# Analytical Solution of Partial Differential Equations I

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Analytical Solution of Partial Differential Equations I
```Week 12
Analytical Solutions of
Partial Differential Equations I
1
Introduction
The method of separation of variables is a powerful approach designed to obtain solutions of
initial and boundary value problems for partial differential equations. The key assumption
made is that the desired solution, being a function of multiple independent variables, can
be represented as a product of simpler functions each of which may depend only on a single
independent variable. As this assumption is introduced into the original partial differential
equation, systems of initial and/or boundary value problems for ordinary differential equations result. These can in turn be readily solved using methods learned earlier and the results
finally substituted back into the original product function to produce the final solution.
2
Elliptic Equations
Laplace’s equation
∇2 φ = 0
and Poisson’e equation
∇2 φ = h
are the most important examples of elliptic equations encountered in applications.
Consider as a first example the problem of steady state heat conduction in a thin rectangular plate of width l and height d. The edges x = 0, x = l and y = 0 are maintained
at T = 0 while at the edge y = d, T (x, d) = f (x). No heat flow along the z direction
perpendicular to the plate. The required temperature T (x, y) satisfies
∂ 2T
∂2T
+
=0
∂x2
∂y 2
1
To find T by separation of variables we assume the a particular solution can be represented
as a product of two functions each depending on a single coordinate, i.e.
Tp (x, y) = X(x)Y (y)
substituting into Laplace’s equation gives
−
1 d2 X
1 d2 Y
=
= k2
X dx2
Y dy 2
where k 2 is a constant called the separation constant. The above must be true since the LHS
is a function of x alone while the RHS a function of y alone. The constant is selected as k 2
in order to obtain a proper Sturm-Liouville problem for X (with real eigenvalues). A clue
about the appropriate choice of the placement of the minus sign above is derived on physical
grounds. Note that for a fixed value of y, the solution must be zero at x = 0, go through a
maximum at x = l/2 and become zero again for x = l. Therefore, a particular solution in
terms of simple trigonometric functions can be expected along the x-direction. On the other
hand, along the y-direction, at constant x, a monotonically increasing/decreasing function
is expected as y changes from zero to d.
With the above the original PDE problem has been transformed into a system of two
ODE’s, i.e.
X 00 + k 2 X = 0
subject to X(0) = X(l) = 0 and
Y 00 − k 2 Y = 0
subject to Y (0) = 0.
The solution for X(x) is
X = Xn = An sin(
nπx
)
l
with eigenvalues
kn =
nπ
l
for n = 1, 2, 3, ....
The solution of Y (y) is
Yn = Bn sinh(
nπy
)
l
so that the particular solution desired is
Tn =
∞
X
n=1
an sin(
nπy
nπx
) sinh(
)
l
l
2
The an ’s are determined by making the above satisfy the nonhomogeneous condition at
y = d, i.e.
f (x) =
∞
X
[an sinh(
n=1
nπx
nπd
)] sin(
)
l
l
which is a Fourier sine series representation of f (x) with coefficients
2
nπd
)=
cn = an sinh(
l
l
Z
0
l
f (x) sin(
nπx
)dx
l
so that the final solution is
∞
X
)
nπx sinh( nπy
l
T (x, y) =
)
cn sin(
nπd
l sinh( l )
n=1
Therefore, as long as f (x) is representable in terms of Fourier series, the obtained solution
coverges to the desired solution. Note also that the presence of homogeneous conditions at
x = 0, x = l made feasible the determination of the required eigenvalues.
