Is this enough to get full marks?

[sa1988]

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.

 

[vela]

Your first boundary condition is redundant, and you're missing a boundary condition. Condition iii doesn't imply A=-B.

 

[sa1988]

Your first boundary condition is redundant, and you're missing a boundary condition. Condition iii doesn't imply A=-B.

So I am missing something then.

I know the first condition was left unused, but I can't see how it was needed.

If condition $iii$ doesn't imply $A=-B$, the only thing I could say is that it 'suggests':

$A=-B$
OR
$m=\frac{n\pi}{L}$
or both.

Similarly condition $i$ can only 'suggest':
$A=-B$
OR
$C=-D$
or both.

Is there some other route I should take to achieve values for A, B, C, D and m with certainty instead of having to use this loose 'trial guess' method?

And you say I'm missing another boundary condition altogether. Which condition is that?

Thanks.

 

[vela]

Reread the problem. You'll see you left out the condition for one of the edges.

 

[sa1988]

Reread the problem. You'll see you left out the condition for one of the edges.

Ha. Christ. I see it now. That changes everything.

$i. T(0,y) = 0$
$ii. T(x,0) = 0$
$iii. T(x,L) = 0$
$iv. T(L,y) = sin(\frac{6\pi y}{L})$

So,

$i. (A+B)(Ce^{imy}+De^{-imy}) = 0$

$\implies B = -A$

$ii. A(e^{mx}-e^{-mx})(C+D) = 0$

$\implies D = -C$

$iii. AC(e^{mx}-e^{-mx})(e^{imL}-e^{-imL}) = 0$

$\implies m = \frac{n\pi}{L}$, $\Big( n\in \mathbb{N}\Big)$

$iv. AC(e^{n\pi}-e^{-n\pi})(e^{i\frac{n\pi y}{L}}-e^{-i\frac{n\pi y}{L}}) = sin(\frac{6\pi y}{L})$

rewrite $iv$ as trig:

$iv. ACsinh(n\pi)(sin(\frac{n\pi y}{L}) = sin(\frac{6\pi y}{L})$
$\implies AC = \frac{1}{sinh(n\pi)}$
and
$n = 6$

Which gives the same answer as before.

But still, my error was only due to a tiny 'typo' so to speak. It still seems like a surprisingly small amount of steps across parts a, b and c to gain the full 30 marks.

 

[vela]

In the last part, I'd say
$$\sum_{n=1}^\infty A_n \sinh n\pi \sin \left(\frac{n \pi y}{L}\right) = \sin\left(\frac{6 \pi y}{L}\right),$$ from which you can conclude $A_6 = \frac{1}{\sinh 6\pi}$ and $A_n=0$ for $n \ne 6$.

 

[Ray Vickson]

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.

You would likely lose marks for using the exact symbol $T$ to stand for three different things in the same problem. Use better notation, such as $T_1(x), T_2(y)$ or $F(x), G(y)$ or something similar. It is never a good idea to make the marker work extra hard trying to decipher what you are doing.

 

[sa1988]

In the last part, I'd say
$$\sum_{n=1}^\infty A_n \sinh n\pi \sin \left(\frac{n \pi y}{L}\right) = \sin\left(\frac{6 \pi y}{L}\right),$$ from which you can conclude $A_6 = \frac{1}{\sinh 6\pi}$ and $A_n=0$ for $n \ne 6$.

You would likely lose marks for using the exact symbol $T$ to stand for three different things in the same problem. Use better notation, such as $T_1(x), T_2(y)$ or $F(x), G(y)$ or something similar. It is never a good idea to make the marker work extra hard trying to decipher what you are doing.

Thanks, duly noted.