SMA 2371: PARTIAL DIFFERENTIAL EQUATIONS
c
⃝Francis
O. Ochieng
francokech@gmail.com
Department of Pure and Applied Mathematics
Jomo Kenyatta University of Agriculture and Technology
Course content
• Surfaces and curves in three dimensions. Simultaneous differential equations of the first order.
dx
dy
dz
Methods of solution of
=
=
. Orthogonal trajectories of system of curves on a surface.
P
Q
R
• Linear partial differential equations of the first order.
• Partial differential equations of the second order: Laplace, Poisson, heat and wave equations.
Methods of the solution by separation of the variables for Cartesian, spherical polar and cylindrical polar coordinates, and by Laplace and Fourier transform.
• Applications to engineering.
References
[1] Elements of PDE by Ian N. Sneddon
[2] Advanced Engineering Mathematics (10th ed.) by Erwin Kreyszig
[3] Schaum’s Outline Series: Theory and problems of PDE
Lecture 1
1
1.1
Review of basic concepts
Partial differentials
Let u = u(x, y), where x = x(s, t) and y = y(s, t). The partial differential of u is defined by
∂u =
∂u
∂u
∂x +
∂y.
∂x
∂y
Therefore, the partial derivative of u with respect to s and t is defined by
∂u
∂u ∂x ∂u ∂y
=
+
∂s
∂x ∂s
∂y ∂s
and
∂u ∂x ∂u ∂y
∂u
=
+
,
∂t
∂x ∂t
∂y ∂t
respectively.
Example(s):
1
1. Given that u = x2 y + y 2 , x = 2st, and y = tes . Find
∂u
∂u
and
.
∂s
∂t
Solution
∂u
∂u ∂x ∂u ∂y
=
+
= (2xy)(2t) + (x2 + 2y)(tes ) =
∂s
∂x ∂s
∂y ∂s
=
∂u
∂u ∂x ∂u ∂y
=
+
= (2xy)(2s) + (x2 + 2y)(es ) =
∂t
∂x ∂t
∂y ∂t
=
2(2st · tes )(2t) + ((2st)2 + 2tes )(tes )
8st3 es + 4s2 t3 es + 2t2 e2s
2(2st · tes )(2s) + ((2st)2 + 2tes )(es )
8s2 t2 es + 4s2 t2 es + 2te2s
Exercise:
2. Given that u is a function of x and y. If x = r cos θ and y = r sin θ, show that
(
)
(
∂u 2
1
+ 2
∂r
r
2
∂ u 1 ∂u
+
(b)
+
∂r2
r ∂r
(a)
1.2
)
(
∂u 2
=
∂θ
1 ∂2u
=
r2 ∂θ2
)
(
∂u 2
∂u
+
∂x
∂y
2
2
∂ u ∂ u
+ 2
∂x2
∂y
)2
Jacobian
⃗ by
Let F and G be functions of x, y and z and define a partial vector differential operator ∇
⃗ = î ∂ + ĵ ∂ + k̂ ∂
∇
∂x
∂y
∂z
Thus,
⃗ = î ∂F + ĵ ∂F + k̂ ∂F
∇F
∂x
∂y
∂z
⃗ = î ∂G + ĵ ∂G + k̂ ∂G
and ∇G
∂x
∂y
∂z
⃗ and ∇G
⃗ is defined by
The cross product of ∇F
î
ĵ
k̂
⃗
⃗
∇F × ∇G = Fx Fy Fz
Gx Gy Gz
= (Fy Gz − Fz Gy )î + (Fz Gx − Fx Gz )ĵ + (Fx Gy − Fy Gx )k̂
=
Fy Fz
F Fx
F
Fy
î + z
ĵ + x
k̂
Gy Gz
Gz Gx
Gx Gy
=
∂(F, G)
∂(F, G)
∂(F, G)
î +
ĵ +
k̂
∂(y, z)
∂(z, x)
∂(x, y)
= P î + Qĵ + Rk̂
The components of the above vector are called direction cosines or Jacobian of F and G.
2
Surfaces and curves in three dimensions
If the rectangular Cartesian coordinates (x, y, z) of a point P in three dimensions are connected by a
relation of the form
F (x, y, z) = 0
then the point P lies on the surface.
2
(1)
We call equation (1) the equation of a surface. For example, the equation x2 + y 2 + z 2 = a2 is the
equation of a sphere centre (0, 0, 0) and radius a. The total differential of equation (1) is given by
dF ≡
∂F
∂F
∂F
dx +
dy +
dz = 0
∂x
∂y
∂z
which can be interpreted in vector form as
(
)
∂F
∂F
∂F
î +
ĵ +
k̂ · (dxî + dy ĵ + dz k̂) = 0
∂x
∂y
∂z
(
(
⇒
∂F ∂F ∂F
,
,
∂x ∂y ∂z
)
· (dx, dy, dz) = 0
)
∂F ∂F ∂F
,
,
are the direction ratios of the normal to the surface at point P while (dx, dy, dz)
∂x ∂y ∂z
are direction ratios of the tangent to the surface at the same point.
Thus,
→ Note: a surface may be envisaged as being generated by a curve. A curve is obtained as the
intersection of any two surfaces. For instance, a conic is a curve obtained as the intersection of the
surface of a cone with a plane. Conics include hyperbola, parabola, and ellipse.
Example(s):
(a) Determine the direction cosines of the tangent at the point (x, y, z) to the conics, x2 y + xz = 0
and x2 + y 2 + z 2 = 0.
Solution
Here, F = x2 y + xz and G = x2 + y 2 + z 2 . So,
P
=
Q =
R =
∂(F, G)
= Fy Gz − Fz Gy = 2x2 z − 2xy = 2x(xz − y)
∂(y, z)
∂(F, G)
= Fz Gx − Fx Gz = 2x2 − 2z(2xy + z)
∂(z, x)
∂(F, G)
= Fx Gy − Fy Gx = 2y(2xy + z) − 2x3
∂(x, y)
(b) Show that the direction cosines of the tangent at the point (x, y, z) to the conics ax2 +by 2 +cz 2 = 1
and x + y + z = 1 are proportional to (by − cz, cz − ax, ax − by).
Solution
Here, F = ax2 + by 2 + cz 2 − 1 and G = x + y + z − 1. So,
P
=
Q =
R =
∂(F, G)
= Fy Gz − Fz Gy = 2by − 2cz
∂(y, z)
∂(F, G)
= Fz Gx − Fx Gz = 2cz − 2ax
∂(z, x)
∂(F, G)
= Fx Gy − Fy Gx = 2ax − 2by
∂(x, y)
Therefore, (P, Q, R) = 2(by − cz, cz − ax, ax − by)
by), where 2 is the constant of proportionality.
3
⇒
(P, Q, R) ∝ (by − cz, cz − ax, ax −
2.1
Simultaneous differential equations of the first order
Consider a curve in space which is the intersection of the surfaces F (x, y, z) = 0 and G(x, y, z) = 0.
It follows that
dG = 0
∂F
∂F
∂F
dx +
dy +
dz = 0
∂x
∂y
∂z
⇒
dF = 0
∂G
∂G
∂G
dx +
dy +
dz = 0
∂x
∂y
∂z
⇒
∂F
∂F
∂F
dx +
dy = −
dz · · · (i)
∂x
∂y
∂z
⇒
∂G
∂G
∂G
dx +
dy = −
dz · · · (ii)
∂x
∂y
∂z
⇒
In matrix form we have

