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Contents
- MATH 421: PARTIAL DIFFERENTIAL E QUATIONS
- Purpose of the Module
- Module Learning Outcomes..........................................................................................................................
- Course Description........................................................................................................................................
- LESSON
- CONCEPTS OF PARTIAL DIFFERENTIAL EQUATIONS
- LESSON
- Formation of Partial Differential Equations................................................................................................
- LESSON THREE.............................................................................................................................................
- SOLUTION OF FIRST ORDER QUASILINEAR PDEs BY THE LAGRANGE’S METHOD OF TYPE ONE AND TWO
- LESSON FOUR..............................................................................................................................................
- FOUR SOLUTION OF FIRST ORDER QUASILINEAR PDEs BY THE LAGRANGE’S METHOD OF TYPES THREE AND
- LESSON FIVE
- SYSTEM OF SURFACES................................................................................................................................. INTEGRAL SURFACES PASSING THROUGH GIVEN CURVES AND SURFACES ORTHOGONAL TO A GIVEN
- LESSON SIX
- THE METHOD OF CANONICAL FORMS........................................................................................................
- LESSON SEVEN
- THE METHOD OF SEPARATION OF VARIABLES
- LESSON EIGHT
- Non-linear Partial Differential Equations of Order One..............................................................................
- LESSON NINE...............................................................................................................................................
- NON-LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF TYPES THREE AND FOUR........................................
- LESSON TEN.................................................................................................................................................
- SOLUTION OF NON-LINEAR PDEs USING TRANSFORMATIONS
- LESSON ELEVEN...........................................................................................................................................
- CHARPIT’S METHOD OF SOLVING FIRST ORDER NON-LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Purpose of the Module
The aim of this course is to introduce the students to the theory of first order partial differential
equations (PDE) and give competence in solving them via analytical methods.
Module Learning Outcomes
By the end of this module the learner will be able to:
i) Classify any first order PDE into linear, semi-linear, quasilinear and nonlinearPDE;
ii) Form PDE by eliminating arbitrary constants and arbitrary functions in given relations;
iii) SolvePDE in Lagrange’s form using auxiliary methods;
iv) Solve first order PDE through the use of characteristics;
v) Find a specific solution of a first order Cauchy problem;
vi) Identify first order non-linearPDE as belonging to some special types and hence solve
them;
vii) Use Charpit’s and Jacobi methods to solve first order non-linearPDEs.
Course Description
Partial differential equations of the first order, first degree. Solutions using Lagrange’s system of
linear equations. Linear, semi-linear and quasi-linear partial differential equations of the first
order. Integral surface passing through a given curve (Cauchy problem). Use of the methods of
Cauchy, Charpit and Jacobi in solving non-linear partial differential equations of the first order.
𝑥𝑦
𝑥
𝑥𝑥
𝑥𝑦
𝑦𝑦
𝑥
2
𝑦
2
𝑥𝑥
𝑦𝑦
are partial differential equations. The functions 𝑢
3
and 𝑢
= sin
are solutions of (∗), as can easily be verified.
Definition 2
The order of a partial differential equation is the order of the highest-ordered partial derivative
appearing in the equation. For example, considering 𝑢 as the dependent variable and 𝑥, 𝑦 as
independent variables,
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
= 𝑢 or 𝑥𝑝 + 𝑦𝑞 = 𝑢 is of order one and
𝜕
2
𝑢
𝜕𝑥
2
𝜕
2
𝑢
𝜕𝑥𝜕𝑦
𝜕
2
𝑢
𝜕𝑦
2
= 0 or 𝑟 + 3 𝑠 + 𝑡 = 0 is of order two.
We have used the standard notation: 𝑝 =
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
𝜕
2
𝑢
𝜕𝑥
2
𝜕
2
𝑢
𝜕𝑥𝜕𝑦
𝜕
2
𝑢
𝜕𝑦
2
The degree of a PDE is the degree of the highest order derivative present in the PDE after
clearing the fractional powers.
