SMA 2371-PARTIAL DIFFRENTIAL
EQUATIONS
January 4, 1980
Course Outline
1. Surfaces and Curves in three dimensions
2. -Simultaneous dierential equations of the rst order.
3. -Methods of solution of
dx
P
=
dy
Q
=
dz
R
4. -Orthogonal trajectories of systems cuves on a surface
5. -Linear partial dierential equationd of the rst order.
6. Partial dierential equations of the second order: Laplace, Poisson,
heat and wave equations.
7. -Methods of solution by separation of variables foe Cartesian, Spherical,
polar and cylindrical polar coordinates and by Laplace and Fourier
Transforms
8. Applications to Engineering
Reference Books
1. Elements of Partial dierential equations by I,N, Sneddon, Mc Graw
Hill(1957)
1
2. p.d.e-Schaum Outline Series
3. p.d.e-Garabedian (1964)
2
Contents
TABLE OF CONTENTS
3
1 INTRODUCTION
5
2 Partial Dierential Equation
6
2.1 Denition . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Formation of Partial Dierential Equations. . . . . . . . . .
2.2.1 Formation of P.D.E by eliminating arbitrary constant
2.2.2 Formation of P.D.E by eliminating arbitrary function
2.3 Derivation of P.D.E equations by eliminating arbitrary functions φ (u, v) = 0 . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
6
6
7
9
. 10
3 LANGRAGE METHOD OF SOLVING THE QUASI LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF ORDER
1 NAMELY; P p + Qq = R
14
3.1 Integral surfaces passing through a given curve . . . . . . . . . 16
3.2 Surfaces Orthogonal to given system of surfaces . . . . . . . . 20
4 SPECIAL TYPES OF FIRST ORDER PARTIAL DIFFERENTAL
EQUATIONS
23
5 SURFACES AND CURVES IN THREE DIMENSIONS:
30
6 METHODS OF SOLVING
33
dx
p
=
dy
q
=
dz
R
=λ
7 ORTHOGONAL TRAJECTORIES OF CURVES:
41
8 METHOD OF SEPARATION OF VARIABLES:
49
9 DERIVATION OF WAVE EQUATIONS
59
9.1 Forces on the bit of the string . . . . . . . . . . . . . . . . . . 60
3
10 HEAT EQUATION
62
11 DERIVATION OF THE WAVE EQUATION
72
12 DERIVATION OF HEAT EQUATION
74
13 OTHER BOUNDARY VALUE PROBLEMS
76
14 METHOD OF LAPLACE TRANSFORMS AND FOURIER
TRANSFORMS:
77
14.1 SINE AND COSINE FOURIER TRANSFORMS . . . . . . . 78
15 FURTHER EXAMPLES ON METHOD OF SEPARATION
VAIABLES
79
4
1 INTRODUCTION
Partial dierential equations are used to formulate problems involving functions of several variables and are either solved by hand, or used to create a
relevant computer model.
Denition
A dierential equation is an equation that relates the derivatives at a scalar
function depending on one or more variables.
Example:
2
δ4 u
+ δδxu2 +u2 = cos(x)is a dierential equation of the function u(x)depending
δx4
2
2
on x, while δu
= δδxu2 + δδyu2 − uis a dierential equation involving a function
δt
u(x, y, t) depending on x, y, t. This equation can be written in subscript
notation as ut = uxx + uyy − u.
Partial dierential operators
(i) δxδ = δx
2
(ii) δxδ 2 = δx2 or δxx
3
(iii) δxδ2 δy = δxxy or δx2 δy
Surfaces and Curves in three dimensions
(i) 3-D coordinate system in often denoted by R3
(ii) 2-D coordinate system in often denoted by R2
(iii) 1-D coordinate system in often denoted by R
(iv) n-D coordinate system in often denoted by Rn
q
Distance between any two points is given by: d (p1 , p2 ) = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 and
the general equation for a sphere with center (h, k, l)and radius r is given by
(x − h)2 + (y − k)2 + (z − l)2 = r2
Examples
1. Graph x = 3in R,R2 and R3
Solution
In Rwe have a single coordinated system and so x = 3is a point in a 1-D
coordinate system
5
In R2 the equation x = 3tells us to graph all the points that are in the form
(3, y).This is a vertical line in a 2-D coordinate system
In R3 the equation x = 3tells us to graph all the points onf the form (3, y, z).This
is similar to the coordinates for the yz plane except this time we have
x = 3instead of x = 0. So in a 3-D system this is a plane that will be
parallel to the yz plane and pass through the x-axis at x = 3.
2 Partial Dierential Equation
2.1 Denition
This is an equation involving partial dierential coecients of functions of
two or more variables is known as partial dierential equation. If a p.d.e
contains nth and lower derivatives of nth order the degree of such equation
is the greatest exponent of the highest order. We adopt the following notion
through out the study of p.d.e's
δz δ 2 z
δ2 z
δ2 z
δz
,q = δy
, δx2 = r, δxδy
= sand δy
p = δx
2 = twhere z is a dependent variable. In
case there are n independent variables we take them to be x1 , x2 , x3 , ..., xn and
z is regarded as dependent variable i.e. p1 = δxδz1 ,p2 = δxδz2 ,....pn = δxδzn ,
Sometimes the partial derivatives are as denoted by making use of suxes
thus
2
δ2 u
ux = δu
,u = δu
,u = δδxu2 ,ut = δu
,uxy = δxδy
.
δx y
δy xx
δt
2.2 Formation of Partial Dierential Equations.
There are two main methods of forming a p.d.e.
(i) Method of elimination of arbitrary constants
(ii)Method of elimination of arbitrary functions
6
2.2.1 Formation of P.D.E by eliminating arbitrary constant
Example
(i) z = ax + by + a2 + b2 .............(i)
Solution
δz
= a.................................(ii)
δx
δz
= b.................................(iii)
δy
Replacinng (ii) and (iii)
in equation
(i), wehave:
z=
δz
δx
x+
δz
δy
y+
δz
δx
2
+
δz
δy
2
z = px + qy + p2 + q 2
(ii) z = (x − a)2 + (y − b)2 ..............(i)
Solution
δz
= 2 (x − a)
δx
δz
= 2 (y − b)
δy
2
2
1 δz
1 δz
z=
+
.
4 δx
4 δy
1
1
z = p2 + q 2
4
4
4z = p2 + q 2
Exercise: Form a P.D.E.
a) z = Aept sin(px)
Solution
δz
= Apept cos(px)
δx
δ2z
= −Ap2 ept sin(px)......................(i)
δx2
7
δz
= Apept sin(px)
δt
δ2z
= Ap2 ept sin(px)......................(ii)
δt2
Ading, (i) and (ii), we have:
δ2z δ2z
+ 2 = −Ap2 ept sin(px) + Ap2 ept sin(px) = 0
2
δx
δt
b) xa2 + yb2 + zc2 = 1..................(i)
Solution
Dierentiting (i) with respect to x,
2
2
2
2x 2z δz
δz
+ 2
= 0 =⇒ c2 x + a2 z
= 0..............(ii)
2
a
c δx
δx
Dierentiting (i) with respect to y,
δz
2y 2z δz
+ 2
= 0 =⇒ c2 y + b2 z
= 0..............(iii)
2
b
c δy
δy
Dierentiting (ii) with respect
to x,
2
c +a
δz
δx
2
2
δ2z
+ a z 2 = 0..................(iv)
δx
2
Dierentiting (iii) with respect
to y,
2
c +b
From (ii)
2
2
δz
δy
+ b2 z
a2 z
c =−
x
2
Putting (vi) into(iv)
a2 z
−
x
δz
δx
+a
2
δz
δx
δz
δx
2
δ2z
= 0..................(v)
δy 2
.....................(vi)
δ2z
+ a z 2 = 0..................(vii)
δx
2
Dividing (vii) by a2 andmultiplying
by x
δ2z
zx 2 + x
δx
δz
δx
2
−z
δz
δx
Similarly from (iii),
8
= 0....................(viii)
b2 z
c =−
y
2
Putting (ix) into (v)
2
bz
−
y
δz
δy
+b
2
δz
δy
δz
δy
2
δ2z
+ b z 2 = 0..................(x)
δy
.....................(ix)
2
Dividing (x) by a2 and multiplying
by
y
δ2z
zy 2 + y
δy
δz
δy
2
−z
δz
δy
= 0....................(xi)
2.2.2 Formation of P.D.E by eliminating arbitrary function
This is done by eliminating the arbitrary function; f or F.
Example:4
a) y = f (x − at) + F (x + at) .............(i)
Solution
δy
= 1 · f 0 (x − at) + 1 · F 0 (x + at)
δx
δy
= −a · f 0 (x − at) + a · F 0 (x + at)
δt
δ2y
= f 00 (x − at) + F 00 (x + at)
2
δx
δ2y
= a2 [f 00 (x − at) + F 00 (x + at)]
δt2
2
δ2y
2δ y
=
a
δt2
δx2
b) z = f (x2 − y) + g (x2 + y) ......(i)
Solution
δz
= 2xf 0 x2 − y + 2xg 0 x2 + y ........(ii)
δx
δ2z
= 2x · 2xf 00 x2 − y + 2f 0 x2 + y + 2x · 2xg 0 x2 + y + 2g 0 x2 + y
2
δx
00 2
0 2
δ2z
2
00
2
0
2
=
4x
f
x
−
y
+
g
x
+
y
+2
g
x
+
y
+
f
x
+
y
........(iii)
δx2
9
δz
= −f 0 x2 − y + g 0 x2 + y .......(iv)
δy
δ2z
00
2
00
2
=
f
x
−
y
+
g
x
+
y
.............(v)
δy 2
1 δz
= f 0 x2 − y + g 0 x2 + y ............(vi)
2x δx
Substituting (vi) and (v) into (iii),
2
δ2z
1 δz
2 δ z
+2
........(vii)
= 4x
δx2
δy 2
2x δx
2 δ2z
δz
3 δ z
x 2 = 4x
+ ........(viii)
2
δx
δy
δx
2.3 Derivation of P.D.E equations by eliminating arbitrary functions φ (u, v) = 0
Consider φ (u, v) = 0...........................(i), where u and v are functions of x,y
and z. This expression represents a solution that has variables u and v. We
take z to be dependent variable while x and y are the independent variables
δy
δz
δz
so that δx
= p, δy
= q, δx
= 0, δx
=0
δy
Dierentiating(i) partially
to xand y in turn we get;
with respect
δu
δv
δu
δφ δv
+p
+p
+
.........................(ii)
δx
δz
δv δx
δz
δφ δu
δu
δv
δφ δv
+q
+
+q
.........................(iii)
δu δy
δz
δv δy
δz
δφ
δu
From (ii),
δφ
δu
− δφ
=
δv
Similarly from (iii)
δφ
δu
− δφ
=
δv
δv
δx
δu
δx
+ p δv
δz
...........................(iv)
δu
+ p δz
δv
δy
δu
δy
+ q δv
δz
+ q δu
δz
.............................(v)
Equating (iv) and (v) and simplifying yields,
10
δv
δv
+p
δx
δz
δu
δu
+q
δy
δz
=
δv
δv
+q
δy
δz
δu
δu
+p
δx
δz
δv δu
δv δu
δv δu
δv δu
δv δu
δv δu
δv δu
δv δu
· +q · +p · +pq
=
· +p · +q · +pq ·
δx δy
δx δz
δz δy
δz δz
δy δx
δy δz
δz δx
δz δz
δv δu δv δu
δv δu
δv δu δv δu
δv δu
δv δu δv δu
·
−
·
+q
·
−
·
= − · −pq
+ · +pq ·
p
δz δy δy δz
δx δz
δz δx
δx δy
δz δz δy δx
δz δz
δv δu δv δu
δv δu δv δu
δv δu δv δu
p
·
−
·
+q
·
−
·
=
·
−
·
δz δy δy δz
δx δz
δz δx
δy δx δx δy
P p + Qq = R
This is a partial dierential equation and can be classied as follows:
(i) If it is a linear p.d.e in p and q(i.e)
P (x, y, z) p + Q (x, y, z) q = R (x, y, z)
(ii) Semi-Linear p.d.e: if it is linear in p and q and the coecient of p and q
are functions of x and y only i.e.