∂F
 ∂x



 ∂G
∂x

∂F
 ∂x

Let A = 

 ∂G

∂F
∂y 

 ∂F 
−
dz 

  ∂z 

=
 ∂G 


∂G  dy
−
∂y
∂F
∂x
∂F
∂y
∂x
∂G
∂y
. Then, Det(A) = |A| =

∂G 
∂G
∂x ∂y
Cramer’s rule we obtain
−
∂F
dz
∂z
−
dx =

∂F  
∂y 
 dx
∂z
(2)
dz
= Fx Gy − Fy Gx . Solving equation (2) using
∂F
∂y
∂G
∂G
dz
F y Gz − F z Gy
∂z
∂y
=
dz
|A|
F x Gy − F y Gx
⇒
dx
dz
=
Fy Gz − Fz Gy
F x Gy − F y Gx
⇒
dy
dz
=
Fz Gx − Fx Gz
Fx Gy − Fy Gx
and
∂F
∂x
dy =
∂G
∂x
−
∂F
dz
∂z
∂G
dz
Fz Gx − Fx Gz
∂z
=
dz
|A|
Fx Gy − Fy Gx
−
Therefore,
dx
dy
dz
=
=
Fy Gz − Fz Gy
Fz Gx − Fx Gz
Fx Gy − Fy Gx
⇒
dx
dy
dz
=
=
P
Q
R
(3)
Equations (3) are called symmetric equations and are the differential equations of the integral curves
(intersection of two surfaces).
Lecture 2
2.2
Methods of solution of
dx
dy
dz
=
=
P
Q
R
The reverse problem is that of finding the surfaces F (x, y, z) = 0 and G(x, y, z) = 0 from the equations
(3). We need to obtain two unique solutions (integral curves), say u(x, y, z) = c1 and v(x, y, z) = c2 ,
where c1 and c2 are arbitrary constants. There are 3 methods of solving equations (3):
1. Method of multipliers (intuition)
2. One variable absent (method of grouping)
3. Exact differential method
4
2.2.1
Method of Multipliers (method of intuition)
This method uses multipliers l, m and n (which are not necessarily constants) so that for a constant
k, equation (3) becomes
ldx + mdy + ndz
=k
lP + mQ + nR
⇒
ldx + mdy + ndz = k(lP + mQ + nR)
These multipliers are chosen such that the condition lP + mQ .+ nR = 0 is satisfied. Hence,
ldx + mdy + ndz = 0,
which on integration yields the first integral curve, u(x, y, z) = c1 . It may be possible to get another set
of multipliers l, m and n to obtain the second integral curve, v(x, y, z) = c2 . Therefore, the integral
surface generated by these curves is given by ϕ(u, v) = 0, where ϕ is an arbitrary function.
Example(s):
Find the integral curves and the integral surface of the following differential equations.
(a)
dx
dy
dz
=
=
x(y − z)
y(z − x)
z(x − y)
Solution
From the given equations we have P = x(y −z), Q = y(z −x) and R = z(x−y). Using the
multipliers l, m and n, we require that lP +mQ+nR = 0 · · · (∗) so that ldx+mdy+ndz = 0.
Now,
i) By intuition, condition (∗) is satisfied if we take l = 1, m = 1 and n = 1. Hence,
dx + dy + dz = 0. Integrating yields
x + y + z = c1
ii) Again by intuition, condition (∗) is satisfied if we take l =
1
1
1
, m = and n = . Hence,
x
y
z
dx dy dz
+
+
= 0. Integrating yields
x
y
z
ln x + ln y + ln z = ln c2
⇒
xyz = c2
Therefore, the required integral curves are given by x + y + z = c1 and xyz = c2 , where
c1 and c2 are parameters. The integral surface generated by these curves is ϕ(c1 , c2 ) = 0.
That is,
ϕ(x + y + z, xyz) = 0
2.2.2
One variable absent
If one of the variables is absent in one of the equations, we can easily derive one of the solutions. Also,
it is possible to use the first solution to obtain a second solution by eliminating one of the variables
in one of the equations.
Example(s):
Find the integral curves of the following equations.
(a)
dy
dz
dx
=
=
x+z
y
z + y2
Solution
Since x is absent in the second and third equations, we take
dy
dz
=
y
z + y2
5
⇒
dz
1
− z=y
dy y
This is a linear ODE in z with integrating factor
∫
−1
dy
y
µ=e
1
d
Multiplying through by we get
y
dy
= e− ln y =
1
y
( )
z
y
z
= y + c1
y
= 1. Integrating yields
⇒
z = y 2 + c1 y
Using the first solution to eliminate z in the first equation, we have
dx
dy
=
x+z
y
dx
dy
=
x + y 2 + c1 y
y
⇒
dx 1
− x = y + c1
dy y
⇒
This is a linear ODE in x with integrating factor
∫
−1
dy
y
µ=e
Multiplying through by
1
d
yields
y
dy
= e− ln y =
( )
x
y
x
= y + c1 ln y + c2
y
=1+
⇒
1
y
c1
. Integrating yields
y
x = c1 y ln y + c2 y + y 2
Therefore, the required integral curves are given by z = c1 y + y 2 and x = c1 y ln y + c2 y + y 2 ,
where c1 and c2 are parameters.
2.2.3
Exact differential method
It is much simpler to try to cast the given differential equations into a form which suggests their
solution. Note that adding/subtracting the numerators and denominators of any two “fractions”
doesn’t alter their value. E.g, we may write
dx
dy
dz
ldx + mdy + ndz
dx + dy
dx − dy − 2dz
=
=
=
=
=
, e.t.c.
P
Q
R
lP + mQ + nR
P +Q
P − Q − 2R
So that if lP + mQ + nR = 0, then ldx + mdy + ndz = 0 is an exact differential. This method is best
illustrated in the following example.
Example(s):
Find the integral curves of the following equations:
(a)
dy
dz
dx
=
=
.
y(x + y) + az
x(x + y) − az
z(x + y)
Solution
We have
dx
dy
dz
dx + dy
xdx − ydy
=
=
=
=
2
y(x + y) + az
x(x + y) − az
z(x + y)
(x + y)
az(x + y)
From
dx + dy
dz
=
(x + y)2
z(x + y)
dx + dy
dz
=
x+y
z
⇒
⇒
From
xdx − ydy
dz
=
az(x + y)
z(x + y)
⇒
⇒
⇒
ln(x + y) = ln z + ln c1
x + y = c1 z
xdx − ydy − adz = 0
x2 − y 2 − 2az = c2
Therefore, the required integral curves are given by x + y = c1 z and x2 − y 2 − 2az = c2 ,
where c1 and c2 are parameters.
6
Exercise:
Use any method to find the integral curves of the following equations.
(a)
dx
dy
dz
=
=
.
2
y
−xy
x(z − 2y)
Solution
dx
dy
From 2 =
y
−xy
⇒
xdx + ydy = 0
⇒
From
dy
dz
=
−xy
x(z − 2y)
⇒
x2 + y 2 = c1
zdy + ydz − 2ydy = 0
⇒
⇒
d(zy) − 2ydy = 0
zy − y 2 = c2
Therefore, the integral curves of the given differential equations are the members of the
two-parameter family x2 + y 2 = c1 and zy − y 2 = c2 .
(b)
x2
dx
dy
dz
=
=
.
2
2
−y −z
2xy
2xz
Solution
dy
dz
From
=
2xy
2xz
From
x(x2
⇒
dy
dz
=
y
z
⇒
ln y = ln z + ln c1
dy
xdx + ydy + zdz
=
2
− − z ) + y(2xy) + z(2xz)
2xy
y2
⇒
ln(x2 + y 2 + z 2 ) = ln y + ln c2
⇒
y = c1 z
2(xdx + ydy + zdz)
dy
=
2
2
2
(x + y + z )
y
⇒
⇒
x2 + y 2 + z 2 = c2 y
Therefore, the integral curves of the given differential equations are the members of the
two-parameter family y = c1 z and x2 + y 2 + z 2 = c2 y.
(c)
dx
dy
dz
=
=
.
xz − y
yz − x
1 − z2
Solution
dx + dy
dz
From
=
xz − y + yz − x
1 − z2
⇒
⇒
From
−dz
dx + dy
=
x+y
1+z
(x + y)(1 + z) = c1
zdz
xdx − ydy
=
x(xz − y) − y(yz − x)
z(1 − z 2 )
⇒
⇒
xdx − ydy
zdz
=
2
2
x −y
1 − z2
(x2 − y 2 )(1 − z 2 ) = c2
Therefore, the integral curves of the given differential equations are the members of the
two-parameter family (x + y)(1 + z) = c1 and (x2 − y 2 )(1 − z 2 ) = c2 .
(d)
dx
dy
dz
=
=
.
x(y 2 − z 2 )
y(z 2 − x2 )
z(x2 − y 2 )
(e)
dy
dz
dx
=
= 2
.
y − xz
x + yz
x + y2
(f)
dx
dy
dz
=
=
.
x(2y 4 − z 4 )
y(z 4 − 2x4 )
z(x4 − y 4 )
(g)
x2
[ans: x2 + y 2 + z 2 = c1 ,
[ans: x2 − y 2 + z 2 = c1 ,
[ans: x4 + y 4 + z 4 = c1 ,
dx
dy
dz
= 2
= 2
.
− yz
y − xz
z − xy
[ans:
7
x−y
= c1 ,
y−z
xyz = c2 ]
xy − z = c2 ]
xyz 2 = c2 ]
xy + yz + xz = c2 ]
(h)
dx
dy
dz
=
=
.
x(x + y)
−y(x + y)
−(x − y)(2x + 2y + z)
[ans: xy = c1 ,
(x + y)(x + y + z) = c2 ]
Lecture 3
2.3
Orthogonal trajectories of a system of curves on a surface
Given a surface
F (x, y, z) = 0
(4)
and a family (or system) of curves on it, we need to find another family of curves each of which lies on
the surface (4) and cuts every curve of the original family at right angles. The new family of curves
is called the family of orthogonal trajectories on the surface (4).
The original family of curves may be thought of as the intersections of the surface (4) with the oneparameter family of surfaces
G(x, y, z) = k
From F (x, y, z) = 0 and G(x, y, z) = k, the tangential direction (dx, dy, dz) to the given curve through
the point (x, y, z) on the surface (4) satisfies the equations
dF = 0 i.e, Fx dx + Fy dy + Fz dz = 0
and
dG = 0 i.e, Gx dx + Gy dy + Gz dz = 0
We solve the above two equations using Cramer’s rule to obtain
dy
dz
∂(F, G)
dx
=
=
, where P =
,
P
Q
R
∂(y, z)
Q=
∂(F, G)
∂(F, G)
, and R =
∂(z, x)
∂(x, y)
The curve through (x, y, z) of the orthogonal family has tangential direction (dx′ , dy ′ , dz ′ )
which lies on the surface (4), so that
Fx dx′ + Fy dy ′ + Fz dz ′ = 0
(5)
and is perpendicular to the original system of curves. Also since the vectors (dx, dy, dz) and (dx′ , dy ′ , dz ′ )
are orthogonal, then
(dx, dy, dz) · (dx′ , dy ′ , dz ′ ) = 0
⇒
dxdx′ + dydy ′ + dzdz ′ = 0
i.e.,
P dx′ + Qdy ′ + Rdz ′ = 0
Solving for dx′ , dy ′ and dz ′ between equations (5) and (6) using Cramer’s rule, we obtain
dy ′
dz ′
dx′
=
=
RFy − QFz
P Fz − RFx
QFx − P Fy
8
(6)
This can be rewritten as:
dx′
Pb
=
dy ′
=
b
Q
dz ′
b
R
F Fz
b = Fz Fx , R
b = Fx Fy
where Pb = y
, Q
Q R
R P
P Q
,
(7)
The solution of equations (7) alongside with (4) gives the system of orthogonal trajectories.
Example(s):
(a) Find the orthogonal trajectories on the surface x2 + y 2 = αz 2 of its intersections with the family
of planes parallel to xy-plane (i.e, z = k)
Solution
Let F (x, y, z) = x2 + y 2 − αz 2 = 0 and G(x, y, z) = z = k. Then,
P
=
Q =
R =
∂(F, G)
= Fy Gz − Fz Gy = 2y − (−2αz)(0) = 2y
∂(y, z)
∂(F, G)
= Fz Gx − Fx Gz = (−2αz)(0) − 2x(1) = −2x
∂(z, x)
∂(F, G)
= Fx Gy − Fy Gx = 2x(0) − 2y(0) = 0
∂(x, y)
Pb =
Fy Fz
2y −2αz
=
= −4αxz
Q R
−2x
0
b =
Q
Fz Fx
−2αz 2x
=
= −4αyz
R P
0
2y
b =
R
Fx Fy
2x 2y
=
= −4x2 − 4y 2
P Q
2y −2x
The symmetric equations for the orthogonal trajectories are given by:
dx
Pb
=
dy
b
Q
=
dx
dy
=
x
y
parameter.
From
From
dz
⇒
b
R
⇒
dy
dz
dx
=
=
2
−4αxz
−4αyz
−4x − 4y 2
dx
dy
=
x
y
αzdz
xdx + ydy
= 2
2
2
x +y
x + y2
⇒
⇒
ln x = ln y + ln c1
xdx + ydy = αzdz
⇒
⇒
dy
αzdz
dx
=
= 2
x
y
x + y2
⇒
x = c1 y, where c1 is a
x2 + y 2 = αz 2 + c2
⇒
c2 = 0
∴ x2 + y 2 = αz 2 . Hence F ≡ 0 is necessarily one of the solutions. Therefore, the orthogonal
trajectories are the curves
x = c1 y and x2 + y 2 = αz 2
(b) Find the equations of the system of curves on the cylinder 2y = x2 orthogonal to its intersections
with the hyperboloids of the one-parameter system xy = z + c, where c is a parameter.
Solution
Let F = 2y − x2 = 0 and G = xy − z = c. The partial derivatives are:
⇒
Fx = −2x,
Fy = 2,
Fz = 0,
9
Gx = y,
Gy = x,
Gz = −1
P
=
Q =
R =
∂(F, G)
= Fy Gz − Fz Gy = −2
∂(y, z)
∂(F, G)
= Fz Gx − Fx Gz = −2x
∂(z, x)
∂(F, G)
= Fx Gy − Fy Gx = −2x2 − 2y = −2(x2 + y)
∂(x, y)
Pb =
Fy Fz
2
0
=
= −4(x2 + y)
Q R
−2x −2(x2 + y)
b =
Q
Fz Fx
0
−2x
=
= −4x(x2 + y)
2
R P
−2(x + y) −2
b =
R
−2x
2
Fx Fy
= 4(x2 + 1)
=
−2 −2x
P Q
The symmetric equations for the orthogonal trajectories are
⇒
From
dx
dy
dz
=
=
2
2
2
−4(x + y)
−4x(x + y)
4(x + 1)
dy
dx
=
x2 + y
x(x2 + y)
⇒
⇒
From
⇒
xdx = dy
c1 = 0
⇒
dx
Pb
=
dy
b
Q
=
dz
b
R
dx
dy
−dz
=
= 2
2
+y
x(x + y)
x +1
x2
x2
c1
=y+
2
2
⇒
x2 = 2y + c1
∴ x2 = 2y
dx
−dz
x2
=
.
Substituting
y
=
yields
x2 + y
x2 + 1
2
dx
x2 +
x2
=
2
−dz
x2 + 1
(
3dz + 2 1 +
⇒
2dx
−dz
= 2
3x2
x +1
⇒
2(x2 + 1)dx
= −3dz
x2
)
(
1
1
dx = 0. Integrating yields 3z + 2 x −
x2
x
)
= c2
Therefore, the orthogonal trajectories are the curves
(
3z + 2 x −
1
x
)
= c1 ,
2y = x2
Exercise:
(a) [Assignment 1 ] Find the orthogonal trajectories on surface of the sphere x2 + y 2 + z 2 = a2 of
its intersections with the paraboloids xy = cz, where c is a parameter.
[ans:
2 − x2
a
x2 + y 2 + z 2 = a2 ,
= c2 ]
a2 − y 2
(b) Show that the orthogonal trajectories on the hyperboloid x2 + y 2 − z 2 = 1 of the conics in
which it is cut by the system of planes x + y = c are its curves of intersection with the surfaces
(x − y)z = k, where k is a parameter.
(c) Find the orthogonal trajectories on the conicoid (x + y)z = 1 of the conics in which it is cut by
the system of planes x − y + z = k, where k is a parameter. [ans: The orthogonal trajectories
1
1
are the curves x = z − 3 +
+ c2 , (x + y)z = 1]
6z
2z
Lecture 4
10
3
Pfaffian differential equations
A differential equation of the form:
n
∑
Fi (x1 , x2 , · · · , xn )dxi = 0
i=1
is called a Pfaffian differential equation in n variables x1 , x2 , · · · , xn . Thus, a Pfaffian differential
equation in 3 variables is an equation of the form
P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz = 0
(8)
If equation (8) has a solution, we say that it is integrable. The following theorem gives the necessary
and sufficient condition for equation (8) to be integrable.
⃗ = (P, Q, R), then equation (8) is integrable if and only if X
⃗ · curlX
⃗ =0
Theorem 3.1. Let X
Example(s):
Verify that the equation (x2 z − y 3 )dx + 3xy 2 dy + x3 dz = 0 is integrable.
Solution
⃗ = (P, Q, R) where P = x2 z − y 3 , Q = 3xy 2 , R = x3
Let X
î
ĵ
∂
∂
⃗ =
curlX
∂x
∂y
x2 z − y 3 3xy 2
k̂
∂
= 0î − (3x2 − x2 )ĵ + (3y 2 + 3y 2 )k̂ = (0, −2x2 , 6y 2 )
∂z
x3
⃗ · curlX
⃗ = (x2 z − y 3 , 3xy 2 , x3 ) · (0, −2x2 , 6y 2 ) = 0 − 6x3 y 2 + 6x3 y 2 = 0, hence integrable.
X
3.1
Methods of solution of P dx + Qdy + Rdz = 0
The following methods are employed in solving Pfaffian differential equations in three variables:
1. Variables separable
2. Re-grouping (inspection)
3. One variable separable
4. Homogeneous equations
5. Natani’s method
3.1.1
Variables separable
If equation (8) can be rearranged to take ∫the form P (x)dx
+ Q(y)dy
∫
∫ + R(z)dz = 0, then we integrate
directly in order to obtain a solution i.e., P (x)dx + Q(y)dy + R(z)dz = c, where c is a constant
of integration.
Example(s):
Solve 2zydx + zxdy + xy(1 + z)dz = 0.
Solution
⃗ · curlX
⃗ = 0.
Clearly, the equation is integrable i.e. X
2dx dy (1 + z)dz
Dividing through by xyz we get
+
+
= 0. Integrating yields
x
y
z
2 ln x + ln y + z + ln z = − ln c
11
⇒
cx2 yz = e−z
3.1.2
Re-grouping (inspection)
This method employs complete methods like.
i) d(xy) = xdy + ydx
( )
ii) d
y
x
=
xdy − ydx
x2
iii) d(ln(x2 + y 2 )) =
2(xdx + ydy)
x2 + y 2
Example(s):
Solve (x2 z − y 3 )dx + 3xy 2 dy + x3 dz = 0.
Solution
⃗ · curlX
⃗ = 0. Now, expanding and regrouping the given
Clearly, the equation is integrable i.e., X
2
3
2
3
equation yields x zdx − y dx + 3xy dy + x dz = 0 ⇒ x2 (zdx + xdz) + 3xy 2 dy − y 3 dx = 0
⇒
Integrating yields zx +
3.1.3
(
3xy 2 dy − y 3 dx
(zdx + xdz) +
=0
x2
y3
d(zx) + d
x
⇒
)
=0
y3
=c
x
One variable separable
Suppose one variable, say z, is separable then equation (8) takes the form
P (x, y)dx + Q(x, y)dy + R(z)dz = 0
(9)
⃗ · curlX
⃗ = 0. Now,
Since the equation is integrable, then X
î
ĵ
k̂
)
)
(
(
∂Q ∂P
∂Q ∂P
∂
∂
∂
⃗ =
curlX
−
−
k̂ = 0, 0,
= (0 − 0)î − (0 − 0)ĵ +
∂x
∂y
∂x
∂y
∂x
∂y
∂z
P (x, y) Q(x, y) R(z)
(
)
(
)
∂Q ∂P
∂Q ∂P
⃗
⃗
X · curlX = (P, Q, R) · 0, 0,
−
=R
−
= 0. But R ̸= 0
∂x
∂y
∂x
∂y
∂Q ∂P
∂Q
∂P
⇒
−
=0 ⇒
=
∂x
∂y
∂x
∂y
Thus, P dx + Qdy is an exact differential i.e., there exists a function ϕ(x, y) such that dϕ = P dx + Qdy.
∂P . ∂Q
∂ϕ
∂ϕ
∂ϕ
∂ϕ
=
This means that
dx +
dy = P dx + Qdy ⇒
= P,
= Q ⇒
.
∂y
∂x
∂x
∂y
∂x
∂y
Therefore, equation (8) reduces to
.
dϕ + R(z)dz
=0
Example(s):
Solve x(y 2 − a2 )dx + y(x2 − z 2 )dy − z(y 2 − a2 )dz = 0.
Solution
⃗ · curlX
⃗ = 0.
Clearly, the given equation is integrable i.e. X
2
2
2
2
Dividing through by (y − a )(x − z ) yields
xdx
ydy
zdz
+
−
=0
x2 − z 2 y 2 − a2 x2 − z 2
⇒
xdx − zdz
y
+ 2
dy = 0
x2 − z 2
y − a2
1
1
1
Integrating yields ln(x2 − z 2 ) + ln(y 2 − a2 ) = ln c
2
2
2
is a constant.
12
⇒
c = (y 2 − a2 )(x2 − z 2 ), where c
Exercise:
1. Solve x(x − z)dx + x2 zdy − x(1 + xy)dz = 0.
[ans: ln x +
2. Solve xz 3 dx − zdy + 2ydz = 0.
z
+ zy = c]
x
[ans: x2 z 2 − 2y = cz 2 ]
Lecture 5
3.1.4
Homogeneous equations
Equation (8) is said to be homogeneous of degree n if P, Q and R are homogeneous and of the same
degree n. Note that a function f (x, y, z) is said to be homogeneous of degree n if there exists a real
number λ such that
f (λx, λy, λz) = λn f (x, y, z)
For example, the function f (x, y, z) = x2 y + xyz + y 2 z is homogeneous of degree 3. If equation (8) is
homogeneous of degree n then we can solve it by making the following substitutions:
y = ux
⇒
dy = udx + xdu
z = vx
⇒
dz = vdx + xdv
Thus, equation (8) becomes
P (x, ux, vx)dx + Q(x, ux, vx)[udx + xdu] + R(x, ux, vx)[vdx + xdv] = 0
⇒ xn P (u, v)dx + xn Q(u, v)[udx + xdu] + xn R(u, v)[vdx + xdv] = 0
⇒ P (u, v)dx + Q(u, v)[udx + xdu] + R(u, v)[vdx + xdv] = 0
