how does the concept of y going to infinity simplify the equation??

[physica]

I’ve been stuck on question D for some time now. I cannot even begin to get my head around the fact that when Y goes to infinity the solution remains finite therefore we can use this to simplify the equation.

 

For X I got X(x)=Ae^(kx)+Be^(-kx)

For Y I got Y(y)=Dcos(ky)+Esin(Ky)

 

Can anyone give me any clues as to how the infinity concept simplifies the Y equation or at least what concepts to look up as we didn’t cover this in class. I may have got the two equations I typed above wrong. In that case I would be very grateful if someone told me as I can’t see it. Any help that stops me chasing my tail would be much appreciated.

[studiot]

 

Can anyone give me any clues as to how the infinity concept simplifies the Y equation

 

So you have the heat equation in a chimney.

 

x is bounded but y is unbounded, but the question helpfully points out that u is bounded everywhere (=remains finite).

 

When you separate the variables, the fact that u is bounded rules out many possible candidate functions for X(x) and Y(y) since

 

X(x) must be bounded in the rannge 0<x<1 and Y(y) must be bounded in the range 0<y<infinity

 

so for instance e^x is admissible but e^y is not since this tends to infinity (is unbounded) as y tends to infinity.

[physica]

thank you for your reply, will think about it some more.... are the equations that I gave in my initial post correct?