Partial Differential Equation

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4. • Definition
• One of the classical partial differential equation of
mathematical physics is the equation describing the
conduction of heat in a solid body (Originated in the
18th century). And a modern one is the space
vehicle reentry problem: Analysis of transfer and
dissipation of heat generated by the friction with
earth’s atmosphere.

5. • A differential equation is an equation for an
unknown function of one or several variables that
relates the values of the function itself and of its
derivatives of various orders.
• Ordinary Differential Equation:
Function has 1 independent variable.
• Partial Differential Equation:
At least 2 independent variables.

6. • General (implicit) form for one function u(x,y) :
• Highest derivative defines order of PDE
• Explicit PDE => We can resolve the equation
to the highest derivative of u.
• Linear PDE => PDE is linear in u(x,y) and
for all derivatives of u(x,y)
• Semi-linear PDEs are nonlinear PDEs, which
are linear in the highest order derivative.
6

7.  Partial Differential Equations are those which involve partial
derivatives with respect to two or more independent variables.
 Partial derivatives
𝜕𝑧
𝜕𝑥
,
𝑑𝑧
𝑑𝑦
,
𝜕2 𝑧
𝜕𝑥2,
𝜕2 𝑧
𝜕𝑦2,
𝜕2 𝑧
𝜕𝑥𝜕𝑦
are denoted by p,q,r,t,s
respectively.
 The ORDER of a PDE is the order of the highest derivatives in
the equation.
 The DEGREE of a PDE is the power of the highest order partial
derivative present in the equation.

8.  Partial differential equations can be formed using the
following methods.
 1) By elimination of arbitrary constants in the equation of
the type f(x, y, z, a, b)=0 where a and b are arbitrary
constants.
 2) By elimination of arbitrary functions in the equation of
the type z = f(u) where u is a function of x, y and z

9. Consider the function f(x, y, z, a, b) = 0 ..(i)
where a and b are two independent arbitrary constants. To
eliminate two constants, we require two more equations.
Differentiating eq. (i) partially with respect to x and y in turn, we
obtain
𝜕𝑓
𝜕𝑥
.
𝜕𝑥
𝜕𝑥
+
𝜕𝑓
𝜕𝑧
.
𝜕𝑧
𝜕𝑥
= 0 ⇒
𝜕𝑓
𝜕𝑥
+
𝜕𝑓
𝜕𝑧
. 𝑝 = 0 ……(ii)
𝜕𝑓
𝜕𝑦
.
𝜕𝑦
𝜕𝑦
+
𝜕𝑓
𝜕𝑧
.
𝜕𝑧
𝜕𝑦
= 0 ⇒
𝜕𝑓
𝜕𝑦
+
𝜕𝑓
𝜕𝑧
. 𝑞 = 0 ……(iii)
Eliminating a and b from the set of equations (i), (ii) and (iii) , we
get a
PDE of the first order of the form F(x, y, z, p, q) = 0. .. (iv)

10. Consider a relation between x, y and z of the type (u; v) = 0 ..(v)
where u and v are known functions of x, y and z and is an
arbitrary function of u and v. Differentiating equ. (v) partially
with respect to x and y, respectively, we get
𝜕𝜑
𝜕𝑢
𝜕𝑢
𝜕𝑥
+
𝜕𝑢
𝜕𝑧
𝑝 +
𝜕𝜑
𝜕𝑣
𝜕𝑣
𝜕𝑥
+
𝜕𝑣
𝜕𝑧
𝑝 = 0 …….(vi)
𝜕𝜑
𝜕𝑢
𝜕𝑢
𝜕𝑦
+
𝜕𝑢
𝜕𝑧
𝑞 +
𝜕𝜑
𝜕𝑣
𝜕𝑣
𝜕𝑦
+
𝜕𝑣
𝜕𝑧
𝑞 = 0 .…..(vii)

11. Eliminating
𝜕𝜑
𝜕𝑢
and
𝜕𝜑
𝜕𝑣
from the eq. (vi) and (vii), we get the
equations
𝜕 𝑢,𝑣
𝜕(𝑦,𝑧)
𝑝 +
𝜕 𝑢,𝑣
𝜕(𝑧,𝑥)
𝑞 =
𝜕 𝑢,𝑣
𝜕(𝑥,𝑦)
…..(viii)
which is a PDE of the type (iv).
since the power of p and q are both unity it is also linear
equation,
whereas eq. (iv) need not be linear.

12. partial differential equation of first order is said to be linear if the dependent variable and its
derivatives are degree of one and the products of the dependent variable and its derivatives do
not appear in the equation .
The equation is said to be quasi-linear if the degree of the highest order derivative is one and
the products of the highest order partial derivatives are not present. A quasi-linear partial
differential equation is represented as
P(x,y,z)*P+Q(X,Y,Z)*q=R(x,y,z)
This equation is known as lagrange’s linear equation.

13.  Working rule
 1. write the given differential equation in the standard form
Pp+Qq=R
 2. Form the lagrange’s auxiliary equation
 d𝑥/P=dy/Q=R (1.1)
3.Solve the simultaneous equation in Eq. to obtain its two
independent solutions u=c1 , v=c2
 4.Write the general solution of the given equation as,
f(u,v) = 0.