⇒ [P (u, v) + uQ(u, v) + vR(u, v)]dx + x[Q(u, v)du + R(u, v)dv] = 0
Q(u, v)du + R(u, v)dv
dx
dx
+
=0 ⇒
+ A(u, v)du + B(u, v)dv = 0
⇒
x
[P (u, v) + uQ(u, v) + vR(u, v)]
x
which can now be solved using the method of one variable separable.
Example(s):
Solve the following equations.
(a) ydx − (x + z)dy + ydz = 0.
Solution
⃗ · curlX
⃗ = 0. The given equation is homogeneous
Clearly, the given equation is integrable i.e., X
of degree 1. Thus, we use the following substitutions
y = ux
⇒
dy = udx + xdu
z = vx
⇒
dz = vdx + xdv
Substituting in the given equation yields
uxdx − (x + vx)[udx + xdu] + ux[vdx + xdv] = 0
⇒
⇒
udx − (1 + v)[udx + xdu] + u[vdx + xdv] = 0
[u + uv − (1 + v)u]dx − x[(1 + v)du − udv] = 0
⇒
(1 + v)du − udv = 0
du
dv
⇒
−
=0
u
1+v
Integrating yields
∫
du
−
u
∫
dv
=
1+v
∫
0
⇒
ln u − ln(1 + v) = ln c
13
⇒
u = c(1 + v)
Back substitution yields
(
z
y
=c 1+
x
x
)
⇒
⇒
y = c(x + z)
x + z = c1 y
(b) yz(y + z)dx + xz(x + z)dy + xy(x + y)dz = 0
Solution
⃗ · curlX
⃗ = 0. It is also homogeneous
Clearly, the given equation is integrable i.e., X
Put y = ux ⇒ dy = udx + xdu and z = vx ⇒ dz = vdx + xdv. Substituting we have
x2 uv(ux + vx)dx + x2 v(x + xv)[udx + xdu] + x2 u(x + ux)[vdx + xdv] = 0
⇒
⇒
uv(u + v)dx + v(1 + v)[udx + xdu] + u(1 + u)[vdx + xdv] = 0
[uv(u + v) + uv(1 + v) + uv(1 + u)]dx + xv(1 + v)du + xu(1 + u)dv = 0
dx
v(1 + v)du
u(1 + u)dv
⇒
+
+
=0
x
2uv(u + v + 1) 2uv(u + v + 1)
2dx
(1 + v)du
(1 + u)dv
⇒
+
+
= 0 · · · · · · · · · (i)
x
u(u + v + 1) v(u + v + 1)
A
B
(1 + v)
= +
u(u + v + 1)
u
(u + v + 1)
A = 1. Putting u = −(v + 1) ⇒
⇒
By partial fractions, let
Putting u = 0
⇒
A(u + v + 1) + Bu = 1 + v
B = −1. Substituting yields
(1 + v)
1
1
= −
u(u + v + 1)
u (u + v + 1)
Similarly,
(1 + u)
1
1
= −
v(u + v + 1)
v (u + v + 1)
Substituting in equation (i) we get
⇒
du
dv
dv
2dx du
+
−
+
−
=0
x
u
(u + v + 1)
v
(u + v + 1)
2dx du dv
du + dv
+
+
−
=0
x
u
v
(u + v + 1)
Integrating yields
2 ln x + ln u + ln v − ln(u + v + 1) = ln c
⇒
x2 uv = c(u + v + 1)
Back substitution yields
2y
(
)
z
y
z
x
=c
+ +1
xx
x x
3.1.5
⇒
xyz = c(x + y + z)
Natani’s method
Step 1: suppose one variable, say z, is a constant ⇒
dz = 0. Thus, equation (8) reduces to
P dx + Qdy = 0
We then solve the above equation and equate the solution, say ϕ(x, y, z), to an arbitrary function of
z i.e.,
ϕ(x, y, z) = f (z) · · · (i), (this is the solution provided f (z) is determine).
Step 2: let either of the remaining variables, say x, to be equal to 1 or 0. For instance, if x = 1,
dx = 0. Thus, equation (8) reduces to
Q(y, z)dy + R(y, z)dz = 0
Let the solution to this equation be
ψ(y, z) = c · · · (ii)
14
⇒
Also put x = 1 in equation (i) to obtain
ϕ(y, z) = f (z) · · · (iii)
Eliminate y between equations (ii) and (iii) to obtain f (z). Substitute the value of f (z) into equation
(i) to obtain the solution for the given differential equation.
Example(s):
Verify that the following equations are integrable and find their primitives (solutions).
(a) (x2 z − y 3 )dx + 3xy 2 dy + x3 dz = 0.
Solution
⃗ · curlX
⃗ = 0. By Natani’s method:
Clearly, the given equation is integrable i.e., X
Step 1: we note that the solution is simpler if we treat x to be a constant, ⇒ dx = 0. So the
given differential equation reduces to
⇒
3xy 2 dy + x3 dz = 0
3y 2 dy + x2 dz = 0
Integrating yields
∫
∫
2
3y dy +
∫
2
x dz =
0
⇒
y 3 + x2 z = f (x) · · · (i)
Equation (i) is the required solution provided that f (x) is determined.
Step 2: put y = 1 in the given differential equation, ⇒ dy = 0. So the differential equation
reduces to
(x2 z − 1)dx + x3 dz = 0
⇒
zdx + xdz −
dx
=0
x2
⇒ d(zx) −
dx
=0
x2
Integrating yields
∫
d(zx) −
∫
dx
=
x2
∫
0
⇒
zx +
1
=c
x
⇒
z=
cx − 1
· · · (ii),
x2
where c is an arbitrary constant. Also, putting y = 1 in equation (i) we get
1 + x2 z = f (x) · · · (iii)
Eliminating z between equations (ii) and (iii) (i.e., substituting equation (ii) in equation (iii))
yields
{
}
cx − 1
1 + x2
= f (x) ⇒ f (x) = cx · · · (iv)
x2
Substituting equation (iv) in equation (i) yields y 3 + x2 z = cx, which is the solution to the given
differential equation.
(b) z(z + y 2 )dx + z(z + x2 )dy − xy(x + y)dz = 0.
Solution
⃗ · curlX
⃗ = 0. By Natani’s method:
Clearly, the given equation is integrable i.e., X
Step 1: we note that the solution is simpler if we treat y to be a constant, ⇒ dy = 0. So the
differential equation reduces to
z(z + y 2 )dx − xy(x + y)dz = 0
⇒
15
dx
ydz
−
= 0 − − − (∗)
x(x + y) z(z + y 2 )
By partial fractions:
Let
⇒
Putting x = 0
1
A
B
= +
x(x + y)
x
x+y
A=
1
. Putting x = −y
y
⇒
⇒
A(x + y) + Bx = 1
1
B = − . Therefore,
y
1
1
1
= −
x(x + y)
x x+y
Similarly,
Let
⇒
Putting z = 0
y
A
B
= +
2
z(z + y )
z
z + y2
A=
1
. Putting z = −y 2
y
⇒
⇒
A(z + y 2 ) + Bz = y
1
B = − . Therefore,
y
y
1
1
= −
2
z(z + y )
z z + y2
Hence, equation (∗) becomes
(
)
1
1
−
dx +
x x+y
(
)
1
1
dz = 0
−
2
z+y
z
Integrating yields
∫ (
)
1
1
−
dx+
x x+y
∫ (
)
1
1
dz =
−
z + y2 z
∫
0
⇒
ln x−ln(x+y)+ln(z+y 2 )−ln z = ln |f (y)|
x(z + y 2 )
= f (y) · · · (i)
z(x + y)
Equation (i) is the required solution provided that f (y) is determined.
Step 2: put z = 1 in the given differential equation, ⇒ dz = 0. So the differential equation
reduces to
dx
dy
(1 + y 2 )dx + (1 + x2 )dy = 0 ⇒
+
=0
2
1+x
1 + y2
Integrating yields
∫
dx
+
1 + x2
Note: tan(A + B) =
(
∫
dy
=
1 + y2
tan A+tan B
1−tan A tan B .
∫
0
⇒
tan−1 x + tan−1 y = tan−1 c − − − (∗)
Taking tan on both sides of equation (∗) yields
)
tan tan−1 x + tan−1 y = c
⇒
x+y
=c
1 − xy
⇒
x=
c−y
· · · (ii),
cy + 1
where c is an arbitrary constant. Also, putting z = 1 in equation (i) we get
f (y) =
x(1 + y 2 )
· · · (iii)
(x + y)
Eliminating x between equations (ii) and (iii) (i.e., substituting equation (ii) in equation (iii))
yields
c−y
1 + y2
}
f (y) =
·{
⇒ f (y) = 1 − c−1 y = 1 − c1 y · · · (iv)
c−y
cy + 1
+y
cy + 1
Substituting equation (iv) in equation (i) yields x(z + y 2 ) = z(x + y)(1 − c1 y).
16
Exercise:
Solve the following differential equations
(a) 2y(a − x)dx + [z − y 2 + (a − x)2 ]dy − ydz = 0.
(b) (1 + yz)dx + x(z − x)dy − (1 + xy)dz = 0.
[ans: (a − x)2 + z = y(c − y)]
[ans: yz + 1 = (xy + 1)(cy + 1)]
(c) yzdx + (x2 y − zx)dy + (x2 z − xy)dz = 0.
[ans: y 2 + z 2 −
(d) (y 2 + yz)dx + (z 2 + xz)dy + (y 2 − xy)dz = 0.
2yz
= c]
x
[ans: y(x + z) = c(y + z)]
(e) (y 2 + yz + z 2 )dx + (z 2 + xz + x2 )dy + (x2 + xy + y 2 )dz = 0. [ans: xy + yz + xz = c(x + y + z)]
(h) zydx + 3zxdy − 2xydz = 0.
[ans: xy 3 = Cz 2 ]
17
Lecture 6
4
Linear partial differential equations of the first order
A PDE is a differential equation which contains derivative(s) of a function with respect to more than
one independent variables. For example,
x
∂2z
∂2z
∂z
∂2z
∂z
+
+
−
y
+
= 2(x + y)
∂x2 ∂x∂x
∂y 2 ∂x ∂y
The order of a PDE is the order of the highest ordered derivative occurring in the equation. For
example, if the dependent variable z is a function of x, y, and t, then the equation
x
∂z
∂z ∂z
+y
+
=0
∂x
∂y
∂t
is a first order equation in three independent variables while the equation
∂z
∂2z
= c2 2
∂t
∂x
is a second order equation in two independent variables. The degree of a PDE is the greatest power
of the highest-ordered derivative occuring in the equation.
First-order PDEs arise in gas flow problems, traffic flow problems, phenomenon of shock waves, the
motion of wave fronts, Hamilton-Jacobi theory, nonlinear continum mechanics and quantum mechanics etc.
A first order PDE in two independent variables x, y and the dependent variable z takes the general
form
f (x, y, z, p, q) = 0, where z = z(x, y),
p=
∂z
,
∂x
q=
∂z
∂y
(10)
The equations of the type (10) arise in many applications in geometry and physics. Equation (10) is
said to be linear if it satisfies the following three conditions:
(i) Dependent variable z and its derivatives are of power 1 only.
(ii) No product of the dependent variable, z, and/or its derivatives are present.
(iii) No transcendental functions of the dependent variable, z, and its derivatives occur. For example,
cos(z), ln z or ez is absent.
For example, the equation
(
cos xy 2
) ∂z
∂x
− y2
∂z
= tan(x2 + y 2 )
∂y
is linear. An equation which is not linear is said to be nonlinear. Examples of nonlinear equations
are:
∂z
∂z
+z
= 0 (shock wave)
∂x
∂y
∂2z
∂2z
−
+ z 3 = 0 (wave with interaction)
∂t2
∂x2
∂z
∂3z
∂z
+z
+ 3 = 0 (dispersive wave)
∂t
∂x ∂x
4.1
Formation of first-order PDEs
A first-order PDE can be formed either by eliminating arbitrary constants or an arbitrary function
involved.
18
4.1.1
Elimination of arbitrary constants
Consider an equation of the form
F (x, y, z, a, b) = 0,
(∗)
where a and b are arbitrary constants. Let z be regarded as a function of x and y. Differentiating
equation (∗) with respect to x and y, respectively, we get
∂F
∂F ∂z
+
=0
∂x
∂z ∂x
∂F
∂F
+p
=0
∂x
∂z
⇒
and
and
∂F
∂F ∂z
+
=0
∂y
∂z ∂y
∂F
∂F
+q
=0
∂y
∂z
(∗∗)
Eliminating a and b between equations (∗) and (∗∗), we obtain an equation of the form
F (x, y, z, p, q) = 0,
which is a PDE of first order. Similarly, if there are more arbitrary constants than the number of
independent variables, the above procedure of elimination will give rise to a PDE of highest order
than the first.
Example(s):
1. Construct the PDE from the following primitives.
(a) z = ax2 + by 3 .
Solution
zx = 2ax
zx
zy
a=
, zy = 3by 2 ⇒ b =
. Substituting a and b in the given
2x
3y 2
zy
zx 2
x + 2 y 3 ⇒ 6z = xzx + yzy ⇒ xp + yq = 6z.
primitive yields z =
2x
3y
(b) z = a(x + y) + b(x − y) + abt + c, where a, b, c are arbitrary constants.
⇒
Solution
zx = a + b, zy = a − b, zt = ab. Now, (a + b)2 − (a − b)2 = 4ab
⇒
(zx )2 − (zy )2 = 4zt
⇒
p2 − q 2 = 4zt
(c) ax2 + by 2 + z 2 = 1.
Solution
Differentiating partially wrt x yields 2ax + 2zzx = 0
⇒
Differentiating partially wrt y yields 2by + 2zzy = 0
⇒
−zzx
x
−zzy
b=
y
a=
Substituting a and b in the given primitive yields
−zzx 2 −zzy 2
x +
y + z2 = 1
x
y
⇒
−z(xzx + yzy ) + z 2 = 1
⇒
zxp + zyq = z 2 − 1
2. Form partial differential equations by eliminating arbitrary constants from the following relations:
(a) z = (x − a)2 + (y − b)2 .
[ans: p2 + q 2 = 4z]
(b) z = a(x + y)2 + b.
[ans: p = q]
(c) z = ax + by + ab.
[ans: z = xp + yq + pq]
+ b.
[ans: q = 2yp2 ]
(e) z = (x + a)(y + b).
[ans: z = pq]
(f) 2z = (ax + y)2 + b.
[answer: xp + yq = q 2 ]
(d) z = ax +
a2 y 2
19
4.1.2
Elimination of arbitrary function
Consider the equation
F (u, v) = 0,
(∗)
where u and v are functions of x, y, z and F is an arbitrary function of u, v. If we treat z as a dependent
variable while x and y as independent variables, we differentiate equation (∗) with respect to x and
y, respectively, to get
∂F ∂u ∂F ∂u ∂z
∂F ∂v
∂F ∂v ∂z
+
+
+
=0
∂u ∂x
∂u ∂z ∂x
∂v ∂x
∂v ∂z ∂x
⇒
⇒
∂F
∂u
(
∂u
∂u
+p
∂x
∂z
)
∂F
+
∂v
(
∂v
∂v
+p
∂x
∂z
)
=0
and
) (
)
(
∂F / ∂F
∂v
∂v / ∂u
∂u
=−
+p
+p
∂u ∂v
∂x
∂z
∂x
∂z
Therefore,
(
∂F ∂u ∂F ∂u ∂z ∂F ∂v ∂F ∂v ∂z
+
+
+
=0
∂u ∂y
∂u ∂z ∂y
∂v ∂y
∂v ∂z ∂y
and
and
∂F
∂u
(
∂u
∂u
+q
∂y
∂z
)
∂F
+
∂v
(
∂v
∂v
+q
∂y
∂z
)
=0
) (
)
(
∂F / ∂F
∂v
∂v / ∂u
∂u
=−
+q
+q
∂u ∂v
∂y
∂z
∂y
∂z
) (
) (
) (
)
∂v
∂v / ∂u
∂u
∂v
∂v / ∂u
∂u
+p
+p
=
+q
+q
∂x
∂z
∂x
∂z
∂y
∂z
∂y
∂z
(
⇒
∂v
∂v
+p
∂x
∂z
)(
∂u
∂u
+q
∂y
∂z
)
(
=
∂u
∂u
+p
∂x
∂z
)(
∂v
∂v
+q
∂y
∂z
)
∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v
+q
+p
+ pq =
+q
+p
+ pq
∂y ∂x
∂z ∂x
∂y ∂z ∂z ∂z
∂x ∂y
∂x ∂z
∂z ∂y ∂z ∂z
⇒
(
⇒
)
∂u ∂v ∂u ∂v
−
p+
∂y ∂z
∂z ∂y
(
)
∂u ∂v
∂u ∂v
∂u ∂v ∂u ∂v
−
−
q=
∂z ∂x ∂x ∂z
∂x ∂y
∂y ∂x
∂(u, v)
∂(u, v)
∂(u, v)
p+
q=
∂(y, z)
∂(z, x)
∂(x, y)
⇒
Example(s):
1. Eliminate the arbitrary function F from the equation F (x2 + y 2 + z 2 , z 2 − 2xy) = 0.
Solution
Let F (u, v) = 0, where u(x, y) = x2 + y 2 + z 2 ,
ux = 2x + 2zzx ,
uy = 2y + 2zzy ,
v(x, y) = z 2 − 2xy and z = z(x, y). Now,
vx = 2zzx − 2y, and vy = 2zzy − 2x
Differentiating partially with respect to x and y in turn we have
Fu ux + Fv vx = 0
⇒
Fu (2x + 2zzx ) + Fv (2zzx − 2y) = 0
⇒
Fu
y − zzx
=
· · · (i)
Fv
x + zzx
⇒
Fu (2y + 2zzy ) + Fv (2zzy − 2x) = 0
⇒
Fu
x − zzy
=
· · · (ii)
Fv
y + zzy
and
Fu uy + Fv vy = 0
From equations (i) and (ii) we have
y − zzx
x − zzy
=
x + zzx
y + zzy
⇒
(y − zzx )(y + zzy ) − (x + zzx )(x − zzy ) = 0
y 2 + yzzy − yzzx − z 2 zx zy − x2 + xzzx − xzzy + z 2 zx zy = 0
⇒
⇒
⇒
y 2 − x2 + yzzy − yzzx + xzzx − xzzy = 0
y 2 − x2 + yz(zy − zx ) − xz(zy − zx ) = 0
⇒
⇒
y 2 − x2 + z(y − x)(zy − zx ) = 0
z(y − x)p − z(y − x)q = y 2 − x2
20
2. Eliminate the arbitrary function f from the equations
[ans: (y − z)p + (z − x)q = x − y]
(a) x + y + z = f (x2 + y 2 + z 2 ).
Solution
Let x + y + z = f (u), where u = x2 + y 2 + z 2 and z = z(x, y). Now,
ux = 2x + 2zzx and uy = 2y + 2zzy
Differentiating partially with respect to x and y in turn we have
1 + zx = fu · ux
⇒
1 + zx = (2x + 2zzx )fu
⇒
fu =
1 + zx
· · · (i)
2x + 2zzx
1 + zy = fu · uy
⇒
1 + zy = (2y + 2zzy )fu
⇒
fu =
1 + zy
· · · (ii)
2y + 2zzy
and
From equations (i) and (ii) we have
1 + zx
1 + zy
=
2x + 2zzx
2y + 2zzy
⇒
(1 + zx )(2y + 2zzy ) − (2x + 2zzx )(1 + zy ) = 0
2y + 2zzy + 2yzx + 2zzx zy − 2x − 2zzx − 2xzy − 2zzx zy = 0
⇒
⇒
⇒
2y + 2zzy + 2yzx − 2x − 2zzx − 2xzy = 0
y + zq + yp − x − zp − xq = 0
⇒
(b) z = f (x2 − y 2 ).
(c) z =
xn f (y/x).
(d) z =
eax+by f (ax
[ans: yp + xq = 0]
[ans: xp + yq = nz]
− by).
[ans: bp + aq = 2abz]
[ans: py − qx = y 2 − x2 ]
(e) z = xy + f (x2 + y 2 ).
[ans: xp − yq = x − y]
(f) z = x + y + f (xy).
(g) z =
(y − z)p + (z − x)q = x − y
f ( xy
z ).
[ans: xp − yq = 0]
Lecture 7
4.2
Lagrange’s equation
This is an equation of the form
P p + Qq = R,
where p =
∂z
∂z
and q =
. Equation (11) is said to be:
∂x
∂y
• Linear if P, Q and R are functions of x and y only.
• Semilinear if P and Q are functions of x and y only but R is a function of x, y and z.
• Quasi-linear if P, Q and R are functions of x, y and z.
→ Note: An equation that does not fit into one of the above categories is called nonlinear.
21
(11)
4.2.1
Formation of Lagrange’s equation
Example(s):
1. Find the Lagrange equation whose solution is given by ϕ(xyz, x2 + y 2 + z 2 ) = 0
Solution
Let ϕ(u, v) = 0 where u = xyz and v = x2 + y 2 + z 2
=
∂(u, v)
u uz
= y
= uy vz − uz vy = (xz)(2z) − (xy)(2y) = 2x(z 2 − y 2 )
vy vz
∂(y, z)
Q =
∂(u, v)
u ux
= uz vx − ux vz = (xy)(2x) − (yz)(2z) = 2y(x2 − z 2 )
= z
vz vx
∂(z, x)
R =
∂(u, v)
u uy
= ux vy − uy vx = (yz)(2y) − (xz)(2x) = 2z(y 2 − x2 )
= x
vx vy
∂(x, y)
P
The Lagrange’s equation is given by
P p + Qq = R
⇒
⇒
2x(z 2 − y 2 )p + 2y(x2 − z 2 )q = 2z(y 2 − x2 )
x(z 2 − y 2 )p + y(x2 − z 2 )q = z(y 2 − x2 )
2. Form a linear partial differential equation whose solution is ϕ(x2 ez , ye−z ) = 0.
[6 Marks]
3. Form a linear partial differential equation whose solution is f (x2 + y 2 + z 2 , 2z + 2xy) = 0. [10
Marks]
4.2.2
Methods of solving P p + Qq = R
There are two methods of solving Lagrange’s equation: (1) method of characteristics and (2) Lagrange’s
method of solution.
2. Lagrange’s method of solution
This method is applicable to all the three cases i.e., linear, semi-linear, and quasi-linear Lagrange’s equations. The method depends on the introduction of subsidiary equations.
Theorem 4.1. The general solution of the Lagrange’s equation P p + Qq = R is ϕ(u, v) = 0,
where u ≡ u(x, y, z) = c1 and v ≡ v(x, y, z) = c2 are the integral curves of the subsidiary
equations
dy
dz
dx
=
=
, − − −(i)
P
Q
R
and ϕ is an arbitrary function.
Proof. The curves u ≡ u(x, y, z) = c1 and v ≡ v(x, y, z) = c2 are solutions of the system of
equations (i). Hence, the equations
du =
∂u
∂u
∂u
dx +
dy +
dz = 0,
∂x
∂y
∂z
dv =
∂v
∂v
∂v
dx +
dy +
dz = 0
∂x
∂y
∂z
are compatible with equations (i). Thus, we must have
P
∂u
∂u
∂u
+Q
+R
=0
∂x
∂y
∂z
and
22
P
∂v
∂v
∂v
+Q
+R
=0
∂x
∂y
∂z
In matrix form, we have