P (x, y) p + Q (x, y) q = R (x, y, z)
Example
a) Form a partial dierential equation whose solution is φ (x2 ez , yez ) = 0
Solution
φ x2 ez , yez = 0.................................(i)
Here u = x2 ez and v = yez
φ (u, v) = 0
δu
δx
δv
= 2xez , δu
= x2 ez , δu
= 0, δx
= 0, δv
= yez
δz
δy
δz
Substituting this in the equation,
δv
δx
δu
δx
+ p δv
δz
=
+ p δu
δz
δv
δy
δu
δy
+ q δv
δz
+ q δu
δz
0 + pyez
ez + qyez
=
2xez + px2 ez
0 + qx2 ez
11
pyez
ez + qyez
=
2xez + px2 ez
qx2 ez
py
1 + qy
=
2
2x + px
qx2
pyqx2 = 2x + px2 (1 + qy)
pyqx2 = 2x + px2 + 2xqy + pqx2 y + x2 p + 2xyq
x2 p + 2xyq = −2x
xp + 2yq = −2
From this, p = x, q = 2y , R = −2
b) Form a p.d.e equation whose solution is φ (x + y + z, x2 + y 2 − z 2 ) = 0
Solution
Here u = x + y + z and v = x2 + y 2 − z 2
φ (u, v) = 0
δu
δx
δv
δv
= 1, δu
= 1, δu
= 1, δx
= 2x, δv
= −2z, δy
= 2y
δz
δy
δz
Substituting this in the equation,
δv
δx
δu
δx
+ p δv
δz
=
+ p δu
δz
δv
δy
δu
δy
+ q δv
δz
+ q δu
δz
2x − 2pz
2y − 2qz
=
1+p
1+q
(2x − 2pz) (1 + q) = (1 + p) (2y − 2qz)
2x + 2xq − 2pz − 2pqz = 2y − 2qz + 2py − 2pqz
x + xq − zp = y − zq + py
(x + z) q − (z + y) p = y − x
wherep = −z − y , q = x + z , R = y − x
(b) Form a partial dierential equation whose solution is φ (x2 + y 2 + z 2 , z 2 − 2xy) =
0
12
Solution
φ x2 ez , yez = 0.................................(i)
Here u = x2 + y 2 + z 2 and v = z 2 − 2xy
φ (u, v) = 0
δu
δx
δv
= 2x, δu
= 2z, δu
= 2y, δx
= −2y, δv
= 2z
δz
δy
δz
Substituting this in the equation,
δv
δx
δu
δx
+ p δv
δz
=
δu
+ p δz
δv
δy
δu
δy
+ q δv
δz
+ q δu
δz
−2y + 2pz
−2x + 2qz
=
2x + 2pz
2y + 2qz
−x + qz
−y + pz
=
x + pz
y + qz
(−y + pz) (y + qz) = (x + pz) (−x + qz)
−y 2 − qyz + pyz + pqz 2 = −x2 − pxz + qxz + pqz 2
(yz + xz) p − (yz + xz) q = y 2 − x2
z (y + x) p − z (y + x) q = (y + x) (y − x)
zp − zq = (y − x)
wherep = z , q = −z , R = y − x
13
3 LANGRAGE METHOD OF SOLVING THE
QUASI LINEAR PARTIAL DIFFERENTIAL
EQUATIONS OF ORDER 1 NAMELY; P p+
Qq = R
The general solution of the linear partial dierential equation are P p + Qq =
R where p,q, R are functions of x,y,z is φ (u, v) = 0where φis an arbitrary
function and u (x, y, z) = c1 )andv (x, y, z) = c2 ) form a solution of equations
dy
dz
dx
=
=
p
q
R
Example:
2
1. Solve yxz p + x2 q = y 2
Solution
This dierential equation is of the form P p + Qq = R
2
wherep = yxz , q = xz , R = y 2
Using langrage auxillary equation,
dx
y2 z
x
Taking the rst two fractions,
=
dz
dy
= 2
xz
y
dy
xdx
=
2
y z
xz
x2 dx − y 2 dy = 0
x3 y 3
c1
−
=
3
3
3
or
x3 − y 3 = c1 =⇒ u
Taking rst and third fraction
xdx
dz
= 2
2
y z
y
xdx − zdz = 0
14
x2 z 2
c2
−
=
2
2
2
x2 − z 2 = c2 =⇒ v
But φ (u, v) = 0, therefore
φ x3 − y 3 , x2 − z 2 = 0
where φis an arbitrary function.
Exercise:
a) p + 3q = 5z + tan (y − 3x)
This dierential equation is of the form P p + Qq = R
wherep = 1, q = 3, R = 5z + tan (y − 3x)
Using langrage auxillary equation,
dx
dy
dz
=
=
1
3
5z + tan (y − 3x)
Using the rst two fractions
dx
dy
=
1
3
3dx = dy
3x − y = c
y − 3x = c1
Using the rst and the third fraction,
dx
dz
=
1
5z + tan (y − 3x)
dx
dz
=
1
5z + tan (c1 )
1
x − ln |5z + tan (c1 )| = c2
5
5x − ln |5z + tan (c1 )| = c2
But φ (u, v) = 0, therefore
φ (y − 3x, 5x − ln |5z + tan (y − 3x)|) = 0
15
thus,
v (x, y, z) = φ(u)
where φis an arbitrary function.
Note:
The possible solutions of p.d.e is φ (u, v) = 0,u = φ(v)or v = φ(u)
3.1 Integral surfaces passing through a given curve
1.Determine the integral surface of the p.d.e.
(x − y) p + (y − x − z) q = z
through the circle z = 1, x + y = 1
Solution
The equation (x − y) p + (y − x − z) q = z is of the form P p + Qq = R
wherep = (x − y), q = (y − x − z), R = z
Using langrage auxillary equation,
2
2
dy
dz
dx
=
= ......(i)
(x − y)
(y − x − z)
z
Using all the fractions
dx + dy + dz
dx + dy + dz
=
x−y+y−x−z+z
0
dx + dy + dz = 0
x + y + z = c1 ...................(ii)
Using the second and the third fraction,
dy
dz
= .
(y − (x + z))
z
From (ii) (x + z) = (c1 − y)
dy
dz
= .
(y − (c1 − y))
z
dy
dz
= .
2y − c1
z
1
ln |2y − c1 | = ln |z| + ln |c|
2
16
(2y − c1 ) = c2 z 2
2y − x − y − z
= c2
z2
y−x−z
= c2 .......................(iii)
z2
Put z =1 in equation (ii) and (iii)
x + y = c1 − 1...................(iv)
y − x = c2 + 1...................(v)
But,
2 x2 + y 2 = (x + y)2 + (y − x)2 ........................(vi)
But x2 + y 2 = 1
2 = (c1 − 1)2 + (c2 + 1)2 ........................(vii)
2 = c21 − 2c1 + 1 + c22 + 2c2 + 1
c21 − 2c1 + c22 + 2c2 = 0
2
2 (y − x − z)
(y − x − z)
2
=0
+
(x + y + z) − 2 (x + y + z) +
2
z
z2
(x + y + z)2 − 2 (x + y + z) +
(y − x − z)2 2 (y − x − z)
+
=0
z4
z2
z 4 (x + y + z)2 + (y − x − z)2 − 2z 4 (x + y + z) + 2z 2 (y − x − z) = 0
2. Find the equation of the surface satisfying the equation 4yzp + q + 2y = 0
and passing through y 2 + x2 = 1, x + z = 2
Solution
The equation (x − y) p + (y − x − z) q = z is of the form P p + Qq = R
wherep = 4yz , q = 1, R = −2y
Using langrage auxillary equation,
dx
dy
dz
=
=
......(i)
4yz
1
−2y
Taking rst and third fractions,
17
dx
dz
dx
dz
1
=
=⇒
= − =⇒ dx = −zdz
4yz
−2y
2z
1
2
1
z2
x+
= c1
2
2
x + z 2 = c1 ..................(ii)
Taking second and third fractions,
dy
dz
=
=⇒ dz + 2ydy = 0
1
−2y
z + y 2 = c2 ..............(iii)
Equations (iii) and (iv) are the integrable surfaces
x + z 2 + z + y 2 = c1 + c2
(x + z) + z 2 + y 2 = c1 + c2
2 + 1 = c1 + c2
3 = c1 + c2
(x + z) + z 2 + y 2 = 3
is the given integral surface.
3) Find the integral surface of x2 p + y 2 q + z 2 = 0 which passes through the
hyperbolar xy = x + y , z = 1
Solution
The dierential equation x2 p + y 2 q + z 2 = 0can be written in the form
P p + Qq = R
wherep = x2 , q = y 2 , R = −z 2
Using langrage auxillary equation,
dy
dz
dx
= 2 =
......(i)
2
x
y
−z 2
Taking rst and third fractions
dx
dz
1 1
=
=⇒ − − = c
2
2
x
−z
x z
18
1 1
+ = c1
x z
Taking second and third fractions
dy
dz
1 1
− =c
=
=⇒
−
y2
−z 2
y z
1 1
+ = c2
y z
Adding (i) and (ii)
1 1 2
+ + = c1 + c2
x y z
y+x 2
+ = c1 + c2
yx
z
2 + 1 = c1 + c2
1 1 2
+ + =3
x y z
yz + xz + 2xy = 3xyz
ASSIGNMENT
Find the integral surface of equation 4yzp − q + 2y = 0which passes through
the hyperbola y 2 = 9 + z and z = 5 − x.
Solution
The dierential equation 4yzp − q + 2y = 0can be written in the form P p +
Qq = R
wherep = 4yz , q = −1, R = −2y
Using langrage auxillary equation,
dy
dz
dx
=
=
......(i)
4yz
−1
−2y
Taking rst and third fractions
dx
dz
dx
dz
1
zdz
1
z2
=
=⇒
=
=⇒ dx = −
=⇒ x +
= c1
4yz
−2y
2z
−1
2
1
2
2
x + z 2 = c1 ...........................(ii)
Taking second and third fractions
19
dy
dz
dz
y2 1
=
=⇒ ydy =
=⇒
− z = c2
−1
−2y
2
2
2
y 2 − z = c2 ..................................(iii)
Adding (ii) and (iii)
x + 2z + y 2 − z = c1 + c2
However there are no combination suiting our given equations, thus;
But y 2 = 9 + z 2 and z = 5 − x.
Rewriting,
9 + z2 = y2
9 + z 2 = z + c2
But z = 5 − xand z 2 = c1 − x
9 + c1 − x = 5 − x + c2
4 = c2 − c1
Replacing c1 and c2 from equations(ii) and (iii),
4 = y2 − z − x + z2
4 = y2 − z − x − z2
3.2 Surfaces Orthogonal to given system of surfaces
..............(i) are
The curves whose equations are the solution of dxp = dyq = dz
R
orthogonal to the system of the surfaces whose equations are pdx + qdy +
Rdz = 0..............(ii).
1. Find the family of surfaces orthogonal to the family of surfaces given by
dierential equations (y + z) p + (z + x) q = x + y
Solution
p = y + z , q = z + x, R = x + y
Then the given dierential equation can be written in the form P p+Qq = R.
20
The dierential equation of the family of the surface is orthogonal to the
given family is given by pdx + qdy + Rdz = 0where (y + z) dx + (z + x) dy +
(x + y) dz = 0
Rearranging,
(y + z) dx + (z + x) dy + (x + y) dz = 0
(ydx + zdx) + (zdy + xdy) + (xdz + ydz) = 0
Rearranging,
(ydx + xdy) + (zdy + ydz) + (xdz + zdx) = 0
d (xy) + d (yz) + d (xz) = 0
Integrate,
xy + yz + xz = c
where c is a constant of integration.
Exercise:
1. Find the family of curves orthogonal to φ z (x + y)2 , x2 − y 2 = 0
Solution
Solution
φ z (x + y)2 , x2 − y 2 = 0.................................(i)
Here u = z (x + y)2 and v = x2 − y 2
φ (u, v) = 0
δv
δv
= 2z (x + y) , δu
= (x + y)2 , δu
= 2z (x + y) , δx
= 2x, δv
= 0, δy
=
δz
δy
δz
δv
−2y, δz = 0
δu
δx
Substituting this in the equation,
δv
δx
δu
δx
+ p δv
δz
=
+ p δu
δz
δv
δy
δu
δy
+ q δv
δz
+ q δu
δz
2x
−2y
2 =
(x + y) + p (x + y)
2z (x + y) + q (x + y)2
2
x [2z + q (x + y)] = −y [2z + p (x + y)]
21
2xz + qx2 + qxy = −2zy − xyp − y 2 p
xy + y 2 p + x2 + xy q = −2z (y + x)
py (x + y) + qx (x + y) = −2z (y + x)
py + qx = −2z
From this, p = y , q = x, R = −2z
Hence the equation is of the form P p + Qq = R and is orthogonal to pdx +
qdy + Rdz = 0.
Thus,
ydx + xdy − 2zdz = 0
But ydx + xdy = d (xy)
Hence,
d (xy) − 2zdz = 0
xy − z 2 = c
22
4 SPECIAL TYPES OF FIRST ORDER PARTIAL DIFFERENTAL EQUATIONS
We discuss special types of rst order partial dierential equations whose
solution can be determined by the Charpit's method.
