Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

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





  1. Determine the general solution of
    ∂u
    ∂y


= 4 ty. [u= 2 ty^2 +f(t)]


  1. Solve


∂u
∂t

= 2 tcosθgiven thatu= 2 twhen
θ= 0. [u=t^2 (cosθ− 1 )+ 2 t]


  1. Verify thatu(θ ,t)=θ^2 +θtis a solution of
    ∂u
    ∂θ


− 2

∂u
∂t

=t.


  1. Verify thatu=e−ycosx is a solution of
    ∂^2 u
    ∂x^2


+
∂^2 u
∂y^2

=0.


  1. 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

]


  1. 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

]
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