∂u
 ∂x

 ∂v
∂x


∂u ( ) 
∂u
−R ∂z 
∂y 
 P
∂v  Q = 
∂v 
−R
∂y
∂z
Solving these equations for P, Q and R using Cramer’s rule, we obtain
P
Q
R
=
=
∂u ∂v ∂u ∂v
∂u ∂v
∂u ∂v
∂u ∂v ∂u ∂v
−
−
−
∂y ∂z
∂z ∂y
∂z ∂x ∂x ∂z
∂x ∂y
∂y ∂x
⇒
P
Q
R
=
=
− − − (a)
∂(u, v)
∂(u, v)
∂(u, v)
∂(y, z)
∂(z, x)
∂(x, y)
Also, we need to eliminate the arbitrary function ϕ from the equation ϕ(u, v) = 0 − − − (∗).
From differential calculus, if ϕ = ϕ(u, v), then
∂ϕ
∂ϕ
du +
dv
∂u
∂v
dϕ =
and z = z(x, y) gives
dz =
∂z
∂z
dx +
dy = pdx + qdy
∂x
∂y
Differentiating equation (∗) with respect to x and y, respectively, we have
∂ϕ du ∂ϕ dv
+
=0
∂u dx ∂v dx
and
∂ϕ du ∂ϕ dv
+
= 0 − − − (∗∗)
∂u dy
∂v dy
Now,
du
∂u
∂u
=
+p ,
dx
∂x
∂z
and
du
∂u
∂u
=
+q
dy
∂y
∂z
dv
∂v
∂v
=
+p ,
dx
∂x
∂z
and
dv
∂v
∂v
=
+q ,
dy
∂y
∂z
Similarly,
where p =
∂z
∂z
and q =
. Substituting in equations (∗∗), we get
∂x
∂y
ϕu (ux + puz ) + ϕv (vx + pvz ) = 0
⇒
ϕu
vx + pvz
=−
ϕv
ux + puz
ϕu (uy + quz ) + ϕv (vy + qvz ) = 0
⇒
ϕu
vy + qvz
=−
ϕv
uy + quz
Therefore,
vx + pvz
vy + qvz
=
ux + puz
uy + quz
⇒
∂(u, v)
∂(u, v)
∂(u, v)
p+
q=
− − − (b)
∂(y, z)
∂(z, x)
∂(x, y)
Using equations (a) and (b), we obtain P p + Qq = R, which shows that ϕ(u, v) = 0 is a solution
to equation (i).
Example(s):
1. (a) Find the general solution of the equation x2 p + y 2 q = (x + y)z.
Solution
23
⇒
x2 p + y 2 q = (x + y)z
are given by
P = x2 , Q = y 2 , R = (x + y)z. The subsidiary equations
dx
dy
dz
=
=
P
Q
R
From
dx
dy
= 2
x2
y
∫
⇒
From
dx
−
x2
∫
dy
=
y2
∫
1 1
− = c1
y x
⇒
0
⇒
x−y
= c1 ≡ u
xy
dx − dy
dz
=
x2 − y 2
z(x + y)
∫
⇒
dx
dy
dz
= 2 =
2
x
y
z(x + y)
⇒
d(x − y)
−
x−y
∫
∫
dz
=
z
⇒
0
ln(x − y) − ln z = ln c2
⇒
x−y
= c2 ≡ v
z
x−y
x−y
and v =
. The general solution is given
xy)
z
(
x−y x−y
ϕ
= 0. Therefore, the general solution may be
,
xy
z
Hence, the integral curves are u =
⇒
by ϕ(u, v) = 0
written as
(
z = (x − y)f
)
x−y
, where f is an arbitrary function.
xy
(b) Find the integral surface of the quasi-linear PDE x(y 2 + z)p − y(x2 + z)q = z(x2 − y 2 )
which contains the straight lines x + y = 0, z = 1.
Solution
Given x(y 2 + z)p − y(x2 + z)q = z(x2 − y 2 ); x + y = 0 and z = 1. Here, P =
x(y 2 + z), Q = −y(x2 + z), and R = (x2 − y 2 )z. The subsidiary equations are
dy
dz
dx
=
=
2
2
2
x(y + z)
−y(x + z)
z(x − y 2 )
From
x2 (y 2
xdx + ydy
dz
=
2
2
2
+ z) − y (x + z)
z(x − y 2 )
⇒
xdx + ydy − dz = 0
xdx + ydy
dz
=
2
2
2
z(x − y )
z(x − y 2 )
⇒
⇒
x2 y 2
c1
+
−z =
2
2
2
∴ x2 + y 2 − 2z = c1 ≡ u
From
1
x dx
(y 2
+ y1 dy
+ z) −
(x2
⇒
+ z)
=
1
z dz
(x2 − y 2 )
⇒
ln x + ln y + ln z = ln c2
dx dy dz
+
+
=0
x
y
z
⇒
∴ xyz = c2 ≡ v
Therefore, the integral curves are u = x2 + y 2 − 2z and v = xyz. So the integral surface
is ϕ(u, v) = 0. Now,
z=1
⇒
u + 2 = x2 + y 2 and v = xy
Also,
x+y =0
⇒
(x + y)2 = 0
⇒
x2 + y 2 + 2xy = 0
⇒
u + 2 + 2v = 0.
Substituting the original values of the integral curves, u and v, we obtain
x2 + y 2 − 2z + 2 + 2xyz = 0
Alternatively, the parametric equations of the curve may be taken to be x = t, y =
−t, z = 1 so that u = 2t2 − 2, v = −t2 ⇒ u = −2v − 2 ⇒ u + 2v + 2 = 0
⇒
x2 + y 2 − 2z + 2xyz + 2 = 0
24
2. Find the general solution of the partial differential equations.
(a) x(x + y)p − y(x + y)q = (y − x)(2x + 2y + z). [ans: xy = c1 , (x + y)(x + y + z) = c2 ]
(b) (xz + y 2 )p + z(y − x)q + 2xy + z 2 = 0.
[ans: F (yz + x2 , 2xz − y 2 ) = 0]
x2 + y 2
(c) (x2 + y 2 + yz)p + (x2 + y 2 − xz)q = z(x + y).
[ans: z − x + y = c1 ,
= c2 ]
z2
(d) z(x + y)p + z(x − y)q = x2 + y 2 .
[ans: 2xy − z 2 = c1 , x2 − y 2 − z 2 = c2 ]
3. Find the equation of the integral surface of the differential equation 2y(z −3)p +(2x −z)q =
y(2x − 3) which passes through the circle z = 0, x2 + y 2 = 2x.
[ans:
2
2
2
x + y − z − 2x + 4z = 0]
4. Find the integral surface of the partial differential equation (x − y)p + (y − x − z)q = z
through the circle z = 1, x2 + y 2 = 1.
[ans:
4
2
2
4
2
z (x + y + z) + (y − x − z) − 2z (x + y + z) + 2z (y − x − z) = 0]
5. Find the integral surface of the equation (x − y)y 2 p + (y − x)x2 q = (x2 + y 2 )z through the
curve xz = a3 , y = 0.
[ans: z 3 (x3 + y 3 )2 = a9 (x − y)3 ]
6. Find the integral surface of the differential equation (x2 − yz)p + (y 2 − xz)q = z 2 − xy which
pass through the line z = 1, y = 0.
[ans: (x − y)(xy + yz + xz) + y − z = 0]
2
2
2
7. Find the integral surface of x p + y q + z = 0 which pass through the hyperbola xy =
x + y, z = 1.
[ans: yz + 2xy + xz = 3xyz]
8. Find the general integral of the equation (2x − y)y 2 p + 8(y − 2x)x2 q = 2(4x2 + y 2 )z and
1
deduce the solution of the Cauchy problem when z(x, 0) =
on a portion of the x-axis.
2x
(
)3
2x − y
[ans: (8x3 + y 3 )2 =
]
z
9. Find the equation of the surface satisfying the equation 4yzp + q + 2y = 0 and passing
through y 2 + z 2 = 1, x + z = 2.
[ans: y 2 + z 2 + x + z − 3 = 0, 8mks]
Lecture 8
5
Partial differential equations of the second order
These types of PDEs arise in connection with various physical problems such as the motion of a
vibrating string, heat flow, electricity, magnetism and fluid dynamics. A second order linear PDE in
two independent variables takes the general form
A
∂2u
∂2u
∂2u
+
C
+
B
= F (x, y, u, p, q),
∂x2
∂x∂y
∂y 2
(12)
∂u
∂u
and q =
.
∂x
∂y
By analogy with the criteria for a conic to be hyperbolic, parabolic, or elliptic, equation (12) is said
to be
where A, B and C are continuous functions of the independent variables x and y, p =
Hyperbolic: if B 2 − 4AC > 0
Parabolic: if B 2 − 4AC = 0
Elliptic: if B 2 − 4AC < 0
The simplest, and in elementary applications, the most important examples of hyperbolic, parabolic,
and elliptic PDEs are:
Wave equation:
Heat equation:
Laplace equation:
∂2u
∂2u
= c2 2 , A = c2 , B = 0, C = −1 ⇒ B 2 − 4AC = 4c2 > 0 (hyperbolic)
2
∂t
∂x
∂u
∂2u
= α 2 , A = k, B = 0, C = 0 ⇒ B 2 − 4AC = 0 (parabolic)
∂t
∂x
∂2u ∂2u
+ 2 = 0, A = 1, B = 0, C = 1 ⇒ B 2 − 4AC = −4 < 0 (elliptic)
∂x2
∂y
25
→ Note: For the heat equation, α > 0 is the diffusivity. For the wave equation, c > 0 is the velocity
of the wave.
5.1
5.1.1
Methods of the solution of PDEs of the second order
Method of separation of variables
This method applies to problems where:
• The PDE is linear and homogeneous (not necessarily constant coefficients)
• The boundary conditions are linear and homogeneous.
→ Note: we are interested in a nontrivial solution. Furthermore, for the case of the wave equation,
we are interested in a solution which has physical significance (i.e., a periodic solution).
Example(s):
(a) Use the method of separation of variables to solve
∂u
∂u
−4
= 0; u(0, y) = 8e−3y .
∂x
∂y
Solution
Let the solution of the given equation be
u(x, y) = X(x)Y (y),
∂u
= Y X ′ and
where X is a function of x alone and Y is a function of y alone. Then,
∂x
∂u
= XY ′ .Substituting in the given equation yields
∂y
Y X ′ − 4XY ′ = 0
⇒
X′
Y′
=
· · · · · · (∗)
4X
Y
The left hand side of equation (∗) is independent of y. So the right hand side, which is identically
equal to the expression on the left, must be independent of y too. Similarly, the right hand side
is independent of x. So the left hand side must be independent of x too. Therefore, both sides
must be independent of both x and y. So each side of equation (∗) must be a constant, say −λ2 .
So we have
X′
Y′
=
=k
4X
Y
X′
= k ⇒ X ′ − 4kX = 0 whose solution is X = A1 e4kx
From
4X
Y′
= k ⇒ Y ′ − kY = 0 whose solution is Y = A2 eky . Therefore,
From
Y
u(x, y) = A1 e4kx · A2 eky = Ae4kx+ky
Applying the condition u(0, y) = 8e−3y , we get u(0, y) = 8e−3y = Aeky
Hence,
u(x, y) = 8e−12x−3y
⇒
A = 8, k = −3.
∂2u
∂u
=
, 0 < x < 2, t > 0
∂t
∂x2
subject to the conditions u(0, t) = 0, u(2, t) = 0 and u(x, 0) = sin(πx).
(b) Solve by the method of separation of variables the heat equation
SolutionUsing the method of separation of variables, we let the solution of the given equation
be
u(x, t) = X(x)T (t)
26
(13)
Hence,
∂u
= XT ′
∂t
∂2u
= T X ′′
∂x2
and
Substituting in the given equation, we obtain
XT ′ = T X ′′
T′
X ′′
=
=k
T
X
⇒
(A)
Since we need to consider only physical solutions, there are three cases to investigate. That is,
case 1 : k < 0, case 2 : k = 0 and case 3 : k > 0.
Case 1: k = −λ2 < 0, where λ is a nonzero real number.
From equations (A), we obtain the following set of ODEs:
X ′′
= −λ2 ⇒ X ′′ + λ2 X = 0
X
T′
= −λ2 ⇒ T ′ + λ2 T = 0
T
Let Dx =
d
d
and Dt = . Then,
dx
dt
X ′′ + λ2 X = 0
and
[
⇒
]
⇒
Dx2 + λ2 X = 0
T ′ + λ2 T = 0
[
⇒
Dx = ±iλ
]
⇒
Dt + λ2 T = 0
⇒
X = A1 cos(λx) + B1 sin(λx)
Dt = −λ2
⇒
T = A2 e−λ
2t
Substituting in equation (13) yields
u(x, t) = e−λ t (A cos λx + B sin λx)
2
(14)
We now determine the value of the λ, and constants A and B. Using the boundary condition
u(0, t) = 0 in equation (14), we get
0 = e−λ t (A)
2
⇒
A = 0 (for non-trivial solution)
Thus, equation (14) reduces to the simpler form
u(x, t) = Be−λ t sin λx
2
(15)
Using the second boundary condition u(2, t) = 0 in equation (15), we get
0 = Be−λ t sin 2λ
2
⇒
sin 2λ = 0 (for non-trivial solution)
nπ
for n = 1, 2, 3, · · · . Here, λn ’s are called the eigenvalues and
2
the function un (x, t) corresponding to any λn is called an eigen function. Hence, equation (15)
reduces to the simpler form (for each eigen function)
Thus, sin 2λ = 0
⇒
λn =
un (x, t) = Bn e−λn t sin(λn x) = Bn e−(n
2
2 π 2 /4)t
(
sin
)
nπ
x , for n = 1, 2, 3, · · ·
2
Since the given equation is linear, then by the principle of superposition, the superposition of
all the linear solutions is a general solution of the given equation. Hence, the general solution is
given by
u(x, t) =
=
∞
∑
n=1
∞
∑
un (x, t)
Bn e−(n
2 π 2 /4)t
n=1
27
(
sin
nπ
x
2
)
(16)
The general solution (16) can be recognized as the Fourier half-range sine series. To determine
the Fourier coefficients, Bn , we apply the initial condition u(x, 0) = sin πx. Thus, we have
sin πx =
(
∞
∑
Bn sin
n=1
(
nπ
x
2
)
)
(
)
π
3π
= B1 sin
x + B2 sin (πx) + B3 sin
x + ···
2
2
{
Comparing the coefficients of like terms, we obtain Bn =
1,
0,
if n = 2
. Substituting in
otherwise
equation (16) we get the solution:
u(x, t) = e−π t sin πx
2
Case 2: k = 0.
From equations (A), we obtain the following set of ODEs:
X ′′
=0
X
⇒
X ′′ = 0
T′
=0
T
⇒
X = A1 x + B1
T′ = 0
⇒
⇒
T =C
Therefore, the general solution is
u(x, t) = (A1 x + B1 ) (C) = Ax + B,
where A and B are arbitrary constants. Using the boundary conditions u(0, t) = 0, we get
0 = A · 0 + B ⇒ B = 0. Hence, the solution reduces to
u(x, t) = Ax
Using the boundary conditions u(2, t) = 0, we get 0 = A · 2
reduces to
u(x, t) = 0,
⇒
A = 0. Hence, the solution
which is a trivial solution. Thus, the possibility that k = 0 leads to no solution of interest to us.
Case 3: k = λ2 > 0, where λ is a nonzero real number.
From equations (A), we obtain the following set of ODEs:
X ′′
= λ2 ⇒ X ′′ − λ2 X = 0
X
T′
= λ2 ⇒ T ′ − λ2 T = 0
T
Thus,
X = A1 e−λx + B1 eλx
and
T = A 2 eλ
2t
Substituting in equation (13) yields
u(x, t) = eλ t (Ae−λx + Beλx ),
2
(17)
where A and B are arbitrary constants. To determine A and B, we use the boundary conditions.
Using the boundary condition u(0, t) = 0 in equation (17), we get
0 = e−λ t (A + B)
2
⇒
A + B = 0 (for non-trivial solution)
Thus, equation (17) reduces to the simpler form
u(x, t) = Aeλ t (e−λx − eλx )
2
28
(18)
