Complete derivation of 2 Dimensional Heat equation or the Laplace equation.

1. MATHEMATICS FOR TECHNOLOGY
Department of Chemical & Biotechnology

2.  Consider the flow of heat in a metal plate ABCD in X-Y
plane.
 The temperature depends upon the 𝑥, 𝑦 and 𝑡 (time).
Given by function 𝑢 𝑥, 𝑡
 Let ∆𝑥 𝑎𝑛𝑑 ∆y be the sides of rectangle and ℎ be the
thickness of plate.
Assumptions:
1. The temperature function 𝑢 𝑥, 𝑡 is constant on
each cross section of rod at each time.
2. There is no heat generation within the metal plate.
A B
D C
O
Y
X
Δx
Δy

3. Rate of heat flow by Fourier law of heat conduction is
given as:
Where,
𝑘 is thermal conductivity of material
𝐴 is cross sectional area of surface.
𝜕 𝑢
𝜕 𝑥
is temperature at distance x at time t
x, y x+Δx, y
x, y+ Δy
Δx
Δy
A B
D C
O
Y
X
x+Δx +, y+ Δy
𝑄 = −𝑘 × 𝐴 ×
𝜕 𝑢
𝜕 𝑥

4.  Heat energy entering the plate from side AB is given as
 Heat energy leaving the plate from side CD is given as
x, y x+Δx, y
x, y+ Δy
Δx
Δy
A B
D C
O
Y
X
x+Δx +, y+ Δy
Q1
Q1
’
𝑄1 = −𝑘 × 𝐴 ×
𝛿 𝑢
𝛿 𝑦 𝑦
𝑄1 = −𝑘 × ∆𝑥 × ℎ ×
𝛿 𝑢
𝛿 𝑦 𝑦
𝑄1
′
= −𝑘 × 𝐴 ×
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
𝑄1
′
= −𝑘 × ∆𝑥 × ℎ ×
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦

5.  Similarly, Heat energy entering the plate from side AD is given as
 Heat energy leaving the plate from side CD is given as
x, y x+Δx, y
x, y+ Δy
Δx
Δy
A B
D C
O
Y
X
x+Δx +, y+ Δy
Q2
’
Q2
𝑄2 = −𝑘 × 𝐴 ×
𝛿 𝑢
𝛿 𝑥 𝑥
𝑄2 = −𝑘 × ∆𝑦 × ℎ ×
𝛿 𝑢
𝛿 𝑥 𝑥
𝑄2
′
= −𝑘 × 𝐴 ×
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
𝑄2
′
= −𝑘 × ∆𝑦 × ℎ ×
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥

6.  Now, the net gain of heat energy by the plate will be:
Using above equations
Substituting the values of 𝑄1, 𝑄2, 𝑄1
′
, 𝑄2
′
, we have
𝑁𝑒𝑡 𝐺𝑎𝑖𝑛 𝑄 = 𝑄1 + 𝑄2 − 𝑄1
′
+ 𝑄2
′
Q = −𝑘 × ∆𝑥 × ℎ ×
𝛿 𝑢
𝛿 𝑦 𝑦
− 𝑘 × ∆𝑥 × ℎ ×
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
−
𝑘 × ∆𝑦 × ℎ ×
𝛿 𝑢
𝛿 𝑥 𝑥
− 𝑘 × ∆𝑦 × ℎ ×
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
Q = 𝑘 × ℎ ∆𝑦
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
−
𝛿 𝑢
𝛿 𝑥 𝑥
+ ∆𝑥
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
−
𝛿 𝑢
𝛿 𝑦 𝑦

7.  Multiply and divide by ∆𝑥 and ∆y for both terms
 Taking common terms, we get,
Q = 𝑘 × ℎ ∆𝑥∆𝑦
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
−
𝛿 𝑢
𝛿 𝑥 𝑥
∆𝑥
+ ∆𝑥∆𝑦
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
−
𝛿 𝑢
𝛿 𝑦 𝑦
∆𝑦
Q = 𝑘 × ℎ × ∆𝑥∆𝑦
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
−
𝛿 𝑢
𝛿 𝑥 𝑥
∆𝑥
+
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
−
𝛿 𝑢
𝛿 𝑦 𝑦
∆𝑦

