- #1

sa1988

- 222

- 23

## Homework Statement

## Homework Equations

## The Attempt at a Solution

I'd first like to apologise in advance because I may be making a few more threads like this. I have an exam on Tuesday and am looking at past papers which don't have any answers for me to check whether I'm doing the right thing, so I'm turning to the heroes of Physics Forums in my hour need.

For the above question I'm pretty sure I've done it correctly but I can't see how the work I've done is even remotely worth 30 marks.

Here's my working:

__a)__##T(x,y) = T(x)T(y)##

Put into Laplace equation:

##\frac{d^2T(x)}{dx^2}T(y) + \frac{d^2T(y)}{dy^2}T(x) = 0##

##\frac{d^2T(x)}{dx^2}\frac{1}{T(x)} + \frac{d^2T(y)}{dy^2}\frac{1}{T(y)} = 0##

##\frac{d^2T(x)}{dx^2}\frac{1}{T(x)} = - \frac{d^2T(y)}{dy^2}\frac{1}{T(y)}##

Then for separation constant ##m^2## :

##\frac{d^2T(x)}{dx^2} = m^2 T(x)##

##\frac{d^2T(y)}{dy^2} = -m^2 T(y)##

Done. Is that really worth 8 marks?

__b)__By inspection I can already see the solutions:

##T(x) = Ae^{mx}+Be^{-mx}##

##T(y) = Ce^{imy}+De^{-imy}##

So,

##T(x,y) = T(x)T(y) = (Ae^{mx}+Be^{-mx})(Ce^{imy}+De^{-imy})##

Done. For 10 marks? I suppose I need to show the step by step working to get to those obvious solutions using the the usual ##f(x) = e^{\lambda x}## trial solution route? But even then it seems a bit simple for 10 marks.

__c)__Boundary conditions are:

##T(0,0) = 0##

##T(x,0) = 0##

##T(x,L) = 0##

##T(L,y) = sin(\frac{6\pi y}{L})##

These conditions give four equations:

##i. (A+B)(C+D) = 0##

##ii. (Ae^{mx}+Be^{-mx})(C+D) = 0##

##iii. (A+B)(Ce^{imL}+De^{-imL}) = 0##

##iv. (Ae^{mL}+Be^{-mL})(Ce^{imy}+De^{-imy}) = sin(\frac{6\pi y}{L})##

Condition ##ii## leads to ##D = -C##

Condition ##iii## leads to ##B= -A##

These can then be put into condition ##iv##:

##AC(e^{mL}-e^{-mL})(e^{imy}-e^{-imy}) = sin(\frac{6\pi y}{L})##

Then rewrite the exponentials in trig form:

##AC sinh(mL)sin(my) = sin(\frac{6\pi y}{L})##

Now this is clearly only possible for ##m = \frac{6\pi}{L}## and ##AC = \frac{1}{sinh(6\pi)}##,

leading to a full solution of:

##T(x,y) = \frac{sinh(\frac{6\pi x}{L})}{sinh(6\pi)}sin(\frac{6\pi y}{L})##

I've checked this against the Laplace equation and it holds so I'm fairly sure it's correct. But it's a pretty easy 15 marks.

So overall I'm wondering if there's something I've done wrong? Should I be showing more step-by-step working or something? The 10-mark question is the biggest worry for me as it seemed far too easy. I'm convinced I missed something, yet I've come to the correct final answer at the end so what could possibly be penalised?

Thanks for any thoughts/opinions.