Page 377 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes respectively. Therefore
x
V ( , ) = (A cos x B sin x )( cos h y D sin h y ) ...(xxi)
C
y
Using boundary conditions (xvii) and (xviii) in (xxi), we get
n
A = 0 and (n 1, 2, 3,...)
a
Hence for each value of n, we have

n y n y n
x
V ( , ) = C n cosh D n sin h sin x ...(xxii)
y
n 1 a a a
Using (xix) in (xxii) we have
n
x
V ( ,0) = C n sin x f ( )
x
n 1 a
Therefore
a
2 n
x
C n = f ( )sin x dx ...(xxiii)
a a
0
Again using (xx) in (xxii), we have
V n b n b n
C m sin h D n cos h sin
y a a a = 0
y b n 1
This will be true for all values of x, if
n b n
C sinh D cos h b = 0
n n
a a
or
n b
D n = C n tanh s ...(xxiv)
a
Therefore (xxii) with coefficients given by (xxiii) and (xxiv) is the solution of the equation (i)
satisfying all the given boundary conditions.
Self Assessment

3. Solve
2 2
U U 0
x 2 y 2

subject to the conditions

y
U (0, ) 0
y
l
U ( , ) 0
n x
x
a
and U ( , ) sin and U ( , 0) 0 for n 1, 2, 3, ...
x
l
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