Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

214_Functions of Several Variables_[C/i. 6
..
dt 2 ~~dt dt~~dadt
Differentiate again using
the same formulas:
d 2 u d 2
u\, x , (d
dx*~dx\dx) da\dx
d z u
_d 2 u d*u d'u
/)ft 8 '
rtn /)ft
'
/}R2 '
2 u d*u
Substituting into the ^iven equation, we will have
2 /d 2 a d 2 u d 2 u\ Jd 2 u d 2 u d*u\
d 2 u r = 0.
Example 5. Transform the equation x2
^- + = y2 z
-g-
2
, taking = ,v, v ~
-=
y x
for the new independent *
variables, and w
z x
for the new
function.
Solution. Let us express the partial derivatives y- and ~ in terms of the
partial derivatives ^~ and ^
ships between the old and new variables:
On the other hand,
Therefore,
or
Whence
and, consequently,
u x t v
dw ,
. To do this, differentiate the given relation-
dx dy dx dz
-^^, oy_
dw dw
dw = :; du ~\ dv
du dv
, dw
, dx dz
-3- du + -5- dv = =
du dv x2 z 2
dw , i^ w _f^ x __m dy\^,^ x . &*_
.
2
^
dw 1 dw\ , . z 2 dw .
dz^_ 2 / J
dx~~ \x z ~~du ~~
dw 1 dw \
x*dv J