Classification of PDE

(BMG2CC1B)

Order and degree of Partial Differential Equations (PDEs)

Partial Differential Equation (PDE) : An equation containing one or more partial derivatives of an unknown function of two or more independent variables is known as a Partial Differential Equation.

Order of a PDE : The order of a PDE is defined as the order of the highest partial derivative occurring in the PDE.

Degree of a PDE : The of a PDE is the degree of the highest order derivative which occurs in it after the equation has been rationalized.

Examples :

𝜕𝑥 𝜕𝑦

(i) ^𝜕𝑧 + ^𝜕𝑧 = 𝑧 + 𝑥𝑦 (1^st order & 1^st degree PDE)

𝜕𝑧

𝜕𝑥

(ii) ( )

𝜕𝑦

3

𝜕𝑧

𝜕𝑥

² 3

+ ^{𝜕 𝑧} = 2𝑥 ( )

(3^rd order & 1^st degree PDE)

𝜕𝑧

(iii) 𝑧 ( ) +

𝜕𝑥 𝜕𝑦

𝜕𝑧 _{= 𝑥}

(1^st order & 1^st degree PDE)

𝜕𝑥 𝜕𝑦 𝜕𝑧

(iv) ^𝜕𝑢 + ^𝜕𝑢 + ^𝜕𝑢 = 𝑥𝑦𝑧

(1^st order & 1^st degree PDE)

(v)

𝜕²𝑧

𝜕𝑥²

= (1 +

𝜕𝑦

1

𝜕𝑧₎2

(2^nd order & 2^nd degree PDE)

𝜕𝑧

𝜕𝑧 𝜕𝑧

𝜕𝑥 𝜕𝑦 𝜕𝑦

(vi) 𝑦 {( ) ( ) } = 𝑧 ( )

2 2

+ (1^st order & 2^nd degree PDE)

Linear and Non-linear PDEs :

A PDE is said to be linear if the dependent variable and its partial derivatives occur only in the first degree and are not multiplied, otherwise it is said to be non-linear.

Examples :

𝜕𝑥 𝜕𝑦

(i) ^𝜕𝑧 + ^𝜕𝑧 = 𝑧 + 𝑥𝑦 (Linear PDE)

𝜕𝑧

𝜕𝑥

(ii) ( )

𝜕𝑦

3

𝜕𝑧

𝜕𝑥

² 3

+ ^{𝜕 𝑧} = 2𝑥 ( )

(Non-linear PDE)

𝜕𝑧

(iii) 𝑧 ( ) +

𝜕𝑥 𝜕𝑦

𝜕𝑧 _{= 𝑥}

(Non-linear PDE)

𝜕𝑥 𝜕𝑦 𝜕𝑧

(iv) ^𝜕𝑢 + ^𝜕𝑢 + ^𝜕𝑢 = 𝑥𝑦𝑧

(Linear PDE)

(v)

𝜕²𝑧

𝜕𝑥²

= (1 +

𝜕𝑦

1

𝜕𝑧₎2

(Non-linear PDE)

𝜕𝑧

𝜕𝑥

𝜕𝑧

𝜕𝑦

2 2

+

𝜕𝑧

𝜕𝑦

(vi) 𝑦 {( ) ( ) } = 𝑧 ( )

(Non-linear PDE)

Notations : (i) When we consider two independent variables 𝑥 and 𝑦 and one dependent variable 𝑧. Then, we use the following notations in the PDEs.

𝜕²𝑧

2

^𝜕𝑧 = 𝑝, ^𝜕𝑧 = 𝑞, ^{𝜕 𝑧} = 𝑟, = 𝑠,

𝜕𝑥 𝜕𝑦 𝜕𝑥² 𝜕𝑥𝜕𝑦 𝜕𝑦²

𝜕²𝑧 _{= 𝑡}

(ii) When we consider 𝑛 independent variables 𝑥₁, 𝑥₂, 𝑥₃, … … , 𝑥_𝑛 and one dependent variable 𝑧. Then, we use the following notations in the PDEs.

