Download Partial Differential Equations

PARTIAL
DIFFERENTIAL
EQUATIONS
A partial differential equation is a
differential equation which involves
partial derivatives of one or more
dependent variables with respect to
one or more independent variables.
 Consider the first-order partial
differential equation
(1)
u
2
2
x y
x
In which
u is the dependent variable and
x and y are independent variables.


The solution is;
u   ( x 2  y 2 )x   ( y)
where indicates a “partial integration”
with respect to “x”, holding y constant,
and is an arbitrary function of y only.
 Thus, solution of Eq (1) is;
x3
u 
 xy 2   ( y )
(2)
3
where is an arbitrary function of y.

As a second example, consider the secondorder2partial differential equation:
 u
 x 3  y (3)
yx

We first write this equation in the form
 u
(
)  x3  y
y x
and integrate partially with respect to y,
holding x constant,
2
u
y
 x3 y 
  ( x)
x
2
where is an arbitrary function of x.

We now integrate this result partially
with respect to x, holding y constant
(4)
x 4 y xy 2
u
4

2
 f ( x)  g ( y )
where f defined by
f(x)=   ( x)dx is an arbitrary function of x
and g is an arbitrary function of y.
The solution of the first-order partial
differential equation contains one
arbitrary function, and the solution of the
second-order partial differential equation
contains two arbitrary functions.
LINEAR PARTIAL DIFFERENTIAL
EQUATIONS OF THE SECOND-ORDER
The general linear partial differential
equation of the second order in two
independent variables x and y is;
 2u
2 y
 2u
u
u
A 2 B
 C 2  D  E  Fu  G
x
xy
y
x
y
(5)
where A, B, C, D, E, F and G are functions
of x and y.
If G(x,y)=0 for all (x,y), the Equation (5)
reduces to;
 2u
2 y
 2u
u
u
(6)
A
B
C
 D  E  Fu  0
x 2
xy
y 2
x
y
HOMOGENOUS LINEAR EQUATIONS
OF SECOND ORDER WITH CONSTANT
COEFFICIENTS
 2u
 2u
 2u
a
b
c
0
2
2
x
xy
y


(7)
where a, b and c are constants.
The word homogenous refers to the fact
that all terms in Equation (7) contain
derivatives of the same order (the second).
u= f(y+mx)
(8)
where f is an arbitrary function and m is a
constant. Differentiating (8);
 2u
2

m
f ''( y  mx )
2
x
 2u
 mf ''( y  mx )
xy
(9)
 2u
 f ''( y  mx )
2
y

Substituting (9) into the Equation (7);
m2 f ''( y  mx)  bmf ''( y  mx)  cf ''( y  mx)  0
f ''( y  mx)[am2  bm  c]  0

Thus f(y+mx) will be a solution of (7) if m
satisfies the quadratic equation
am2+bm+c=0
(10)
we now consider the following four cases
of Equation (7)
1. a  0 , and the roots of the quadratic
Equation (10) are distinct.
2. a  0 ,the roots of the Equation (10) are
equal.
3. a  0, b  0
4. a  0, b  0, c  0
In case (I) a  0 , and the roots of
Equation(10) are distinct.
Let the distinct roots of Equation
(10) m1 and m2.
Then the Equation (7) has the
solutions:
f(y+m1x)+g(y+m2x)
In case (ii), a,  0 and the roots of
Equation (10) are equal.
Let the double root of Equation (10) be
m1.
The Equation (7) has the solution
f(y+m1x), where f is an arbitrary function
of its argument.
The Equation (7) also has the solution
xg(y+m1x), where g is an arbitrary
function of its argument.
“f(y+m1x)+ xg(y+m1x)”
is a solution of Eq.7.
In case (iii), a  0, b  0
 The quadratic Equation (10) reduces to
bm+c=0 and has only one root.
 Denoting this root by m1, Equation (7) has
the solution f(y+m1x) where f is an
arbitrary function of its argument.
 g(x) is an arbitrary function of x only, is
also a solution of Equation (7).

f(y+m1x)+g(x)

is a solution of Equation (7).

