BVP for two-dimensional Lap lace Equation:

We consider the cases of semi-infinite Lamina & Finite Lamina respectively.

BVP for steady Heat flow in a semi-infinite Lamina:

We rewrite BVP as

(1) Uxx + Uyy = 0 for 0<x<a, 0<y<∞

(2) U(0,y) = 0 = U(a,y) for 0<y<∞

(3) U(x, ∞)=0 for 0<x<a

(4) U(x,0)=f(x)

Assuming a separation of variables solution of the form

---------------------------------------------------------------(5)

Where F(x) is only a function of x and G(y) is only a function of y only. On

differentiating equation (1), we obtain

( ) ( )yH xFx

U "2

2

=∂∂

& ( ) ( )yH xFy

U "2

2

=∂∂

On substituting of (1) & above in Lap lace’s equation & division by FH gives

.H

H

F

F ""

λ=−= . & Hence we get two ordinary differential equations as

-------------------------------

(6)

-------------------------------------------------

(7)

& The infinitely many solutions of (6) are given by

-----------------------------------------------------(8)

1

Lap lace Equation

U(x, y)=F(x)H(y)

( ) } ( ) ( ) ( )aF00F Nn a

n- where0xFF

2" ==∈

π=λ=λ−

( ) } ( )Nn a

n- where0yHH

2" ∈

π=λ=λ+

( ) ( ) Nn wherea

xnsinBxF nn ∈

π=

& The general solution of (7) is given by ( ) ( )Nn eEeDyH a

yn

na

yn

nn ∈

+=

π

−

π

.

Applying the principle of superposition the general solution is given by

--------------(9)

Where an=BnDn and bn=BnEn. Applying boundary conditions we get

( ) .0a .0a

xnsinea,xU n

a

n

1nn =⇒=

π=∞

∞π∞

=∑

( ) ( ) ( ) ( )

π=⇒=

π+= ∑∑

∞

=

∞

= a

xnsinbxf .xf

a

xnsinba0,xU

1nn

1nnn

& Hence from (5) the general solution is given by

----------------

(10)

BVP for steady Heat flow in a finite Lamina:

We rewrite BVP as

(1) Uxx + Uyy = 0 for 0<x<a, 0<y<b

(2) U(0,y) = 0 = U(a,y) for 0<y<b

(3) U(x, b)=0 for 0<x<a

(4) U(x,0)=f(x)

Assuming a separation of variables solution of the form

---------------------------------------------------------------(5)

Where F(x) is only a function of x and G(y) is only a function of y only. As in the

previous case the general solution of (1) is given by

--------

------(6)

2

( ) ( ) ( ) ( ) ∑∑∑∞

=

π

−

π∞

=

∞

=

π

+===

1n

a

yn

na

yn

nn1n

n1n

n a

xnsinebeayHxFy,xUy,xU

( ) ( ) ( ) .dxa

xnsinxf

a

2b with Nn ;

a

xnsineay,xU

a

0

n1n

a

yn

n

π=∈

π== ∫∑

∞

=

π

U(x, y)=F(x)H(y)

( ) ( ) ( ) ( ) ∑∑∑∞

=

π

−

π∞

=

∞

=

π

+===

1n

a

yn

na

yn

nn1n

n1n

n a

xnsinebeayHxFy,xUy,xU

Where an=BnDn and bn=BnEn are arbitrary constants. For ease in evaluating the constants

rewriting equation (6) as

-----------------------------------(7)

For combining two hyperbolic functions in the square bracket, with any lose of

generality, we may write,

π

=

π

=a

yncoshFb &

a

ynsinhFa 0

n'n

0n

'n

Where Fn and 0y are arbitrary constants. Substituting above values in equation (7) we get,

( ) ( )Nn a

xnsin

a

ynsinh

a

yncosh

a

yncosh

a

ynsinhFy,xU

1n

00n ∈

π

π

π

+

π

π

= ∑∞

=

--------------------------------------------------(8)