Case 1:
A rst order partial dierential equation is said to be of the form Charpit's
form, if it can be written in the form:
z = px + qy + f (p, q) ................(i)
Solution
Let F (x, y, z, p, q) = z − px − q(y) − f (p, q).................(ii)
Therefore the Charpit's auxillary equation is:
δF
δx
dp
=
+ p δF
δZ
δF
δy
dq
dz
dx
dy
=
= δF = δF
δF
δF
δF
+ q δZ
−p δp − q δq
− δp
− δq
Using the rst and the second fractions,
dq
dp
=
−p + p
−q + q
This implies that
dp = 0 =⇒ p = aand dq = 0 =⇒ q = b.............................(iii)
Substituting equation(iii) into equation (i)
z = ax + by + f (a, b) ................(iv)
Equation (iv) gives the complete integral of equation (i)
Example:
a) Find the complete integral(C.I) and the Singular Integral (S.I) of
(i) z = px + qy + log(pq)..................(i)
Solution
Comparing this with the given equation of Charpit form, the complete integral is given by
z = ax + by + log(ab)
z = ax + by + log(a) + log(b)......................(ii)
23
To obtain the singular integral we dierentiate (ii) partially with respect to
a and b to obtain (iii) and (iv),
1
1
=⇒ a = − ..................(iii)
a
x
1
1
0 = y + =⇒ b = − ..................(iv)
b
y
0=x+
We then substitute equations (iii) and (iv) in (ii) to obtain the singular
integral.
z=
−
1
x
x+ −
1
y
y + log −
1
x
+ log −
1
y
1
1
z = −1 − 1 + log −
+ log −
x
y
1
z = −2 + log
xy
√
(ii) z = px + qy − 2 pq..................(i)
Solution
Comparing this with the given equation of Chaurat's form, the complete
integral is given by
√
z = ax + by − 2 ab..................(ii)
To obtain the singular integral we dierentiate (ii) partially with respect to
a and b to obtain (iii) and (iv),
r
2b
b
b
0 = x − √ = x − √ =⇒ x = √ =
2 ab
ab
ab
b
.........................(iii)
a
r
2a
a
a
a
0 = y − √ = y − √ =⇒ y = √ =
.........................(iv)
b
2 ab
ab
ab
r
r
r
r
√
b
b
a
b
x−z =
−a
−b
+ 2 ab =
...............................(v)
a
a
b
a
r
r
r
r
√
a
b
a
a
y−z =
−a
−b
+ 2 ab =
...............................(vi)
b
a
b
b
Multiplying (v) and (vi)
24
r
(x − z) (y − z) =
b
×
a
r
a
=1
b
(x − z) (y − z) = 1
pq
(iii) Prove that z = px + qy + pq−p−q
represents all planes such that the
algebraic sum of the intercepts on three coordinate axes is unity.
Solution
Comparing this with the given equation of Charpit's form, the complete
integral is given by
z = ax + by +
ab
.................(ii)
ab − a − b
Rearranging equation (ii) in order that outlines the intercepts,
ax + by − z =
ax
−ab
ab−a−b
x
−b
ab−a−b
+
y
−a
ab−a−b
+
+
by
−ab
ab−a−b
z
ab
ab−a−b
−ab
ab − a − b
−
z
−ab
ab−a−b
=1
= 1.....................(iii)
−b
−a
From equation (iii), the x intercept is given by: ab−a−b
, y- intercept by ab−a−b
and
ab
z-intercept ab−a−b
.
The sum of the intercepts is given by
z−a
ab
−b − a + ab
−b
+
+
=
=1
ab − a − b ab − a − b ab − a − b
ab − a − b
(iv) Find the C.F and S.I. of 4xyz = pq + 2px2 y + 2qxy 2 ......................(i)
Solution
This equation can only be compared to Charpit's form if it is transformed
rst,
Tranforming equation (i) to Charpit's form
δz
δz
Recall p = δx
, q = δy
But 4xy = (2x) (2y) ............................(ii)
Divide (i) by (ii),
25
1 δz
1 δz
2x2 y δz
2xy 2 δz
z=
·
·
+
+
2x δx
2y δy
4xy δx
4xy δy
1 δz
1 δz
1 δz
1 δz
2
2
·
·
+x
·
+y
·
z=
2x δx
2y δy
2x δx
2y δy
Let 2xδx = δX and 2yδy = δY
let x2 = X and y 2 = Y z=
p=
δz
,q
δX
=
δz
δX
δz
δY
+X
δz
δX
+Y
δz
δY
δz
δY
Z = pq + pX + qY
which is of Charpit's form
Comparing this with the given equation of Charpit's form, the complete
integral is given by
z = aX + bY + ab.................(iii)
But x2 = X and y 2 = Y
Replacing yields;
z = ax2 + by 2 + ab.................(iv)
For the singular integral,we dierentiate (iv) partially with respect to a and
b to obtain equations (v) and (vi)
0 = x2 + b =⇒ b = −x2 ........................(v)
0 = y 2 + a =⇒ a = −y 2 ...........................(vi)
z = −y 2 x2 − x2 y 2 + x2 y 2 .................(vii)
z = −y 2 x2 ........(viii)
1
(v) Show that the complete integral of the equation z = px+qy+(p2 + q 2 + 1) 2 ..........(i)represents
all planes at a unit distance from the origin.
Solution
This equation is of Chaurit's form, thus
26
z = ax + by + a2 + b2 + 1
12
............................(ii)
Rearranging equation (ii) in order that outlines the intercepts,
ax + by − z = − a2 + b2 + 1
21
............(iii)
1
Dividing through by − (a2 + b2 + 1) 2 , we get
ax
− (a2 + b2 + 1)
x
1
2
+
a
1
−(a2 +b2 +1) 2
by
−
z
1 = 1............(iv)
− (a2 + b2 + 1)
− (a2 + b2 + 1) 2
y
z
+
+
= 1............(v)
1
b
1
2
1
1
(a2 +b2 +1) 2
−(a2 +b2 +1) 2
Squaring the intercepts and adding them to get the distance,
2
b2
1
a2 + b 2 + 1
a
+
+
=
=1
a2 + b 2 + 1 a2 + b 2 + 1 a2 + b 2 + 1
a2 + b 2 + 1
Case 11: A dierentail equation with only p,q,z in the form:
f (p, q, z) = 0..................................(i)
Solution
Apply Charpit's auxillary equation
dq
dz
dx
dy
dp
=
= 0
=
=
0
0
0
0
0 + pf (p, q, z)
0 + qf (p, q, z)
pf (p, q, z)
−f (p, q, z)
−qf (p, q, z)
pf 0
dp
dq
dz
dx
dy
= 0
= 0
=
=
0
0
(p, q, z)
qf (p, q, z)
pf (p, q, z)
−f (p, q, z)
−qf (p, q, z)
dp
dq
= δf
δf
p δz
q δz
dp
dq
=
p
q
lnp = lnq + lna
p = qa or q = pa.................................(ii)
dz = pdx + qdy
dz = p (dx + ady) = pd (x + ay) ...............................(iii)
Letu = x + ay
27
dz = pdu
p=
dz
............................(iv)
du
Thus by equation (ii),
q=a
dz
............................(v)
du
Replacing back in equation
(i)
f
dz dz
, a , 0 = 0......................(vi)
du du
Equation (vi) is a p.d.e of rst order. Solving (vi), we get z as a function of
U. The Complete Integral is then obtained by replacing u with x+ay.
Example:
a) Find the Complete Integral and the Singular Integral of q 2 = z 2 p2 (1 − p2 ) .....................(i)
Solution
Step 1:
dz
dz
We take p = du
,q = a du
and u = x + ay and substitute in
! (i)
dz
a
du
2
2
2
dz
dz
=z
1−
du
du
"
2 #
dz
a2 = z 2 1 −
du
2
2
2
a =z −z
2
dz
du
2
2
dz
z 2 − a2
=
du
z2
r
√
dz
z 2 − a2
z 2 − a2
=
=
du
z2
z
ˆ
ˆ
z
√
du =
dz
2
z − a2
Integrating by substitution yields,
u+b=
√
28
z 2 − a2
Squaring both sides,
(u + b)2 = z 2 − a2
Replacing u = x + ay
(x + ay + b)2 = z 2 − a2 ..............................(ii)
Dierentiatiate (ii) with respect to a
2y (x + ay + b) = −2a..................................(iii)
Dierentiatiate (ii) with respect to b
2 (x + ay + b) = 0..................................(iv)
From (iii) and (iv) we get,
z 2 = 0 =⇒ z = 0.............................(v)
Equation (v) is an xy plane.
(Incomplete)
29
5 SURFACES AND CURVES IN THREE DIMENSIONS:
In rectangular co-ordinates, the equation of a surface is expressed in the form
f (x, y, z) = 0..........................(i). For example, the equation x2 + y 2 + z 2 =
a2 is the equation of a sphere centered at the origin (0, 0, 0)radius r. From
dx + δf
dy + δf
dz = 0showing that
equation (i) the total dierential of F is δf
δx
δy
δz
δf δf δf
the triad δx , δy , δz are the directional rations normal to the surface at a
given point while (dx, dy, dz)are the direction ratios tangent to the surface
of the same point. Consider a curve which is the intersection of the surfaces
f (x, y, z)and g (x, y, z) = 0. With the two equations of surfaces satisfying
their total dierentials i.e. df = 0and dg = 0or
δf
δf
δf
dx + dy + dz = 0
δx
δy
δz
δg
δg
δg
dx + dy + dz = 0
δx
δy
δz
which on solving for dx, we have;
δf
δf
δf
dx + dy = − dz
δx
δy
δz
δg
δg
δg
dx + dy = − dz
δx
δy
δz
4=
δf
δx
δg
δx
δf
δy
δg
δy
δf δg δf δg
=
−
=J
δx δy δy δx
δf
δy
δg
δy
− δf
dz
δz
dx =
− δg
dz
δz
δf
δx
δg
δx
δf
δy
δg
δy
δz
=
J
30
δf
δy
δg
δy
δf
δz
δg
δz
f,g
x,y
f, g
x, y
or
dx =
f,g
x,y
δzJ
dx
dz
or = .....................(A)
f,g
f,g
J x,y
J f,g
J x,y
y,z
Next solving for dy, we have;
δf
δf
δf
dx + dy = − dz
δx
δy
δz
δg
δg
δg
dx + dy = − dz
δx
δy
δz
δf
δx
δg
δx
4=
δf
δy
δg
δy
δf
δx
δg
δx
dy =
or
δf δg δf δg
=
−
=J
δx δy δy δx
− δf
dz
δz
− δg
dz
δz
δf
δx
δg
δx
δz
=
δf
δy
δg
δy
J
δf
δz
δg
δz
δf
δx
δg
δx
f,g
x,y
f, g
x, y
f,g
δzJ x,y
dy
dz
or = .........................(B)
dx =
f,g
f,g
f,g
J x,y
J z,x
J x,y
From A and B,
J
dx
=
f,g
y,z
which is a system of the form,
where
p=J
f, g
y, z
J
dy
=
f,g
z,x
J
dz
f,g
x,y
dy
dz
dx
=
=
p
q
R
or q = J
31
f, g
z, x
or R =
J
dz
f,g
x,y
p=
δf
δy
δg
δy
δf
δz
δg
δz
or q =
δf
δz
δg
δz
δf
δx
δg
δx
or R =
δf
δx
δg
δx
δf
δy
δg
δy
The reverse problem is that of nding the curves of intersection for these
surface i.e. We show that the surface f (x, y, z) = c1 will involve only one
arbitrary constant c1 and f (x, y, z) = c2 will involve only one arbitrary constant c2 . Therefore, the curve of intersection will involve two parameters and
we call them a two parameters and we call them a two parameter family of
curves.
32
6 METHODS OF SOLVING
dx
p
=
dy
q
=
dz
R
=λ
(i) The Spot Method
..........................(i)
Suppose that U (x, y, z)is a solution of dxp = dyq = dz
R
δu
δu
δu
δu
δu
dx
+
dy
+
dz
=
0
hence
λ
dx
+
dy
+
dz
= 0where λ 6=
then δu
δx
δy
δz
δx
δy
δz
0hence p du
+ q du
+ R du
= 0. We then try to spot functions p0 , q 0 and R0 such
dx
dy
dz
that pp0 + qq 0 + RR0 = 0implying that p0 dx + q 0 dy + R0 dz = 0will be an exact
dierential equation with the solution U (x, y, z) = c1 . Repeats the same
process with the function p00 , q 00 and R00 to get V (x, y, z) = c2 which is the
required two parameter family of curves c1 and c2 .
