An introduction to partial differential equations 517
Problem 3. Verify that
φ(x,y,z)=
1
√
x^2 +y^2 +z^2
satisfies the partial
differential equation:
∂^2 φ
∂x^2
+
∂^2 φ
∂y^2
+
∂^2 φ
∂z^2
=0.
The partial differential equation
∂^2 φ
∂x^2
+
∂^2 φ
∂y^2
+
∂^2 φ
∂z^2
=0 is calledLaplace’s equation.
Ifφ(x,y,z)=
1
√
x^2 +y^2 +z^2
=(x^2 +y^2 +z^2 )−
1
2
then differentiating partially with respect toxgives:
∂φ
∂x
=−
1
2
(x^2 +y^2 +z^2 )−
(^32)
( 2 x)
=−x(x^2 +y^2 +z^2 )−
3
2
and
∂^2 φ
∂x^2
=(−x)
[
−
3
2
(x^2 +y^2 +z^2 )−
5
(^2) ( 2 x)
]
+(x^2 +y^2 +z^2 )−
3
(^2) (− 1 )
by the product rule
3 x^2
(x^2 +y^2 +z^2 )
5
2
−
1
(x^2 +y^2 +z^2 )
3
2
( 3 x^2 )−(x^2 +y^2 +z^2 )
(x^2 +y^2 +z^2 )
5
2
Similarly, it may be shown that
∂^2 φ
∂y^2
( 3 y^2 )−(x^2 +y^2 +z^2 )
(x^2 +y^2 +z^2 )
5
2
and
∂^2 φ
∂z^2
( 3 z^2 )−(x^2 +y^2 +z^2 )
(x^2 +y^2 +z^2 )
5
2
Thus,
∂^2 φ
∂x^2
- ∂^2 φ
∂y^2
∂^2 φ
∂z^2
( 3 x^2 )−(x^2 +y^2 +z^2 )
(x^2 +y^2 +z^2 )
(^52)
( 3 y^2 )−(x^2 +y^2 +z^2 )
(x^2 +y^2 +z^2 )
5
2
( 3 z^2 )−(x^2 +y^2 +z^2 )
(x^2 +y^2 +z^2 )
5
2
⎛
⎜
⎜
⎝
3 x^2 −(x^2 +y^2 +z^2 )
- 3 y^2 −(x^2 +y^2 +z^2 )
- 3 z^2 −(x^2 +y^2 +z^2 )
⎞
⎟
⎟
⎠
(x^2 +y^2 +z^2 )
5
2
= 0
Thus,
1
√
x^2 +y^2 +z^2
satisfies the Laplace equation
∂^2 φ
∂x^2
∂^2 φ
∂y^2
∂^2 φ
∂z^2
= 0
Now try the following exercise
Exercise 199 Further problemson the
solution of partial differential equations by
directpartial integration
- Determine the general solution of
∂u
∂y
= 4 ty. [u= 2 ty^2 +f(t)]
- Solve
∂u
∂t
= 2 tcosθgiven thatu= 2 twhen
θ= 0. [u=t^2 (cosθ− 1 )+ 2 t]
- Verify thatu(θ ,t)=θ^2 +θtis a solution of
∂u
∂θ
− 2
∂u
∂t
=t.
- Verify thatu=e−ycosx is a solution of
∂^2 u
∂x^2
+
∂^2 u
∂y^2
=0.
- Solve
∂^2 u
∂x∂y
=8eysin2xgiven that aty=0,
∂u
∂x
=sinx,and atx=
π
2
,u= 2 y^2.
[
u=−4eycos 2x−cosx+4cos2x
+ 2 y^2 −4ey+ 4
]
- Solve
∂^2 u
∂x^2
=y( 4 x^2 − 1 )given that atx=0,
u=sinyand
∂u
∂x
=cos2y.
[
u=y
(
x^4
3
−
x^2
2
)
+xcos 2y+siny
]