8. We know that, Heat energy gained is given by:
Where,
𝑚 is mass,
𝐶𝑝 is the specific heat
∆𝑡 is temperature gradient
In our case, above heat gain equation will become,
𝑄 = 𝑚 × 𝐶𝑝 × ∆𝑡
𝑄𝑔 = 𝑚 × 𝐶𝑝 ×
𝜕𝑢
𝜕𝑡
𝑄𝑔 = 𝑉𝑜𝑙𝑢𝑚𝑒 × 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 × 𝐶𝑝 ×
𝜕𝑢
𝜕𝑡
𝑄𝑔 = (∆𝑥 × ∆𝑦 × ℎ) × 𝜌 × 𝐶𝑝 ×
𝜕𝑢
𝜕𝑡

9. Where,
∆𝑥 × ∆𝑦 × ℎ represents volume of plate and
𝜌 be the density of material of plate.
 As the equations 𝑄𝑔 and Q both represents the heat energy gained by the system
Hence,
Now evaluating for above equation, we have values of Q and 𝑄𝑔
𝑄𝑔 = ∆𝑥 × ∆𝑦 × ℎ × 𝜌 × 𝐶𝑝 ×
𝜕𝑢
𝜕𝑡
𝑄𝑔 = Q

10.  Cancelling the common terms we have,
𝑄𝑔 = Q
∆𝑥 × ∆𝑦 × ℎ × 𝜌 × 𝐶𝑝 ×
𝜕𝑢
𝜕𝑡
=
𝑘 × ℎ × ∆𝑥∆𝑦
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
−
𝛿 𝑢
𝛿 𝑥 𝑥
∆𝑥
+
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
−
𝛿 𝑢
𝛿 𝑦 𝑦
∆𝑦
𝜌 × 𝐶𝑝 ×
𝜕𝑢
𝜕𝑡
=
𝑘
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
−
𝛿 𝑢
𝛿 𝑥 𝑥
∆𝑥
+
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
−
𝛿 𝑢
𝛿 𝑦 𝑦
∆𝑦

11.  Taking limits,
 Solving the limits, we get
 Let the constant term,
𝑘
𝜌×𝐶𝑝
be represented as 𝑐2
𝜕𝑢
𝜕𝑡
=
𝑘
𝜌 × 𝐶𝑝
lim
𝛿𝑥→0
𝛿 𝑢
𝛿 𝑥 𝑥+∆𝑥
−
𝛿 𝑢
𝛿 𝑥 𝑥
∆𝑥
+ lim
𝛿𝑦→0
𝛿 𝑢
𝛿 𝑦 𝑦+∆𝑦
−
𝛿 𝑢
𝛿 𝑦 𝑦
∆𝑦
𝜕𝑢
𝜕𝑡
=
𝑘
𝜌 × 𝐶𝑝
𝛿2
𝑢
𝛿 𝑥2
+
𝛿2
𝑢
𝛿 𝑦2
𝜕𝑢
𝜕𝑡
= 𝑐2
×
𝛿2
𝑢
𝛿 𝑥2
+
𝛿2
𝑢
𝛿 𝑦2

12. The equation,
Represents the temperature of the rectangular plate in transient state.
 For a steady state where u is independent of time i.e.
𝜕𝑢
𝜕𝑡
= 0
Hence equation for steady state becomes,
Which is the heat flow equation in 2 Dimension. Also called as the Laplace Equation
𝝏𝒖
𝝏𝒕
= 𝒄𝟐 ×
𝜹𝟐
𝒖
𝜹 𝒙𝟐
+
𝜹𝟐
𝒖
𝜹 𝒚𝟐
𝜹𝟐
𝒖
𝜹 𝒙𝟐
+
𝜹𝟐
𝒖
𝜹 𝒚𝟐
= 𝟎