Example:
dy
dz
dx
= x(x+y)−az
= z(x+y)
1. Find the integral curve of the equations y(x+y)+az
Solution
p = y (x + y) + az , q = x (x + y) − az and R = z (x + y)
By intution take p0 = x, q 0 = y thus pp0 + qq 0 + RR0 will become exact;
xy (x + y) + xaz + yx (x + y) − yaz + R0 z (x + y) = 0
x2 y + xy 2 + xaz + yx2 + y 2 x − yaz + R0 z (x + y) = 0
az (x + y) + R0 z (x + y) = 0 =⇒ az (x + y) = −R0 z (x + y) = 0
Thus
R0 = −a
Then pp0 +qq 0 +RR0 = 0is now exact implying that p0 dx+q 0 dy +R0 dz = 0will
be an exact dierential equation with the solution U (x, y, z) = c1
xdx − ydy − adz = 0
On integration,
x2 y 2
−
− az = c or x2 − y 2 − 2az = c1
2
2
where c1 = 2c
Again by intuition, take p00 = 1, q 00 = i thus pp00 + qq 00 + RR00 = 0
33
y (x + y) + az + x (x + y) − az + R00 z (x + y) = 0
(x + y)2 + R00 z (x + y) = 0
R00 =
− (x + y)
z
Thus p00 dx + q 00 dy + R00 dz = 0will be an exact dierential equation with the
solution V (x, y, z) = c2
(x + y)
dz = 0
z
dx + dy 1
− dz = 0
x+y
z
dx + dy −
d (x + y) 1
− dz = 0
x+y
z
ln |x + y| − ln |z| = lnc2
x+y
= c2
z
Thus the integral curves are given by U (x2 − y 2 − 2az) = c1 and V x+y
= c2
z
Note: This method can be simplied and shortened by the following properties:
0 dx+q 0 dy+R0 dz
dx
= dy
= dz
= p pp
= dx−dy−2dz
so that if pp0 + qq 0 + RR0 = 0 then
0 +qq 0 +RR0
p
q
R
p−q−2R
p0 dx + q 0 dy + R0 dz = 0is exact.
(ii) Method 2
Given that
dx+3dy−dz
pp0 +3qq 0 −RR0
dx
p
=
e.t.c
dy
q
=
dz
R
, It should be noted that
dx
p
=
dy
q
Example:
dy
dx
dz
Find the integral curves of y(x+y)+az
= x(x+y)−az
= z(x+y)
Solution
Adding the rst two and equating to the third equation,
dz
dx + dy
2 =
z (x + y)
(x + y)
dx + dy
dz
=
x+y
z
34
=
dz
R
=
p0 dx+q 0 dy
pp0 +qq 0
=
ln |x + y| = ln |z| + lnc1
x+y
= c1
z
Also,
xdx
ydy
dz
=
=
xy (x + y) + azx
xy (x + y) − azy
z (x + y)
x2 y
Therefore,
+
xy 2
xdx − ydy
xdx − ydy
=
2
2
+ azx − xy − x y + azy
az (x + y)
xdx − ydy
dz
=
az (x + y)
z (x + y)
xdx − ydy = adz
x2 y 2
−
= az + c or x2 − y 2 − 2az = c2
2
2
where c2 = 2c
(iii) Method 3:
It is possible to use the rst solution to obtain the second solution by eliminating one of the variables in one of the fractions:
dy
dx
dz
Example: Obtain integral curve of y(x+y)+az
= x(x+y)−az
= z(x+y)
...............(i)
Solution
Adding the rst two and equating to the third equation,
dx + dy
dz
2 =
z (x + y)
(x + y)
dx + dy
dz
=
x+y
z
ln |x + y| = ln |z| + lnc1
x+y
= c1
z
Eliminating z from the rst and second fraction using the rst solution, we
have;
z=
x+y
c1
35
Replacing in equation (i),
dx
y (x + y) + a
x+y
c1
=
dy
x (x + y) − a
x+y
c1
=
x+y
c1
dz
(x + y)
Using the rst two fractions,
dx
dy
a =
y + c1
x − ca1
a
a
x−
dx = y +
dy
c1
c1
x2
azx
y2
azy
−
=
+
+c
2
x+y
2
x+y
x2 y 2 az (x + y)
−
−
= c or x2 − y 2 − 2az = c2
2
2
(x + y)
where c2 = 2c
(iv) Method 4
If one of the variables is absent in one of the equations, one of the solutions
can be denied easily.
dz
dx
= dy
= z+y
Example: Find the integral curves of the equations x+z
2 ...............(i)
y
Solution
dz
x is absent from the equation dyy = z+y
2 and this equation is linear with z and
´ 1
the dependent variable with I.F = e − y dy = e−lny = y1 . Therefore
dz
z + y2
dz z
=
or
− =y
dy
y
dy y
1 dz
z
− 2 =1
y dy y
z
d
=1
y
z
= y + c1 or z = c1 y + y 2
y
36
Comparing the rst and the second fractions,
dx
dy
=
2
x + (c1 y + y )
y
This is a linear equation with x as the dependent variable
dx
x + c1 y + y 2
=
dy
y
dx
x
= + c1 + y
dy
y
dx x
− = c1 + y
dy
y
´
I.F = e
− y1 dy
= e−lny = y1 . Therefore
1
x
= (y + c1 )
d
y
y
c1
x
d
=1+
y
y
x
= y + c1 lny + c2
y
Replacing c1 =
z−y 2
y
x
z − y2
=y+
lny + c2
y
y
Example:
Find the integral curves of the equation
dx
dz
a) xy
= dy
= zxy−2x
2
y2
The rst two fractions do not have the variable z
dy
dx
= 2
xy
y
dx
dy
=
x
y
lnx = lny + lnc =⇒
Also,
37
x
= c1
y
dy
dz
=
2
y
zxy − 2x2
But x = c1 y
dy
dz
=
2
2
y
zc1 y − 2c21 y 2
dy =
b)
dx
xz(z 2 +xy)
=
dy
−yz(z 2 +xy)
dz
zc1 − 2c21
y=
1
ln zc1 − 2c21 + c2
c1
y=
y
xz 2x2
ln
− 2 + c2
x
y
y
=
dz
z4
−dy
dx
=
x
y
lnx = −lny + lnc =⇒ xy = c1
Also,
dx
dz
= 4
2
xz (z + xy)
z
z (z 2 + xy)
dx
=
dz
x
z4
dx
1 c1
=
+
dz
x
z z3
c1
lnx = lnz − 2 + c2
2z
xy
lnx − lnz + 2 = c2
2z
x
xy
+ 2 = c2
z 2z
dy
dx
dz
c) x(y−z)
= y(z−x)
= z(x−y)
Solution
Adding the rst two fractions and equating with the third,
38
dx + dy
dz
=
−z (x + y)
z (x − y)
dx + dy + dz = 0
Integrating,
x + y + z = c1
Also,
ydx + xdy
dz
=
yx (y − z) + xy (z − x)
z (x − y)
dz
ydx + xdy
=
yx (x − y)
z (x − y)
dx dy dz
+
+
=0
x
y
z
lnx + lny + lnz = lnc2
xyz = c2
d) x2 (ydx
3 −z 3 ) =
Solution
dy
y 2 (z 3 −x3 )
x3
=
(y 3
dz
z 2 (x3 −y 3 )
zdz
xdx + ydy
= 3 3
3
3
3
3
− z ) + y (z − x )
z (x − y 3 )
Considering the denominator,
x3 y 3 − x3 z 3 + y 3 z 3 − x3 y 3 = z 3 x3 − z 3 y 3
z 3 y 3 − x3 = −z 3 x3 − y 3
Therefore,
zdz
xdx + ydy
= 3 3
3
3
3
−z (x − y )
z (x − y 3 )
xdx + ydy + zdz = 0
x2 y 2 z 2
+
+
= c or x2 + y 2 + z 2 = c1
2
2
2
y3 −
dx
x2
x3
+
+
dy
y2
z3
− x3
39
=
dz
z2
x3 − y 3
dx
x2
y3
+
+
dy
y2
z3
=
dz
z2
x3 − y 3
dx dy dz
+ 2 + 2 =0
x2
y
z
1 1 1
− − − =c
x y z
1 1 1
+ + = c2
x y z
40
7 ORTHOGONAL TRAJECTORIES OF CURVES:
p=
δf
δy
δg
δy
δf
δz
δg
δz
,q =
δf
δz
δg
δz
δf
δx
δg
δx
,R =
δf
δx
δg
δx
δf
δy
δg
δy
In three dimensions, given a surface f (x, y, z) = 0..............................(i)and
a system of curves, we have to nd a system of curves each of which lies on
the surface (i) and cuts every curve of the given system at right angles. The
new system of curve is called the system of orthogonal trajectories on the
surface of the given system of curves. The original system of curves may be
thought of as the intersection of the surface (i) with the parameter family of
curves g (x, y, z) = c1 ..............................(ii)
41
In general, the tangential direction (dx, dy, dz)to the given curve through the
points x,y,z on the surface (i) satises the equations δf
dx + δf
dy + δf
dz = 0
δx
δy
δz
δg
δg
and δx
dx+ δy
dy+ δg
dz = 0. Which on solving gives dx
= dy
= dz
.............(iii)
δz
p
q
R
where
δf
δf
δf
δf
fy fz
fz fx
fx fy
δx
δz
δx
=
,q = δg δg =
,R = δg δy
=
p=
δg
gy gz
gz gx
gx gy
δz
δx
δx
δy
The curve through point (x, y, z)of the orthogonal system has tangential
direction (dx0 , dy 0 , dz 0 )which lies on the surface (i) such that δf
dx0 + δf
dy 0 +
δx
δy
δf
dz 0 = 0...................(iv)
δz
δf
δy
δg
δy
δf
δz
δg
δz
42
This implies that dxdx0 + dydy 0 + dzdz 0 = 0
Hence from (iii), we have pdx0 + qdy 0 + Rdz 0 = 0...........................(v)
Solving (iv) and (v) we have,
0
0
dx0
= dy
= dz
.........................(vi)where p0 = Rfy − qfz , q 0 = pfz − Rfx and
p0
q0
R0
R0 = qfx − pfy .........................(vii)
Note that:
δf
dx0
δx
0
+ δf
dy 0 = − δf
dz 0
δy
δz
pdx + qdy 0 = −Rdz 0
dx0 =
fz δz 0 fy
−Rδz 0 q
=
fx fy
p q
dx0
fz fy
−R q
δz 0
fx fy
p q
=
fx fy
p q
dz 0
fx fy
p q
43
fx fz δz 0
p −Rδz 0
dy 0 =
δz 0
=
fx fy
p q
dy 0
fx fz
p −R
fx fy
p q
dz 0
=
fx fy
p q
fx fy
p q
Example:
Find the orthogonal trajectories of the sphere x2 + y 2 + z 2 = a2 at its intersection with a parabola xy = cz where c is a parameter.
Solution
Let f (x, y, z)be identiacal to x2 + y 2 + z 2 = a2
Let g (x, y, z) = xyz = c
Then (dx, dy, dz)satises
2y
x
z
,q =
and
R=
dx
p
2z
2xy 2
=
−
− 2x =
z2
− xy
2
z
δf
δz
δg
δz
δf
δx
δg
δx
δf
δx
δg
δx
δf
δy
δg
δy
=
=
=
dy
q
=
dz
R
where p =
δf
δy
δg
δy
δf
δz
δg
δz
=
fy fz
=
gy gz
−2xy 2 −2xz 2
z2
fz fx
2z 2x
=
=
y
gz gx
− xy
z2
z
2yz
z
+
2x2 y
z2
=
2yz 2 +2x2 y
z2
=
2y (z 2 +x2 )
z2
fx fy
2x 2y
2
2
2(x2 −y 2 )
= y x = 2x −2y
=
z
z
gx gy
z
z
dx
dy
dz
2 2 = 2y(z2 +x2 ) = 2(x2 −y2 )
−2x y z+z
2
z2
z
dx
dy
dz
=
=
2
2
2
2
2
−x (y + z )
y (z + x )
z (x − y 2 )
Then the triad (dx0 , dy 0 , dz 0 )satises
p0 =
dx0
p0
=
dy 0
q0
=
dz 0
R0
where
fy fz
2y
2z
=
= 2yz (x2 − y 2 ) − 2zy (x2 + z 2 ) =
2
2
2
q R
y (x + z ) z (x − y 2 )
44
−2yz (y 2 + z 2 )
fz fx
2z
2x
q0 =
=
= −2zx (y 2 + z 2 )−2xz (x2 − y 2 ) =
2
2
2
2
R p
z (x − y ) −x (y + z )
2
2
−2xz (x + z )
fx fy
2x
2y
R0 =
=
= 2xy (x2 + z 2 )+2xy (y 2 + z 2 ) =
2
2
2
p q
−x (y + z ) y (x + z 2 )
2xy (x2 + 2z 2 + y 2 )
Hence,
dx0
dy 0
dz 0
=
=
−2yz (y 2 + z 2 )
−2xz (x2 + z 2 )
2xy (x2 + 2z 2 + y 2 )
dx0
dy 0
dz 0
=
=
−yz (y 2 + z 2 )
−xz (x2 + z 2 )
xy (x2 + 2z 2 + y 2 )
For the rst two fractions,
dy 0
dx0
=
−yz (y 2 + z 2 )
−xz (x2 + z 2 )
dx0
dy 0
=
y (y 2 + z 2 )
x (x2 + z 2 )
ydy 0
xdx0
= 2
(y 2 + z 2 )
(x + z 2 )
From;x2 + y 2 + z 2 = a2 ,x2 + z 2 = a2 − y 2 and y 2 + z 2 = a2 − x2
xdx0
ydy 0
=
(a2 − x2 )
(a2 − y 2 )
1
1
− ln a2 − x2 = − ln a2 − y 2 + lnk
2
2
a2 − y 2
= k −2 = c
a2 − x 2
−y
2
2
2
= a2 are the
Thus aa2 −x
2 = cand the given surface equation is x + y + z
systems of orthogonal trajectories.
