Multivariate Calculus - Bruce E. Shapiro

LECTURE 9. THE PARTIAL DERIVATIVE 63
f yy = ∂ ∂y
We observe in passing that f xy = f yx .
Example 9.6 The Heat Equation is
∂f
∂y = ∂ ∂y (2x3 y + 21y 2 ) = 2x 3 + 42y
f yx = ∂ ∂f
∂x ∂y = ∂
∂x (2x3 y + 21y 2 ) = 6x 2 y.
∂u
∂t = u
c∂2 ∂x 2
where u(x, t) gives the temperature of an object as a function of position and time.
Show that
u = 1 √
t
e −x2 /(4ct)
satisfies the heat equation (i.e., that it is a solution of the heat equation).
Solution. By the product rule:
By the chain rule,
Furthermore
∂u
∂t
( 1√t
e −x2 /(4ct))
= ∂ ∂t
= √ 1 ∂ /(4ct)
t ∂t e−x2 + e −x2 /(4ct) ∂ 1
√
∂t t
∂
∂t e−x2 /(4ct)
( ) −x
2
= e −x2 /(4ct) ∂ ∂t 4ct
( )
= e −x2 /(4ct) −x
2 ∂
4c ∂t (t−1 )
( )
= e −x2 /(4ct) −x
2
(−t −2 )
4c
= x2
4ct 2 e−x2 /(4ct)
Hence
∂u
∂t
∂ 1
√ = ∂ ∂t t ∂t (t−1/2 ) = (−1/2)t −3/2
= √ 1 ( ) x
2
t 4ct 2 e −x2 /4ct + e −x2 /4ct (−1/2)t −3/2
(
= e −x2 /4ct x
2
4ct 5/2 − 1 )
2t 3/2
= e−x2 /4ct
√
t
( x
2
4ct 2 − 1 2t
)
Math 250, Fall 2006 Revised December 6, 2006.