PARTIAL DIFFERENTIAL EQUATIONS Example #1.

Solve the differential equation

𝜕𝜕2 𝑢𝑢
= 6𝑥𝑥 2 (2𝑦𝑦 − 1) given the boundary conditions that at
At the end of this chapter, the student should be able to: 𝜕𝜕𝑥𝑥 2
𝜕𝜕𝜕𝜕
1. Recognize some important engineering partial differential 𝑥𝑥 = 0, = sin 2𝑦𝑦 and 𝑢𝑢 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐.
𝜕𝜕𝜕𝜕
equations. 𝜕𝜕2 𝑢𝑢
Solution: Since = 6𝑥𝑥 2 (2𝑦𝑦 − 1) then integrating partially
2. Solve a partial differential equation by direct partial 𝜕𝜕𝑥𝑥 2

integration with respect to x gives:

𝜕𝜕𝜕𝜕
3. Solve differential equations by separating the variables. = � 6𝑥𝑥 2 (2𝑦𝑦 − 1)𝑑𝑑𝑑𝑑 = (2𝑦𝑦 − 1) � 6𝑥𝑥 2 𝑑𝑑𝑑𝑑
𝜕𝜕2 𝑢𝑢 1 𝜕𝜕2 𝑢𝑢 𝜕𝜕𝜕𝜕
4. Solve the wave equation = 6𝑥𝑥 3
𝜕𝜕𝑥𝑥 2 𝑐𝑐 2 𝜕𝜕𝑡𝑡 2
𝜕𝜕2 𝑢𝑢 1 𝜕𝜕𝑢𝑢 = (2𝑦𝑦 − 1) + 𝑓𝑓(𝑦𝑦)
5. Solve the heat conduction equation = 3
𝜕𝜕𝑥𝑥 2 𝑐𝑐 2 𝜕𝜕𝑡𝑡 3 (2𝑦𝑦
𝜕𝜕2 𝑢𝑢 𝜕𝜕2 𝑢𝑢
= 2𝑥𝑥 − 1) + 𝑓𝑓(𝑦𝑦)
6. Solve Laplace’s equation + =0 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑓𝑓(𝑦𝑦)𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓.
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2
𝜕𝜕𝜕𝜕
From the boundary condition when x = 0, = sin 2𝑦𝑦
𝜕𝜕𝜕𝜕
Definition: A partial differential equation is an equation that
contains one or more partial derivatives. Examples include: Hence; sin 2𝑦𝑦 = 2(0)3 sin(2𝑦𝑦 − 1) + 𝑓𝑓(𝑦𝑦)
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
(i.) 𝑎𝑎 + 𝑏𝑏 = 𝑐𝑐 From which 𝑓𝑓(𝑦𝑦) = sin 2𝑦𝑦
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑢𝑢 1 𝜕𝜕𝑢𝑢
(ii.) = (known as Heat Conduction Equation) 𝜕𝜕𝜕𝜕
𝜕𝜕𝑥𝑥 2 𝑐𝑐 2 𝜕𝜕𝑡𝑡 Now = 2𝑥𝑥 3 (2𝑦𝑦 − 1) + 𝑓𝑓(𝑦𝑦)
𝜕𝜕2 𝑢𝑢 𝜕𝜕2 𝑢𝑢 𝜕𝜕𝜕𝜕
(iii.) + =0 (known as Laplace Equation)
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2
Integrating partially with respect to x gives:
Equation (i) is a first-order partial differential equation, and
𝑢𝑢 = �[2𝑥𝑥 3 (2𝑦𝑦 − 1) + sin 2𝑦𝑦] 𝑑𝑑𝑑𝑑
equations (ii) and (iii) are second-order partial differential
equations since the highest power of the differential is 2. 2𝑥𝑥 4
= (2𝑦𝑦 − 1) + 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 2𝑦𝑦 + 𝐹𝐹(𝑦𝑦)
4
To be able to solve simple partial differential equations
knowledge of the following is required: From the boundary condition when 𝑥𝑥 = 0, 𝑢𝑢 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ; hence;
a. Partial Integration 04
cos(𝑦𝑦) = (2𝑦𝑦 − 1) + (0)𝑠𝑠𝑠𝑠𝑠𝑠2𝑦𝑦 + 𝐹𝐹(𝑦𝑦)
b. First and Second-Order Partial Differential Equation 2
c. The solution of Ordinary Differential Equations
From which 𝐹𝐹(𝑦𝑦) = cos 𝑦𝑦
A. Partial Integration
Integration is the reverse process of differentiation 𝜕𝜕2 𝑢𝑢
Hence; the solution of = 6𝑥𝑥 2 (2𝑦𝑦 − 1) for the given
𝜕𝜕𝜕𝜕 𝜕𝜕𝑥𝑥 2
For example: = 5 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is integrated partially with boundary condition is:
𝜕𝜕𝜕𝜕
respect to t , then the 5cosx term is considered as constant, and; 𝑥𝑥 4
𝑢𝑢 = (2𝑦𝑦 − 1) + 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑢𝑢 = � 5𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑 = 5𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 � 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑑𝑑 2