14. A homogeneous linear partial differential equation of the nth order is of the form
homogeneous because all its terms contain derivatives of the same order.

15. Equation (1) can be expressed as
As in the case of ordinary linear equations with constant coefficients the complete solution of (1)
consists of two parts, namely, the complementary function and the particular integral.
The complementary function is the complete solution of f (D,D') z = 0-------
(3), which must contain n arbitrary functions as the degree of the polynomial
f(D,D'). The particular integral is the particular solution of equation (2).

16. Finding the complementary function
Let us now consider the equation f(D,D') z = F (x,y)
The auxiliary equation of (3) is obtained by replacing D by m and D' by 1.
Solving equation (4) for „m‟, we get „n‟ roots. Depending upon the nature of the roots, the Complementary
function is written as given below:

17. Finding the particular Integral

19. Expand [f (D,D')]-1 in ascending powers of D or D' and operate on xm yn term by term.
Case (iv) : When F(x,y) is any function of x and y.
into partial fractions considering f (D,D') as a function of D alone. Then operate each
partial fraction on F(x,y) in such a way that
where c is replaced by y+mx after integration

20. Non Homogeneous Linear PDES
If in the equation )1...()........,(),( yxFzDDf 
the polynomial expression is not hom
ogeneous, then (1) is a non- homogeneous linear
partial differential equation
),( DDf 
Complete Solution
= Complementary Function + Particular Integral
To find C.F., factorize
into factors of the form
),( DDf 
)( cDmD 
yx
ezDDDD 3222
)43( 
Ex

21. If the non homogeneous equation is of the form
)()(.
),())((
21
2211
21
xmyexmyeFC
yxFzcDmDcDmD
xcxc



1.Solve
22
)( xzDDDD 
Solution
)1(),( 2
 DDDDDDDDDf
)()(. 21 yxyeFCx
 

22. 
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
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
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 




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



6.5.125.4.34.3123
1
......
)1()1(1
)1(
1
1
.
65443
2
2
2
2
22
2
2
1
22
2
xxxxx
x
D
x
D
D
x
D
D
x
D
x
D
D
DDDDD
x
IP

23. 2.Solve 4)32)(1(  zDDDD
Solution
3
4
)2()( 1
3
1  xyexyez xx


24. Linear Second Order P.D.E :
Examples (Classification)
Hyperbolicis
ACBCBcA
t
txu
x
txu
c
Parabolicis
ACBCBA
t
txu
x
txu
EquationWave
041,0,0
0
),(),(
EquationWave
______________________________________
EquationHeat
040,0,
0
),(),(
EquationHeat
22
2
2
2
2
2
2
2
2



















25. • Consider an elastic string stretched to a length l along
the x-axis with its two fixed ends at x=0 and x=l.
• Assumption made:
i. The string is homogeneous with constant density ρ.
ii. The string is perfectly elastic and offers no
resistance to bending.
iii. The tension in the string is so large that the force
due to weight of the string can be neglected.
• Consider the motion of the small portion PQ of length
δx of the string. Since the string produces no
resistance to bending, the tension T₁ and T₂ at points P
and Q will act tangentially at P and Q respectively.
• Assuming that the point on the string move only in the
vertical direction, there is no motion in the horizontal
direction.
• Hence, the sum of the forces in the horizontal
direction must be zero.

26. • −𝑇₁cos 𝛼 + 𝑇₂ sin 𝛽 = 0
• 𝑇₁cos 𝛼 = 𝑇₂ sin 𝛽 = 𝑇(constant ),say ………..(1)
• The forces acting vertically on the string are the vertical components of tension at points P and Q.
Thus, the resultant vertical force acting on PQ is 𝑇₂ sin 𝛽- 𝑇₁ sin 𝛼.
• By newtons second law of motion,
• Resultant Force=Mass×Accelaration
• −𝑇₁sin 𝛼 + 𝑇₂ sin 𝛽 = 𝜌𝛿𝑥
𝜕2 𝑦
𝜕𝑡2 ………(2)
• Dividing by T,
•
𝑇₂ sin 𝛽
T
−
𝑇₁sin 𝛼
T
= 𝑃𝛿𝑥
𝜕2 𝑦
𝜕𝑡2
•
𝑇₂ sin 𝛽
𝑇₂ cos 𝛽
−
𝑇₁sin 𝛼
𝑇₁cos 𝛼
= 𝑃𝛿𝑥
𝜕2 𝑦
𝜕𝑡2
• tan 𝛽 − tan 𝛼 =
𝑃𝛿𝑥
𝑇
𝜕2 𝑦
𝜕𝑡2 ……….(3)
• Since tan 𝛼 and tan 𝛽 are the slopes of the curve at point P and Q respectively.
• tan 𝛼 =
𝜕𝑦
𝜕𝑥 𝑃
=
𝜕𝑦
𝜕𝑥 𝑥
• tan 𝛽 =
𝜕𝑦
𝜕𝑥 𝑄
=
𝜕𝑦
𝜕𝑥 𝑥+𝛿𝑥