Applying conditions (3) & (4) we get,

( ) ( ) ( )0 F since ; 0

a

bynsinh 0

a

xnsin

a

bynsinhFb,xU n

0

1n

0n ≠=

−π

⇒=

π

−π

= ∑∞

=by 0 =⇒

( ) ( ) ( ) ∑∑∞

=

∞

=

π

π=⇒=

π

π

=1n

n1n

0n a

xnsin

a

b nsinhFxf xf

a

xnsin

a

ynsinhF0,xU

So, f(x) is half-range Fourier series with Fourier coefficient

π

a

bnsinhFn . Finally the

generally is given by

3

( ) ∑∞

=

π

π+

π==

1n

'n

'n a

xnsin

a

ynsinhb

a

yncoshay,xU

( ) ( )∑∞

=

π

−π

==1n

0n a

xnsin

a

yynsinhFy,xU

( ) ( ) ( ) dxa

xnsinxf

a

bnsinha

2F with

a

xnsin

a

ybnsinhFy,xU

a

0

n1n

n

π

π

=

π

−π== ∫∑

∞

=

Examples

Solve the following examples.

1. Solve 0y

u

x

u2

2

2

2

=∂∂+

∂∂

which satisfy the conditions ( ) ( ) ( ) 00,xuy,luy,0u === &

( ) Nn,l

xnsina,xu ∈

π=

2. A rectangular plate with insulated surface is 10 cm wide and so long compared to

its width it may be considered infinite in the length without introducing an

appreciable error. If the temperature along one short edge y=0 is given by

( ) ( )

<<−<<

=10.x5 ,x1020

5.x0 ,x200,xu While the two long edge x=0 & x=10 as well as

the other short edge are kept at 00 C. Prove that the steady-state temperature u(x,t)

at any point (x,y) is given by ( ) ( )( )

( ) ( )10

y1n2

12

1n

2e.

l

x1n2sin

1n2

1800y,xu

π−−∞ + π−−

−π

= ∑ .

3. An infinitely long uniform plate is bounded by two parallel edges and end at right

angle to them. The breadth is π ,this end is maintained at a temperature u0 at all

points and other edges are at zero temperature. Determine the temperature at any

point of the plate in the steady-state.

4. Solve 0y

u

x

u2

2

2

2

=∂∂+

∂∂

for 0<x< π , 0<y< π with conditions given

( ) ( ) ( ) 0,xuy,uy,0u =π=π= & ( ) xsin0,xu 2=

5. A long rectangular plate of width a cm with insulated surface has its temperature

v equal to zero on both the long sides and one of the short sides so that

4

( ) ( ) ( ) ( ) .kx0,xv,0,xv,0y,av,0y,0v ==∞== Show that the steady-state

temperature within the plate is ( ) ( ).

a

xnsine

n

1ak2y,xv

1

a

yn1n

∑∞ π−+ π−

π=

6. A rectangular plate with insulated surfaces is 8 cm wide and so long compared to

its width that it may be considered infinite in length without introducing an

appreciable error. If the temperature along one short edge y=0 is given by

( ) 8x0 ,8

xsin1000,xu <<π= while the two long edges x=0 & x=8 as well as the

other short edge are kept at 00C, show that the steady-state temperature at any

point is given by ( ) .8

xsine100y,xu 8

y π=π−

7. Solve 0y

u

x

u2

2

2

2

=∂∂+

∂∂

within the rectangle by0,ax0 ≤≤≤≤ given that

( ) ( ) ( ) 0b,xuy,auy,0u === & ( ) ( )xax0,xu −= .

8. A square plate is bounded by the lines x=0,y=0,x=20 & y=20. its faces are

insulated. The temperature along the upper horizontal edges is given

u(x,20)=x(20-x), when 0<x<20, while other edges are kept at 00C. Find the steady

state temperature in the plate.

9. A rectangular plate has sides a & b. Let the side of length a be taken along OX

and that of length b along OY and other sides along x=a & y=b. The sides

x=0,x=a,y=b are insulated and the edge y=0 is kept at the temperature .a

xcosu 0

π

Find the steady-state temperature at any point (x,y).

10. The temperature u is maintained at 00 along three edges of a square plate of length

100 cm and the fourth edge is maintained at 1000 until steady-state conditions

prevail. Find an expression for the temperature u at any point (x,y). Hence show

5

that the temperature at the centre of the plate is

−−−−−−−−π

+π

−ππ

2

5cosh5

1

2

3cosh3

1

2cosh

1200

6