= 5𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐(−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) + 𝑐𝑐 Example #2. Solve the differential equation:

= −5𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑡𝑡 + 𝑓𝑓(𝑥𝑥) 𝜕𝜕2 𝑢𝑢 𝜕𝜕𝜕𝜕
= 𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥 + 𝑦𝑦) given that = 2 𝑤𝑤ℎ𝑒𝑒𝑒𝑒 𝑦𝑦 = 0 and 𝑢𝑢 = 𝑦𝑦 2
𝜕𝜕𝑥𝑥𝑥𝑥𝑥𝑥 𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑢𝑢 when 𝑥𝑥 = 0.
Similarly, if = 6𝑥𝑥 2 𝑐𝑐𝑐𝑐𝑐𝑐2𝑦𝑦 is integrated partially with 𝜕𝜕2 𝑢𝑢
𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕
Solution: Since = 𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥 + 𝑦𝑦) then integrating partially
respect to y, then: 𝜕𝜕𝑥𝑥𝑥𝑥𝑥𝑥
𝜕𝜕𝜕𝜕 with respect to y gives:
= � 6𝑥𝑥 2 𝑐𝑐𝑐𝑐𝑐𝑐2𝑦𝑦𝑦𝑦𝑦𝑦 = (6𝑥𝑥 2 ) � 𝑐𝑐𝑐𝑐𝑐𝑐2𝑦𝑦𝑦𝑦𝑦𝑦 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕 = � 𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥 + 𝑦𝑦)dy = sin(𝑥𝑥 + 𝑦𝑦) + 𝑓𝑓(𝑥𝑥)
1 𝜕𝜕𝜕𝜕
= (6𝑥𝑥 2 ) � 𝑠𝑠𝑠𝑠𝑠𝑠2𝑦𝑦� + 𝑓𝑓(𝑥𝑥) 𝜕𝜕𝜕𝜕
2 From the boundary condition when = 2, y = 0
𝜕𝜕𝜕𝜕
= 3𝑥𝑥 2 𝑠𝑠𝑠𝑠𝑠𝑠2𝑦𝑦 + 𝑓𝑓(𝑥𝑥)
𝜕𝜕𝜕𝜕
and integrating
𝜕𝜕𝜕𝜕
partially with respect to x gives: Hence; 2 = sin 𝑥𝑥 + 𝑓𝑓(𝑥𝑥)
From which 𝑓𝑓(𝑥𝑥) = 2 − sin 𝑥𝑥
𝑢𝑢 = �[3𝑥𝑥 2 𝑠𝑠𝑠𝑠𝑠𝑠2𝑦𝑦 + 𝑓𝑓(𝑥𝑥)]𝑑𝑑𝑑𝑑 𝜕𝜕𝑢𝑢
Then = sin(𝑥𝑥 + 𝑦𝑦) + 2 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
2 𝜕𝜕𝜕𝜕
= 𝑥𝑥 𝑠𝑠𝑠𝑠𝑠𝑠2𝑦𝑦 + (𝑥𝑥)𝑓𝑓(𝑥𝑥) + 𝑔𝑔(𝑦𝑦)
Integrating partially with respect to x gives:
f (x) and g(y) are functions that may be determined if extra
information, called boundary conditions or initial conditions, 𝑢𝑢 = �[ sin(𝑥𝑥 + 𝑦𝑦) + 2 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠]𝑑𝑑𝑑𝑑
are known. = −𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥 + 𝑦𝑦) + 2𝑥𝑥 + +𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑓𝑓(𝑦𝑦)
From the boundary conditions, 𝑢𝑢 = 𝑦𝑦 2 , when 𝑥𝑥 = 0 ; hence
A.1 Solution of Partial Differential Equation by Direct Partial
𝑦𝑦 2 = −𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 0 + 𝑐𝑐𝑐𝑐𝑐𝑐(0) + 𝑓𝑓(𝑦𝑦)
Integration
= 1 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑓𝑓(𝑦𝑦)
The simplest form of partial differential equations occurs when
From which 𝑓𝑓(𝑦𝑦) = 𝑦𝑦 2 − 1 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
a solution can be determined by direct partial integration. 𝜕𝜕2 𝑢𝑢