𝜕𝑧 𝜕𝑧

= 𝑝 , = 𝑝 ,

𝜕𝑧

𝜕𝑥₁ 𝜕𝑥₂ 𝜕𝑥₃

1 2 3

= 𝑝 , …………………,

𝜕𝑧

𝜕𝑥_𝑛

= 𝑝

𝑛

(iii) Partial differentiations are also denoted as :

𝜕𝑢

_𝜕𝑥 = 𝑢_𝑥,

𝜕𝑢

_𝜕𝑦 = 𝑢_𝑦,

𝜕²𝑢

_𝜕𝑥2 = 𝑢_𝑥𝑥,

𝜕²𝑢

_𝜕𝑦2 = 𝑢_𝑦𝑦,

𝜕²𝑢

𝜕𝑥𝜕𝑦 ^{= 𝑢𝑥𝑦}

Classification of First Order PDEs :

(i) Linear PDE : A first order equation 𝑓⁽𝑥, 𝑦, 𝑧, 𝑝, 𝑞⁾ = 0 is said to be Linear PDE if it is linear in 𝑝, 𝑞 𝑎𝑛𝑑 𝑧, that is, if the given equation is of the form

𝑃⁽𝑥, 𝑦⁾𝑝 + 𝑄⁽𝑥, 𝑦⁾𝑞 = 𝑅⁽𝑥, 𝑦⁾𝑧 + 𝑆⁽𝑥, 𝑦⁾.

Example : (i) 𝑥²𝑦𝑝 + 𝑥𝑦²𝑞 = 𝑥𝑦𝑧 + 𝑥²𝑦²

(ii) 𝑝 + 𝑞 = 𝑧 + 𝑥𝑦

(iii) 𝑥𝑝 + 𝑦𝑞 = 𝑥²𝑦²𝑧 + ^𝑥

𝑦

(ii) Semi-linear PDE : A first order equation 𝑓⁽𝑥, 𝑦, 𝑧, 𝑝, 𝑞⁾ = 0 is said to be Semi-linear PDE if it is linear in 𝑝 𝑎𝑛𝑑 𝑞, the co-efficients 𝑝 𝑎𝑛𝑑 𝑞 are functions

of 𝑥 𝑎𝑛𝑑 𝑦 only, that is, if the given equation is of the form 𝑃⁽𝑥, 𝑦⁾𝑝 + 𝑄⁽𝑥, 𝑦⁾𝑞 =

𝑅⁽𝑥, 𝑦, 𝑧⁾.

Example : (i) 𝑥²𝑦𝑝 + 𝑥𝑦²𝑞 = 𝑥²𝑦²𝑧²

2

(ii) 𝑦𝑝 + 𝑥𝑞 = ^𝑧

𝑥𝑦

2 2

(iii) 𝑥𝑝 + 𝑦𝑞 = ^{𝑥 𝑦}

_𝑧2

3. Quasi-linear PDE : A first order equation 𝑓⁽𝑥, 𝑦, 𝑧, 𝑝, 𝑞⁾ = 0 is said to be Quasi-linear PDE if it is linear in 𝑝 𝑎𝑛𝑑 𝑞, that is, if the given equation is of the form 𝑃⁽𝑥, 𝑦, 𝑧⁾𝑝 + 𝑄⁽𝑥, 𝑦, 𝑧⁾𝑞 = 𝑅⁽𝑥, 𝑦, 𝑧⁾.

Example : (i) 𝑥²𝑧𝑝 + 𝑧𝑦²𝑞 = 𝑥𝑦

(ii) ⁽𝑥² − 𝑦𝑧⁾𝑝 + ⁽𝑦² − 𝑧𝑥⁾𝑞 = ⁽𝑧² − 𝑥𝑦⁾

(iii) 𝑥𝑦²𝑧𝑝 + 𝑥²𝑦𝑧𝑞 = ⁽𝑥²𝑦²𝑧² − 1⁾

4. Non-linear PDE : A first order equation 𝑓⁽𝑥, 𝑦, 𝑧, 𝑝, 𝑞⁾ = 0 is said to be Non- linear PDE if it is not linear in 𝑝 𝑎𝑛𝑑 𝑞.

