- Show thatf(x,y)=ln(x^2 +y^2 )satisfies thepartial differential equation:
∂^2 f
∂x^2
+
∂^2 f
∂y^2
= 0
This is called theLaplace equationin two-dimensional rectangular coordinates. It is
very important in fluid mechanics, electromagnetism, and many other areas of science
and engineering, as well as being a key equation in pure mathematics.
Answers
- (i)fx= 3 x^2 y^2 + 4 y^4 ,fy= 2 x^3 y+ 16 xy^3 ,
fxx= 6 xy^2 ,fyy= 2 x^3 + 48 xy^2 ,fxy= 6 x^2 y+ 16 y^3
(ii) fx=exy(ycos(x+y)−sin(x+y)),
fy=exy(xcos(x+y)−sin(x+y)),
fxx=exy((y^2 − 1 )cos(x+y)− 2 ysin(x+y)),
fyy=exy((x^2 − 1 )cos(x+y)− 2 xsin(x+y)),
fxy=exy(xycos(x+y)−(x+y)sin(x+y))
16.5 The total differential
∂z
∂x
,
∂z
∂y
give the rates of increase ofz=f(x,y)in thexandy directions respec-
tively – what about the increase in a general direction? I.e. ifx,yincrease byδx,δy,
by how much doeszincrease? From Figure 16.6 we can see that the increase inz,δz,is
given approximately by
δz
δz
δx
δx+
δz
δy
δy
Such diagrams are not to everyone’s taste, so a few examples might help.
Problem 16.3
Obtain expressions fordzin the cases
(i) z=xYy (ii) z=xy (iii) z=x^2 Yy^2
What happens ifdx,dyare so small that products of them can be neglected?
(i) is not so bad:
δz=x+δx+y+δy−(x+y)=δx+δy
This result is exact. There are no products ofδx,δy.
(ii) is a little more complicated:
δz=(x+δx)(y+δy)−xy
=yδx+xδy+δxδy