518 Higher Engineering Mathematics
7. Solve
∂ 2 u
∂x∂t
= sin(x + t) given that
∂u
∂x
= 1
when t = 0, and when u =2t when x = 0.
[u =−sin(x + t) + x + sin x + 2t + sin t ]
8. Show that u(x, y) = x y +
x
y
is a solution of
2x
∂ 2 u
∂x∂ y
+ y
∂ 2 u
∂ y 2 = 2x.
9. Find the particular solution of the differential
equation
∂ 2 u
∂x∂ y
= cos x cos y given the initial conditions that when y =π,
∂u
∂x
= x, and
when x = π, u =2 cos y.
u = sin x sin y +
x 2
2
+ 2 cos y −
π 2
2
10. Verify that φ(x, y) = x cos y + e x sin y satisfies the differential equation
∂ 2 φ
∂x 2 +
∂ 2 φ
∂ y 2 + x cos y = 0.
53.4 Some important engineering
partial differential equations
There are many types of partial differential equations. Some typically found in engineering and science
include:
(a) The wave equation, where the equation of motion
is given by:
∂ 2 u
∂x 2 =
1
c 2
∂ 2 u
∂t 2
where c 2 =
T
ρ
, with T being the tension in a string
and ρ being the mass/unit length of the string.
(b) The heat conduction equation is of the form:
∂ 2 u
∂x 2 =
1
c 2
∂u
∂t
where c 2 =
h
σρ
, with h being the thermal conductivity of the material, σ the specific heat of the
material, and ρ the mass/unit length of material.
(c) Laplace’s equation, used extensively with electrostatic fields is of the form:
∂ 2 u
∂x 2 +
∂ 2 u
∂ y 2 +
∂ 2 u
∂z 2 = 0.
(d) The transmission equation, where the potential u in a transmission cable is of the form:
∂ 2 u
∂x 2 = A
∂ 2 u
∂t 2 + B
∂u
∂t
+ Cu where A, B and C are
constants.
Some of these equations are used in the next sections.
53.5 Separating the variables
Let u(x, t ) = X (x)T (t ), where X (x) is a function of x
only and T (t ) is a function of t only, be a trial solution to
the wave equation
∂ 2 u
∂x 2 =
1
c 2
∂ 2 u
∂t 2 . If the trial solution is
simplified to u = XT, then
∂u
∂x
= X T and
∂ 2 u
∂x 2 = X T .
Also
∂u
∂t
= XT and
∂
2 u
∂t 2 = XT .
Substituting into the partial differential equation
∂ 2 u
∂x 2 =
1
c 2
∂ 2 u
∂t 2 gives:
X
T =
1
c 2 XT
Separating the variables gives:
X
X
=
1
c 2
T
T
Let μ =
X
X
=
1
c 2
T
T
where μ is a constant.
Thus, since μ =
X
X
(a function of x only), it must be
independent of t ; and, since μ =
1
c 2
T
T
(a function of t
only), it must be independent of x.
If μ is independent of x and t , it can only be a constant. If μ =
X
X
then X = μX or X − μX = 0 and if
μ =
1
c 2
T
T
then T = c 2 μT or T − c 2 μT = 0.
Such ordinary differential equations are of the form
found in Chapter 50, and their solutions will depend
on whether μ >0, μ = 0 or μ <0.
7. Solve
∂ 2 u
∂x∂t
= sin(x + t) given that
∂u
∂x
= 1
when t = 0, and when u =2t when x = 0.
[u =−sin(x + t) + x + sin x + 2t + sin t ]
8. Show that u(x, y) = x y +
x
y
is a solution of
2x
∂ 2 u
∂x∂ y
+ y
∂ 2 u
∂ y 2 = 2x.
9. Find the particular solution of the differential
equation
∂ 2 u
∂x∂ y
= cos x cos y given the initial conditions that when y =π,
∂u
∂x
= x, and
when x = π, u =2 cos y.
u = sin x sin y +
x 2
2
+ 2 cos y −
π 2
2
10. Verify that φ(x, y) = x cos y + e x sin y satisfies the differential equation
∂ 2 φ
∂x 2 +
∂ 2 φ
∂ y 2 + x cos y = 0.
53.4 Some important engineering
partial differential equations
There are many types of partial differential equations. Some typically found in engineering and science
include:
(a) The wave equation, where the equation of motion
is given by:
∂ 2 u
∂x 2 =
1
c 2
∂ 2 u
∂t 2
where c 2 =
T
ρ
, with T being the tension in a string
and ρ being the mass/unit length of the string.
(b) The heat conduction equation is of the form:
∂ 2 u
∂x 2 =
1
c 2
∂u
∂t
where c 2 =
h
σρ
, with h being the thermal conductivity of the material, σ the specific heat of the
material, and ρ the mass/unit length of material.
(c) Laplace’s equation, used extensively with electrostatic fields is of the form:
∂ 2 u
∂x 2 +
∂ 2 u
∂ y 2 +
∂ 2 u
∂z 2 = 0.
(d) The transmission equation, where the potential u in a transmission cable is of the form:
∂ 2 u
∂x 2 = A
∂ 2 u
∂t 2 + B
∂u
∂t
+ Cu where A, B and C are
constants.
Some of these equations are used in the next sections.
53.5 Separating the variables
Let u(x, t ) = X (x)T (t ), where X (x) is a function of x
only and T (t ) is a function of t only, be a trial solution to
the wave equation
∂ 2 u
∂x 2 =
1
c 2
∂ 2 u
∂t 2 . If the trial solution is
simplified to u = XT, then
∂u
∂x
= X T and
∂ 2 u
∂x 2 = X T .
Also
∂u
∂t
= XT and
∂
2 u
∂t 2 = XT .
Substituting into the partial differential equation
∂ 2 u
∂x 2 =
1
c 2
∂ 2 u
∂t 2 gives:
X
T =
1
c 2 XT
Separating the variables gives:
X
X
=
1
c 2
T
T
Let μ =
X
X
=
1
c 2
T
T
where μ is a constant.
Thus, since μ =
X
X
(a function of x only), it must be
independent of t ; and, since μ =
1
c 2
T
T
(a function of t
only), it must be independent of x.
If μ is independent of x and t , it can only be a constant. If μ =
X
X
then X = μX or X − μX = 0 and if
μ =
1
c 2
T
T
then T = c 2 μT or T − c 2 μT = 0.
Such ordinary differential equations are of the form
found in Chapter 50, and their solutions will depend
on whether μ >0, μ = 0 or μ <0.