Using the second boundary condition u(2, t) = 0 in equation (18), we get
0 = Aeλ t (e−2λ − e2λ )
2
⇒
since (e−2λ − e2λ ) ̸= 0 as λ ̸= 0.
A = 0,
Thus, equation (18) reduces to the simpler form
u(x, t) = 0,
which is a trivial solution. Thus, there is no nontrivial solution to equation (A) for the case
k > 0.
Exercise:
∂2u
∂2u
(a) Use the method of separation of variables to solve the 1D wave equation 2 = c2 2 −−−−−(∗)
∂t
∂x
subject to the following conditions:
Boundary conditions:
Initial conditions:
u(0, t) = 0 and u(50, t) = 0
ut (x, 0) = 0 and u(x, 0) = f (x)
Solution
Using the method of separation of variables, we let the solution of equation (∗) to be
u(x, t) = X(x)T (t) − − − − − (∗∗)
Thus,
∂2u
= XT ′′
∂t2
∂2u
= T X ′′
∂x2
and
Substituting in equation (∗) we get
XT ′′ = c2 T X ′′
⇒
T ′′
X ′′
=
= −λ2 ,
c2 T
X
(for periodic solution)
This leads to the following two ODEs:
X ′′
= −λ2
X
T ′′
= −λ2
c2 T
Let Dx =
X ′′ + λ2 X = 0
⇒
T ′′ + (cλ)2 T = 0
⇒
d
d
and Dt = . Then,
dx
dt
X ′′ + λ2 X = 0
⇒
(Dx2 + λ2 )X = 0
⇒
(Dt2 + (cλ)2 )T = 0
⇒
Dx = ±iλ
⇒
X = A1 cos λx + B1 sin λx
and
T ′′ + (cλ)2 T = 0
⇒
Dt = ±icλ
⇒
T = A2 cos cλt + B2 sin cλt
Substituting in equation (∗∗) yields
u(x, t) = (A1 cos λx + B1 sin λx)(A2 cos cλt + B2 sin cλt) − − − −(∗ ∗ ∗)
Applying the boundary condition u(0, t) = 0 to equation (∗ ∗ ∗), we get
A1 (A2 cos cλt + B2 sin cλt) = 0
⇒
A1 = 0
(for non-trivial solution)
Thus, equation (∗ ∗ ∗) reduces to the simpler form
u(x, t) = sin(λx) [A cos(cλt) + B sin(cλt)] − − − −(∗ ∗ ∗)′
29
Applying the boundary condition u(50, t) = 0 to equation (∗ ∗ ∗), we get
⇒
sin(50λ) [A cos(cλt) + B sin(cλt)] = 0
sin(50λ) = 0
(for non-trivial solution)
From sin(50λ) = 0, we obtain
λn =
nπ
50
for n = 1, 2, 3, · · ·
Hence, equation (∗ ∗ ∗)′ reduces to the simpler form
un (x, t) = (sin
nπ
nπc
nπc
x)(An cos
t + Bn sin
t), for n = 1, 2, 3, · · ·
50
50
50
Now, by the principle of superposition we have
∞
∑
u(x, t) =
(sin
n=1
nπc
nπc
nπ
x)(An cos
t + Bn sin
t) − − − −(a)
50
50
50
Differentiating equation (a) partially wrt t, we obtain
ut (x, t) =
∞
∑
(sin
n=1
nπc
nπc
nπ
nπc
nπc
x)(−An
sin
t + Bn
cos
t)
50
50
50
50
50
Applying the initial velocity condition ut (x, 0) = 0, we get
∞
∑
(sin
n=1
nπ ( nπc )
x) Bn
=0
50
50
⇒
Bn = 0
for n = 1, 2, 3, · · ·
Thus, equation (a) reduces to the simpler form
u(x, t) =
∞
∑
An sin(
n=1
nπ
nπc
x) cos(
t) − − − − − (b)
50
50
Applying the initial condition u(x, 0) = f (x) to equation (b), we get
f (x) =
∞
∑
An sin(
n=1
nπ
x)
50
This is Fourier sine expansion of f (x) over the interval [0, 50]. Thus, the Fourier coefficient An
is given by
2
An =
50
∫50
f (x) sin(
nπ
x)dx
50
0
Substituting in equation (b) we get the solution
∞
2 ∑
u(x, t) =
50 n=1
[ ∫50
]
nπ
nπc
nπ
t)
f (x) sin( x)dx sin( x) cos(
50
50
50
0
(b) A rod whose surface is insulated has a length of 3 units and deformity of 2 units. The ends
of the rod are kept at 0o C and its initial temperature at any point x, 0 ≤ x ≤ 3, is given by
u(x, 0) = 5 sin 4πx − 3 sin 8πx + 2 sin 10πx. Find the temperature of the rod at any time t [ans:
2
2
2
u(x, t) = 5e−(4πc) t sin(4πx) − 3e−(8πc) t sin(8πx) + 2e−(10πc) t sin(10πx)]
Solution
The PDE to be solved is the 1D heat equation
ut = c2 uxx − − − (∗),
30
where u(x, t) is the temperature of the rod at time t, subject to the boundary conditions
u(0, t) = 0 and u(3, t) = 0 and the initial condition u(x, 0) = 5 sin 4πx − 3 sin 8πx + 2 sin 10πx.
Using the method of separation of variables, we let the solution of equation (∗) to be
u(x, t) = X(x)T (t) − − − (∗∗)
Thus,
ut = X(x)T ′ (t),
uxx = X ′′ (x)T (t)
and
Substituting in equation (∗) we get
X(x)T ′ (t) = c2 X ′′ (x)T (t)
T ′ (t)
X ′′ (x)
=
= −λ2 ,
c2 T (t)
X(x)
⇒
(for periodic solution)
This leads to the following two ODEs:
X ′′ (x)
= −λ2
X(x)
T ′ (t)
= −λ2
c2 T (t)
Let Dx =
⇒
X ′′ (x) + λ2 X(x) = 0
⇒
T ′ (t) + (cλ)2 T (t) = 0
d
d
and Dt = . Then,
dx
dt
X ′′ + λ2 X = 0
⇒
⇒
(Dx2 + λ2 )X = 0
Dx = ±iλ
⇒
X = A1 cos λx + B1 sin λx
and
T ′ + (cλ)2 T = 0
⇒
⇒
(Dt + (cλ)2 )T = 0
Dt = −(cλ)2
T = A2 e−(cλ)
2t
⇒
Substituting in equation (∗∗) we get the solution to be
u(x, t) = (A1 cos λx + B1 sin λx)A2 e−(cλ) t = (A cos λx + B sin λx)e−(cλ) t − − − (∗ ∗ ∗)
2
2
Applying the boundary condition u(0, t) = 0 to equation (∗ ∗ ∗), we get
0 = Ae−(cλ)
2t
⇒
A=0
(for non-trivial solution)
Thus, equation (∗ ∗ ∗) reduces to the simpler form
u(x, t) = Be−(cλ) t sin λx − − − (∗ ∗ ∗)′
2
Applying the boundary condition u(3, t) = 0 to equation (∗ ∗ ∗)′ , we get
0 = Be−(cλ) t sin 3λ
2
⇒
sin 3λ = 0
(for non-trivial solution)
From sin 3λ = 0, we obtain
nπ
, for n = 1, 2, 3, · · ·
3
Thus, equation (∗ ∗ ∗)′ reduces to the simpler form
nπ 2
nπ
un (x, t) = Bn e−( 3 c) t sin( x), for n = 1, 2, 3, · · ·
3
Now, by the principle of superposition we have
λn =
u(x, t) =
∞
∑
Bn e−(
nπ 2
c) t
3
sin(
n=1
nπ
x) − − − (a)
3
Applying the initial condition u(x, 0) = 5 sin 4πx − 3 sin 8πx + 2 sin 10πx, we get
∞
∑
Bn sin(
n=1
nπ
x) = 5 sin 4πx − 3 sin 8πx + 2 sin 10πx
3
Comparing the coefficients of like terms, we obtain B12 = 5, B24 = −3, B30 = 2 and Bn = 0 for
n∈
/ {12, 24, 30}. Therefore, equation (a) reduces to
u(x, t) = 5e−(4πc) t sin(4πx) − 3e−(8πc) t sin(8πx) + 2e−(10πc) t sin(10πx)
2
2
31
2
(c) Use the method of separation of variables to solve
∂2u
∂2u
=
4
; u(0, y) = 0, u(1, y) = 0.
∂x2
∂y 2
(d) Use the method of separation of variables to determine the general solution to the one di∂2u
1 ∂u
mensional heat equation
, 0 ≤ x ≤ L, t > 0 subject to the boundary condition
=
2
∂x
λ ∂t
u(0, t) = u(L, t) = 0 and the initial conditions u(x, 0) = f (x). Hence, determine the particular
solution when f (x) = x2 .
(e) Use the method of separation of variables to determine the general solution to the one dimen2
∂2u
2 ∂ u , 0 ≤ x ≤ L, t > 0 subject to the boundary conditions
sional wave equation
=
c
∂t2
∂tx2
u(0, t) = u(L, t) = 0 and the initial conditions u(x, 0) = f (x) and ut (x, 0) = g(x). Hence,
determine the particular solution given that f (x) = 0 and g(x) = x.
Lecture 9
Reduction to cylindrical coordinate system
If the domain under consideration is cylindrical, then it is appropriate to first transform the independent variables from Cartesian coordinates (x, y, z) to cylindrical coordinates (r, θ, z).
→ Note: In the polar system, the solution must be periodic in θ, with period 2π (this means that
u(r, θ + 2π) = u(r, θ) -i.e., the solution repeats itself every θ = 2π). Also, u(r, θ) should remain finite
as r → 0.
Example(s):
∂2u
∂2u
+
= 0
(∗), where u(x, y) represents the velocity
∂x2
∂y 2
potential of a fluid particle in a certain domain. For example, we need to determine u(x, y)
inside a circular disk x2 + y 2 < a2 , when its values on the circumference and at the center are
finite.
1. Consider the Laplace equation
Solution
In obtaining the solution of equation (∗), we make the following transformations of the independent variables:
x = r cos θ, y = r sin θ, and u = u(r, θ)
⇒
√
r=
x2 + y 2 ,
and θ = tan−1 (y/x)
Thus,
x
∂r
y
∂r
r cos θ
r sin θ
√
=√ 2
=
cos
θ,
=
= sin θ
=
=
∂x
r
∂y
r
x + y2
x2 + y 2
−y
−r sin θ
1
∂θ
x
r cos θ
1
∂θ
= 2
=
= − sin θ,
= 2
=
= cos θ
2
2
2
2
∂x
x +y
r
r
∂y
x +y
r
r
Using chain rule, we know that:
∂u
∂u ∂r
∂u ∂θ
∂u 1
∂u
=
+
= cos θ
− sin θ
∂x
∂r ∂x ∂θ ∂x
∂r
r
∂θ
32
∂u
∂u ∂r ∂u ∂θ
∂u 1
∂u
=
+
= sin θ
+ cos θ
∂y
∂r ∂y
∂θ ∂y
∂r
r
∂θ
Thus, we observe that
∂
∂
1
∂
= cos θ
− sin θ
∂x
∂r r
∂θ
and
∂
∂
1
∂
= sin θ
+ cos θ
∂y
∂r r
∂θ
Therefore, by chain rule and product rule for partial derivatives, we have
∂2u
∂x2
∂
∂x
(
)
(
)(
{
1
∂2u
∂u
+ 2 sin θ sin θ 2 + cos θ
r
∂θ
∂θ
= cos2 θ
∂2u
∂y 2
)
∂u
∂
1
∂
∂u 1
∂u
= cos θ
− sin θ
cos θ
− sin θ
∂x
∂r r
∂θ
∂r
r
∂θ
(
)
(
)
∂
∂u 1
∂u
1
∂
∂u 1
∂u
= cos θ
cos θ
− sin θ
− sin θ
cos θ
− sin θ
∂r
∂r
r
∂θ
r
∂θ
∂r
r
∂θ
{
}
{
}
2
2
2
1 ∂ u
1 ∂u
1
∂ u
∂u
2 ∂ u
−
− sin θ cos θ
− sin θ
= cos θ 2 − cos θ sin θ
∂r
r ∂r∂θ r2 ∂θ
r
∂r∂θ
∂r
=
(
}
2
∂2u
1
2
∂u
∂2u
1
∂u
2
2 ∂ u
+
sin
θ
−
cos
θ
sin
θ
+ sin2 θ
+ 2 cos θ sin θ
∂r2
r2
∂θ2
r
∂r∂θ r
∂r
r
∂θ
)
(
)(
(i)
)
∂u
∂
1
∂
∂u 1
∂u
= sin θ
+ cos θ
sin θ
+ cos θ
∂y
∂r r
∂θ
∂r
r
∂θ
(
)
(
)
∂u
∂
∂u
∂
∂u 1
1
∂u 1
+ cos θ
+ cos θ
= sin θ
sin θ
+ cos θ
sin θ
∂r
∂r
r
∂θ
r
∂θ
∂r
r
∂θ
}
{
}
{
2
2
2
1 ∂u
1
∂ u
∂u
1 ∂ u
2 ∂ u
= sin θ 2 + cos θ sin θ
−
+ cos θ sin θ
+ cos θ
∂r
r ∂r∂θ r2 ∂θ
r
∂r∂θ
∂r
=
∂
∂y
{
1
∂2u
∂u
+ 2 cos θ cos θ 2 − sin θ
r
∂θ
∂θ
= sin2 θ
}
2
∂2u
∂2u
1
∂u
2
1
2
∂u
2 ∂ u
cos
θ
sin
θ
+ cos2 θ
− 2 cos θ sin θ
+
cos
θ
+
2
2
2
∂r
r
∂θ
r
∂r∂θ r
∂r
r
∂θ
(ii)
Hence, substituting in the given PDE (i.e., add equations (i) and (ii)), we get the transformed
equation:
∂ 2 u 1 ∂u
1 ∂2u
+
+
=0
(∗∗)
∂r2
r ∂r
r2 ∂θ2
The boundary condition for equation (∗∗) is given by u(1, θ) = f (θ) for 0 ≤ θ ≤ 2π. By using
the method of separation of variables, we let the solution of equation (∗∗) to be
u(r, θ) = R(r)Θ(θ),
from which we obtain
∂u
= R′ Θ,
∂r
∂2u
= R′′ Θ,
∂r2
and
∂2u
= RΘ′′
∂θ2
Substituting in equation (∗∗), we get
1
1
R′′ Θ + R′ Θ + 2 RΘ′′ = 0
r
r
⇒
(
)
r2 R′′ + rR′ Θ = −RΘ′′
Separating the variables yields
(
)
r2 R′′ + rR′
Θ′′
=−
= k,
R
Θ
where is a constant.
33
(∗ ∗ ∗)
Case 1: k = 0.
From equations (∗ ∗ ∗), we obtain the following set of ODEs:
1
R′′ + R′ = 0
r
and
Θ′′ = 0
Solving yields
R(r) = a ln r + b and Θ(θ) = Cθ + D
Therefore,
u(r, θ) = (a ln r + b) (Cθ + D)
The solution has to be periodic in θ, with period 2π. This implies that C = 0. Hence,
u(r, θ) = A ln r + B
Also, the solution has to remain finite as r → 0, then A = 0. Hence, u(r, θ) = B,
For convenience, we write this solution in the form
u0 (r, θ) =
a constant.
A0
2
Case 2: k = n2 > 0, where n = 1, 2, 3, · · · .
From equations (∗ ∗ ∗), we obtain the following set of ODEs:
r2 R′′ + rR′ − n2 R = 0
(i) and
Θ′′ + n2 Θ = 0
(ii)
Equation (i) is a Cauchy-Euler type equation and its solution can be obtained by making the
dR
dR
d2 R
d2 R
dR
following substitution: r = et
⇒ t = ln r, r
=
, and r2 2 =
.
−
2
dr
dt
dr
dt
dt
Hence, equation (i) becomes
d2 R
− n2 R = 0,
dt2
whose general solution is given by
R(t) = En e−nt + Fn ent
R(r) = En e−n ln r + Fn en ln r
⇒
⇒
R(r) = En r−n + Fn rn ,
where En and Fn are arbitrary constants. The solution of equation (ii) is given by
Θ(θ) = Cn cos nθ + Dn sin nθ,
where Cn and Dn are arbitrary constants. Therefore,
(
)
u(r, θ) = En r−n + Fn rn (Cn cos nθ + Dn sin nθ)
(iii)
From physical ground, we are looking for a solution which is periodic in θ, with period 2π.
Clearly, equation (iii) is periodic in θ, with period 2π. Also, the solution has to remain finite as
r → 0. This is only possible if En = 0. Therefore, equation (iii) reduces to
un (r, θ) = Fn rn (Cn cos nθ + Dn sin nθ) = rn (An cos nθ + Bn sin nθ) ,
where u0 (r, θ) =
A0
. By the principle of superposition, we have
2
u(r, θ) =
∞
A0 ∑
+
rn (An cos nθ + Bn sin nθ)
2
n=1
Applying the boundary condition u(a, θ) = f (θ), we get
f (θ) =
∞
A0 ∑
+
an (An cos nθ + Bn sin nθ)
2
n=1
34
(iv)
n = 1, 2, 3, · · · ,
This is a Fourier series with cosine coefficients an An and sine coefficients an Bn , over the interval
[0, 2π]. The Fourier coefficients are given by
1
An =
πan
∫2π
⇒
f (θ) cos(nθ)dθ
1
Bn =
πan
∫2π
0
f (θ) sin(nθ)dθ
0
Substituting in equation (iv), we get the solution:
u(r, θ) =
1
2π
∫2π
f (θ)dθ +
0
∞
1 ∑
π
n=1

