\theta _{t} = \theta _{tt} = 0$$ . Using the fact that $$u_{st} = - \frac{1}{2}x + 2$$ leads to $$\frac{\partial^{2} u_{st}}{\partial x^{2}} = 0$$ and therefore $$theta_{xx} = 0$$
which doesn't affect the PDE for u, only changes u to theta.
Oh and both boundary conditions become 0.

3) Inserting
$$\theta (x,t) = f(t)g(x)$$ into the PDE leads to;
$$f''(t)g(x) = 4f(t)g''(x) - 2g(x)f'(t)$$
reranging leads to;
$$\frac{f''(t) + 2f(t)}{4f(t)} = \frac{g''(x)}{g(x)}$$
And each side is a function of its argument only and are equal to each other, so each side equals the same constant, which for the sake of this question must be $$-\Lambda$$ to create a trig solution for g(x) (need trig solution as boundary conditions require that g must equal 0 in 2 places g(0) = g(2) = 0) .

4) this is where i start to struggle, the ODE for g(x) is;
(note; i have used $$\Lambda = \lambda ^{2}$$
$$g''(x) + \Lambda g(x) = 0$$ which leads to $$g(x) = Acos(\lambda x) + Bsin(\lambda x)$$ with boundary conditions g(0) = g(2) = 0.
Using those boundary conditions, A = 0 and B = 0 OR $$sin(2 \lambda) = 0$$
B=0 is trivial case, so it is ignored, if $$sin(2 \lambda) = 0$$ then; $$2 \lambda = n \pi$$ which leads to $$\lambda = \frac{n \pi}{2}$$

this is all fine and dandy, but then when i look at the ODE for f(t)
$$f''(t) + 2f'(t) + 4 \Lambda f(t) = 0$$ and applying the value for $$\Lambda = \lambda ^{2}$$ leads to

$$f''(t) + 2f'(t) + 4 \frac{n^{2} \pi ^{2}}{4} f(t) = 0$$

Once i cancel the 4 I am left with an ODE, i try using $$A e^{mx}$$ as a solution but it leaves me with;
$$m^{2} + 2m = -n^{2} \pi ^{2}$$ which has no solutions for m unless m = 0, so I am stuck there, i need to find the separation constant such that i can get solutions to both of these ODE's :(

Help is really appreciated, i think its the fact that there is an extra partial u by t in there, only examples we have really are when both partials are second order.

 

[Spoony]

Anyoen have any ideas? I'll take anything atm.. I've come to a stand still on this one

 

[hgfalling]

That equation with the m's in it is just a quadratic. I think you can solve it.

 

[Spoony]

the lowest point of m^2 + 2m is 1 and n is an integer therefore -n^2 pi^2 > 1 for all n... unless m can take imaginary values... oh hell i haven't been stupid enough to look at that and think "no value of m can surely satisfy that equation m^2 is always positive.." if this what it is i want to be euthinised to better preserve mankinds intelligence

 

[Spoony]

but m can't take imaginary values as the value for 2m would give an imaginary part, and -n^2 pi^2 is a real number

 

[hgfalling]

$$m^2 + 2m + n^2\pi^2 = 0$$

$$m = \frac { -2 \pm \sqrt{4 - 4n^2\pi^2}}{2}$$

etc...

 

[Spoony]

aye true, but you know that 4*(n^2)*(pi^2) is greater than 4 for all n > 0 (n is an integer) except when n = 0 but if n = 0 then this is the only solution which leads to g(x) = 0 which then leads to theta = 0 and u = -(0.5)x + 2, ie u is only a function of x then and not of t.

 

[hgfalling]

m is complex for n > 0. That's ok, it's going to be exponentiated in the answer, right? So sines and cosines ensue.

 

[Spoony]

but then it wouldn't satisfy $$m^{2} + 2m + n^{2}\pi^{2} = 0$$ as 2m would be a complex number... literally just occoured to me that m^2 will then have a complex part to cancel out the 2m part.. d'oh!
cheers mate :D

 

[Spoony]

why would i need sines and cosines btw? would there be more steps before i have the solution to theta?

 

[hgfalling]

Not really, you're almost done.

If f(t) = Ae^mt and you know g(t) all you have to do is write the whole thing down as an infinite sum over all the lambdas. You just will want to move the imaginary part of m inside the summation because it depends on n.

 

[Spoony]

edit: wrong

 

[Spoony]

Do end up with
$$f(t) = Ae^{-1+(1-n^{2}\pi^{2})^{1/2}} + Ae^{-1-(1-n^{2}\pi^{2})^{1/2}}$$
Or is there individual constants for the positive and minus square root bits?

 

[hgfalling]

What you want to do is factor out a $$\sqrt{-1}$$ of the expression under the square root sign. Then you have the very basic second-order ODE with constant coefficients where the roots are $$a \pm bi$$. You should know from some other class or something that the solution to this is

$$f(t) = e^a (c_1 cos bt + c_2 sin bt)$$

It's no different here except recall that you're not really solving for general f(t) but for f_n(t), so the constants should be indexed by n as well. The constants, suppose we call them a_n (on cos) and b_n on sin, will be the Fourier coefficients of u(x,0).