Hence, the solution of = 𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥 + 𝑦𝑦) is given by:
𝜕𝜕𝑥𝑥𝑥𝑥𝑥𝑥
𝑢𝑢 = −𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥 + 𝑦𝑦) + 2𝑥𝑥 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑦𝑦 2 − 1 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
 1 𝜕𝜕2 𝑢𝑢 𝜕𝜕𝑢𝑢
Example #3. Verify that ∅(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = satisfy the 7. Solve = sin(𝑥𝑥 + 𝑡𝑡) given that = 1 when 𝑡𝑡 = 0, and
�𝑥𝑥 2 +𝑦𝑦 2 +𝑧𝑧 2 𝜕𝜕𝑥𝑥𝜕𝜕𝑡𝑡 𝜕𝜕𝑥𝑥
𝜕𝜕2 ∅ 𝜕𝜕2 ∅ 𝜕𝜕2 ∅ when 𝑢𝑢 = 2𝑡𝑡 when 𝑥𝑥 = 0.
differential equation: + + =0 𝑥𝑥
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2 𝜕𝜕𝑧𝑧 2
8. Show that 𝑢𝑢(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥𝑥𝑥 + is a solution of
Solution: The partial differential equation: 𝑦𝑦
𝜕𝜕2 ∅ 𝜕𝜕2 ∅ 𝜕𝜕2 ∅ 𝜕𝜕2 𝑢𝑢 𝜕𝜕2 𝑢𝑢
+ + = 0 is called a Laplace’s Equation. 2𝑥𝑥 − 𝑦𝑦 = 2𝑥𝑥 .
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2 𝜕𝜕𝑧𝑧 2 𝜕𝜕𝑥𝑥𝜕𝜕𝑦𝑦 𝜕𝜕𝑦𝑦 2
9. Find the particular solution of the differential equation
1 𝜕𝜕2 𝑢𝑢
If ∅(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) =
1
= (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )−2 ; = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 given that the initial conditions that
�𝑥𝑥 2 +𝑦𝑦 2 +𝑧𝑧 2 𝜕𝜕𝑥𝑥𝜕𝜕𝑦𝑦
𝜕𝜕𝑢𝑢
Then differentiating partially with respect to x gives: when 𝑦𝑦 = 𝜋𝜋, = 𝑥𝑥, and when 𝑥𝑥 = 𝜋𝜋, 𝑢𝑢 = 2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐.
𝜕𝜕𝑥𝑥
10. Verify that the ∅(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 + 𝑒𝑒 𝑥𝑥 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 satisfies the
𝜕𝜕∅ 1 3 𝜕𝜕2 ∅ 𝜕𝜕2 ∅
= − (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )−2 (2𝑥𝑥) differential equation
𝜕𝜕𝑥𝑥 2
+
𝜕𝜕𝑦𝑦 2
+ 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 = 0
𝜕𝜕𝑥𝑥 2
3
= −𝑥𝑥(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )−2
Answers:
And;
𝜕𝜕 2 ∅ 3 5
2
= (−𝑥𝑥) [− (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )−2 (2𝑥𝑥)]
𝜕𝜕𝑥𝑥 2
3
+ (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )−2 (−1)
by the product rule
3𝑥𝑥 2 1
= 5− 3
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2 (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2
(3𝑥𝑥 2 ) − (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )
= 5
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2
Similarly, it may be shown that