 2π


 2π


∫
∫
f (θ) cos(nθ)dθ cos(nθ) +
f (θ) sin(nθ)dθ sin(nθ)
rn 




0
0
Case 3: k = −n2 < 0, where n = 1, 2, 3, · · · .
From equations (∗ ∗ ∗), we obtain the following set of ODEs:
r2 R′′ + rR′ + n2 R = 0
and
Θ′′ − n2 Θ = 0
Solving yields
R(r) = An r−n + Bn rn
Therefore,
and Θ(θ) = Cn cosh nθ + Dn sinh nθ
(
)
u(r, θ) = An r−n + Bn rn (Cn cosh nθ + Dn sinh nθ)
The solution has to be periodic in θ, with period 2π. This implies that Cn = Dn = 0. Hence,
u(r, θ) = 0 (trivial solution.)
Exercise:
2. The electrostatic potential of electrical charges in the domain D that is free of these charges is
governed by the Laplace equation
∂2u ∂2u
+ 2 =0
∂x2
∂y
Use the method of separation of variables to show that the potential u(x, y) in a cylindrical
capacitor defined by D = {0 ≤ r ≤ a, 0 ≤ θ ≤ 2π} when the upper half of the capacitor is
kept at the potential U (i.e., u(r, θ) = U for 0 ≤ θ ≤ π) and the lower half of the capacitor is
grounded (i.e., u(r, θ) = 0 volts for π < θ ≤ 2π).
Solution
From example 1, the polar form of equation (∗) is given by
∂ 2 u 1 ∂u
1 ∂2u
+
+
=0
∂r2
r ∂r
r2 ∂θ2
35
(∗∗)
The boundary condition for equation (∗∗) is given by u(1, θ) = f (θ) for 0 ≤ θ ≤ 2π. By using
the method of separation of variables, we let the solution of equation (∗∗) to be
u(r, θ) = R(r)Θ(θ),
from which we obtain
∂2u
= R′′ Θ,
∂r2
∂u
= R′ Θ,
∂r
∂2u
= RΘ′′
∂θ2
and
Substituting in equation (∗∗), we get
1
1
R′′ Θ + R′ Θ + 2 RΘ′′ = 0
r
r
⇒
(
)
r2 R′′ + rR′ Θ = −RΘ′′
Separating the variables yields
(
)
r2 R′′ + rR′
Θ′′
=−
= k,
R
Θ
(∗ ∗ ∗)
where is a constant.
Case 1: k = n2 > 0.
From equations (∗ ∗ ∗), we obtain the following set of ODEs:
r2 R′′ + rR′ − n2 R = 0
Θ′′ + n2 Θ = 0
(i) and
(ii)
Equation (i) is a Cauchy-Euler type equation and its solution can be obtained by making the
dR
d2 R
dR
d2 R
dR
=
, and r2 2 =
.
following substitution: r = et
⇒ t = ln r, r
−
dr
dt
dr
dt2
dt
Hence, equation (i) becomes
d2 R
− n2 R = 0,
dt2
whose general solution is given by
R(t) = En e−nt + Fn ent
⇒
R(r) = En e−n ln r + Fn en ln r
⇒
R(r) = En r−n + Fn rn ,
where En and Fn are arbitrary constants. The solution of equation (ii) is given by
Θ(θ) = Cn cos nθ + Dn sin nθ,
where Cn and Dn are arbitrary constants. Therefore,
(
)
u(r, θ) = En r−n + Fn rn (Cn cos nθ + Dn sin nθ)
(iii)
From physical ground, we are looking for a solution which is periodic in θ, with period 2π.
Clearly, equation (iii) is periodic in θ, with period 2π. Also, the solution has to remain finite as
r → 0. This is only possible if En = 0. Therefore, equation (iii) reduces to
un = Fn rn (Cn cos nθ + Dn sin nθ) = rn (An cos nθ + Bn sin nθ) ,
n = 1, 2, · · ·
By the principle of superposition, we have
u(r, θ) = g +
∞
∑
rn (An cos nθ + Bn sin nθ)
(iv)
n=1
{
Applying the voltage boundary condition u(a, θ) =
g+
∞
∑
U
0
if 0 ≤ θ ≤ π,
we get
if π < θ ≤ 2π
{
n
a (An cos nθ + Bn sin nθ) =
n=1
36
U
0
if 0 ≤ θ ≤ π,
if π < θ ≤ 2π
Integrating both sides from 0 to 2π yields:
∫2π
g+
∞
∑
an (An cos nθ + Bn sin nθ) dθ = πU
⇒
2πg = πU
⇒
n=1
0
Therefore,
g=
U
2
∞
∑
U
u(a, θ) =
+
an (An cos nθ + Bn sin nθ)
2 n=1
We apply the orthogonality relationships:
∫2π
∫2π
U
u(a, θ) cos mθdθ =
2
cos mθdθ +
0
∫2π ∑
∞
0 n=1
0
∫π
⇒
cos mθdθ = 0 + Am am π
U
an (An cos nθ + Bn sin nθ) cos mθdθ
⇒
0 = Am am π
⇒
Am = 0, for all m
0
Similarly,
∫2π
U
u(a, θ) sin mθdθ =
2
⇒
U
sin mθdθ +
[
U cos mθ
sin mθdθ =
2
m
0
∫2π ∑
∞
an (An cos nθ + Bn sin nθ) sin mθdθ
0 n=1
0
0
∫π
∫2π
]2π
+ B m am π
⇒
0
2U
= Bm am π, for odd m = (2n − 1)
m
Therefore, equation (iv) becomes
u(r, θ) =
∞
∑
U
2U.
+
r(2n−1) sin(2n − 1)θ
2 n=1 (2n − 1)πa(2n−1)
Case 2: k = 0.
From equations (∗ ∗ ∗), we obtain the following set of ODEs:
1
R′′ + R′ = 0
r
and
Θ′′ = 0
Solving yields
R(r) = a ln r + b and Θ(θ) = Cθ + D
Therefore,
u(r, θ) = (a ln r + b) (Cθ + D)
The solution has to be periodic in θ, with period 2π. This implies that C = 0. Hence,
u(r, θ) = A ln r + B
Also, the solution has to remain finite as r → 0, then A = 0. Hence,
u(r, θ) = B,
a constant
Case 3: k = −n2 < 0.
From equations (∗ ∗ ∗), we obtain the following set of ODEs:
r2 R′′ + rR′ + n2 R = 0
and
Θ′′ − n2 Θ = 0
Solving yields
R(r) = An r−n + Bn rn
Therefore,
and Θ(θ) = Cn cosh nθ + Dn sinh nθ
(
)
u(r, θ) = An r−n + Bn rn (Cn cosh nθ + Dn sinh nθ)
The solution has to be periodic in θ, with period 2π. This implies that Cn = Dn = 0. Hence,
u(r, θ) = 0 (trivial solution.)
37
Lecture 10
5.1.2
Laplace transform
The methods of Laplace and Fourier transforms are powerful whenever a BVP is to be solved over an
infinite or semi-infinite domain. The philosophy in these methods lies in the fact that by taking the
transform of the given PDE, we can reduce the number of independent variables by 1. The formal
step is that the given PDE with its boundary and initial conditions is transformed with respect to one
of its independent variables. In doing so, we assume that the differential operator can be interchanged
with the Laplace transform operator L or Fourier transform operator F .
The Laplace transform is very useful for semi-infinite range, i.e., 0 < x < ∞ and 0 < t < ∞. The
Laplace transform method is as follows:
Suppose a function f (t) is defined for t ≥ 0. Also, let L be the Laplace transform operator, then the
Laplace transform, with respect to t, of f (t) is a function defined by
∫∞
L [f (t)] =
e−st f (t)dt = F (s), Re(s) > 0, where s is a dummy variable.
0
Similarly, the Laplace transform, with respect to t, of u(x, t) is defined by
∫∞
e−st u(x, t)dt = U (x, s)
L [u(x, t)] =
t=0
Example(s):
Find the Laplace transform of the following functions:
(a) f (t) = 1
(b) f (t) = eat
(c) f (t) = t2
Exercise:
Verify the following properties of the Laplace transform.
[ ]
1
1
L[1] =
L eat =
s
s−a
[ at n ]
n!
n!
n
L [t ] = n+1 , n ≥ 0 L e t =
s
(s − a)n+1
[ at
]
s
s−a
L [cos bt] = 2
L e cos bt =
2
s +b
(s − a)2 + b2
[ at
]
b
b
L [sin bt] = 2
L e sin bt =
2
s +b
(s − a)2 + b2
[ at
]
s
s−a
L [cosh bt] = 2
L e cosh bt =
2
s −b
(s − a)2 − b2
[
]
b
b
L [sinh bt] = 2
L eat sinh bt =
s − b2
(s − a)2 − b2
Laplace transform of partial derivatives
To obtain the Laplace transform of partial derivatives, say ux , uxx , ut and utt , we simply take the
Laplace transform as follows:
[
]
∂u
=
L[ux ] = L
∂x
[
]
∫∞
e
∂
dt =
∂x
∂x
−st ∂u
t=0
∫∞
∂2u
L[uxx ] = L
=
∂x2
∫∞
e−st u(x, t)dt =
d
L [u(x, t)]
dx
t=0
e
∂2
dt
=
∂x2
∂x2
−st ∂
2u
t=0
∫∞
t=0
38
e−st u(x, t)dt =
d2
L [u(x, t)]
dx2
and
[
L[ut ]
=
]
∂u
L
=
∂t
∫∞
e−st
∂u
dt.
∂t
Integration by parts yields:
t=0
z = e−st , dv =
put
∫∞
e
−st ∂u
∂t
dt
=
[zv]∞
0
∫∞
−
t=0
∂u
dt
∂t
⇒ dz = −se−st dt, v = u.
[
vdz = u(x, t)e
−st
]∞
t=0
∫∞
+s
t=0
∴
L[ut ]
=
=
e−st u(x, t)dt = −u(x, 0) + sL[u(x, t)]
t=0
sL[u(x, t)] − u(x, 0)
[
L[utt ]
Substituting yields
]
∫∞
∂2u
L
=
∂t2
e−st
∂2u
dt.
∂2t
Integration by parts yields:
t=0
z = e−st , dv =
put
∫∞
e
−st ∂
2u
∫∞
∂2u
dt
∂2t
⇒ dz = −se−st dt, v =
[
dt
=
L[utt ]
=
sL[ut ] − u(x, 0)
=
s sL[u(x, t)] − u(x, 0) − ut (x, 0)
=
s2 L[u(x, t)] − su(x, 0) − ut (x, 0)
∂t2
−
t=0
vdz = ut (x, t)e
−st
]∞
[zv]∞
0
t=0
t=0
∴
∫∞
+s
∂u
= ut .
∂t
Substituting yields
e−st ut dt = −ut (x, 0) + sL[ut ]
t=0
[
]
Generally,
[
]
∂nu
∂ n−1 u(x, 0)
n
n−1
n−2
n−3
L
=
s
L[u(x,
t)]
−
s
u(x,
0)
−
s
u
(x,
0)
−
s
u
(x,
0)
−
·
·
·
−
t
tt
∂tn
∂tn−1
Example(s):
∂u
∂2u
=
, 0 < x < 2, t > 0
∂t
∂x2
subject to the conditions u(0, t) = 0, u(2, t) = 0 and u(x, 0) = sin(πx).
(a) Use Laplace transform to solve the 1D heat conduction equation
Solution
Taking the Laplace transform of the given equation, with respect to t, we get
[
]
[
]
sL [u(x, t)] − u(x, 0) =
d2
L [u(x, t)]
dx2
Applying the initial condition, we obtain sL [u(x, t)] − sin(πx) =
d2
L [u(x, t)].
dx2
∂u
∂2u
L
=L
∂t
∂x2
(
⇒
⇒
)
d2
− s L [u(x, t)] = − sin(πx)
dx2
(19)
Equation (23) is a non-homogeneous second order linear constant coefficient equation. Let
d
D=
. Then, equation (23) becomes
dx
(
)
D2 − s L [u(x, t)] = − sin(πx)
(20)
(
)
√
The auxiliary equation is D2 − s = 0 ⇒ D = ± s. Thus the complimentary solution of
equation (27) is
√
√
Lc [u(x, t)] = A(s)e− sx + B(s)e sx ,
39
where A(s) and B(s) are constants to be determined. The particular solution of equation (27)
is given by
1
1
Lp [u(x, t)] = 2
[− sin(πx)] =
sin(πx).
D −s
s + π2
Therefore, the general solution of equation (27) is L [u(x, t)] = Lc [u(x, t)] + Lp [u(x, t)] i.e.,
L [u(x, t)] = A(s)e−
√
sx
+ B(s)e
√
sx
+
1
sin(πx)
s + π2
(21)
⇒
L[u(0, t)] = 0. Substituting in equation (28) yields
1
+
sin(0) ⇒ A(s) + B(s) = 0. Therefore, equation (28)
s + π2
From the condition u(0, t) = 0
0=
√
A(s)e− s·0
+ B(s)e
√
s·0
reduces to
L [u(x, t)] = A(s)e−
√
sx
− A(s)e
√
sx
+
1
sin(πx)
s + π2
(22)
⇒
L[u(2, t)] = 0. Substituting in equation (29) yields
(
√
√ )
1
−2 s − e2 s = 0
⇒ A(s) = 0.
0=
−
+
sin(2π)
⇒
A(s)
e
s + π2
1
Therefore, equation (29) reduces to L [u(x, t)] =
sin(πx). Taking the inverse Laplace
s + π2
transform we have
From the condition u(2, t) = 0
√
A(s)e−2 s
√
A(s)e2 s
−1
u(x, t) = sin(πx)L
[
]
1
2
= e−π t sin(πx)
2
s+π
∂2u
∂2u
=
4
using Laplace transform method subject to the conditions
∂t2
∂x2
u(0, t) = 0, u(5, t) = 0, u(x, 0) = 0 and ut (x, 0) = 5 sin(πx).
(b) Solve the wave equation
Solution
Taking the Laplace transform, with respect to t, on both sides yields
[
]
[
∂2u
∂2u
= 4L
L
2
∂t
∂x2
]
⇒
s2 L [u(x, t)] − su(x, 0) − ut (x, 0) = 4
Applying the initial conditions, we obtain sL [u(x, t)] − 5 sin(πx) = 4
(
d2
L [u(x, t)]
dx2
d2
L [u(x, t)].
dx2
)
d2
4 2 − s2 L [u(x, t)] = −5 sin(πx)
dx
⇒
(23)
Equation (23) is a non-homogeneous linear second order with constant coefficient.
d
∂2
Let D =
⇒ D2 =
. Then in terms of the D operator, equation (23) becomes
dx
∂x2
(
(
)
4D2 − s2 L [u(x, t)] = −5 sin(πx)
)
The auxiliary equation is 4D2 − s2 = 0
equation (27) is
⇒
(24)
s
D = ± . Thus the complementary solution of
2
Lc [u(x, t)] = A(s)e−sx/2 + B(s)esx/2 ,
where A(s) and B(s) are constants to be determined. The particular solution of equation (27)
is given by
1
5
Lp [u(x, t)] =
[−5 sin(πx)] = 2
sin(πx).
4D2 − s2
s + 4π 2
Therefore, the complete solution of equation (27) is L [u(x, t)] = Lc [u(x, t)] + Lp [u(x, t)] i.e.,
L [u(x, t)] = A(s)e−sx/2 + B(s)esx/2 +
40
5
sin(πx)
s2 + 4π 2
(25)
⇒
L[u(0, t)] = 0. Substituting in equation (28) yields
5
sin(0) ⇒ A(s) + B(s) = 0. Therefore, equation (28)
s2 + 4π 2
From the condition u(0, t) = 0
0=
A(s)e−s·0/2 + B(s)es·0/2 +
reduces to
L [u(x, t)] = A(s)e−sx/2 − A(s)esx/2 +
5
sin(πx)
+ 4π 2
s2
(26)
From the condition u(5, t) = 0
⇒
L[u(5, t)] = 0. Substituting in equation (29) yields
5s
5s
)
(
−
5
−5s/2 − e5s/2 = 0
0 = A(s)e 2 − A(s)e 2 + 2
sin(5π)
⇒
A(s)
e
⇒ A(s) = 0.
s + 4π 2
5
Therefore, equation (29) reduces to L [u(x, t)] = 2
sin(πx). Taking the inverse Laplace
s + 4π 2
transform we have
−1
[
u(x, t) = sin(πx)L
=
]
[
5
5
2π
= sin(πx)L−1
·
s2 + 4π 2
2π s2 + (2π)2
]
5
sin(πx) sin(2πt)
2π
Exercise:
∂u
∂2u
= c2 2 subject to the initial and
∂t
∂x
u(x, t) → 0 as x → ∞.
(a) Use Laplace transform to solve the 1D heat equation
boundary conditions u(x, 0) = 5;
u(0, t) = 10;
Solution
∂2u
∂u
= c2 2 . Applying the Laplace Transform we get
Given:
∂t
∂x
[
∂u
L
∂t
⇒
Let D =
c2
[
∂2u
= c L
∂x2
]
2
d2
L [u(x, t)]
dx2
d2
⇒ sL [u(x, t)] − 5 = c2 2 L [u(x, t)]
dx
)
⇒
(
]
sL [u(x, t)] − u(x, 0) = c2
d2
− s L [u(x, t)] = −5
dx2
(
)
d
. Then, c2 D2 − s L [u(x, t)] = −5.
dx
(
(27)
)
i) Complementary solution: we √
write c2 D2 − s Lc [u(x, t)] = 0. The auxiliary equation is
( 2 2
)
s
c D −s =0 ⇒ D =±
. Thus CF of equation (27) is
c
Lc [u(x, t)] = A(s)e−
ii) Particular integral: we write Lp [u(x, t)] =
√
s
x
c
√
+ B(s)e
1
(c2 D2
− s)
s
c
x
(−5) =
5
s
Hence, the general solution of equation (27) is
L [u(x, t)] = A(s)e−
From the condition
u(∞,
t) = 0
√
√
s
s
5
0 = A(s)e− c ∞ + B(s)e c ∞ +
s
√
√
s
x
c
+ B(s)e
s
c
x+
5
s
(28)
⇒
L[u(∞, t)] = 0. Substituting in equation (28) yields
⇒
B(s) = 0. Therefore, equation (28) reduces to
L [u(x, t)] = A(s)e−
41
√
s
x
c
+
5
s
(29)
10
. Substituting in equation (29)
L[u(0, t)] = L[10] =
s
5
A(s) = . Therefore, equation (29) reduces to
s
⇒
From the condition u(0, t) = 10
√
s
10
5
yields
= A(s)e− c 0 +
⇒
s
s

