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 called Laplace’s equation.
If φ(x, y, z) =
1
x 2 + y 2 + z 2
= (x
2
+ y
2
+ z
2
)
−
1
2
then differentiating partially with respect to x gives:
∂φ
∂x
= −
1
2
(x
2
+ y
2
+ z
2
)
− 3
2 (2x)
= −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 (2x )
+ (x
2
+ y
2
+ z
2
)
− 3
2 (−1)
by the product rule
=
3x 2
(x 2 + y 2 + z 2 )
5
2
−
1
(x 2 + y 2 + z 2 )
3
2
=
(3x 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 =
(3y 2 ) − (x 2 + y 2 + z 2 )
(x 2 + y 2 + z 2 )
5
2
and
∂ 2 φ
∂z 2 =
(3z 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 =
(3x 2 ) − (x 2 + y 2 + z 2 )
(x 2 + y 2 + z 2 )
5
2
+
(3y 2 ) − (x 2 + y 2 + z 2 )
(x 2 + y 2 + z 2 )
5
2
+
(3z 2 ) − (x 2 + y 2 + z 2 )
(x 2 + y 2 + z 2 )
5
2
=
⎛
⎜
⎜
⎝
3x 2 − (x 2 + y 2 + z 2 )
+ 3y 2 − (x 2 + y 2 + z 2 )
+ 3z 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 problems on the
solution of partial differential equations by
direct partial integration
1. Determine the general solution of
∂u
∂ y
= 4t y.
[u =2t y 2 + f (t )]
2. Solve
∂u
∂t
= 2t cos θ given that u = 2t when
θ = 0.
[u =t 2 (cos θ − 1) + 2t ]
3. Verify that u(θ, t ) =θ 2 + θt is a solution of
∂u
∂θ
− 2
∂u
∂t
= t .
4. Verify that u = e
−y cos x is a solution of
∂ 2 u
∂x 2 +
∂ 2 u
∂ y 2 = 0.
5. Solve
∂ 2 u
∂x∂ y
= 8e y sin 2x given that at y = 0,
∂u
∂x
= sin x, and at x =
π
2
, u =2y 2 .
u =−4e y cos 2x − cos x + 4 cos 2x
+ 2y 2 − 4e y + 4
6. Solve
∂ 2 u
∂x 2 = y(4x 2 − 1) given that at x = 0,
u =sin y and
∂u
∂x
= cos 2y.
u = y
x 4
3
−
x 2
2
+ x cos 2y + sin y