Mathematical Tools for Physics

10—Partial Differential Equations 294

Does the exact answer look like this? It doesn’t seem to, but look closer. Ifbathen because 0 < y < byou
also haveya. The hyperbolic function factors in Eq. ( 23 ) will have very small arguments, proportional tob/a.
Recall the power series expansion of the hyperbolic sine:sinhx=x+···. These factors become approximately

sinh

(

(2`+ 1)πy/a

)

sinh

(

(2`+ 1)πb/a

) ≈

(2`+ 1)πy/a

(2`+ 1)πb/a

=

y

b

The temperature solution is then

T(x,y)≈

4

π

T 0

∑∞

`=0

1

2 `+ 1

y

b

sin

(2`+ 1)πx

a

=T 0

y

b

Where did that last equation come from? The coefficient ofy/bis just the Fourier series of the constantT 0 in
terms of sines on 0 < x < a.
What about the opposite extreme, for whichba? This is the second picture just above. Instead of
being short and wide it is tall and narrow. For this case, look again at the arguments of the hyperbolic sines.
Nowπb/ais large and you can approximate the hyperbolic functions by going back to their definition.

sinhx=

ex+e−x

2

≈

1

2

ex, forx 1

The denominators in all the terms of Eq. ( 23 ) are large,≈eπb/a(or larger still because of the(2`+ 1)). This
will make all the terms in the series extremely smallunlessthe numerators are correspondingly large. This means
that the temperature stays near zero unlessyis large. That makes sense. It’s only forynear the top end that
you are near to the wall with temperatureT 0.
You now have the case for whichbaandya. This means that I can use the approximate form of
the hyperbolic function for large arguments.

sinh

(

(2`+ 1)πy/a

)

sinh

(

(2`+ 1)πb/a

)≈

e(2`+1)πy/a

e(2`+1)πb/a

=e(2`+1)π(y−b)/a

The temperature distribution is now approximately

T(x,y)≈

4

π

T 0

∑∞

`=0

1

2 `+ 1

e−(2`+1)π(b−y)/asin

(2`+ 1)πx

a

(24)