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Laplace EquationSeparation of Variables in Three Dimensions (3D)

A two-dimensional (2D) example

The method of separation of variables applied to

Laplace Equation

The method of separation of variables applied to Laplace
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Outline

1 Laplace Equation

2 Separation of Variables in Three Dimensions (3D)

3




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Laplace Equation

We know that the Laplace equation is2V=0, (1)

where, in 3D Cartesian coordinates

2

2

x2+ 2

y2+ 2

z2

withV(r) V(x,y,z)

specified on various boundaries. Boundary conditions may be

of the Dirichlet type (Vspecfied) or of the Neumann kind ( Vnspecified).

Our aim is to develop a method based on the concept of theseparation of variables to solve the Laplace equation,

consistent with the boundary conditions.The method of separation of variables applied to Laplace


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The method of separation of variables is based upon theconjecture (guess, tukka,...) that the solution can be written

in the formV(x,y,z) =X(x)Y(y)Z(z), (2)

where X(x), Y(y), and Z(z) are, respectively, functions of thevariables, x, y, and z, only.

Separation of Variables implies the product form ofV(x,y,z).

Substituting Eq. 2, the Laplace equation, we get

(

2

x2+

2

y2+

2

z2)X(x)Y(y)Z(z) =0 (3)

=

Y(y)Z(z)d2X

dx2 +X(x)Z(z)

d2Y

dy2 +X(x)Y(y)

d2Z

dz2 =0 (4)

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3D Separation of Variables, contd.

Note that partial derivatives have been replaced by totalderivatives, why?

This is because functions X(x), Y(y), and Z(z) are functionsof one variable only.

On dividing Eq. 4byV(x,y,z) =X(x)Y(y)Z(z), we obtain

1

X

d2X

dx2 +

1

Y

d2Y

dy2 +

1

Z

d2Z

dz2 =0

=

1

X

d2X

dx2 =

1

Y

d2Y

dy2

1

Z

d2Z

dz2 (5)

Note that LHS of this equation depends only on x, while RHSon y and z. What does it mean?


L l i
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3D separation of variables, contd.

Eq. 5can be satisfied only if both sides are equal to the sameconstant, say, l2

=1

X

d2X

dx2 =l2

and

1

Y

d2Y

dy2

1

Z

d2Z

dz2 =l2

Xequation becomes

d2X

dx2 + l2

X=0

and Y and Zequation can be rewritten as

1

Y

d2Y

dy2 =

1

Z

d2Z

dz2 l2


L l E i


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Separation of variable argument leads to

1

Y

d2Y

dy2 =m2

and 1

Z

d2Z

dz2 l2 =m2,

where m2 is another constant. So that

d

2

Ydy2

+m2Y=0

andd2Z

dz2 (l2 +m2)Z=0


L l E ti
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Defining n2

=(l2

+m2

) we obtain

d2Z

dz2 +n2Z=0

Finally, we get three ordinary differential equations in X, Y,and Z, in place of a partial differential equation (PDE)

d2X

dx2 + l2X = 0

d2Y

dy2 +m2

Y = 0

d2Z

dz2 +n2Z = 0

where l2 +m2 +n2 =0.


Laplace Equation
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These three ordinary differential equations (ODEs) can be solved,

in conjunction with the boundary conditions.


A E l i 2D
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An Example in 2D

Figure : 2D Boundary conditions

We aim to solve for V=V(x,y) satisfying the 2D Laplaceequation

2V

x2 +

2V

y2 =0,

subject to boundary conditions above.

Laplace Equation
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2D Laplace eqn.

Use separation of variables conjecture V(x,y) =X(x)Y(y) in2D Laplace equation to obtain

1

X

d2X

dx2

=1

Y

d2Y

dy2

=k2 (say),

where k2 is a constant.

=

d2X

dx2 k2X= 0 (6)

d2Y

dy2 +k2Y= 0 (7)


Laplace Equation
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Eqs. 6and7subject to boundary conditions

X(x=0) =V0, X(x ) =0

Y(y=0) = Y(y=a) =0

Eqs. 6and7have solutions

X(x) = Aekx+Bekx

Y(y) = Csin(ky) +Dcos(ky)

X(x) =0 = B=0Y(y=0) =0 = D=0

Y(y=a) =0 = sin(ka) =0= ka=n

k kn=n/a


Laplace Equation
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p qSeparation of Variables in Three Dimensions (3D)


So most general solution satisfying the given BCs

V(x,y) =X(x)Y(y) =

n=1Anenx

/a

sin(ny/a),

where An are constants to be determined. Use the BCV(x=0,y) =V0, for 0y a.

=n=1

An sin(ny/a) =V0

Multilply both sides by sin(my/a) (m is an integer) andintegrated for 0y a

n=1

An

a

0

sin(my/a) sin(ny/a)dy=V0 a

0

sin(my/a)dy

Usinga

0sin(my/a) sin(ny/a)dy= (a/2)m,n, and

a0sin(my/a)dy= (1 cos(m))(a/m)The method of separation of variables applied to Laplace

Laplace Equation
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Separation of Variables in Three Dimensions (3D)A two-dimensional (2D) example

Where m,n is called Kronecker delta and defined as

m,n

= 1, form

=n

= 0, form =n

Using this, we get

An=2V0

n

(1 cosn)

So that

An =0 for even values ofn

An = 4V0n

for odd n

Thus, the final solution is

V(x,y) =4V0 n=1,3,5,...

1

nenx/a sin(ny/a)

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Date post:	13-Apr-2018
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