For example
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
𝜕𝑢
𝜕𝑡
2
𝜕
2
𝑢
𝜕𝑥
2
𝜕
2
𝑢
𝜕𝑦
2
𝜕
2
𝑢
𝜕𝑥𝜕𝑦
𝜕𝑢
𝜕𝑦
3
4
Equations (1) and (2) are of first degree whereas equation (3) is of 4
th
degree.
Classification of first order PDEs
Partial differential equations can be classified in at least three ways. They are
- Order of the PDE
- Linear, Semi-linear, Quasi-linear and fully non-linear
- Scalar equation, system of equations
NOTE
a) A PDE of order 𝑚 is called quasi-linear if it is linear in the derivative of order 𝑚 with
coefficients that depend on the independent variables and derivatives of the unknown
function of order strictly less than 𝑚.
b) A quasilinear PDE where the coefficients of the derivatives of order 𝑚 are functions of
the independent variables alone is called a semi-linear PDE.
c) A PDE which is linear in the unknown function and all its derivatives with coefficients
depending on the independent variables alone is called a linear PDE.
d) A PDE which is not quasi-linear is called a fully non-linear PDE.
We have the following picture
Linear PDE⋤ Semi-linear PDE ⋤ Quasi-linear PDE⋤ Fully non-linear PDE
Remark
a) A single first order quasi-linear PDE must be of the form
𝑥
𝑦
b) A single quasi-linear PDE where 𝑎, 𝑏 are functions of 𝑥 and 𝑦 alone is a semi-linear PDE.
c) A single semi-linear PDE where 𝑐
0
1
(𝑥, 𝑦) is a linear PDE.
d) Linear PDEs can further be classified into two: homogeneous and nonhomogeneous.
Every linear PDE can be written in the form 𝐿[𝑢] = 𝑓, where 𝑢 → 𝐿[𝑢] is a linear map,
and 𝑓 is a function of independent variables only. This linear PDE is said to be
homogeneous if 𝑓 is the zero function; otherwise it is called a non-homogeneous linear
PDE.
Examples
i. 𝑥𝑢
𝑥
𝑦
− 2 𝑢 = 0 is a linear homogenous PDE
ii. 𝑦𝑢
𝑥
𝑦
− 𝑥𝑦 = 0 is a linear non-homogeneous PDE
iii. 𝑥𝑢
𝑥
𝑦
2
2
= 0 is a semi-linear PDE
iv.
2
2
𝑥
𝑦
= 𝑥𝑢 is a quasilinear
v. 𝑢
𝑥
2
𝑦
= 𝑥𝑦 is a fully non-linear PDE
vi. {
𝑥
𝑦
𝑥
𝑦
is a system of first order PDEs
E-tivity1.2: Concepts of partial differential equations
Numbering and Pacing and
Sequencing
Title Partial Differential Equations
Purpose To introduce you to the language and concepts of partial
differential equations.
Brief summary of overall task Watch the PDEs by MATHITUPCANADA
and then attempt the given questions.
Spark
Solutions of PDEs may look like
this.
Individual contribution
- Watch the video on PDEs
- Classify the following PDEs
a) 𝑥𝑢
𝑥
𝑦
2
b) 𝑢𝑢
𝑥
2
𝑦
3
c) 𝑢
𝑥
𝑦
Interaction begins
- Post your answers on the discussion forum 1.2.
- Read what your colleagues have posted.
- In a sentence or two, comment on what your
colleagues have posted.
E-moderator interventions
- Focussing group discussion
- Encouraging quiet ones to contribute
- Providing feedback/ teaching points
- Summarising key points
- Closing the discussion
Schedule and time This activity should take one hour
Next Formation of PDEs
1.4 Assessment
- We know the following classification of PDEs
Linear PDE⋤ Semi-linear PDE ⋤ Quasi-linear PDE⋤ Fully non-linear PDE
Each of the above inclusions is a strict inclusion. Justify the statement by giving
examples.
- Give at least three examples of fifth order PDE belonging to each of the above (i.e
exer.1.) classes.
- Classify the following equations by all the three ways of classification.
i. (
𝜕𝑢
𝜕𝑦
2
𝜕
2
𝑢
𝜕𝑥
2
ii. 𝑠𝑖𝑛 ( 1 +
𝜕𝑢
𝜕𝑥
2
3
iii.