Example : (i) 𝑝² + 𝑞² = 1

(ii) 𝑝𝑞 = 𝑧

(iii) 𝑥²𝑝² + 𝑦²𝑞² = 𝑧²

Formation of first order partial differential equations

Rule I : Derivation of PDE by eliminating constants

Example 1. Find a PDE by eliminating a and b from

𝑧 = 𝑎𝑥 + 𝑏𝑦 + 𝑎² + 𝑏²

Solution : Given, 𝑧 = 𝑎𝑥 + 𝑏𝑦 + 𝑎² + 𝑏² ………… (i) Differentiating (i) partially with respect to x, we get

𝜕𝑧 _{= 𝑎}

𝜕𝑥

​

⇒ 𝑝 = 𝑎 ……………. (ii)

Differentiating (i) partially with respect to y, we get

𝜕𝑧 _{= 𝑏}

𝜕𝑦

⇒ 𝑞 = 𝑏 ……………. (iii)

Now, putting 𝑎 = 𝑝 and 𝑏 = 𝑞 in (i), we get

𝑧 = 𝑝𝑥 + 𝑞𝑦 + 𝑝² + 𝑞², which is the required PDE.

Answer

EXERCISE 1 (A)

1. 𝒛 = 𝑨 𝒆^𝒑𝒕 𝐬𝐢𝐧 𝒑𝒙 ⁽𝒑 𝒂𝒏𝒅 𝑨⁾

Solution : Given, 𝑧 = 𝐴 𝑒^𝑝𝑡 sin 𝑝𝑥 ………… (i) Differentiating (i) partially with respect to x, we get

𝜕𝑥

^𝜕𝑧 = 𝐴 𝑒^𝑝𝑡 cos 𝑝𝑥 . 𝑝

𝜕𝑥

⇒ ^𝜕𝑧 = 𝐴𝑝𝑒^𝑝𝑡 cos 𝑝𝑥 ………………. (ii)

Again, differentiating (ii) partially with respect to x, we get

𝜕𝑥²

^𝜕2𝑧 = 𝐴𝑝𝑒^𝑝𝑡(−sin 𝑝𝑥⁾. 𝑝

𝜕𝑥²

2

⇒ ^{𝜕 𝑧} = −𝐴𝑝²𝑒^𝑝𝑡𝑠𝑖𝑛𝑝𝑥 ………….. (iii)

Differentiating (i) partially with respect to t, we get

𝜕𝑡

^𝜕𝑧 = 𝐴 𝑝𝑒^𝑝𝑡 𝑠𝑖𝑛 𝑝𝑥 …………………. (iv)

Again, differentiating (iv) partially with respect to t, we get

𝜕𝑡²

^𝜕2𝑧 = 𝐴𝑝²𝑒^𝑝𝑡𝑠𝑖𝑛𝑝𝑥 ………………….(v)

Adding (iii) & (v), we get

𝜕𝑥² 𝜕𝑡²

^𝜕2𝑧 + ^𝜕2𝑧 = 0, which is the required PDE.

Answer

2. 𝒛 = 𝑨 𝒆^{−𝒑𝟐𝒕} 𝐜𝐨𝐬 𝒑𝒙 ⁽𝒑 𝒂𝒏𝒅 𝑨⁾

Solution : Given, 𝑧 = 𝐴 𝑒^−𝑝2𝑡 cos 𝑝𝑥 ………… (i) Differentiating (i) partially with respect to x, we get

𝜕𝑥

^𝜕𝑧 = −𝐴 𝑝𝑒^−𝑝2𝑡 sin 𝑝𝑥 …………… (ii)

Again, differentiating (i) partially with respect to x, we get

𝜕𝑥²

^𝜕2𝑧 = −𝐴 𝑝²𝑒^−𝑝2𝑡 cos 𝑝𝑥 ……………(iii)

Differentiating (i) partially with respect to t, we get

𝜕𝑡

^𝜕𝑧 = −𝐴 𝑝²𝑒^−𝑝2𝑡 cos 𝑝𝑥 ……………(iv)

From (iii) & (iv), we get,

𝜕𝑥² 𝜕𝑡

^𝜕2𝑧 = ^𝜕𝑧, which is the required PDE.

