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Deathslayer7
2013-10-19, 01:47 AM
I'm not sure what the question is asking for.

Given:

Geometry: a square box with the top and right side equal to T1

k[T_xx+T_yy]=pcT_t

where T=Temp and t=time and T_xx (read: partial squared T with respect to partial x squared) is the 2nd derivative with respect to those variables.

Given the boundary conditions:

T_x(0,y,t)=0
T(L,y,t)=T1
T_y(x,0,t)=0
T(x,L,t)=T1

and the initial condition:
T(x,y,0)=t0(x,y)

a) Determine a(x0 and b(y) that will give a homogeneous problem for u(x,y,t). State the differential equation, boundary conditions and initial condition for u(x,y,t). Done

b) Under what conditions will you be able to solve for u(x,y,t) by assuming that u(x,y,t)=v(x,t)*w(y,t)?

c) Determine the problems that must be solved for v(x,t) and w(y,t)

Solution:

The professor forces us to normalize the equation before even starting the problem and I thank him for that. That being said, part a) is done. I can do that part. But I do not understand what part b) is asking for.

I have the following for a) after normalizing.

a(x)=0
b(y)=0

u_xx+u_yy=u_t

u_x(0,y,t)=0
u(L,y,t)=0
u_y(x,o,t)=0
u(x,L,t)=0

u(x,y,0)=1

So what exactly is part b) asking for? I don't understand. Note this is a Heat Conduction class, not a math class. This is the first time I ever encountered separation of variables and I am teaching myself because the teacher isn't that great at it. I got the knack of it down, but the wording is what trips me up.

Please note: I know I can normally go straight to separation of variables, but the problem specifically asks for this step.

Teddy
2013-10-19, 07:14 AM
My initial thoughts without actually doing the separation would be that there's one or more values for one or more of the variables (I suspect t) that makes the equation unsolvable, and that the question is asking what conditions must be true for that variable (for example t >= 0).

Deathslayer7
2013-10-19, 01:20 PM
I guess my question is this part:

u(x,y,t)=v(x,t)*w(y,t)?

Why do this? Why not just go to separation of variables since it is homogenous and linear?

Teddy
2013-10-19, 01:27 PM
That is a separation of variables, namely the variables x and y. I guess you could argue that you should separate it as X(x)Y(y)T(t) (I'm skipping the multiplication signs here, because I'd invariably start to mix my '*'s up with convolutions), and by this point, I'm starting to draw blank as well...

Deathslayer7
2013-10-19, 01:57 PM
X(x)Y(y)T(t)

That's how I am used to seeing it as well. Anyway, thanks for the help.

theKOT
2013-10-20, 01:05 AM
So it's been a few years since I've done this but I hope I can explain in a sufficient and accurate manner.
Separation of variables inherently relies on homogeneous boundary conditions in order to be applied. Without homogeneous boundary conditions, we cannot be certain that our dependence on x and y will be able to be expressed as a product function. So if it is impossible to write things in such a way that the BCs are homogeneous, you cannot separate the variables.
That's what half of PDE classes amounts to: torturing your eqn and function until they are homogeneous. For further reading and examples, look here (http://tutorial.math.lamar.edu/Classes/DE/SeparationofVariables.aspx).
Hope this helps.
edit: also this (http://tutorial.math.lamar.edu/Classes/DE/PDESummary.aspx) is a pretty nifty summary of the methods/limitations of separation of variables.
edit #2: I'm pretty sure the question was just asking where you were allowed to assume that the spatial part is separable, so I wouldn't worry about why the time portion was not also separated.

Heliomance
2013-10-20, 02:46 AM
Euch, I hate looking at equations in typable format. So hard to decipher!

Gray Mage
2013-10-20, 02:47 PM
Ok, it has been a while, but I think this is more or less what you need to do.

You know that u(x,y,t)=v(x,t)*w(y,t).

And you know the boundary conditions of u(x,y,t). v(x,t)*w(y,t) must also have the same boundary conditions.

So:

u(x,y,0) = v(x,0)*w(y,0) = 1
u_x(0,y,t) = v(0,t)_x*w(y,t) = 0
u(L,y,t) = v(L,t)_x*w(y,t) = 0
u_y(x,0,t) = v(x,t)*w(0,t)_y = 0
u(x,L,t) = v(x,t)*w(L,t)_y = 0
u_xx + u_yy = v(x,t)_xx*w(y,t) + v(x,t)*w(y,t)_yy = v(x,t)_t*w(y,t) + v(x,t)*w(y,t)_t = u_t

From the above you can conclude (among other things) that:

v(0,t)_x = v(L,t)_x = 0
w(0,t)_y = w(L,t)_y = 0

The above is probably what b) means by conditions.

And I think you can't separe as X(x)Y(y)T(t) because heat equations depend on time as well as space.

Kato
2013-10-23, 04:58 AM
Yeah, I'll try to make this short... the thing is that time dependent and space dependent variables in many physical applications are treated very differently.

Let's consider a simple example of f(x,y,t)=sin(x+t)*cos(y+t).
Basically, it's already in the v(x,t)*w(y,t) format, but there could be (slightly) more complex problems. But it's often much harder to separate the time variable in the equation from the spatial variables and often it's not necessary for the solution. If I recall correctly the assumptions is: If you can solve v and w separately, you can also solve u (if u=v*w).

(It's been way too long since I actually did solve such a PDE and I feel I really need some exercising. :smallfrown:)