Finally in case (iv), a  0, b  0, c  0
 The equation (10) reduces to c=0, which is impossible.
 There exist no solution of the form (Equation 8).
 The differential Equation (7) is;
 2u
c
0
2
y
or
 u
(
)0
y y



Integrating partially with respect to y twice, we obtain
u= f(x)+ yg(x),
where f and g are arbitrary functions of x only.
f(x)+yg(x)
is a solution of Equation (7).
Example 1: Find a solution of
 2u
 2u
 2u
5
6 2  0
2
x
xy
y
(Equation 15)
which contains two arbitrary functions.
Solution: The quadratic Equation (10)
corresponding to the differential equation
(Equation 15) is;
m2-5m+6=0
and this equation has the distinct roots m1=2,
m2=3.
Using Equation (11);
u= f(y+2x)+g(y+3x)
is a solution of Equation (15) which contains
two arbitrary functions.
Example 2: Find a solution of
 2u
 2u
 2u
4
4 2 0
2
x
xy
y
(Equation 16)
which contains two arbitrary functions.
Solution: The quadratic Equation (14.10)
corresponding to the differential equation
(Equation 16) is;
m2-4m+4=0
and this equation has the double root m1=2,
Using Equation (12);
u= f(y+2x)+xg(y+2x)
is a solution of Equation (16) which contains
two arbitrary functions.
The second-order linear partial
differential equation
 2u
2 y
 2u
u
u
A 2 B
 C 2  D  E  Fu  0
x
xy
y
x
y
where A, B, C, D, E and F are real
constants is said to be
i) hyperbolic if B2-4AC>0
ii) parabolic if B2-4AC=0
iii) elliptic if B2-4AC<0
(6)
The equation
 2u
 2u

0
2
2
x
y
(10)
(WAVE EQUATION, a homogenous linear
equation with constant coefficients)
This equation is hyperbolic since A=1, B=0,
C=-1 and B2-4AC>0. Equation (10) is a
special case of the one-dimensional wave
equation, which is satisfied by the small
transverse displacements of the points of a
vibrating string. It has the solution
u=f(y+x)+g(y-x)
where f and g are arbitrary functions.

The equation;
 2u
u

0
2
x
y
(11)
(HEATOR DIFFUSION EQUATION, is
not homogenous)
 This equation is parabolic, since A=1,
B=C=0, and B2-4AC=0. Equation (11) is a
special case of the one-dimensional heat
equation (or diffusion equation), which is
satisfied by the temperature at a point of
a homogenous rod.

The equation;
 2u
 2u

 0
2
2
x
y



(12)
(LAPLACE EQUATION, Homogenous linear
equation with constant coefficients)
This equation is elliptic, since A=1, B=0, C=1
and B2-4AC=-4 <0.
Equation (12) is the so-called twodimensional Laplace equation, which is
satisfied by the steady-state temperature at
points of a thin rectangular plate.
It has the solution;
u= f (y+ix)+ g (y-ix),
where f and g are arbitrary functions.
THE METHOD OF
SEPERATION OF VARIABLES
In this lesson, we introduce the so-called
method of separation of variables. This is a
basic method which is very powerful for
obtaining solutions of certain problems
involving partial differential equations.
Example: Heat or Diffusion Problem
We now illustrate the method of separation
of variables by applying it to obtain a formal
solution of the so-called heat problem.
The Physical Problem:
The temperature of an infinite horizontal slab
of uniform width h is everywhere zero. The
temperature at the top of the slab is then set
and maintained at Th, while the bottom
surface is maintained at zero. Determine the
temperature profile in the slab as a function
of position and time.
Solution:
This problem is solved in rectangular
coordinates.
Differential equation is;
2
(1)
T

T
2

2
t
z
The separation of variables method is now
applied. We assume that the solution for T
may be separated into the product of one
function ψ(z) that depends solely on z, and
by a second function θ(t) that depends only
on t.
T= T(z,t)
T= ψ(z) θ(t) = ψ.θ

The left hand side of Equation (1) becomes
T

(2)

( )   '
t
t
The right hand side of Equation (1) is given by
2
(3)

T
2
2

   
2
z
 Combining equations (2) and (3)
    2 
(4)

   
2




 2 
where –λ2 is a constant. The boundary conditions
(BC) and initial condition (IC) are now written.