 

[Spoony]

can the cosntants c_1 and c_2 be complex here? because i think that's what I am struggling with, also do i end up with;
$$f(t) = e^{-t}(Ae^{+(n^{2}\pi^{2}-1)^{1/2}} + Be^{-(n^{2}\pi^{2}-1)^{1/2}} )$$
Leading to;
$$f(t) = e^{-t} (c_1 cos bt + c_2 sin bt)$$
$$c_1 = A+B$$ and $$c_2 = i(A-B)$$
$$b = (n^{2}\pi^{2}-1)^{1/2}$$
not sure if this is right

 

[ipreferforks]

haha looks a lot like the DEF coursework in for tomorrow :P

 

[Spoony]

LOL hello fellow uni dude, you finished this yet? I think I've cracked it now but not too sure of all the final steps.

EDIT: keep the name of our uni private please, reasons of privacy and this could be, but i highly doubt it viewed as cheating (im of the mind that I am being helped with and not told awnsers :) )

 

[hgfalling]

No, remember that $sin u = \frac {e^u - e^{0u}}{2i}$. (Note the i in the denominator).

So $c_2$ won't be complex.

 

[Spoony]

so;
$$sin(t) = \frac{e^{it} - e^{-it}}{2i}$$
(rewrote it because i wasnt sure of the 0)
Now;
$$f(t) = A(cos(t\sqrt{n^{2}\pi^{2}-1}) + \frac{e^{it\sqrt{n^{2}\pi^{2}-1}} - e^{-it\sqrt{n^{2}\pi^{2}-1}}}{2})) + B(cos(t\sqrt{n^{2}\pi^{2}-1}) - \frac{e^{it\sqrt{n^{2}\pi^{2}-1}} - e^{-it\sqrt{n^{2}\pi^{2}-1}}}{2}))$$

which doesn't seem to have got me anywhere :S
also i think you mean that 0 to be a minus sign, i changed it :)edit; lol i don't know why i/2i = 1/i, its probably something to do with how tired i am... corrected now.

 

[Spoony]

OH unless you also call for cos to be rewritten in terms of exponentials?

 

[hgfalling]

Sorry, that's supposed to be -u, not 0u.

If you have roots of $a \pm bi$, then you have this:

$$f(t) = Ae^{(a + bi)t} + Be^{(a - bi)t}$$

This is the same as saying that $e^{(a + bi)t}$ and $e^{(a - bi)t}$ form a basis for the solution space of this ODE.

But let's change the basis as follows. Factoring out the $e^a$, let:
$$u_1 = \frac{e^{bix} + e^{-bix}}{2}$$
$$u_2 = \frac{e^{bix} - e^{-bix}}{2i}$$

This is also a basis. For any $A, B \in \mathbb C$ you can rewrite $Ae^{(a + bi)t} + Be^{(a - bi)t}$ as a linear combination of $u_1$ and $u_2$. Try it and see.

Now
$$u_1 = \cos {bx}$$
$$u_2 = \sin {bx}$$

So the solution can just be written as a combination of sines and cosines with real coefficients (multiplied by the exponential we factored out earlier, of course).

This is standard stuff from a first ordinary differential equations class. You'll use this over and over as you solve PDEs, so best make sure you understand it well.

 

[Spoony]

Sorry do you mean factor out the;
$$e^{at}$$ instead of $$e^{a}$$ ?

 

[hgfalling]

Yes that's what i meant.

 

[Spoony]

Ok so;
$$f(t) = e^{at}(Ae^{ibt} + Be^{-ibt})$$

now; do i use a direct substitution for u_1 and u_2 (ie just rearanging the e^(ibt) and e^(-ibt) to be in terms of u_1 and u_2 or is it a bit more sneaky than that?)
(I checked all my lecture notes as well just to make sure i hadnt missed this, and from what i can see we haven't touched this idea of writing an imaginary solution in this way, :( although we haven't done an enitre module on ODE's and this module is on Fourier transforms, Fourier series and PDE's oh and series solutions to ODE's)

thanks for all your help mate, i really would be stuffed otherwise :D

edit: I think that we wern't meant to solve this in this way tbh, i think there's a shortcut i probably missed... oh well, this way is just as right

 

[hgfalling]

I don't think you really need to show the substitution. I mean A and B are just arbitrary constants, so the c_1 and c_2 that go with u_1 and u_2 are just arbitrary too.

So as far I would be concerned, you can skip directly from the equation you just wrote down:
$$f(t) = e^{at}(Ae^{ibt} + Be^{-ibt})$$
to
$$f(t) = e^{at} (c_1 \cos {bt} + c_2 \sin {bt})$$.