b) Find the orthogonal trajectories on the cone x2 + y 2 = z 2 tan2 αat its
intersection with the family of planes z = c
Solution
2
2
45
The triad (dx, dy, dz)satises the equation
p=
,q =
R=
δf
δy
δg
δy
δf
δz
δg
δz
δf
δx
δg
δx
δf
δz
δg
δz
δf
δx
δg
δx
δf
δy
δg
δy
dx
p
=
=
dz
R
=
fy fz
2y −2ztan2 α
=
= 2y
gy gz
0
1
=
fz fx
−2ztan2 α 2x
=
= −2x and
gz gx
1
0
=
fx fy
2x 2y
=
=0
gx gy
0 0
dx
dy
dz
dx
dy
dz
=
=
=⇒
=
=
2y
−2x
0
y
−x
0
Then the triad (dx0 , dy 0 , dz 0 )satises
dx0
p0
=
dy 0
q0
=
p0 =
fy fz
2y −2ztan2 α
=
= −4xztan2 α
q R
−2x
0
q0 =
fz fx
−2ztan2 α 2x
=
= −4yztan2 α
R p
0
2y
R0 =
dy
q
dz 0
R0
where
fx fy
2x 2y
=
= −4 (x2 + y 2 )
p q
2y −2x
dy 0
dz 0
dx0
=
=
−4xztan2 α
−4yztan2 α
−4 (x2 + y 2 )
dy 0
dz 0
dx0
=
=
xztan2 α
yztan2 α
(x2 + y 2 )
Comparing the rst two fractions,
dx0
dy 0
=
x
y
lnx = lny + lnc
x
=c
y
Therefore the required system of orthogonal trajectories are given by x2 +y 2 =
z 2 tan2 αand xy = c.
c) Find the orthogonal trajectories on the conicoid (x + y) z = 1of the conics
in which its cut by the system of planes x − y + z = k where k is a parameter.
46
Solution
The triad (dx, dy, dz)satises the equation
dx
p
=
dy
q
=
dz
R
f = xz + yz − 1 = 0
g =x−y+z−k
δf
δf
fy fz
z x+y
δy
δz
p = δg δg
=
=
=z+x+y
gy gz
−1
1
δy
δz
,q =
R=
δf
δz
δg
δz
δf
δx
δg
δx
δf
δx
δg
δx
δf
δy
δg
δy
=
fz fx
x+y z
=
= x + y − z and
gz gx
1
1
fx fy
z z
=
= −z − z = −2z
gx gy
1 −1
dy
dz
dx
dy
dz
dx
=
=
=⇒
=
=
x+y+z
x+y−z
−2z
y
−x
0
=
Then the triad (dx0 , dy 0 , dz 0 )satises
dx0
p0
=
dy 0
q0
=
dz 0
R0
where
p0 =
fy fz
z
y+x
=
= −2z 2 − (x + y) (x + y − z)
q R
x + y − z −2z
q0 =
fz fx
x+y
z
=
= (x + y) (x + y + z) + 2z 2
R p
−2z x + y + z
R0 =
−2z 2
fx fy
z
z
=
= z (x + y − z) − z (x + y + z) =
p q
x+y+z x+y−z
dy 0
dz 0
dx0
=
=
−2z 2 − (x + y) (x + y − z)
(x + y) (x + y + z) + 2z 2
−2z 2
Adding the rst two fractions,
dx0 + dy 0
dz 0
=
−2z 2 − (x + y) (x + y − z) + (x + y) (x + y + z) + 2z 2
−2z 2
dx0 + dy 0
dz 0
=
2z (x + y)
−2z 2
dz 0
dx0 + dy 0
=
(x + y)
−z
ln |x + y| = −ln |z| + lnc
47
(x + y) z = c
If c = 1,
(x + y) z = 1
Using x + y = z1 in the rst and the third fractions,
dx0
dz 0
=
−2z 2 − (x + y) (x + y − z)
−2z 2
dx0
−2z 2 − z1
1
z
−z
=
dz 0
−2z 2
dz 0
dx0
=
−2z 2
−2z 2 − z12 + 1
1
dz 0
1
1
0
0
2
dx =
−2z − 2 + 1 = dz 1 + 4 − 2
−2z 2
z
2z
2z
On integrating,
x=z−
1
1
+
+c
3
6z
2z
d) Find the orthogonal trajectories on the surface f = x2 +y 2 +2p (yz)+d = 0
of its intersections with the one family of planes parallel toz = c whre c is a
parameter.
48
8 METHOD OF SEPARATION OF VARIABLES:
Example 1:
Displacement of vibrating spring or wave equation.
The partial dierential equation governing the motion of the vibrating string
2
2
ension
is given as δδt2u = c2 δδxu2 where c2 = TM
ass
Therefore our task is to determine u (x, t)under the boundary conditions
u (0, t) = 0,u (1, t) = 0 and the initial conditions u (x, 0) = φ(x)and ut (x, 0) =
ϕ(x). Here u (x, t)represents displacement of the string at any position and
at any time t. u(t)represents the velocity of the wave produced by the
string at time t. By the method of separation of variables, the dependent variable u (x, t)may be expressed as u (x, t) = X(x)T (t)where X(x)is
a function of x alone and T (t)is a function of t alone. Then it follows that
uxx = X 00 (x)T (t)andutt = X 00 (x)T (t)
Given,
2
δ2u
2δ u
=
c
....................(i)
δt2
δx2
49
Let U = XT........................(ii)
δX
δT
δX
δu
=
T +X
=
T = X 0 T.........................(iii)
δx
δx
δx
δx
δ2u
δX 0
δT
δ2X
δ
0
(X
T
)
=
T
+
X
=
=
T = X 00 T.............................(iv)
2
2
δx
δx
δx
δx
δx
Similarly,
δ2u
= T 00 X.........................(v)
δt2
Substituting (v) and (iv) into (i)
XT 00 = c2 X 00 T
c2 X 00
T 00
=
=k
T
X
These two functions could only be equal to a constant, say k as shown above,
T 00
= k =⇒ T 00 − kT = 0
T
and
k
c2 X 00
= k =⇒ X 00 − 2 X = 0
X
c
Assuming that we need to consider only the physical solution, there are three
cases to be be investigated.
Case 1:
k=0
If k=0, then T 00 = 0, T 0 = A,T = At + B and X 00 = 0, X 0 = C ,X = Cx + D
Therefore, U (x, t) = X(x)T (t) =⇒ U (x, t) = (Cx + D) (At + B)
These solutions cannot describe the vibtrating of a system because it is not
periodic.
Applying U (0, t) = 0,U (1, t) = 0
D (At + B) = 0
Dt = 0, t 6= 0
hence D = 0
U (x, t) = Cx (At + B)
0 = Ct
50
hence C = 0
This shows that U (x, t) = 0has trivial solutions.
Case 2:k = λ2
Then the two dierential equations are
T 00
= λ2 =⇒ T 00 − λ2 T = 0
T
and
c2 X 00
λ2
= λ2 =⇒ X 00 − 2 X = 0
X
c
T = Aeλt + Be−λt
λ
λ
X = Ce c x + De− c x
Thus
U (x, t) = Ce
λ
x
c
+ De
− λc x
Aeλt + Be−λt
This solution cannot describe the oscillating motion of a system because it
is not periodic. THerefore we reject the solution;
U (0, t) = (C
C = −D,
D = −C λ
+λD) t =⇒ λC + D = 0 =⇒
λ
λ
x
−cx
x
−cx
x
− λc x
c
c
c
t=C e −e
t =⇒ t 6= 0, e − e
6=
U (x, 0) = Ce − Ce
0, C = 0hence the solution is trivial
Case 3:k = −λ2
Then the two dierential equations are
T 00
= λ2 =⇒ T 00 + λ2 T = 0
T
and
c2 X 00
λ2
= λ2 =⇒ X 00 + 2 X = 0
X
c
T = Acosλt + Bsinλt
λ
λ
X = Ccos
x + Dsin
x
c
c
λ
λ
U (x, t) = Ccos
x + Dsin
x [Acosλt + Bsinλt]
c
c
51
. However, we need to determine
This solution is periodic with a period of 2π
λ
λ, A, B, C, andD from the rst boundary condition.
U (0, t) = 0 =⇒ 0 = C (Acosλt + Bsinλt) =⇒ 0 = CT, T 6= 0 =⇒ C = 0
Hence U (x, t) = Dsin λc x [Acosλt + Bsinλt]
The second boundary condition gives
U (l, t) = 0 =⇒ 0 = Dsin λc l [Acosλt + Bsinλt] or 0 = Dsin λc l · T
l 6= 0and we cannot choose D = 0because it gives a trivial solution.
Solution
sin λlc = 0, only possible solution is when λlc = nπ =⇒ λ = nπ
c or λn =
l
nπ
c, n = 1, 2, 3, ...
l
λn is called the eigen solution. Then, we therefore have innite number of
λn for which the product solution of the wave equation exists./ Therefore we
write the solution as
Un (x, t) = sin
λn
x (An cosλnt + Bn sinλnt)
l
or
nπ nπc
nπc x An cos
t + Bn sin
t
l
l
l
where An and Bn are denes as fourier coecients. To obtain An , we use the
Un (x, t) = sin
initial displacement conditions
U (x, 0) = φ(x) =
∞
X
An sin
n=1
nπ
x
l
This series can be organized as the half range since the expansion of a periodic function φ(x)is denes in the range (0, l). Now An can be obtained by
multiplying the series by sin nπl xand integrating with respect to x from 0 to
´
l which gives An = π2 0l φ(x)sin nπl xdx
∞
a0 X
F (x) =
+
(an cos(nx) + bn sin(nx))
2
n=1
To determine Bn a, we use the fourth condition
U (x, 0) = ψ(x, t) =
∞ n
X
nπ
nπc
nπc
nπc
nπc o
sin x −
An sin
t+
Bn cos
t
l
l
l
l
l
n=1
52
ψ(x, t) =
∞
X
nπc
l
n=1
Bn sin
nπc
x
l
Then multiplying both sides of this equation by sin nπl xand integrating with
respect to x, we get
ˆ
Bn =
2
nπc
l
ψ(x)sin
0
nπ
xdx
l
Therefore the equation :
nπ nπc
nπc x An cos
t + Bn sin
t
l
l
l
is the required solution of the wave equation with An and Bn as shown above.
Un (x, t) = sin
Example:
a) A string is stretched and fastened to two points. The motion is started
by displacing the string in the form y (x, t) = asin πxl from which it is
released at time t = 0. Show that the displacement of any point at y (x, t) =
asin
πx
l
Solution
cos
πct
l
The equation of a vibration of a string is given by:
53
2
δ2u
2δ u
=
c
....................(i)
δt2
δx2
Boundary conditions are y (0, t) = 0and y (l, t) = 0........................(ii)
The string is in the shape of y (x, t) = asin πxl .........................(iii)at rest
i.e whent = 0. Also dy
=0
dt
The objective isto solve equation (i) subject to conditions given by equations
(ii) and (ii). As discussed in the previous section the solution to(i) is given
by Y = XT
Y (x, t) = (a1 coskx + a2 sinkx) (a3 coskct + a4 sinkct) .........................(iv)
Using the rst boundary condition in equation (ii)
0 = (a1 cos0 + a2 sin0) (a3 coskct + a4 sinkct)
0 = a1 (a3 coskct + a4 sinkct)
Therefore a1 = 0since the solution must be true for all values of t, hence
equation (iv) reduces to
Y (x, t) = a2 sinkx (a3 coskct + a4 sinkct)
Y (x, t) = sinkx (a2 a3 coskct + a2 a4 sinkct)
Let a2 a3 = b1 and a2 a4 = b2
Y (x, t) = sinkx (b1 coskct + b2 sinkct) .....................(v)
Using equation (ii) in (v),y (l, t) = 0
0 = sinkl (b1 coskct + b2 sinkct)
sinkl = 0
k=
Using (vi) in (v),
Y (x, t) = sin
Using
nπ
.........................(vi)
l
nπ nπ
nπ x b1 cos ct + b2 sin ct .....................(vii)
l
l
l
= 0 at t= 0
nπc
nπ
nπ
nπc
nπ 0 = sin x −b1
sin ct + b2
cos ct .....................(viii)
l
l
l
l
l
dy
dt
54
nπc nπ x 0 + b2
...................(ix)
l
l
Since the answer must be true for all values of x, b2 = 0 and equation (vii)
0 = sin
becomes;
nπ nπ x b1 cos ct
l
l
nπ
nπ
Y (x, t) = b1 sin xcos ct....................(x)
l
l
Applying the initial conditiont = 0 in (x),
nπ
nπ
Y (x, 0) = b1 sin xcos(0) = b1 sin x.......................(xi)
l
l
Comparing equation (xi) with te givenequation
a = b1 , n = 1
πx πct
cos
Y (x, t) = asin
....................(xii)
l
l
Y (x, t) = sin
b) A tightly stretched spring with xed ends at x = 0and x = l is initially
in equilibrium position. It is vibrating by giving each of the poins an initial
3 πx
=
bsin
. Find the dispacement Y (x, t).
velocity, Y (x, 0) = 0, ∂y
∂t t=0
l
Solution
55
Equation of a vibrating string is
2
δ2u
2δ u
=
c
....................(i)
δt2
δx2
We solve equation (i) subject to the bondary conditionsy (0, t) = 0................(ii)and
y (l, t) = 0........................(iii)
Initial conditions are y (x, 0) = 0................(iv)and
The solution to equation (i) is of the form
∂y
∂t t=0
= bsin3
πx
l
....................(v).