Answer

3. 𝒛 = 𝒂𝒙^𝟑 + 𝒃𝒚^𝟑 ; ( 𝒂, 𝒃 )

Solution : Given equation is,

𝑧 = 𝑎𝑥³ + 𝑏𝑦³ ……………. (i)

Differentiating (i) partially with respect to x, we get

𝜕𝑥

^𝜕𝑧 = 3𝑎𝑥² ………………. (ii)

Differentiating (i) partially with respect to y, we get

𝜕𝑦

^𝜕𝑧 = 3𝑏𝑦² ………………. (iii)

From (ii) and (iii), we get

𝜕𝑧 𝜕𝑧

𝑥 + 𝑦 = 3𝑎𝑥³ + 3𝑏𝑦³

𝜕𝑥 𝜕𝑦

​

⇒ 𝑥 ^𝜕𝑧 + 𝑦 ^𝜕𝑧 = 3⁽𝑎𝑥³ + 𝑏𝑦³⁾

𝜕𝑥 𝜕𝑦

⇒ 𝑥 ^𝜕𝑧 + 𝑦 ^𝜕𝑧 = 3𝑧 ^[𝐹𝑟𝑜𝑚 (𝑖)^]

𝜕𝑥 𝜕𝑦

​

which is the required PDE.

Answer

𝒂

𝟐

4. 𝟒𝒛 = [𝒂𝒙 + ^𝒚 + 𝒃] ; ⁽𝒂, 𝒃⁾

Solution : Given equation is,

𝑦

𝑎

2

4𝑧 = [𝑎𝑥 + + 𝑏] ……………. (i)

Differentiating (i) partially with respect to x, we get

𝜕𝑧 𝑦

4 _𝜕𝑥 = 2 (𝑎𝑥 + _𝑎 + 𝑏) . 𝑎

⇒ ^𝜕𝑧 = ¹.𝑎. (𝑎𝑥 + ^𝑦 + 𝑏) ………….. (ii)

𝜕𝑥 2 𝑎

​

Differentiating (i) partially with respect to y, we get

₄ 𝜕𝑧

𝜕𝑦

𝑦

𝑎

= 2. (𝑎𝑥 + + 𝑏) .

1

𝑎

_⇒ 𝜕𝑧

1 1 𝑦

𝜕𝑦 2 𝑎 𝑎

= . (𝑎𝑥 + + 𝑏) ……………..(iii)

Multiplying (ii) by (iii), we get

𝜕𝑧 𝜕𝑧 1 𝑦

𝜕𝑥 𝜕𝑦 4 𝑎

. = . [𝑎𝑥 + + 𝑏]

2

_⇒ 𝜕𝑧 _. 𝜕𝑧 1

𝜕𝑥 𝜕𝑦 4

= . 4𝑧 ^[𝐹𝑟𝑜𝑚 (𝑖)^]

𝜕𝑥 𝜕𝑦

⇒ ^𝜕𝑧 . ^𝜕𝑧 = 𝑧, which is the required PDE.

Answer

5. 𝒛 = 𝒂𝒙^𝟐 + 𝒃𝒙𝒚 + 𝒄𝒚^𝟐, ⁽𝒂, 𝒃, 𝒄⁾

Solution : Given equation is,

𝑧 = 𝑎𝑥² + 𝑏𝑥𝑦 + 𝑐𝑦² …………………….. (i) Differentiating (i) partially with respect to x, we get

𝜕𝑥

^𝜕𝑧 = 2𝑎𝑥 + 𝑏𝑦 …………………… (ii)

Differentiating (ii) partially with respect to x, we get

​

^𝜕2𝑧 = 2𝑎 ……………….. (iii)

𝜕𝑥²

Differentiating (ii) partially with respect to y, we get

𝜕 𝜕𝑧

𝜕𝑦 𝜕𝑥

( ) = 𝑏

⇒

𝜕²𝑧

𝜕𝑥𝜕𝑦

= 𝑏 ………………………(iv)

Differentiating (i) partially with respect to y, we get

𝜕𝑦

^𝜕𝑧 = 𝑏𝑥 + 2𝑐𝑦 …………………… (v)