𝜕
2
𝑢
𝜕𝑥
2
𝜕
2
𝑢
𝜕𝑦
2
iv. 𝑒
𝜕
2
𝑢
𝜕𝑥
2
𝜕
2
𝑢
𝜕𝑦
2
v.
𝜕
2
𝑢
𝜕𝑡
2
𝜕
2
𝑢
𝜕𝑥
2
1.5 References
- Tyn Myint-U and Lokenath Debnath (1992). Linear Partial Differential Equations for
Scientists and Engineers, 4
th
Ed. Birkhauser publishers.
- Frank, Ayres (1952). Theory and problems of Differential Equations. McGRAW-HILL
Publishers.
In general, the arbitrary constants may be eliminated from
yielding a partial
differential equation of order one 𝑓(𝑥, 𝑦, 𝑢, 𝑝, 𝑞) = 0.
Example 2.
Form a PDE by eliminating the arbitrary constants 𝑎 and 𝑏 from
i. 𝑢 = 𝑎𝑥 + 𝑏𝑦 + 𝑎𝑏
ii. 𝑢 = 𝑎𝑥 + 𝑎
2
2
iii. 𝑢 = (𝑥
2
2
iv. 𝑢 = (𝑥 − 𝑎)
2
2
v. 𝑢 = 𝑎𝑥 + ( 1 − 𝑎)𝑦 + 𝑏
Solutions
i. 𝑝 =
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
Substituting for 𝑎 and 𝑏 in 𝑢 = 𝑎𝑥 + 𝑏𝑦 + 𝑎𝑏 we get 𝑢 = 𝑝𝑥 + 𝑞𝑦 + 𝑝𝑞, a linear PDE
of first order.
ii. 𝑝 =
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
2
𝑦 Eliminating 𝑎 from these results, we get
2
2
𝑦 which is the required non-linear PDE of first order.
iii. 𝑝 =
𝜕𝑢
𝜕𝑥
2
𝜕𝑢
𝜕𝑦
2
2
𝑝
2 𝑥
and
2
𝑞
2 𝑦
. Substituting these in 𝑢 = (𝑥
2
2
𝑝
2 𝑥
𝑞
2 𝑦
𝑝𝑞
4 𝑥𝑦
or 4 𝑥𝑦 = 𝑝𝑞, a non-linear first order PDE.
iv. 𝑝 =
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
𝑝
2
𝑞
2
𝑝
2
4
𝑞
2
4
or 4 𝑢 = 𝑝
2
2
, a non-linear first order PDE
v. 𝑝 =
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
Example 2.
Form a PDE by eliminating the arbitrary constants 𝑎, 𝑏 and 𝑐 from the relation
𝑥
2
𝑎
2
𝑦
2
𝑏
2
𝑢
2
𝑐
2
Solution
Note that 𝑎, 𝑏 and 𝑐 are arbitrary constants and 𝑢 is the dependent variable, depending
on 𝑥 and 𝑦. We can write the given relation as 𝑓(𝑥, 𝑦, 𝑢) = (
𝑥
2
𝑎
2
𝑦
2
𝑏
2
𝑢
2
𝑐
2
Then differentiating ( 1 ) partially w.r.t 𝑥 and 𝑦 respectively, we have
𝜕𝑓
𝜕𝑥
𝜕𝑓
𝜕𝑢
𝜕𝑢
𝜕𝑥
= 0 and
𝜕𝑓
𝜕𝑦
𝜕𝑓
𝜕𝑢
𝜕𝑢
𝜕𝑦
2 𝑥
𝑎
2
2 𝑢
𝑐
2
𝜕𝑢
𝜕𝑥
= 0 and
2 𝑦
𝑏
2
2 𝑢
𝑐
2
𝜕𝑢
𝜕𝑦
= 0 or 𝑐
2
2
and 𝑐
2
2
Differentiating ( 2 ) w.r.t 𝑥 we get 𝑐
2
2
𝜕𝑢
𝜕𝑥
2
2
𝜕
2
𝑢
𝜕𝑥
2
On substituting
𝑐
2
𝑎
2
𝑢
𝑥
𝜕𝑢
𝜕𝑥
from ( 2 ) in the above equation, we get
2
2
2
Or 𝑥𝑢
𝜕
2
𝑢
𝜕𝑥
2
𝜕𝑢
𝜕𝑥
2
𝜕𝑢
𝜕𝑥
Similarly differentiating ( 3 ) w.r.t y we get 𝑐
2
2
𝜕𝑢
𝜕𝑦
2
2
𝜕
2
𝑢
𝜕𝑦
2
= 0 and
substituting
𝑐
2
𝑏
2
𝑢
𝑦
𝜕𝑢
𝜕𝑦
from ( 3 ) in the above equation, we get
𝜕
2
𝑢
𝜕𝑦
2
𝜕𝑢
𝜕𝑦
2
𝜕𝑢
𝜕𝑦
Thus equations ( 4 ) and ( 5 ) are partial differential equations of first degree and second order.