Y (x, t) = (a1 coskx + a2 sinkx) (a3 coskct + a4 sinkct) .........................(vi)
Using condition (ii),
0 = (a1 cos0 + a2 sin0) (a3 coskct + a4 sinkct)
0 = a1 (a3 coskct + a4 sinkct) ..................(vii)
Therefore a1 = 0since the solution must be true for all values of t, hence
equation (vii) reduces to
Y (x, t) = a2 sinkx (a3 coskct + a4 sinkct)
Y (x, t) = sinkx (a2 a3 coskct + a2 a4 sinkct)
Let a2 a3 = b1 and a2 a4 = b2
Y (x, t) = sinkx (b1 coskct + b2 sinkct) .....................(viii)
Using equation (ii) in (v),y (l, t) = 0
0 = sinkl (b1 coskct + b2 sinkct)
sinkl = 0
k=
nπ
.........................(ix)
l
Using (ix) in (viii),
nπ nπ
nπ x b1 cos ct + b2 sin ct .....................(x)
l
l
l
Using initial condition, y (x, 0) = 0
nπ
0 = sin x (b1 cos0 + b2 sin0) .....................(xi)
l
Y (x, t) = sin
56
nπ
x (b1 ) ...................(xii)
l
Since the answer must be true for all values of x, b1 = 0 and equation (x)
0 = sin
becomes;
nπ nπ Y (x, t) = sin x b2 sin ct
l
l
nπ
nπ
Y (x, t) = b2 sin xsin ct....................(xiii)
l
l
where n=1,2,3,...
Therefore the sum of this solution is also a solution by superposition principle
of linear solutions
∞
Y (x, t) =
X
An sin
n=0
nπ
nπ
xsin ct....................(xiv)
l
l
Dierentiating (xiv) with repect to t
∞
δy X nπc
nπ
nπc
=
An
sin xcos
t
δt
l
l
l
n=0
3 πx
=
bsin
Applying the initial condition in ∂y
,
∂t t=0
l
∞
X
nπc
nπ
nπc
δy
=
An
sin xcos
(0)
δt t=0 n=0
l
l
l
bsin
3
πx l
=
∞
X
n=0
An
nπc
nπ
sin x..............(xv)
l
l
3
From trigonometric identities
sin3θ = 3sinθ − 4sin θ
πx πx 3πx
= 3sin
− 4sin3
l
l
l
3
πx 1
3πx
3 πx
= sin
− sin
sin
l
4
l
4
l
X
∞
πx 1
3
3πx
nπ
nπc
− bsin
bsin
=
An
sin x
4
l
4
l
l
l
n=0
πx 1
3πx
πc
π
2πc
2π
3πc
3π
4πc
4π
3
bsin
− bsin
= 0+A1 sin x+A2
sin x+A3
sin x+A4
sin x+.....
4
l
4
l
l
l
l
l
l
l
l
l
sin
57
Equating coecient of sin
πx
l
3
πc
3bl
b = A1
=⇒ A1 =
4
l
4πc
Equating coecient of sin 2πl x
Equating coecient of sin
2πc
=⇒ A2 = 0
0 = A2
l
3πx
l
1
3πc
−bl
− b = A3
=⇒ A3 =
4
l
12πc
Equating coecient of sin 4πl x
0 = A4
4πc
=⇒ A4 = 0
l
Substitututing A1 , A2 , A3 , A4 into equation (xvii)
Y (x, t) =
3bl
πx
πct
bl
3πx
3πct
sin sin
−
sin
sin
4πc
l
l
12πc
l
l
Equation (vii) represents displacement of the wave.
58
9 DERIVATION OF WAVE EQUATIONS
The wave equation is derived with respect to travelling transverse waves on
a taut string. The following assumptions are made:
(i) The amplitude of the wave is always small.
(ii) the string has a constant linear mass density
(iii) The string is under a constant tension of magnitude T.
(iv) The string is not made of atoms or molecules. It is compresses of continuous non-quantized matter.
(v) There is no damping(no loss of energy due to friction)
(vi) Gravity is ignored in the treatment.
INTRODUCTION
Through out the derivation, we will focus on a tiny bity of strings, so small
that we can pretend that it is almost straight line. The bit of a string which
will have a tension pulling at each end(with magnitude T ). Length of the
bit of string will be 4s.
Assuming that the amptitude of the wave is extremely small, then the ap59
proximation of the string length is 4x. Dening a linear mass density as the
mass of the string in kilograms per meter of string to be µ, then the mass of
bit of string is given by: m = µ4x.........................................(i).
9.1 Forces on the bit of the string
Since we are looking at transverse travelling waves, we worry about the net
force in the vertical direction.
X
Fnet =
Fy
Note that: Deing vertical coordinates above, onm the string to be positive.
The net force on the bit of the string is:
Fnet = T sin (θ2 ) − T sin (θ1 )
Converting sine functons into tangent functions,
60
Since we are dealing only with small amplitudes, we use this approximation
4x = 4s. Therefore, sinθ = tanθ
Then the new net force is expressed as
Fnet = T tan (θ2 ) − T tan (θ1 )
4y
looking at small and smaller
Using tan is more convenient because tanθ = 4x
δy
bits of string letting 4xapproach zero then tanθ = δx
(tangent to the string
at a point)
Letting the string bit smaller
smaller,
the net force is
and Fnet = T
δy
δx
−
R
δy
δx
Employing Newton's Second Law
...........................(ii)
Looking at the actual motion of the bit of string. Newton's second law tells
us that Fnet = ma...............(iii) where Fnet is the net force, m is the mass
while a is the acceleration. Puting vertical acceleration as the second time
61
derivative of vertical displacements
Fnet = m
δ2y
δt2
............................(iv)
Substituting values of mass from
(i)in equation (iv)
Fnet = µ4x
Equating equation (v)
to (ii)
T
"
T
δy
δx 2
−
4x
δ2y
δt2
2 δy
δ y
= µ4x
δx 1
δt2
2 δ y
=µ
............................(vi)
δt2
δy
−
δx 2
#
δy
δx 1
............................(v)
From denition of a derivative
f (x + 4x) − f (x)
δf
= lim
δx 4x→0
4x
If 4xgets smaller and smaller,
equation
(vi) becomes:
T
δ2y
δx2
=µ
δ2y
δt2
............................(vii)
δ y
δ y
where T (Tension)and µ (mass)are constants. δx
2 and δt2 are second derivatives of vertical displacement.
Writing
both constants as a single constant,
2
δ2y
δx2
c2 =
δ2y
δt2
2
........................(viii)
where c2 = Tµ . Equation (viii) is a p.d.e since vertical diplacement depends
on both position (x) and time (t)
10 HEAT EQUATION
Consider the heat conduction equation in a thin metal bar of legth l with
insulated sides. Let us assume that the ends are x = 0and x = l and
62
are hels at temperature 00 cfor all timet > 0. In addition, let us assume
that the temperature distribution at t=0 is U (x, 0) = f (x). Determine the
temperature distribution in a bar of some subsequent time t > 0.
Solution
The heat equation is given by Ut = αUxx i.e.
δu
δ2u
=α 2
δt
δx
where U (x, t)is the heat distribution subject to boundary conditions U (0, t) =
0and U (1, t) = 0and the initial conditions U (x, 0) = f (x).
By method of separation of variables,
Let U = XT........................(ii)
δX
δT
δX
δu
=
T +X
=
T = X 0 T.........................(iii)
δx
δx
δx
δx
δ2u
δ
δX 0
δT
δ2X
0
=
(X
T
)
=
T
+
X
=
T = X 00 T.............................(iv)
δx2
δx
δx
δx
δx2
Similarly,
δu
= T 0 X.........................(v)
δt
Substituting (v) and (iv) into (i)
T 0 X = αX 00 T
T0
X 00
=
=k
αT
X
These two functions could only be equal to a constant, say k as shown above,
T0
= k =⇒ T 0 − αkT = 0
αT
and
X 00
= k =⇒ X 00 − kX = 0
X
Assuming that we need to consider only the physical solution, there are three
cases to be be investigated.
Case 1:
k=0
If k=0, then T 0 = 0, T = Aand X 00 = 0, X 0 = C ,X = Cx + D
63
Therefore, U (x, t) = X(x)T (t) =⇒ U (x, t) = (Cx + D) (A) = CAx + DA =
Ex + F
These solutions cannot describe the vibtrating of a system because it is not
periodic.
Applying U (0, t) = 0,U (1, t) = 0
F = 0 T 6= 0
hence F = 0
U (x, t) = Ex X 6= 0
hence E = 0
This shows that U (x, t) = 0has trivial solutions.
Case 2:k = λ2
Then the two dierential equations are
T 0 X = αX 00 T
X 00
T0
=
= λ2
αT
X
These two functions could only be equal to a constant, say k as shown above,
T0
= λ2 =⇒ T 0 − αλ2 T = 0
αT
δT
δT
2
= αλ2 T =⇒
= αλ2 δt =⇒ lnT = αλ2 t + A ⇒ T = Aeαλ t
δt
T
and
X 00
= k =⇒ X 00 − λ2 X = 0
X
T = Aeαλ
2t
X = Ceλx + De−λx
Thus
2
U (x, t) = Ceλx + De−λx Aeαλ t
2
U (x, t) = Beλx + F e−λx eαλ t
but when t increases the temperature in this case increases too much instead
64
of decreasing i.e. physical condition does not allow. Therefore we reject the
solution;
Case 3
k = −λ2
Then the two dierential equations are
T 0 X = αX 00 T
T0
X 00
=
= −λ2
αT
X
These two functions could only be equal to a constant, say k as shown above,
T0
= λ2 =⇒ T 0 + αλ2 T = 0
αT
δT
δT
2
= −αλ2 T =⇒
= −αλ2 δt =⇒ lnT = −αλ2 t + A ⇒ T = Ae−αλ t
δt
T
and
X 00
= λ2 =⇒ X 00 + λ2 X = 0
X
T = Ae−αλ
2t
X = Ccosλx + Dsinλx
Thus
U (x, t) = (Ccosλx + Dsinλx) Ae−αλ
U (x, t) = (Bcosλx + F sinλx) e−αλ
2t
2t
This is quite possible, applying the boundary conditions
a) U (0, t) = 0
2t
0 = (B + 0) e−αλ
2t
0 = Be−αλ
B=0
U (x, t) = (F sinλx) e−αλ
b) U (l, t) = 0
65
2t
0 = (F sinλl) e−αλ
2t
Here, F 6= 0, e−αλ t =
6 0. Thus,
2
sinλl = 0
λl = nπ =⇒ λ =
Then
nπ
l
nπ −α( nπl )2 t
U (x, t) = F sin x e
l
λn are dened to be the eigen values and the function Un (x, t)corresponds to
any λn is called the eigen function. Thus the general solution becomes
nπ 2
nπ Un (x, t) = Fn sin x e−α( l ) t
l
From the superposition principle all the linear solutions, we get the general
solutions as
∞
∞
U (x, t) =
X
Un (x, t) =
X
n=1
n=1
nπ −α( nπl )2 t
Fn sin x e
l
This can be recognized as the fourier half range sine series and Fn is the
fourier coecient which can be determined from the initial condition as follows:
∞
U (x, t) =
X
Fn sin
n=1
nπ
x
l
Multiplying both sides by sin nπl xand integrating, we obtain:
2
Fn =
l
f (x) =
ˆ
l
f (x)sin
0
nπ
xdx
l
∞
X
a0
nπx
nπx
+
an cos
+ bn sin
2
l
l
n=1
Assignment
1. Use the method of separation of variables to solve the heat equation.
66
2. By stating the necessary assumption show that the one dimensional wave
2
δ2 y
µ
T ension
2
equation is given by δδt2y = c2 δx
.