Differentiating (v) partially with respect to y, we get

​

^𝜕2𝑧 = 2𝑐 …………………………(vi)

𝜕𝑦²

​

From (iii), (iv) and (vi), we get

𝑥² + 2𝑥𝑦

𝜕²𝑧 𝜕²𝑧

𝜕𝑥² 𝜕𝑥𝜕𝑦

𝜕²𝑧

𝜕𝑦²

+ 𝑦² = 2𝑎𝑥² + 2𝑏𝑥𝑦 + 2𝑐𝑦²

2 2

⇒ 𝑥^{2 𝜕 𝑧} + 2𝑥𝑦 ^{𝜕 𝑧}

𝜕𝑥² 𝜕𝑥𝜕𝑦

2

_{+ 𝑦2} 𝜕 𝑧

𝜕𝑦²

= 2⁽𝑎𝑥² + 𝑏𝑥𝑦 + 𝑐𝑦²⁾

​

= 2𝑧 ^[𝐹𝑟𝑜𝑚 (𝑖)^]

2 2

⇒ 𝑥^{2 𝜕 𝑧} + 2𝑥𝑦 ^{𝜕 𝑧}

𝜕𝑥² 𝜕𝑥𝜕𝑦

2

_{+ 𝑦2} 𝜕 𝑧

𝜕𝑦²

which is the required PDE.

Answer

6. 𝒛^𝟐 = 𝒂𝒙^𝟑 + 𝒃𝒚^𝟑 + 𝒂𝒃, ⁽𝒂, 𝒃⁾

Solution : Given equation is,

𝑧² = 𝑎𝑥³ + 𝑏𝑦³ + 𝑎𝑏 …………….. (i) Differentiating (i) partially with respect to x, we get

𝜕𝑥

2𝑧 ^𝜕𝑧 = 3𝑎𝑥² …………………… (ii)

Differentiating (i) partially with respect to y, we get

𝜕𝑦

2𝑧 ^𝜕𝑧 = 3𝑏𝑦² …………………… (iii)

𝜕𝑧 𝜕𝑧

𝜕𝑥 𝜕𝑦

(ii) x (iii) ⇒ 4𝑧² ( ) ( ) = 9𝑎𝑏𝑥²𝑦² ………………….. (iv)

From (ii) and (iii), we get

6𝑥³𝑦²𝑧 ^𝜕𝑧 + 6𝑥²𝑦³𝑧 ^𝜕𝑧 = 9𝑎𝑥⁵𝑦² + 9𝑏𝑥²𝑦⁵ …………… (v)

𝜕𝑥 𝜕𝑦

​

Adding (iv) and (v), we get

​

6𝑥³𝑦²𝑧 ^𝜕𝑧 + 6𝑥²𝑦³𝑧 ^𝜕𝑧 + 4𝑧² (^𝜕𝑧) (^𝜕𝑧) = 9𝑎𝑥⁵𝑦² + 9𝑏𝑥²𝑦⁵ + 9𝑎𝑏𝑥²𝑦²

𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

𝜕𝑧 𝜕𝑧

𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

⇒ 6𝑥³𝑦²𝑧 ^𝜕𝑧 + 6𝑥²𝑦³𝑧 ^𝜕𝑧 + 4𝑧² ( ) ( ) = 9𝑥²𝑦²⁽𝑎𝑥³ + 𝑏𝑦³ + 𝑎𝑏⁾

𝜕𝑧

𝜕𝑧

𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

⇒ 𝑧 [6𝑥³𝑦^{2 𝜕𝑧} + 6𝑥²𝑦^{3 𝜕𝑧} + 4𝑧 ( ) ( )] = 9𝑥²𝑦²𝑧²

^[𝐹𝑟𝑜𝑚 (𝑖)^]

𝜕𝑧

𝜕𝑧

𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

⇒ 6𝑥³𝑦^{2 𝜕𝑧} + 6𝑥²𝑦^{3 𝜕𝑧} + 4𝑧 ( ) ( ) = 9𝑥²𝑦²𝑧, which is required PDE.

Answer