BC (1)
At z=0, T=0 for t ≥0
BC (2)
At z=h, T=Th for t>0
IC
At t=0, T=0 for 0≤z≤h
The following solution results if –λ2 is zero.
T0=C1+C2z
(5)
Substituting BC(1) and BC(2) into Equation (5)
gives
T0=Th (z/h)
(6)
The equation (6) is the steady-state solution to
the partial differential equation (PDE).
If the constant is nonzero, we obtain
a  t
(7)
T e
(a sin   b cos  )
2 2

2
z

z
Equations (6) and (7) are both solutions
to (1). Since (1) is a linear PDE, the sum of
Equations (6) and (7) also is a solution,
i.e.,
T  To  T– 
(8)
2
z
 a 2  2t
T  Th ( )  e
(a sin  z  b cos z )
h
Resubstitution of the BC gives (9)
z
n z (9)
 a 2 ( n / h )2 t
T  Th ( )  an e
sin( )
h
h
The constant an is evaluated ısing the IC.
2Th
an 
(1)n
n
z 2Tn  (1)n  a2 ( n / h )2 t
n z
T  Tn ( ) 

e
sin(
)
h  n1 n
h
Steady-state
solution
Transient
solution
(10)
Consider the two-dimensional problem of
a very thin solid bounded by the y-axis
(z=0), the lines y=0 and y=l, and extending
to infinity in the z direction. The
temperature of the vertical edge at y=0
and y=l is maintained at zero. The
temperature at z=0 is To. Determine the
steady-state temperature profile in the
solid. Assume both plane surfaces of the
solid are insulated.
The two-dimensional view of this system
is presented in the Figure.
z
T=0
T=0
T=To
y
z=0
y=0
y=L
Based on the problem statement T is not a function of x.
Based on the problem statement T is not a
function of x.
 2T
 2T

0
2
2
y
z
(1)
This PDE is solved using the
separation-of-variables method.
Since the temperature of the solid
is a function of y and z, we assume
the solution can be separated into
the product of one function Φ (z)
that depends only on z.
T=T(y,z)
T=ψ (y) Φ (z)
(2)
T=ψ Φ
The first term of Equation (1) is
 2T
=Φ ψ’’
2
y
The second term makes the form
 2T
=
ψ
Φ’’
(3)
2
z
Substituting Equations (2) and (3) into
Equation (1) gives
Φ ψ’’+ ψ Φ’’ =0
 ''
 ''
(4)


Function of y

Function of z
Both terms in Equation (4) are equal to the
same constant –λ2.
 ''
 ''

 – 2


(5)
A positive constant or zero does not
contribute to the solution of the problem.
  a sin  y  b cos  y
a and b are constants that depend on
the value of λ.
  Ce z  De  z
T  
  (a sin  y  b cos  y )(Ce  De
z
 z
)
(6)
What are the BC?
 BC(1)
T=0
at y=0
 BC(2)
T=0
at y=R
 BC(3)
T=To
z=0
 BC(4)
T=0
z=∞
Based on physical grounds, BC(4) gives,
C=0
T  (a sin  y  b cos  y) De z
T  (a sin  y  b cos  y)e
 z

BC(1) gives
bλ =0
 T  (a sin  y )e   z

(7)
BC (2) gives
0  (a sin  L)e  z
The RHS of this equation is zero if aλ=0
or sin λL=0. If aλ is zero, no solution
results. Therefore
 sin λL=0
 λL=nπ; n=1,2,3,4,…
 λ=( nπ)/L

The equation (7) now becomes
T  an e
 ( n / L ) z
n y
sin(
)
L
The constant an is evaluated from BC(3).
To  an sin(
n y
)
L
The Fourier series analysis gives
an 
2To
[( 1) n 1  1]
n
So that;
T   n 1

2To
n y
n 1
 ( n / L ) z
[(1)  1]e
sin(
)
n
L