L [u(x, t)] = 5 
Recall: L−1
e
−
√
s
x
c
s


1
+ 
s
u(x, t) = 5L−1 
⇒
[
[ ]
√
1
e−a
= 1 and L−1
s
s
s
]
[
= 1 − erf
(
e
−

√
s
x
c
[
2
→ Note: the error function is defined by erf (z) = √
π
−
√
s
x
c

 + L−1
[ ]
, where erf is the error function.
x
√
2c t
∫z
)]
e−x dx.
2
0
(
)
(
)
(b) Using Laplace transform, show that y(x, t) = b cos λ t − xc H t − xc is the solution to the 1D
wave equation ytt = c2 yxx (∗) subject to the following initial and boundary conditions:
Boundary conditions:
y(0, t) = b cos(λt),
Initial conditions:
y(x, 0) = 0,
y(∞, t) < ∞
yt (x, 0) = 0
Solution
Taking the Laplace transform of equation (∗) wrt t we have L[ytt ] = c2 L[yxx ]
d2
⇒ s2 L[y(x, t)] − sy(x, 0) − yt (x, 0) = c2 2 L[y(x, t)]
dx
Using the Initial Conditions we have
d2
L[y(x, t)] −
dx2
( )2
s
c
⇒
L[y(x, t)] = 0
s
s
(D − )(D + )L[y(x, t)] = 0
c
c
The general solution of this ODE is given by
L[y(x, t)] = A(s)e c x + B(s)e− c x ,
s
(∗)′
s
where A(s) and B(s) are arbitrary functions. To determine A(s) and B(s), we apply the Boundary Conditions as follows:
From y(∞, t) < ∞: taking Laplace transform wrt t yields L[y(∞, t)] < ∞. Putting x = ∞ in
equation (∗)′ , we get
L[y(∞, t)] = A(s)e c ∞ + B(s)e− c ∞ = Ae c ∞ < ∞
s
s
s
⇒
A(s) = 0
So equation (∗)′ reduces to
L[y(x, t)] = B(s)e− c x
(∗)′′
s
From y(0, t) = b cos(λt): taking Laplace transform wrt t yields
L[y(0, t)] = bL[cos(λt)] =
s2
bs
+ λ2
Putting x = 0 in equation (∗)′′ , we get
L[y(0, t)] = B(s)e− c ·0 = B(s)
s
42
⇒
B(s) =
s2
bs
+ λ2

1 
s
)]
(
∴ u(x, t) = 5 2 − erf

1
e
+  = 5 L−1 
s
s
s
a
√
2 t

Substituting in equation (∗)′′ we have
L[y(x, t)] =
s2
s
bs
e− c x − − − (i)
2
+λ
Recall from the second shifting theorem that if L [f (t)] = F (s), then
L [f (t − a) · H(t − a)] = F (s)e−sa − − − (ii)
{
where the unit step function (Heaviside function) is defined as H(t − a) =
1,
0,
if t > a
.
if t < a
Comparing equations (i) and (ii) we have
y(x, t) = f (t − a) · H(t − a),
F (s) =
s2
x
bs
and a =
2
+λ
c
So,
f (t) = L−1 [F (s)] = L−1
[
]
bs
= b cos λt
2
s + λ2
Therefore,
(
(
⇒
f (t − a) = f t −
)
(
x
x
y(x, t) = b cos λ t −
H t−
c
c
x
c
)
(
= b cos λ t −
x
c
)
)
This result represents a wave traveling to the right along the string with velocity c.
∂y
∂y
+x
= 0 subject to the following initial and boundary
∂x
∂t
y(0, t) = t.
(c) Use Laplace transform to solve
conditions: y(x, 0) = 0;
Solution
Taking the Laplace transform of the given equation wrt t we have
[
]
[
]
∂y
∂y
L
+ xL
=0
∂x
∂t
[
]
∂
:0
L [y(x, t)] + x sL [y(x, t)] − y(x, 0) = 0
⇒
∂x
d
⇒
L [y(x, t)] + xsL [y(x, t)] = 0
dx
Let z = L [y(x, t)]. Then,
ln z + s
∂z
+ sxz = 0
∂x
x2
= A(s)
2
⇒
(30)
∂z
+ sx∂x = 0. Integrating yields
z
⇒
ln z = A(s) − s
x2
2
⇒
z = A(s)e−s
x2
2
,
where A(s) is a constant of integration. Hence, the general solution of equation (30) is
L [y(x, t)] = A(s)e−s
x2
2
(31)
1
From the condition y(0, t) = t ⇒ L[y(0, t)] = L[t] = 2 . Substituting in equation (31) yields
s
02
1
1
= A(s)e−s 2
⇒ A(s) = 2 . Therefore, equation (31) reduces to
2
s
s
L [y(x, t)] =
1 −s x2
e 2 − − − (i)
s2
Recall from the second shifting theorem that if L [f (t)] = F (s), then
L [f (t − a) · H(t − a)] = F (s)e−sa − − − (ii)
43
{
where the unit step function (Heaviside function) is defined by H(t − a) =
1,
0,
if t ≥ a
.
if t < a
Comparing equations (i) and (ii) we have
y(x, t) = f (t − a) · H(t − a),
So,
−1
f (t) = L
[F (s)] = L
−1
[
Therefore,
(
y(x, t) =
1
x2
and
a
=
s2
2
(
]
1
=t
s2
F (s) =
⇒
x2
t−
2
x2
f (t − a) = f t −
2
)
(
x2
H t−
2
)
(
=
x2
t−
2
)
)
∂u
∂2u
= 4 2 subject
∂t
∂x
to the conditions: u(0, t) = u(3, t) = 0 and u(x, 0) = 10 sin(2πx) − 6 sin(4πx).
(d) [Assignment 2 ] Use Laplace transform to solve the boundary value problem
∂u ∂u
+
= x, x > 0, t > 0 with initial and
∂x
∂t
(t − x)2
t2
[ans: u(x, t) =
H (t − x) − + xt]
2
2
(e) Use Laplace transform method to solve the equation
boundary condition: u(x, 0) = 0 and u(0, t) = 0.
∂u
∂2u
=
, 0 < x < 2, t > 0
∂t
∂x2
subject to the conditions u(0, t) = 0, u(2, t) = 0 and u(x, 0) = 3 sin(2πx).
[ans:
2
u(x, t) = 3e−4π t sin(2πx)]
(f) Use the method of Laplace transform to solve the heat equation
∂2u
∂2u
=
+ sin(πx), 0 < x < 1, t > 0 using Laplace transform
∂t2
∂x2
subject to the following initial and boundary conditions: u(x, 0) = 0; ut (x, 0) = 0; u(0, t) =
1
0; u(1, t) = 0.
[ans: u(x, t) = 2 [1 − cos(πt)] sin(πx)]
π
(g) Solve the 1D wave equation
(h) Solve by Laplace transform;
i) ux + ut + u = 0, x > 0, t > 0 subject to u(0, t) = 0 t > 0 and u(x, 0) = 0 x > 0.
u(x, t) = e−t [cos t sin x − cos x sin t − sin(t − x) · H(t − x)]]
ii) ux + 2xut = 2x,
u(x, 0) = 1,
u(0, t) = 1.
[ans:
[ans: u(x, t) = 1 + t − (t − x2 ) · u(t − x2 )]
(i) Using the Laplace transform method, solve the partial differential equation
to the initial condition u(x, 0) = e−7x , x > 0, t > 0.
∂u
∂u
= u− . Subject
∂t
∂x
[15 mks]
Lecture 11
5.1.3
Fourier transform
Fourier transform is useful when the domain of definition with respect to x and t are −∞ < x < ∞
and −∞ < t < ∞. The Fourier integral of a function f (x) defined on the interval (−∞, ∞) is given
by:
1
f (x) =
π
∫∞
[A(ω) cos(ωx) + B(ω) sin(ωx)] dω,
0
where the Fourier coefficients are defined by
∫∞
A(ω) =
∫∞
f (t) cos(ωt)dt
and B(ω) =
−∞
f (t) sin(ωt)dt
−∞
44
(32)
Substituting the Fourier coefficients in equation (32) yields:
1
f (x) =
π
∫∞ ∫∞
f (t) [cos(ωt) cos(ωx) + sin(ωt) sin(ωx)] dtdω − − − (I)
0 −∞
In order to motivate the definition of Fourier transform, we first express complex form of the Fourier
integral as follows. Using the identity cos(AB) = cos A cos B + sin A sin B, we write the integral (I)
as.
∫∞ ∫∞
1
f (x) =
f (t) [cos ω(x − t)] dtdω
π
0 −∞
Since cos θ =
eiθ
+
2
e−iθ
, we obtain
f (x) =
=
1
2π
1
2π
∫∞ ∫∞
[
]
f (t) eiω(x−t) + e−iω(x−t) dtdω
0 −∞
∫∞ ∫∞
f (t)e
iω(x−t)
0 −∞
1
dtdω +
2π
∫∞ ∫∞
f (t)e−iω(x−t) dtdω
0 −∞
Replacing ω by −ω in the second term on the right hand side and adjusting limits from −∞ to 0, we
obtain
f (x) =
1
2π
∫∞ ∫∞
1
2π
f (t)eiω(x−t) dtdω =
−∞ −∞

∫∞  ∫∞
−∞



e−iωt f (t)dt eiωx dω

−∞
which is the complex form of the Fourier integral of f . This leads to following pair of transforms.
Definition 5.1.1 (Fourier transform). The Fourier transform of a function f (x), with respect to x,
is defined by
∫∞
F [f (x)] =
e−iωx f (x)dx .= U (ω),
(−∞ < ω < ∞)
−∞
provided this integral exists.
Similarly, the Fourier transform of a function u(x, t) for −∞ < x < ∞, with respect to x is given by
∫∞
F [u(x, t)] =
e−iωx u(x, t)dx = U (ω, t),
(−∞ < ω < ∞ and t > 0)
−∞
Example(s):
Find the Fourier transform of the function defined by
{
f (x) =
1 for |x| ≤ L,
0 for |x| > L
Solution
Using the definition of Fourier transform, we have
∫∞
F [f (x)] =
e
−∞
−iωx
∫L
f (x)dx =
e−iωx dx,
(since |x| ≤ L)
−L
2
1 [ −iωx ]L
e
=
= −
−L
iω
ω
(
eiωL − e−iωL
2i
)
=
2
sin(ωL)
ω
→ Note: only functions that tend to zero sufficiently fast as x → ±∞ (rapidly decreasing to zero
functions) have Fourier transform. Thus, the nonzero constant function C, sin x, ex and x2 do not
have Fourier transform.
45
Definition 5.1.2 (Inverse Fourier transform). The inverse Fourier transform of a function U (ω), for
−∞ < ω < ∞, is defined as
F
−1
1
[U (ω)] = f (x) =
2π
∫∞
U (ω)eiωx dω,
(−∞ < x < ∞)
−∞
provided this integral exists.
→ Note: ω = 2πf represents the angular frequency, f being the oscillation frequency.
Theorem 5.1 (Convolution theorem). If F −1 [U (ω)] = f (x) and F −1 [G(ω)] = g(x), then
F
−1
∫∞
[U (ω)G(ω)] =
f (s)g(x − s)ds
−∞
Fourier transform of partial derivatives
Let u = u(x, t) be a function defined for −∞ < x < ∞ and t ≥ 0. If u(x, t) → 0 as x → ±∞, and
F [u(x, t)] = U (ω, t), then the Fourier transform of ux (x, t), with respect to x, is
∫∞
F [ux (x, t)] =
e−iωx
−∞
[
=
u(x, t)e
∂u
dx.
∂x
−iωx
Integration by parts yields:
∫∞
]∞
+ iω
−∞
e−iωx u(x, t)dx = iωU (ω, t).
−∞
If, in addition, ux (x, t) → 0 as x → ±∞, then the Fourier transform of uxx (x, t), with respect to x, is
∫∞
F [uxx (x, t)] =
e
−∞
−iωx
[
=
[
]
∂ ∂u
dx. Integration by parts yields:
∂x ∂x
ux (x, t)e
−iωx
]∞
−∞
∫∞
+ iω
e−iωx ux (x, t)dx
−∞
= iω {F [ux (x, t)]} = iω {iωU (ω, t)]} = −ω 2 U (ω, t).
Similarly, the Fourier transform of ut (x, t) and utt (x, t), with respect to x, are given by
∫∞
F [ut (x, t)] =
e
−∞
∫∞
F [utt (x, t)] =
e
−∞
d
dx =
∂t
dt
−iωx ∂u
∫∞
−∞
d2
dx
=
∂t2
dt2
−iωx ∂
2u
e−iωx u(x, t)dx =
∫∞
d
d
F [u(x, t)] = U (ω, t)
dt
dt
e−iωx u(x, t)dx =
−∞
d2
d2
F
[u(x,
t)]
=
U (ω, t)
dt2
dt2
Example(s):
1. An infinitely long string extending from −∞ < x < ∞ under tension is displaced into a curve
u = f (x) and let go from rest. Find the displacement u(x, t) of the string at any point at any
subsequent time.
Solution
The PDE to be solved is the 1D wave equation
2
∂2u
2∂ u
=
c
, −∞ < x < ∞, t > 0 − − − (∗)
∂t2
∂x2
46
subject to the initial conditions:
u(x, 0) = f (x) and ut (x, 0) = 0, −∞ < x < ∞
with u(x, t) → 0, ux (x, t) → 0 as x → ±∞, t > 0.
Taking the Fourier transform of both sides of equation (∗) with respect to x, we obtain
]
[
[
∂2u
∂2u
2
F
=
c
F
∂t2
∂x2
]
or
d2
F [u(x, t)] = c2
dt2
2
= c
∫∞
e
−iωx
−∞
[



ux (x, t)e
2
= ic ω
−iωx
∫∞
∞
+ iω
−∞



]
∂ ∂u
dx. Use integration by parts
∂x ∂x
u(x, t)e
−iωx
∞
−∞
e
−iωx ∂u
∂x
−∞
∫∞
+ iω
e
−iωx


dx

∫∞
2
= ic ω


u(x, t)dx

−∞
e−iωx
−∞
= −c ω
∂u
dx
∂x
∫∞
2 2
e−iωx u(x, t)dx
−∞
d2
F [u(x, t)] = −c2 ω 2 F [u(x, t)]
dt2
⇒
d
Let the differential operator D = . Then, in terms of the D operator, the above differential
dt
equation becomes
[D2 + c2 ω 2 ]F [u(x, t)] = 0
The root of the auxiliary equation are D = ±ikω. Thus, the general solution is given by
F [u(x, t)] = A(ω) cos(kωt) + B(ω) sin(kωt) − − − (∗∗)
Applying the initial conditions u(x, 0) = f (x)
U (ω) = A(ω). Hence, equation (∗∗) becomes
⇒
F [u(x, 0)] = F [f (x)] = U (ω), yields
F [u(x, t)] = U (ω) cos(cωt) + B(ω) sin(cωt) − − − (∗ ∗ ∗)
Applying the initial conditions ut (x, 0) = 0 ⇒ F [ut (x, 0)] = 0, yields 0 = B(ω). Hence,
equation (∗ ∗ ∗) becomes
F [u(x, t)] = U (ω) cos(cωt)
Taking the inverse Fourier transform, we get
u(x, t) = F
=
=
=
−1
1
2π
1
4π
1
[U (ω) cos(cωt)] =
2π
∫∞
−∞
∫∞
(
U (ω) cos(cωt)eiωx dω
−∞
eicωt + e−icωt
U (ω)
2
(
)
eiωx dω
)
U (ω) eiω(x+ct) + eiω(x−ct) dω
−∞