Formation of PDEs by elimination of arbitrary functions
Let 𝑈 = 𝑈(𝑥, 𝑦, 𝑢) and 𝑉 = 𝑉(𝑥, 𝑦, 𝑢) be the dependent variables of the variables 𝑥, 𝑦, 𝑢 and 𝑢 is
a function of 𝑥 and 𝑦. Let
be an arbitrary relation between them. Partially differentiating w.r.t 𝑥 and 𝑦, we obtain
𝜕𝜑
𝜕𝑈
𝜕𝑈
𝜕𝑥
𝜕𝑈
𝜕𝑢
𝜕𝜑
𝜕𝑉
𝜕𝑉
𝜕𝑥
𝜕𝑉
𝜕𝑢
And
𝜕𝜑
𝜕𝑈
𝜕𝑈
𝜕𝑦
𝜕𝑈
𝜕𝑢
𝜕𝜑
𝜕𝑉
𝜕𝑉
𝜕𝑦
𝜕𝑉
𝜕𝑢
Eliminating
𝜕𝜑
𝜕𝑈
and
𝜕𝜑
𝜕𝑉
from
and
, we have
𝜕𝑈
𝜕𝑥
𝜕𝑈
𝜕𝑢
𝜕𝑉
𝜕𝑥
𝜕𝑉
𝜕𝑢
𝜕𝑈
𝜕𝑦
𝜕𝑈
𝜕𝑢
𝜕𝑉
𝜕𝑦
𝜕𝑉
𝜕𝑢
𝜕𝑈
𝜕𝑥
𝜕𝑈
𝜕𝑢
𝜕𝑉
𝜕𝑦
𝜕𝑉
𝜕𝑢
𝜕𝑈
𝜕𝑦
𝜕𝑈
𝜕𝑢
𝜕𝑉
𝜕𝑥
𝜕𝑉
𝜕𝑢
E-tivity 2.2: Formation of PDEs
Numbering and Pacing and
Sequencing
Title Formation of PDEs
Purpose To expose you to formation of PDEs through
elimination of arbitrary constants and arbitrary
functions..
Brief summary of overall task Watch the video formation of PDEs by MATHSOLVES
ZONE
and then attempt the questions given.
Spark
FORMATION OF PDEs
Individual contribution • Watch the video on formation of PDEs
a) 𝑧 = 𝑓(𝑥 − 𝑦)
b) (𝑥 − 𝑎)
2
2
2
Interaction begins
- Post your answers on the discussion forum 2.2.
- Read what your colleagues have posted.
- In a sentence or two, comment on what two of
your colleagues have posted.
E-moderator interventions • Focussing group discussion
- Encouragingquiet ones to contribute
- Providing feedback/ teaching points
- Summarising key points
- Closing the discussion
Schedule and time This activity should take one hour
Next Solution of quasilinear PDEs
2. 4 Assessment
- Form PDEs by eliminating arbitrary constants from the following relations
(note that 𝑧 = 𝑧(𝑥, 𝑦))
a) 𝑧 = 𝑎𝑥 + 𝑏𝑦 + 𝑎
2
2
b) 𝑧 = 𝑎𝑥𝑦 + 𝑏
c) 𝑧 = 𝑎
2
d)
2
2
2
e) 𝑧 = 𝑥𝑦 + 𝑦√𝑥
2
2
f) 𝑧 = 𝑎𝑒
−𝑏
2
𝑡
g) 𝑎𝑥
2
2
2
- Form PDEs by eliminating arbitrary functions
a) 𝑓
2
2
2
b) 𝑓
2
c) 𝑧 = 𝑓(
𝑦
𝑥
2. 5 References
- Tyn Myin-U and Lokeath Debnath (1992). Linear Partial Differential Equations for
Scientists and Engineers, 4
th
Ed. Birkhauser publishers.