3
Steady State Temperature in a Brick
Consider now the problem of steady state heat conduction in a rectangular parallelepiped
(length l1 , width l2 , height d) subject to zero temperature on all five faces except that at
z = d which is maintained at T (x, y, d) = f (x, y). Laplace’s equation is
∂2T
∂ 2T
∂ 2T
+
+
=0
∂x2
∂y 2
∂z 2
Assume a particular solution is of the form
Tp = X(x)Y (y)Z(z)
With the above one has
d2 X
∂ 2T
=
Y
Z
= Y ZX 00
∂x2
dx2
d2 Y
∂ 2T
=
XZ
= XZY 00
2
2
∂y
dy
and
d2 Z
∂ 2T
=
XY
= XY Z 00
∂z 2
dz 2
3
Substituting into the original PDE yields, after division by XY Z,
1 00 1 00 1 00
X + Y + Z =0
X
Y
Z
which can be rearranged to
1
1 00 1 00
Y + Z = − X 00
Y
Z
X
But the left hand side is a function of y and z only while the right hand side is a function of
x only, therefore, necessarily
1
1 00 1 00
Y + Z = − X 00 = k12
Y
Z
X
Now, rearrangement of the term on the left hand side yields
1
1 00
Z − k12 = − Y 00
Z
Y
but the term on the left hand side is a function of z only while the one on the right hand
side is a function of y only, thus, necessarily
1
1 00
Z − k12 = − Y 00 = k22
Z
Y
Two separation constants are needed in this case giving
X 00 + k12 X = 0
Y 00 + k22 Y = 0
and
Z 00 − (k12 + k22 )Z = 0
Note that homogeneous conditions occur at both boundaries in both the x and y directions
Xm = Am sin(k1 x) = Am sin(
Yn = Bn sin(k2 y) = Bn sin(
nπy
)
l2
and
Zmn = Cmn sinh(kmn z)
4
mπx
)
l1
2
where kmn
= k12 + k22 . The desired particular solution is then
Tp =
∞
∞ X
X
m=1 n=1
amn sin(
mπx
nπy
) sin(
) sinh(kmn z)
l1
l2
where the coefficients amn must be determined by incorporating the nonhomogeneous condition. This leads to a double Fourier series representation and the coefficients
amn =
4
1
mπx
nπy
4 Z l1 Z l2
f (x, y) sin(
) sin(
)dydx
sinh(kmn d) l1 l2 0 0
l1
l2
Nonhomogeneous Problems
Heat is generated internally inside a rectangular plate of dimensions a × b at an rate of Q
units of energy per unit time per unit volume. Two of the plate’s edges are insulated while
the other two are maintained at zero temperature. The mathematical formulation of the
problem consists of the statement of Poisson’s equation
∂2T
Q
∂ 2T
+
=−
2
2
∂x
∂y
K
The (homogeneous) boundary conditions are, at x = 0 for all y
∂T (0, y)
=0
∂x
at y = 0 for all x
∂T (x, 0)
=0
∂y
at x = a for all y
T (a, y) = 0
and at y = b for all x
T (x, b) = 0
Here we assume a particular solution exists with the following special form
T (x, y) = ψ(x, y) + φ(x)
such that the original problem can be reformulated as a superposition of two simpler problems; the one-dimensional, nonhomogeneous problem given by
d2 φ Q
=0
+
dx2 K
5
subject to
dφ(0)
=0
dx
and
φ(a) = 0
with the solution
Qa2
x
[1 − ( )2 ]
2K
a
And the two-dimensional homogeneous problem given by
φ(x) =
∂ 2ψ ∂ 2ψ
+ 2 =0
∂x2
∂y
subject to
∂ψ(0, y)
=0
∂x
at y = 0 for all x
∂ψ(x, 0)
=0
∂y
at x = a for all y
ψ(a, y) = 0
and at y = b for all x
ψ(x, b) = −φ(x)
which can readily by solved by separation of variables assuming a particular solution of the
form
ψ(x, y) = X(x)Y (y)
to give
ψ(x, y) = −
∞
2Q X
(−1)n cos(λn x) cosh(λn y)
aK n=0 λ3n
cosh(λn b)
where the eigenvalues λn are given by
λn =
(2n + 1)π
2a
with n = 0, 1, 2, ...
Therefore the required solution of the original problem is
T (x, y) =
∞
x
2 X
(−1)n cos(λn x) cosh(λn y)
Qa2
{[1 − ( )2 ] − 3
}
K
a
a n=0 λ3n
cosh(λn b)
6
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