∫∞
∫∞

1 1
1
U (ω)eiω(x+ct) dω +
U (ω)eiω(x−ct) dω

2  2π
2π
−∞
=
∫∞
−∞
1
{f (x + ct) + f (x − ct)}
2
47
In obtaining this result, we have used the fact that the Fourier integral representation of the
∫∞
1
function f (x) can be written as: f (x) =
U (ω)eiωx dω.
2π
−∞
Exercise:
∫∞
1
2
f (s)e−(x−s) /4kt ds
2. Use the complex form of the Fourier transform to show that u(x, t) = √
2 πkt
−∞
is a solution of the BVP below, governing the heat conduction in a very long metal bar which
extends from −∞ to ∞.
ut = kuxx ; u(x, 0) = f (x), −∞ < x < ∞.
Solution
The 1D heat conduction equation is subject to the boundary conditions:
u(−∞, t) = u(∞, t) = 0 and ux (−∞, t) = ux (∞, t) = 0, t > 0
and initial conditions:
u(x, 0) = f (x) and ut (x, 0) = 0, −∞ < x < ∞
Taking the Fourier transform of the given equation with respect to x, we get
[
[
]
∂u
∂2u
F
= kF
∂t
∂x2
]
or
d
F [u(x, t)] = k
dt
∫∞
−∞
e−iωx
[


= k
= k
⇒
]
∂ ∂u
dx.
∂x ∂x
ux (x, t)e−iωx
∫∞
∞
+ iω



∞

−∞
−∞
ux (x, t)e−iωx
Use integration by parts
−∞

∂u 
e−iωx dx
∂x 

+ iω u(x, t)e−iωx
∞
−∞
∫∞
+ iω
−∞


e−iωx u(x, t)dx

{[
}
]∞
d
2
−iωx
F [u(x, t)] = k {ux (x, t) + iωu(x, t)} e
− ω F [u(x, t)]
−∞
dt
Applying the boundary conditions yields
d
F [u(x, t)] = −kω 2 F [u(x, t)]
dt
d
Let the differential operator D = . Then, in terms of the D operator, the above differential
dt
equation becomes
[D + kω 2 ]F [u(x, t)] = 0
The root of the auxiliary equation is D = −kω 2 . Thus, the general solution is given by
F [u(x, t)] = A(ω)e−kω t − − − (∗)
2
Applying the initial conditions u(x, 0) = f (x)
U (ω) = A(ω). Hence, equation (∗) becomes
⇒
F [u(x, 0)] = F [f (x)] = U (ω), yields
F [u(x, t)] = U (ω)e−kω
48
2t
Taking the inverse Fourier transform, we get
[
u(x, t) = F −1 U (ω)e−kω
2t
]
If we can find a function g(x, t) whose Fourier transform with respect to x is G(ω, t) = e−kω t ,
then we can write
2
u(x, t) = F −1 [U (ω)G(ω, t)],
where U (ω) = F [f (x)]
and conclude from the convolution theorem that
∫∞
f (s)g(x − s, t)ds − − − (i)
u(x, t) =
−∞
Now, we find g(x, t) by evaluating the integral
g(x, t) =
=
=
=
1
2π
1
2π
1
2π
∫∞
G(ω, t)e
−∞
∫∞
−∞
∫∞
∫∞
1
dω =
2π
e−kω t · eiωx dω
2
−∞
e−kω t [cos ωx + i sin ωx] dω
2
e−kω t cos ωxdω
(since e−kω t sin ωx is an odd function of ω).
2
2
−∞
∫∞
1
π
iωx
e−kω t cos ωxdω
2
(since e−kω t cos ωx is an even function of ω).
2
0
We now introduce a new variable of integration by making the substitution
z 2 = kω 2 t
z
ω=√
kt
⇒
⇒
1
dω = √ dz
kt
Hence, we get
g(x, t) =
∫∞
where I =
1
√
π kt
∫∞
e
0
∫∞
(
)
xz
cos √
dz
kt
−z 2
=
1
√
π kt
=
1
√ I, − − −(ii)
π kt
e−z cos(uz)dz,
2
0
x
where u = √ .
kt
e−z cos(uz)dz. This is an elementary function which can be determined explicitly.
2
0
Thus,
dI
=−
du
∫∞
ze−z sin(uz)dz
2
0
−z dz. Thus, dp = u cos(uz)dz and
Using
∫ integration by parts: put p = sin(uz) and dv = ze
1
2
2
v = ze−z dz = − e−z . Hence,
2
2
dI
du


∞

0
= − pv
= −
u
2
−
∫∞
0
∫∞




]∞
∫∞
[ 1

u
2
2
vdp = − − e−z sin(uz)
+
e−z cos(uz)dz

 2

2
0
0
u
2
e−z cos(uz)dz = − I
2
0
49
dI
u
= − I, whose general solution is
du
2
√
∫∞
π
2
−z 2
−u
/4
given by I = Ce
, where C is an arbitrary constant. Now, I(0) =
e dz =
.
2
0
√
π
. Hence,
Applying this initial condition, we get C =
2
√
π −u2 /4
I=
e
2
Therefore, we obtain the ordinary differential equation
Substituting in equation (ii), we obtain
√
π −u2 /4
1
1
2
g(x, t) = √ ·
e
= √
e−x /4kt
π kt 2
2 πkt
Therefore,
1
2
e−(x−s) /4kt
g(x − s, t) = √
2 πkt
Substituting in equation (i), we get
1
u(x, t) = √
2 πkt
∫∞
f (s)e−(x−s)
2 /4kt
ds
−∞
3. A bar whose surface is insulated has a length of 3 units and diffusivity 2 units. If the ends are
kept at temperature zero units and initial temperature is u(x, 0) = 5 sin(4πx) − 3 sin(8πx), find
the temperature distribution at any point x at any time t using Fourier transform.
5.1.4
Fourier sine and cosine transforms
When f (x) is an even function on the interval (−∞, ∞), then the product f (x) cos(ωx) is also an even
function whereas f (x) sin(ωx) is an odd function. As a consequence B(ω) = 0 and
∫∞
A(ω) = 2
f (t) cos(ωt)dt
0
Hence, the Fourier integral (32) reduces to the simpler form
f (x) =
1
π
∫∞
A(ω) cos(ωx)dω =
2
π
0

∫∞ ∫∞
0



f (t) cos(ωt)dt cos(ωx)dω

0
Similarly, when f (x) is an odd function on the interval (−∞, ∞), then the product f (x) cos(ωx) is
also an odd function whereas f (x) sin(ωx) is an even function. Thus, A(ω) = 0 and
∫∞
B(ω) = 2
f (t) sin(ωt)dt
0
Hence, the Fourier integral (32) reduces to the simpler form
f (x) =
1
π
∫∞
B(ω) sin(ωx)dω =
0
2
π

∫∞ ∫∞
0



f (t) sin(ωt)dt sin(ωx)dω
0

From these Fourier integrals, we define the following Fourier transform pairs.
Definition 5.1.3 (Fourier cosine transform). The Fourier transform of an even function f is called
Fourier cosine transform of f , and is defined as:
∫∞
Fc [f (x)] =
f (x) cos(ωx)dx = U (ω),
0
50
(0 ≤ ω < ∞)
The inverse Fourier cosine transform of a function U (ω), for 0 ≤ ω < ∞, is defined as
Fc−1 [U (ω)]
2
= f (x) =
π
∫∞
(0 ≤ x < ∞)
U (ω) cos(ωx)dω,
0
Definition 5.1.4 (Fourier sine transform). The Fourier transform of an odd function f is called
Fourier sine transform of f , and is defined as:
∫∞
Fs [f (x)] =
f (x) sin(ωx)dx = U (ω),
(0 ≤ ω < ∞)
0
The inverse Fourier sine transform of a function U (ω), for 0 ≤ ω < ∞, is defined as
Fs−1 [U (ω)]
2
= f (x) =
π
∫∞
U (ω) sin(ωx)dω,
(0 ≤ x < ∞)
0
→ Note: the Fourier cosine and sine transforms are appropriate for problems over semi-infinite intervals (i.e., when the range of the variables selected extends from 0 to ∞) in a spatial variable in which
the function or its derivative are prescribed on the boundary.
Example(s):
Find the Fourier sine transform and Fourier cosine transform of the function defined by
{
f (x) =
1 if 0 ≤ x ≤ 2,
0 if x > 2
Solution
i) Using the definition of Fourier sine transform of f (x) with respect to x, we have
∫∞
Fs [f (x)] =
∫2
f (x) sin(ωx)dx =
0
sin(ωx)dx,
(since 0 ≤ x ≤ 2)
0
]2
1[
1
= − cos(ωx) = [1 − cos(2ω)]
ω
ω
0
ii) Using the definition of Fourier cosine transform of f (x) with respect to x, we have
∫∞
Fc [f (x)] =
∫2
f (x) cos(ωx)dx =
0
cos(ωx)dx,
(since 0 ≤ x ≤ 2)
0
]2
sin(2ω)
1[
sin(ωx) =
ω
ω
0
=
Fourier cosine and sine transform of partial derivatives
Let u = u(x, t) be a function defined for 0 ≤ x < ∞ and t ≥ 0. If u(x, t) → 0 as x → ∞, then
i) Fourier sine transform of ux (x, t), with respect to x, is
∫∞
Fs [ux (x, t)] =
sin(ωx)
∂u
dx.
∂x
Integration by parts yields:
0
[
=
]∞
u(x, t) sin(ωx)
0
−ω
∫∞
0
51
u(x, t) cos(ωx)dx = −ωFc [ux (x, t)]
If, in addition, ux (x, t) → 0 as x → ∞, then the Fourier sine transform of uxx (x, t), with respect
to x, is written as follows:
∫∞
Fs [uxx (x, t)] =
[
]
∂ ∂u
sin(ωx)
dx. Integration by parts yields:
∂x ∂x
0
[
=
]∞
ux (x, t) sin(ωx)
0
−ω

∂u
cos(ωx) dx = −ω
∂x
0

[
= −ω
∫∞
]∞
u(x, t) cos(ωx)
0
∫∞
cos(ωx)
0
∫∞
+ω
∂u
dx. Integrate by parts.
∂x


u(x, t) sin(ωx)dx

0
= ωu(0, t) − ω 2 Fs [u(x, t)]
Similarly, the Fourier sine transform of ut (x, t) and utt (x, t), with respect to x, are given by
∫∞
Fs [ut (x, t)] =
∂u
d
sin(ωx) dx =
∂t
dt
0
∫∞
u(x, t) sin(ωx)dx =
d
d
Fs [u(x, t)] = Us (ω, t)
dt
dt
0
∫∞
Fs [utt (x, t)] =
sin(ωx)
∂2u
d2
∂t
dt2
dx =
2
0
∫∞
u(x, t) sin(ωx)dx =
−∞
d2
d2
F
[u(x,
t)]
=
Us (ω, t)
s
dt2
dt2
ii) Fourier cosine transform of ux (x, t) and uxx (x, t), with respect to x, are given by
∫∞
Fc [ux (x, t)] =
cos(ωx)
∂u
dx. Integration by parts yields:
∂x
0
[
=
u(x, t) cos(ωx)
]∞
0
∫∞
+ω
u(x, t) sin(ωx)dx = −u(0, t) + ωFs [ux (x, t)]
0
∫∞
Fc [uxx (x, t)] =
[
]
[
]∞
∂ ∂u
dx = ux (x, t) cos(ωx) + ω
cos(ωx)
∂x ∂x
0
∫∞
sin(ωx)
∂u
dx
∂x
0
0
= −ux (0, t) + ω
= −ux (0, t) + ω
∫∞
sin(ωx)
∂u
dx. Integrate by parts again.
∂x
0

[

u(x, t) sin(ωx)
]∞
0
−ω


∫∞
u(x, t) cos(ωx)dx
0

= −ux (0, t) − ω 2 Fc [u(x, t)]
Similarly, the Fourier cosine transform of ut (x, t) and utt (x, t), with respect to x, are given by
∫∞
Fc [ut (x, t)] =
∂u
d
cos(ωx) dx =
∂t
dt
0
cos(ωx)
0
u(x, t) cos(ωx)dx =
d
d
Fc [u(x, t)] = Uc (ω, t)
dt
dt
0
∫∞
Fc [utt (x, t)] =
∫∞
d2
∂2u
dx
=
∂t2
dt2
∫∞
u(x, t) cos(ωx)dx =
−∞
d2
d2
F
[u(x,
t)]
=
Uc (ω, t)
c
dt2
dt2
→ Note: The choice of cosine or sine transform is decided by the form of the boundary conditions at
the lower limit of the variable selected. For instance, if u(0, t) is given as a boundary condition, the
Fourier sine transform is appropriate. On the other hand, if ux (0, t) is given as a boundary condition,
the Fourier cosine transform is suitable.
Example(s):
52
1. consider the 1D heat conduction equation in a semi-infinite metal bar 0 ≤ x < ∞. The PDE
which governs this process is written as
ut = kuxx , 0 ≤ x < ∞, t > 0,
where k is the thermal diffusivity of the metal bar, with boundary conditions u(0, t) = u0 , t > 0
and initial condition u(x, 0) = 0, x > 0. Find the temperature at any time t and position x.
Solution
Since the temperature u(0, t) = u0 is given when x = 0, the Fourier sine transform is appropriate
for this problem. We define
∫∞
Fs [u(x, t)] =
u(x, t) sin(ωx)dx = Us (ω, t)
0
Taking the Fourier sine transform of the given equation, we get
Fs [ut ] = kFs [uxx ]
⇒
∫∞
∫∞
ut sin(ωx)dx = k
0
uxx sin(ωx)dx
0
[
]
d
Fs [u(x, t)] = k ωu(0, t) − ω 2 Fs [u(x, t)]
dt
d
⇒
Fs [u(x, t)] + kω 2 Fs [u(x, t)] = kωu0 ,
dt
d
which is an ODE. Let D = . Then in terms of the D operator, the above ODE becomes
dt
⇒
(D + kω 2 )Fs [u(x, t)] = kωu0
The general solution is given by
Fs [u(x, t)] = A(ω)e−kω t +
2
⇒
Applying the initial condition u(x, 0) = 0
Fs [u(x, t)] =
u0
ω
Fs [u(x, 0)] = 0 yields A(ω) = −
)
u0 (
2
1 − e−kω t
ω
u0
. Hence,
ω
Taking the inverse Fourier sine transform, we obtain
u(x, t) =
Fs−1
[
∫∞
)]
) sin(ωx)
u0 (
2u0 (
2
−kω 2 t
1−e
=
dω
1 − e−kω t
ω
π
ω
0
=
2u0
π
∫∞
sin(ωx)
2u0
dω −
ω
π
0
=
∫∞
e−kω
2t
sin(ωx)
dω
ω
0
2u0 π 2u0
· −
π 2
π
∫∞
e
0
−kω 2 t


sin(ωx)
2
dω = u0 1 −

ω
π

∫∞
e
0
−kω 2 t
sin(ωx) 
dω

ω
The infinite integral can be expressed in terms of the error function by using the well-known
result
∫∞
π −u2 /λ2
2 2
e
e−λ x cos(2ux)dx =
2λ
0
and integrating with respect to u from 0 to u. Therefore,
[
u(x, t) = u0 1 − erf
53
(
x
√
2 kt
)]
2. The steady-state temperature in a semi-infinite plate is determined from
∂2u ∂2u
+ 2 = 0, y > 0
∂x2
∂y
subject to the conditions u(0, y) = 0, u(π, y) = e−y and uy (x, 0) = 0. Solve for u(x, y).
Solution
The domain of the variable y and the prescribed condition at y = 0 indicate that the Fourier
cosine transform is suitable for the problem. We define
∫∞
Fc [u(x, y)] =
u(x, y) cos(ωy)dy = Uc (x, ω)
0
Taking the Fourier cosine transform of the given equation, we get
[
Fc
⇒
]
[
]
∂2u
∂2u
+
F
= Fc [0]
c
∂x2
∂y 2
d2
Fc [u(x, y)] − ω 2 Fc [u(x, t)] − uy (x, 0) = 0
dx2
d2
Fc [u(x, y)] − ω 2 Fc [u(x, t)] = 0
dx2
Since the domain of x is a finite interval, we choose to write the solution of the above ODE as
⇒
Fc [u(x, y)] = C1 cosh(ωx) + C2 sinh(ωx)
⇒
Applying the boundary condition u(0, y) = 0
Fc [u(0, y)] = 0, we obtain C1 = 0. Hence,
Fc [u(x, y)] = C2 sinh(ωx)
Applying the boundary condition u(π, y) = e−y .
⇒
[
Fc [u(π, y)] = Fc e
−y ]
∫∞
=
e−y cos(ωy)dy =
1
1 + ω2
0
Thus, we obtain C2 =
1
(1 +
ω 2 ) sinh ωπ
. Hence,
Fc [u(x, y)] =
sinh(ωx)
(1 + ω 2 ) sinh ωπ
Taking the inverse Fourier cosine transform, we obtain
u(x, t) =
Fc−1
[
]
sinh(ωx)
2
=
2
(1 + ω ) sinh ωπ
π
∫∞
0
54
sinh(ωx) cos ωy
dω
(1 + ω 2 ) sinh ωπ