- Frank Ayres (1952). Shaum’s outline of theory and problems of differential equations.
McGRAW-HILL publishers
Step 2. Write down Lagrange’s auxiliary system, namely;
𝑑𝑥
𝑃
𝑑𝑦
𝑄
𝑑𝑢
𝑅
Step 3. Solve ( 2 ) to get 𝑈(𝑥, 𝑦, 𝑢) = 𝑐
1
and 𝑉(𝑥, 𝑦, 𝑢) = 𝑐
2
taken as two independent solutions
of ( 2 ).
Step 4. The general solution (or integral of ( 1 ) is then written in one of the following three
equivalent forms:
𝜙(𝑈, 𝑉) = 0 , 𝑈 = 𝜙(𝑉) or 𝑉 = 𝜙(𝑈).
Complete solutions. If 𝑈 = 𝑎 and 𝑉 = 𝑏 are two independent solutions of
𝑑𝑥
𝑃
𝑑𝑦
𝑄
𝑑𝑢
𝑅
and if
𝛼 and 𝛽 are arbitrary constants, 𝑈 = 𝛼𝑉 + 𝛽 is called the complete solution of 𝑃𝑝 + 𝑄𝑞 = 𝑅.
Type 1. The PDE whose auxiliary system,
𝒅𝒙
𝑷
𝒅𝒚
𝑸
𝒅𝒖
𝑹
is such that one of the variables is
either absent or cancels out from any two fractions of the given equations is said to be of type 1.
The general solution can be obtained by grouping two fractions.
Example 3.
Solve 𝑦𝑢𝑝 + 𝑥𝑢𝑞 = 𝑥𝑦
Solution
Here 𝑃 = 𝑦𝑢, 𝑄 = 𝑥𝑢 and 𝑅 = 𝑥𝑦
The Lagrange’s auxiliary system for the given PDE are
𝑑𝑥
𝑦𝑢
𝑑𝑦
𝑥𝑢
𝑑𝑢
𝑥𝑦
Grouping the first and second fractions, we have
𝑑𝑥
𝑦𝑢
𝑑𝑦
𝑥𝑢
⇒𝑥𝑑𝑥 − 𝑦𝑑𝑦 = 0 , now by integrating
term by term we get
𝑥
2
2
𝑦
2
2
= 𝑎 or 𝑥
2
2
1
. Similarly grouping the first and third
fractions, we have
𝑑𝑥
𝑦𝑢
𝑑𝑢
𝑥𝑦
⇒𝑥𝑑𝑥 − 𝑢𝑑𝑢 = 0. Integrating term by term we get
2
2
2
. Hence the required general solution is 𝜙
2
2
2
2
Example 3.
Solve 𝑦
2
Solution
Here, 𝑃 = 𝑦
2
Lagrange’s system are
𝑑𝑥
𝑦
2
𝑑𝑦
−𝑥𝑦
𝑑𝑢
𝑥(𝑢− 2 𝑦)
. Grouping the first two fractions, we have
2
2
1
. Now grouping second and third fractions, we have
𝑑𝑦
−𝑥𝑦
𝑑𝑢
𝑥(𝑢− 2 𝑦)
𝑑𝑢
𝑑𝑦
(𝑢− 2 𝑦)
𝑦
𝑑𝑢
𝑑𝑦
𝑢
𝑦
= 2 which is linear in 𝑢 and 𝑦. Its integrating
factor is 𝑒
𝑙𝑛𝑦
= 𝑦. Hence 𝑦
𝑑𝑢
𝑑𝑦
𝑑
𝑑𝑦
2
2
2
2
. Hence the required general solution is 𝜙(𝑥
2
2
2
Type 2. Suppose one integral is known by the method of type 1 and suppose also that
another integral cannot be obtained by using the method of type 1. Then if one integral known to
us is used to find another integral, then the corresponding PDE is said to be of type 2.