27. • Substitute in Eq.(3)
•
𝜕𝑦
𝜕𝑥 𝑥+𝛿𝑥
−
𝜕𝑦
𝜕𝑥 𝑥
=
𝑃𝛿𝑥
𝑇
𝜕2 𝑦
𝜕𝑡2
• Dividing by 𝛿𝑥 and taking limit 𝛿𝑥 → 0,
• lim
𝛿𝑥→0
𝜕𝑦
𝜕𝑥 𝑥+𝛿𝑥
−
𝜕𝑦
𝜕𝑥 𝑥
𝛿𝑥
=
𝑃𝛿𝑥
𝑇
𝜕2 𝑦
𝜕𝑡2
•
𝜕2 𝑦
𝜕𝑥2 =
1
𝑐2
𝜕2 𝑦
𝜕𝑡2 where 𝑐2
=
𝑇
𝑃
•
𝜕2 𝑦
𝜕𝑡2 = 𝑐2 𝜕2 𝑦
𝜕𝑥2
• This equation is known as one dimensional wave equation.

28. Solution of One –Dimensional Wave Equation
• The one dimensional wave equation is
•
𝜕2 𝑦
𝜕𝑡2 = 𝑐2 𝜕2 𝑦
𝜕𝑥2 ………………(4)
• Let 𝑦 = 𝑋 𝑥 𝑇 𝑡 be a solution of Eq.(4)
•
𝜕2 𝑦
𝜕𝑡2 = 𝑋𝑇′′ ,
𝜕2 𝑦
𝜕𝑥2 = 𝑋′′ 𝑇
• Substituting in Eq.(4)
• 𝑋𝑇′′ = 𝑐2 𝑋′′ 𝑇
•
𝑋``
𝑋
=
1
𝑐2
𝑇``
𝑇
• Since 𝑋 and 𝑇 are only the functions of x and t respectively, this equation
holds good if each term is constant.
• Let
𝑋``
𝑋
=
1
𝑐2
𝑇``
𝑇
= 𝑘,say
• Considering
𝑋``
𝑋
= 𝑘,
𝑑2 𝑋
𝑑𝑥2 − 𝑘𝑋 = 0 …….(5)

29. • Considering
1
𝑐2
𝑇``
𝑇
= 𝑘,
𝑑2 𝑇
𝑑𝑡2 − 𝑘𝑐2
𝑇 = 0 ………(6)
• Solving Equations (5) & (6), the following cases arise:
I. When k is positive
• Let 𝑘 = 𝑚2
•
𝑑2 𝑋
𝑑𝑥2 − 𝑚2 𝑋 = 0 and
𝑑2 𝑇
𝑑𝑡2 − 𝑚2 𝑐2 𝑇 = 0
• 𝑋 = 𝑐1 𝑒 𝑚𝑥
+ 𝑐2 𝑒−𝑚𝑥
and T = 𝑐3 𝑒 𝑚𝑐𝑡
+ 𝑐4 𝑒−𝑚𝑐𝑡
• Hence , the solution of Eq.(4) is
• 𝑦 = (𝑐1 𝑒 𝑚𝑥 + 𝑐2 𝑒−𝑚𝑥)(𝑐3 𝑒 𝑚𝑐𝑡 + 𝑐4 𝑒−𝑚𝑐𝑡) ……(7)
• II. When k is negative
• Let 𝑘 = −𝑚2
• 𝑋 = 𝑐1cos 𝑚𝑥 + 𝑐2 sin 𝑚𝑥 and T = 𝑐3cos 𝑚𝑐𝑡 + 𝑐4 sin 𝑚𝑐𝑡
Hence the solution of Eq.(4) is
𝑦 = 𝑐1cos 𝑚𝑥 + 𝑐2 sin 𝑚𝑥 𝑐3cos 𝑚𝑐𝑡 + 𝑐4 sin 𝑚𝑐𝑡 ……(8)

30. III. When k=0
𝑋 = 𝑐1 𝑥 + 𝑐2 and T = 𝑐3 𝑡 + 𝑐4
Hence, the solution of Eq.(4) is
• 𝑦 = (𝑐1 𝑥 + 𝑐2)(𝑐3 𝑡 + 𝑐4) ……(9)
• Out of these three solution , we need to choose that solution which is consistent
with the physical nature of the problem. Since we are dealing with problems on
vibrations, y must be a periodic function of x and t. Thus , the solution must involve
trigonometric terms.
• Hence, the solution of the form given by Eq.(8)
• 𝑦 = 𝑐1cos 𝑚𝑥 + 𝑐2 sin 𝑚𝑥 𝑐3cos 𝑚𝑐𝑡 + 𝑐4 sin 𝑚𝑐𝑡

31. (
1
)
(
2
)
(
3
)
(
4
)
(
5
)
d'Alembert's Solution
The method of d'Alembert provides a solution to the one-dimensional wave equation
that models vibrations of a string.
The general solution can be obtained by introducing new variables and ,
and applying the chain rule to
Using (4) and (5) to compute the left and right sides of (3) then gives