𝜕𝜕 2 ∅ (3𝑦𝑦 2 ) − (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )
= 5
𝜕𝜕𝑦𝑦 2
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2
and;
𝜕𝜕 2 ∅ (3𝑧𝑧 2 ) − (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )
= 5
𝜕𝜕𝑧𝑧 2
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2
Thus;
𝜕𝜕 2 ∅ 𝜕𝜕 2 ∅ 𝜕𝜕 2 ∅
+ +
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2 𝜕𝜕𝑧𝑧 2
(3𝑥𝑥 2 ) − (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )
= 5
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2
(3𝑦𝑦 ) − (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )
2
+ 5
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2
(3𝑧𝑧 ) − (𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )
2
+ 5
(𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 )2
(3𝑥𝑥 2 )−�𝑥𝑥 2 +𝑦𝑦 2 +𝑧𝑧 2 �+(3𝑦𝑦 2 )−(𝑥𝑥 2 +𝑦𝑦 2 +𝑧𝑧 2 )+(3𝑧𝑧 2 )−(𝑥𝑥 2 +𝑦𝑦 2 +𝑧𝑧 2 )
= 5 =0
(𝑥𝑥 2 +𝑦𝑦 2 +𝑧𝑧 2 )2
1
Thus; satisfies the Laplace Equation
�𝑥𝑥 2 +𝑦𝑦 2 +𝑧𝑧 2
𝜕𝜕2 ∅ 𝜕𝜕2 ∅ 𝜕𝜕2 ∅
+ + =0
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2 𝜕𝜕𝑧𝑧 2

Practice Exercises: Verify the solution of the following

problems. Write your solution/answer in clean long bond paper.
𝜕𝜕𝑢𝑢
1. Determine the general solution of = 4𝑡𝑡𝑡𝑡.
𝜕𝜕𝑦𝑦
𝜕𝜕𝑢𝑢
2. Solve = 2𝑡𝑡𝑡𝑡𝑜𝑜𝑜𝑜𝜃𝜃 given that 𝑢𝑢 = 2𝑡𝑡 when 𝜃𝜃 = 0.
𝜕𝜕𝑡𝑡
3. Verify that 𝑢𝑢(𝜃𝜃, 𝑡𝑡) = 𝜃𝜃 2 + 𝜃𝜃𝑡𝑡 is a solution of
𝜕𝜕𝑢𝑢 𝜕𝜕𝑢𝑢
−2 = 𝑡𝑡
𝜕𝜕𝜃𝜃 𝜕𝜕𝑡𝑡
4. Verify that 𝑢𝑢 = 𝑒𝑒 −𝑦𝑦 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is a solution of
𝜕𝜕2 𝑢𝑢 𝜕𝜕2 𝑢𝑢
+ =0
𝜕𝜕𝑥𝑥 2 𝜕𝜕𝑦𝑦 2
𝜕𝜕2 𝑢𝑢
5. Solve =8𝑒𝑒 𝑦𝑦 𝑠𝑠𝑠𝑠𝑠𝑠2𝑥𝑥 given that at 𝑦𝑦 = 0,
𝜕𝜕𝑥𝑥𝜕𝜕𝑦𝑦
𝜕𝜕𝑢𝑢 𝜋𝜋
= 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, and at 𝑥𝑥 = , 𝑢𝑢 = 2𝑦𝑦 2 .
𝜕𝜕𝑥𝑥 2
𝜕𝜕2 𝑢𝑢 2
6. Solve = 𝑦𝑦(4𝑥𝑥 − 1) given that at 𝑥𝑥 = 0, 𝑢𝑢 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝜕𝜕𝑥𝑥 2
𝜕𝜕𝑢𝑢
and = 𝑐𝑐𝑐𝑐𝑐𝑐2𝑦𝑦
𝜕𝜕𝑥𝑥