Example 3.
Solve
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
Solution
The Lagrange auxiliary equations are
Grouping the first two fractions:
𝑑𝑥
1
𝑑𝑦
1
1
so that
1
Grouping the last two fractions:
𝑑𝑦
1
𝑑𝑢
𝑥+𝑦+𝑢
𝑑𝑦
1
𝑑𝑢
2 𝑦+𝐶 1
+𝑢
or
𝑑𝑢
𝑑𝑦
1
𝑑𝑢
𝑑𝑦
1
which is linear in 𝑢 and 𝑦. Its integrating factor is 𝑒
−𝑦
and hence
−𝑦
𝑑𝑢
𝑑𝑦
1
−𝑦
𝑑
𝑑𝑦
−𝑦
1
−𝑦
−𝑦
1
−𝑦
−𝑦
2
by chain rule.
−𝑦
2
and hence the required general solution is
ϕ(𝑥 + 𝑦, 𝑒
−𝑦
Example 3.
Solve 𝑥𝑢
2
2
4
3. 4 Assessment
Find the general solution of each of the following using the appropriate method
a) 𝑢𝑢
𝑥
𝑦
b)
2
𝑥
𝑦
c)
𝑦
2
𝑢
𝑥
𝑥
𝑦
2
d) 𝑥𝑦𝑝 + 𝑦
2
2
e) 𝑝 + 3 𝑞 = 5 𝑢 + tan (𝑦 − 3 𝑥)
f) 𝑝 − 2 𝑞 = 3 𝑥
2
sin (𝑦 + 2 𝑥)
3. 5 References
- Tyn Myin-U and Lokeath Debnath (1992). Linear Partial Differential Equations for
Scientists and Engineers, 4
th
Ed. Birkhauser publishers.
- Frank Ayres (1952). Shaum’s outline of theory and problems of differential equations.
your colleagues have posted.
E-moderator interventions
- Focussing group discussion
- Encouraging quiet ones to contribute
- Providing feedback/ teaching points
- Summarising key points
- Closing the discussion
Schedule and time This activity should take 1 hour
Next end
LESSON FOUR
SOLUTION OF FIRST ORDER QUASILINEAR PDEs BY THE LAGRANGE’S
METHOD OF TYPES THREE AND FOUR
4.1 Introduction
In this lesson we will study two other methods of solutions of quasilinear PDES. We will
specifically consider the method of solutions of quasilinear PDEs of types 3 and 4.
4 .2 Learning Outcomes
By the end of this lesson the learner will be able to:
4.2.1 Solve quasilinear PDEs using the Lagrange’s method, type 3;
4.2.2 Solve quasilinear PDEs using the Lagrange’s method, type 4
- 3 Quasilinear PDEs of Type 3 (Existence of multipliers 𝑷 𝟏
𝟏
and 𝑹
𝟏
such that
𝟏
𝟏
𝟏
The class of quasilinear PDEs 𝑃𝑝 + 𝑄𝑞 = 𝑅, such that there exist 𝑃
1
1
and 𝑅
1
so that
1
1
1
𝑅 = 0 , is said to be of type 3.
Method of solution.
Given the PDE 𝑃𝑝 + 𝑄𝑞 = 𝑅, the auxiliary system is
𝑑𝑥
𝑃
𝑑𝑦
𝑄
𝑑𝑢
𝑅
and if
1
1
1
𝑅 = 0 then 𝑃
1
1
1
𝑑𝑢 = 0 which can be integrated to give
1
1
. This method may be repeated to get another integral 𝑈
2
2
1
1
and 𝑅
1
are called multipliers. As a special case, these can be constants also.
Sometimes only one integral is possible by use of multipliers. In such cases second integral
should be obtained by using type 1 or type 2 methods as the case may be.
Example 4.
Solve
2
2
