518 Higher Engineering Mathematics
- Solve
∂^2 u
∂x∂t
=sin(x+t)given that
∂u
∂x
= 1
whent=0, and whenu= 2 twhenx= 0.
[u=−sin(x+t)+x+sinx+ 2 t+sint]
- Show thatu(x,y)=xy+
x
y
is a solution of
2 x
∂^2 u
∂x∂y
+y
∂^2 u
∂y^2
= 2 x.
- Find theparticular solutionof thedifferential
equation
∂^2 u
∂x∂y
=cosxcosygiven the ini-
tial conditions that when y=π,
∂u
∂x
=x,and
whenx=π,u=2cosy.
[
u=sinxsiny+
x^2
2
+2cosy−
π^2
2
]
- Verify thatφ(x,y)=xcosy+exsinysatis-
fies the differential equation
∂^2 φ
∂x^2
+
∂^2 φ
∂y^2
+xcosy=0.
53.4 Some important engineering
partial differential equations
There are many types of partial differential equa-
tions. Some typically found in engineering and science
include:
(a) Thewave equation, where the equation of motion
is given by:
∂^2 u
∂x^2
=
1
c^2
∂^2 u
∂t^2
wherec^2 =
T
ρ
, withTbeing the tension in a string
andρbeing the mass/unit length of the string.
(b) Theheat conduction equationis of the form:
∂^2 u
∂x^2
=
1
c^2
∂u
∂t
wherec^2 =
h
σρ
, withhbeing the thermal conduc-
tivity of the material,σthe specific heat of the
material, andρthe mass/unit length of material.
(c) Laplace’s equation, used extensively with elec-
trostatic fields is of the form:
∂^2 u
∂x^2
+
∂^2 u
∂y^2
+
∂^2 u
∂z^2
= 0.
(d) Thetransmission equation, where the poten-
tialu in a transmission cable is of the form:
∂^2 u
∂x^2
=A
∂^2 u
∂t^2
+B
∂u
∂t
+CuwhereA,BandCare
constants.
Some of these equations are used in the next sections.
53.5 Separating the variables
Letu(x,t)=X(x)T(t),whereX(x)is a function ofx
onlyandT(t)is a functionoftonly, be a trial solutionto
the wave equation
∂^2 u
∂x^2
=
1
c^2
∂^2 u
∂t^2
. If the trial solution is
simplified tou=XT,then
∂u
∂x
=X′Tand
∂^2 u
∂x^2
=X′′T.
Also
∂u
∂t
=XT′and
∂^2 u
∂t^2
=XT′′.
Substitutinginto thepartial 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
Letμ=
X′′
X
=
1
c^2
T′′
T
whereμis a constant.
Thus, sinceμ=
X′′
X
(a function ofxonly), it must be
independent oft; and, sinceμ=
1
c^2
T′′
T
(a function oft
only), it must be independent ofx.
Ifμis independent ofx and t, it can only be a con-
stant. Ifμ=
X′′
X
thenX′′=μXorX′′−μX=0andif
μ=
1
c^2
T′′
T
thenT′′=c^2 μTorT′′−c^2 μT=0.
Such ordinary differential equations are of the form
found in Chapter 50, and their solutions will depend
on whetherμ>0,μ=0orμ<0.