2 where c = m =
M ass
3. Derive the heat equation.
Example 2:
A rod whose surface is measured has a length of 3 units. The end of the rod
is kept at 00 cand its initial temperature at any point x, 0 < x < 3is given
by U (x, 0) = 5sin4πx − 3sin8πx + 2sin10πx. Find the temperature at any
given time t.
Solution
Let U (x, t)be temperature at any given time t.
The boundary conditions are x = 3and x = 0
Initial conditions are U (x, 0) = 5sin4πx − 3sin8πx + 2sin10πx.
The heat equation is one dimension
δ2u
δu
=α 2
δt
δx
Using case 3 wherek = −λ2 since case 1 and case 2 give trivial solutions;
Then the two dierential equations are
T 0 X = αX 00 T
X 00
T0
=
= −λ2
αT
X
These two functions could only be equal to a constant, say k as shown above,
T0
= λ2 =⇒ T 0 + αλ2 T = 0
αT
δT
δT
2
= −αλ2 T =⇒
= −αλ2 δt =⇒ lnT = −αλ2 t + A ⇒ T = Ae−αλ t
δt
T
and
X 00
= λ2 =⇒ X 00 + λ2 X = 0
X
T = Ae−αλ
2t
X = Ccosλx + Dsinλx
67
Thus
U (x, t) = (Ccosλx + Dsinλx) Ae−αλ
U (x, t) = (Bcosλx + F sinλx) e−αλ
2t
2t
This is quite possible, applying the boundary conditions
a) U (0, t) = 0
2t
0 = (B + 0) e−αλ
2t
0 = Be−αλ
B=0
U (x, t) = (F sinλx) e−αλ
2t
b) U (3, t) = 0
0 = (F sinλ3) e−αλ
2t
Here, F 6= 0, e−αλ t =
6 0. Thus,
2
sinλ3 = 0
λl = nπ =⇒ λ =
Then
nπ
3
2
nπ −α( nπ
t
3 )
U (x, t) = F sin x e
3
λn are dened to be the eigen values and the function Un (x, t)corresponds to
any λn is called the eigen function. Thus the general solution becomes
nπ 2
nπ Un (x, t) = Fn sin x e−α( 3 ) t
3
From the superposition principle all the linear solutions, we get the general
solutions as
∞
∞
U (x, t) =
X
n=1
Un (x, t) =
X
n=1
68
2
nπ −α( nπ
t
3 )
Fn sin x e
3
U (x, 0) =
∞
X
Un (x, 0) =
n=1
∞ X
n=1
Fn sin
nπ π
2π
3π
4π
5
x = F1 sin x+F2 sin x+F3 sin x+F4 sin x+F5 sin
3
3
3
3
3
3
Using the initial conditions and comparing,
(a) U (x, 0) = 5sin4πx − 3sin8πx + 2sin10πx.
π
2π
3π
4π
5π
5sin4πx−3sin8πx+2sin10πx = F1 sin x+F2 sin x+F3 sin x+F4 sin x+F5 sin x+...
3
3
3
3
3
F1 = 5, n3 = 4 =⇒ n = 12
= 8 =⇒ n = 24
F2 = −3, 2n
3
F3 = 2, 3n
= 10 =⇒ n = 10
3
F4 = 0
F5 = 0
Therefore,
2
2
U (x, t) = 5sin4πxe−α(4π) t − 3sin8πxe−α(8π) t + 2sin10πxe−α(10π)
2
t
Example:
A rod of length l and with insulated side is initially at a uniform temperature U0 . It ends are suddenly cooled at 00 cand kept at that temperature. Find the temperature distribution function U (x, t). Ans: U (x, t) =
U0 sin
nπ
l
xe−
c 2 n2 π 2
t
l
Example:
A bar whose surface is insulated and has a length of 3 units and diusivity
2 units. If the ends are kept at temperature 0 units and initial temperature is U (x, 0) = 5sin4πx − 3sin8πx + 2sin10πx. Find the sistribution of
temperature at any point x at any time t.
Solution
69
Let the bar be placed on the x-ais such that one end is at the origin and the
other end is at x = 3. The distribution of heat is generally given by
Let U (x, t)be temperature at any given time t.
The boundary conditions are x = 3and x = 0
Initial conditions are U (x, 0) = 5sin4πx − 3sin8πx + 2sin10πx.
The heat equation is one dimension
δ2u
δu
=2 2
δt
δx
Using case 3 wherek = −λ2 since case 1 and case 2 give trivial solutions;
Then the two dierential equations are
T 0 X = 2X 00 T
T0
X 00
=
= −λ2
2T
X
These two functions could only be equal to a constant, say k as shown above,
T0
= λ2 =⇒ T 0 + 2λ2 T = 0
αT
δT
δT
2
= −2λ2 T =⇒
= −2λ2 δt =⇒ lnT = −2λ2 t + A ⇒ T = Ae−2λ t
δt
T
70
and
X 00
= λ2 =⇒ X 00 + λ2 X = 0
X
2t
T = Ae−2λ
X = Ccosλx + Dsinλx
Thus
U (x, t) = (Ccosλx + Dsinλx) Ae−2λ
2t
2t
U (x, t) = (Bcosλx + F sinλx) e−2λ
This is quite possible, applying the boundary conditions
a) U (0, t) = 0
0 = (B + 0) e−2λ
0 = Be−2λ
2t
2t
B=0
U (x, t) = (F sinλx) e−2λ
2t
b) U (3, t) = 0
2t
0 = (F sinλ3) e−2λ
Here, F 6= 0, e−αλ t =
6 0. Thus,
2
sinλ3 = 0
λl = nπ =⇒ λ =
Then
nπ
3
nπ 2
nπ U (x, t) = F sin x e−2( 3 ) t
3
λn are dened to be the eigen values and the function Un (x, t)corresponds to
71
any λn is called the eigen function. Thus the general solution becomes
nπ 2
nπ Un (x, t) = Fn sin x e−2( 3 ) t
3
From the superposition principle all the linear solutions, we get the general
solutions as
∞
∞
U (x, t) =
X
Un (x, t) =
n=1
U (x, 0) =
∞
X
Un (x, 0) =
n=1
Fn sin
n=1
∞ X
n=1
X
Fn sin
2
nπ −2( nπ
t
3 )
x e
3
nπ π
2π
3π
4π
5
x = F1 sin x+F2 sin x+F3 sin x+F4 sin x+F5 sin
3
3
3
3
3
3
Using the initial conditions and comparing,
(a) U (x, 0) = 5sin4πx − 3sin8πx + 2sin10πx.
5sin4πx−3sin8πx+2sin10πx = F1 sin
F1
F2
F3
F4
F5
n1 π
n2 2π
n3 3π
n4 4π
n5 5π
x+F2 sin
x+F3 sin
x+F4 sin
x+F5 sin
3
3
3
3
3
= 5, n31 = 4 =⇒ n1 = 12
= −3, 2n3 2 = 8 =⇒ n2 = 24
= 2, 3n3 3 = 10 =⇒ n3 = 10
=0
=0
Therefore,
U (x, t) = 5sin4πxe−
288 2
π t
9
− 3sin8πxe−
2576 2
π t
9
+ 2sin10πxe−
1800
(10π)2 t
9
11 DERIVATION OF THE WAVE EQUATION
Let the string be elastic with linear density ρstretched to length l with tension
T and xed at 0 and A then allowed to vibrate:
72
Let the tension at P and θbe T1 and T2 respectively.
Horizontal T along PA is −T1 cosθ1 , Vertical T along PA is −T1 sinθ1 .
Horizontal T along θB is −T2 sinθ2 , Vertical T along PA is −T2 cosθ2 .
PQ does not move horizontally. This implies that T1 cosθ1 = T2 cosθ2 =
T.......................(i)
Now, mass of PQ is given by linear denity multiplied by length. 4s · ρ
and accelLet the displacement of PQ be u1 . This implies that velocity = δu
δt
2
eration is δδt2u . Then from Newton's second law of motion F = ma.
4s · ρ
δ2u
= −T1 sinθ1 + T2 sinθ2
δt2
4s · ρ
δ2u
= T2 sinθ2 − T1 sinθ1
δt2
Dividing left hand side and right hand side by T
2
4s · ρ δδt2u
T2 sinθ2 T1 sinθ1
=
−
T
T
T
73
2
4s · ρ δδt2u
T2 sinθ2 T1 sinθ1
=
−
T
T2 cosθ2 T1 cosθ1
2
4s · ρ δδt2u
= T anθ2 − T anθ1
T
Now T anθ2 = δu
|at Q (slope or gradient)
δx x+4x
Similarly T anθ1 = δu
|at P (slope or gradient)
δx x
4s · ρ δ 2 u
δu
δu
|x+4x −
|x
=
2
T δt
δx
δx
δ2u
T
=
2
δt
ρ
δu
|
δx x+4x
− δu
|
δx x
4x
δ2u
T δ2u
=
δt2
ρ δx2
where 4x = 4s
12 DERIVATION OF HEAT EQUATION
The diusivity of material is given by
This can be expressed as α = sρk
74
T hermal conductivity
Specif ic heat capacity×density of the material
.
Let U (x, l)be the temperature at any point x at any time t. Let one end
be the origin and the bar be kept on x axis so that the other end is kept at
x = l. At A consider the cross-sectional area a at planepand another plane
p0 at x + 4x
Let pp0 = 4x → 0. Let temperature atpq beu and atp0 q 0 be u + 4u. The
4t |at x . a is the area while k is
heat ow from pq to p0 q 0 is given by −ka 4u
4x
the constant of proportionality(thermal conductivity).
By fouries law, the heat ux is directly proportional to the temperature
gradient.
4t |at x+4x .
The heat fow from p0 q 0 to the next string is given by −ka 4u
4x
Therefore the heat retained is given by:
ka
4u
4u
4t |x+4x −ka
4t |x
4x
4x
The amount of heat retained by the small element Also, heat gained or
lost frompqp0 q 0 is given by M sθwhere M is the mass (volume multiplied by
density), s is the specic heat capacity while θis the temperature change.
= (a4x × ρ) × s4u
75
ρas4u4x = ka
4u
4u
4t |x+4x −ka
4t |x
4x
4x
With the limit 4x → 0, 4u −→ 0 in time 4t
Dividing both sides by 4tthen by 4x
|
−ka 4u
|
ka 4u
4u
4x x+4x
4x x
ρas
=
4t
4x
ka δu
δu
δu
=
|x+4x −
|x
δt
ρas δx
δx
δu
k δ2u
=
δt
ρs δx2
By denition of a p.d.e
zero.
k
ρs
= c2 . c2 is a positive quantity and is not equal to
δu
δ2u
= c2 2
δt
δx
13 OTHER BOUNDARY VALUE PROBLEMS
Solve the boundary value problem δu
= 4 δu
, where U (0, y) = 5e−7y
δx
δy
Solution
By method of separation of variables
U (x, y) = X(x)Y (y)
δu
= X 0Y
δx
δu
= Y 0X
δy
X 0 Y = 4Y 0 X
X0
Y0
=
=c
4X
Y
Taking
X0
4X
= cand
Y0
Y
=c
X 0 − 4cX = 0
Y 0 − cY = 0
76
Integrating factor is given by
´
I.F = e
−4cdx
´
I.F = e
−cdy
= e−4cx
= e−cy
X = Ae4cx
Y = Becy
Since the solution is a product of X and Y i.e.U (x, y) = X(x)Y (y)
= Ae4cx (Becy )
= Dec(4x+y)
If compared to the boundary conditionsU (0, y) = 5e−7y ,
5e−7y = Decy
c = −7, D = 5
Thus,
U (x, y) = 5e−7(4x+y)
Exercise:
Solve the following boundary value problems:
(i) δu
= 2 δu
+ u ; U (x, 0) = 6e−3x
δx
δy
(ii) 3 δu
+ 2 δu
= 0 ; U (x, 0) = 4ex
δx
δy
Answers
(i) U (x, t) = 6e−3x−2t
3
(ii)U (x, t) = 4e−x− 2 y
14 METHOD OF LAPLACE TRANSFORMS
AND FOURIER TRANSFORMS:
−
The Laplace transorm of the function f(t) is Lf (t) = f (s) = f (t) =
77
´∞
0
e−st f (t)dt
ˆ
0
ˆ
∞
L [Ux ] =
ˆ
∞
U (x, t) dt
L [U (x, t)] =
0
δu −st
d
e dt =
δx
dx
ˆ
∞
e−st U (x, t) dt
0
ˆ ∞
δ u −st
d
e−st U (x, t) dt
L [Uxx ] =
e dt = 2
2
δx
dx 0
0
ˆ ∞
δu −st
δu
L [Ut ] = L
=
e dt
δt
δt
0
∞
2
2
u = e−st =⇒
δu
= −se−st
δt
δv
δu
=
=⇒ v = u
δt
δt
ˆ ∞
δu
−st ∞
L [Ut ] = L
= U e |0 +s
U e−st dt
δt
0
−
= 0 − u(0) + su
−
= u(0) + su
14.1 SINE AND COSINE FOURIER TRANSFORMS
The sine and cosine fourier transforms can be employed when the range of
integration varies from 0 to ∞. The choice of sine and cosine transform is
decided by the four of the boundary conditions. Now if a sine transform is
being used, we multiply the dierential equation by sinγxand integrate with
respect to x from 0 to ∞.Thus the sine transform of the term Uxx can be
written as follows:
ˆ ∞
ˆ ∞
sinγxUxx dx = Ux sinγx |∞
0 −γ
0
cosγxUx dx
0
The rst term of the right hand side vanishes at the lower limit through the
sine term. It also vanishes at the upper limit if U (x, t)is such that Ux → ∞as
x → ∞. Thus
78
ˆ
∞
sinγxUxx dx = γ [u(0, t)] − γ 2 Fs {u}
´∞
0
where Fs {u} = 0 U (x, t)sinγxdxand U (0, t)is the value of U at x = 0.
Example
If U (x, t)is the temperature at time t and γ the thermal diusivity of the
metal bar nd U (x, t) from the partial dierential equation. Ut = αUxx ,
x > 0 and t > 0with the boundary conditions U (0, t) = U0 , t > 0 and initial
condition U (x, 0) = 0, x > 0 .
Solution
Here we use the sine transform. Thus, taking the sine transform of Ut = αUxx ,
x > 0 and t > 0, weˆhave:
ˆ ∞
∞
Uxx sinγxdx
Ut sinγxdx = α
0
0
or
−
du
2
= α γU (0, t) − γ U
dt
−
−
or ddtu + αγ 2 u = αγU0 using equation (ii) which is an ordinary dierential
−
−
equation. The soluion is given by U = Ae−αγ t + γ1 U0 at t = 0, U = 0.
−
Therefore U =
U0
γ
2
i
h
2
1 − e−αγ t . The inversion is
2U0
U (x, t) =
π
ˆ
∞
1 − e−αγ
0
2t
sinγx
γ
dx
15 FURTHER EXAMPLES ON METHOD OF
SEPARATION VAIABLES
1. Consider Laplace's equation Uxx +Uyy = 0..........................(i)where U (x, y)epresents
the velocity potential of a uid particle in a certain domain. For example we
need to determine U (x, y)inside a unit circle x2 + y 2 < 1, when its values on
the circumference x2 + y 2 = 1are described.
79
Solution
This boundary value problem for Laplace's equation is dened as the dirichilet
problem. In obtaining the solution of (i), we make the following transformation. x = Rcosθand y = Rsinθ. Then the laplacee's equation can be reduced
to:
URR +
where U = U (R, θ).
R2 = x2 + y 2 , θ = tan−1 xy
− y2
dR
x
δθ
x
=
=
=
cosθ
,
=
y2
dx
R
δx
1+
dR
dy
=
y
R
δθ
= sinθ, δy
=
We know that
x2
1
x
2
1+ y 2
x
=
1
1
UR + 2 Uθθ = 0
R
R
−Rsinθ
R2
x
R2
=
= − R1 sinθ
Rcosθ
R2
=
cosθ
R
and
δ
δ δR
δθ
δ
δ
= δR
+ δθδ δx
= cosθ δR
− R1 sinθ δθδ
=
δx
δx
δy
δ
1
δ
sinθ δR + R cosθ δθ
δ
δu
δ
δ
δ
δ
δ2 u
= δx
= δx
u = cosθ δR
− R1 sinθ δθδ cosθ δR
−
δx2
δx
δx
δu
1
δu
δu
1
δu
cosθ δR− R sinθ
δθ cosθ δR − R sinθ
δθ
δ δR
δR δy
δθ
+ δθδ δy
=
1
sinθ δθδ
R
u =
1
δ
δu
1
δu
= cosθ
− sinθ
cosθ
− sinθ
R
δθ
δR R
δθ
1
δu
δ2u
1
δ2u
1
δu
δ2u
1
δu
1
= cosθ cosθ 2 + 2 sinθ − sinθ
− sinθ −sinθ
+ cosθ
− cosθ
− s
δR
R
δθ R
δuδθ
R
δR
δRδθ R
δR R
δ
δR
= cos2 θ
δu
1
δu
cosθ
− sinθ
δR R
δθ
δ2u 1
δu 1
δ2u 1
1
δ2u
1
δu 1
2
+
cosθsinθ
−
sinθcosθ
+
sin
θ−
sinθcosθ
+ 2 sinθcosθ + 2 sin
2
2
δR R
δθ R
δuδθ R
R
δRδθ R
δR R
80
2
δ2u 2
δ2u
1
1
δu 2
2 δu
2 δ u
−
sinθcosθ
+
+
+
cosθsinθ
sin
θ
sin
θ
.........................(a)
δR2 R2
δθ R
δRδθ R2
δR R2
δθ2
δu
δ
δ
δ
δ
δ2 u
δ
=
u = sinθ δR
=
+ R1 cosθ δθδ sinθ δR
+ R1 cosθ δθδ u =
δy 2
δy
δy
δy δy
δu
δu
1
δu
1
δu
sinθ δR
sinθ
+
cosθ
+
cosθ
δR
R δθ
R δθ
δ
δu
1
δu
1
δ
δu
1
δu
= sinθ
sinθ
+ cosθ
+ cosθ
sinθ
+ cosθ
δR
δR R
δθ
R
δθ
δR R
δθ
= cos2 θ
= sin2 θ
2
1
δ2u 2
δu 2
δ2u 1
2 δu
2 δ u
−
+
cos
θ
+
+
sinθcosθ
sinθcosθ
cos
θ
..........................(b)
δR2 R2
δθ R2
δRδθ R
δR R2
δθ2
Adding (a) and (b),
δ2u
1 δu
1 δ2u
δ2u δ2u
+
=
+
+
.
δx2 δy 2
δR2 R δR R2 δθ2
This can be written as:
URR +
1
1
UR + 2 Uθθ = 0
R
R
where U = U (R, θ).
The boundary condition for equation (i) is given by U (1, θ) = f (θ). By
using method of separation of variables, we have U (R, θ) = U (R)θ(θ),
R2 R00 + RR0 − U 2 = 0.............................(iii)
θ4 + n2 θ = 0 ; n = 0, 1, 2, .........................(iv)
Since
R00 θ +
1 0
1
R θ + 2 θ00 R = 0
R
R
1 0
1
R + R θ + 2 θ00 R = 0
R
R
00
1 0
1
R + R θ = − 2 θ00 R
R
R
00
81
or
R00 + R1 R0
1 θ00
=− 2
= n2
R
R θ
Equation (iii) is a cauchy euler type of equation and its solution can be
obtained by changing the independet variables. R = ez or logR = z
Now
dR dz
dR
=
·
dR
dz dR
=
1 dz
·
R dR
=
where D =
1
DR
R
d
dt
d
d2 R
=
2
dR
dR
1
DR
R
1 dR 1 d
=− 2
+
R dz
x dz
dR
dz
1 dR
1 dz d
=− 2
+ ·
·
R dz
R dR dz
=−
=−
dR
dz
1 dy
1 d2 R
+
·
R2 dt R2 dz 2
1
1
1 2
DR + 2 DR
= 2 D2 − D R
2
R
R
R
=
1
D (D − 1) R
R2
82
Thus,
R
R2
dR
dR
= DR =
dR
dz
d2 R
d2 R dR
=
D
(D
−
1)
=
−
R
dR2
dz 2
dz
On substituting these values in equation (iii),
d2 R dR dR
+
− n2 R = 0
−
dz 2
dz
dz
The solution is:
R = Aenz + Be−nz = ARn + BR−n
The solution of equation (iv) is given by θ = Ccosnθ+Dsinnθ........................(vi)
and hence U (R, θ) = (ARn + BR−n ) (Ccosnθ + Dsinnθ) .............................(vii)
From physical ground, we ae looking for a bounded solution, equation (vii)
is bounded provided Un = (An cosnθ + Bn sinnθ) Rn or n = 0, 1, 2, 3...where
´ 2π
´ 2π
An = π1 0 f (θ)cos(θ)dθand Bn = π1 0 f (θ)sin(θ)dθ
Example 2:
Consider the heat conduction in a thin metal bar of length l with insulated
sides. Let us assume that the end x = 0 is held at U0 degree celcius and the
end x = lis held at U1 degree celcius for all time t > 0. Let us suppose that
the temperature distribution at t = 0is (x, 0) = f (x) 0 ≤ x ≤ l. Determine
the temperature distribution in the bar at any position at any time t > 0.(10
marks)
Solution
The mathematical problem is the following:
The heat conduction equaiton Ut = αUxx ............................(i)is to be
solved subject to the boundary conditions x = 0 ,U (0, t) = U0 and x = l,
U (l, t) = Ul
and the initial conditions U (x, 0) = f (x). Use the method of separation
83
of variables for the solution of equation (i).
Example 3:
A transmission line of negligible resistance and capacitance has its sending
end at x = 0and receiving end at x = l. A constant voltage E0 is applied to
the sending end while an open circuit is maintained at the receiving end so
that the current there is zero. Assuming that the initial voltage and current
are zero, determine the voltage at any position x at any time t > 0
Solution
The telephone equation can be written as
d2 E
dE
d2 E
=
LC
+ (RC + LG)
+ RGE
2
2
dx
dt
dT
It is given that R = 0, C = 0, hence the equation reduces to
d2 E
dE
= LG ...................................(i)
2
dx
dT
with boundary conditions given as:
dE
x = 0, E (0, t) = E0 and x = L, dX
(L, t) = 0........................................(ii)
The initial conditions are given as
t = 0, U (x, 0) = 0
Taking Laplace transforms of equation (i), we have
−
d2 E
L 2 = LG S E − E (0, 0)
dx
d2 E
− LGSE = 0
dx2
√
E = Ae
LGSx
+ Be−
√
LGSx
A and B are constants to be determined.
To determine these constants, we rst need to transform the boundary
conditions as follows:
84
x = 0, L E (0, t) = E (0, t) = Es0
x = L, L E (L, t) = E (0) = 0
The application of the above boundary conditions yields
x = 0, A + B = Es0
√
√
x = l, Ae LGSl + Be− LGSl = 0
Solving the two equations, we have;
√
E0
E0 e2L Las
√
√
and B =
A=
s 1 + e2L Las
s 1 + e2L Las
Then substituting A and B we have;
√
√
√
E0 e2L Las
E0
LGSx
√
√
e
e− LGSx
+
E=
s 1 + e2L Las
s 1 + e2L Las
√
E0 cosh Las (L − x)
√
E=
scosh Las · L
Using the inverse laplace transforms, we have;
"
E (x, L) = E0
∞
4 X (−1)n (2n−1)2 π2
(2n − 1) (L − x) π
1+ y
e 4LGR cos
π n=1 2n − 1
2L
#
Example 2: Unsteady ow of a at plate
Lets have a uid of uid density ρconned over half plane y = 0. Let at
t = 0, the plate y = 0(which is initially at rest) start moving with velocity u
along the x-axis, then the uid above it will start moving because of velocity.
Since the plate is taken to be innite, we shall assume that the variables are
2
functions of y and t only. The equation of motion is δu
= γ δδyu2 .......(i). It
δt
resembles the heat equation.
Its boundary conditions are
a) t < 0, U (y, t) = 0
b) t > 0, U (0, t) = U
85
c) y −→ ∞, U (∞, t) = 0
Its initial conditions are
a) t = 0, U (y, 0) = 0
The equation of motion U given by (i) which on taking laplace transforms
d2 u s
d2 u
s
u
or
=
− u=0
dy 2
γ
dy 2 γ
u = Ae
√s
γ
y
+ Be
−
√s
γ
y
where A and B are constants of integration. The transformed boundary
conditions are :
a) u (0, s) = us
b) u (∞, s) = 0
Using the boundary conditions, we have
A = 0and B = us
Thus
√
u=
u −
e
s
s
y
γ
The inverse transform is
1
u=
2πi
ˆ
k+i∞
st
u (y, s) e dt = u 1 − ekf
k−i∞
y
√
2 st
The Fourier transform of U (x, t), Ux (x, t), Uxx (x, t)with respect to t are
given below:
ˆ
+∞
U (x, t) e−ist dt
F {Uxx } =
−∞
ˆ
+∞
F {Ux } =
−∞
ˆ
+∞
F {Uxx } =
−∞
δ
d
U (x, t) e−ist dt =
F [U (x, t)]
δx
dx
δ2
d2
−ist
U
(x,
t)
e
dt
=
F [U (x, t)]
δx2
dx2
86
and
ˆ
+∞
F {Ut } =
−∞
ˆ
=
e−iγt |+∞
−∞
δ −iγt
e dt
δt
+∞
U e−iγt dt
+ir
−∞
If U (±∞) = 0
F {Ut } = irF (u)
87