MAT4847
Assignment 2
DUE : 7 SEPTEMBER 2026
, ASSIGNMENT 02
Chapters 1–3 — Applied Partial Differential Equations (Haberman, 4th Edition)
Unique Number: 196817
Closing Date: 07 September 2026
Question 1 — Exercise 1.4.1(h) [7 Marks]
Problem (PB, p. 19): Equilibrium temperature of a rod, Q = 0, with
du du
( 0 ) − [ u ( 0 ) −T ) = 0 ,− K0 ( L ) = α [ u ( L ) −u ∞ )
dx dx
Solution
2
d u
With Q = 0at equilibrium, 2
= 0 (Sec. 1.3), so
dx
u ( x ) = c1 + c2 x .
Left BC: u ′ ( 0 ) = c2, u ( 0 ) = c1, so
c 2 − ( c 1 −T ) = 0 ⇒c 2 = c1 −T ( ⋆)
Right BC: u ′ ( L ) = c2, u ( L ) = c1 + c2 L, so
− K0 c 2 = α ( c 1 + c2 L−u ∞ )
Substituting ( ⋆):
− K0 ( c 1 −T ) = α c1 + α L( c 1 −T ) − α u∞
c 1 [ α ( 1+ L ) + K0 ) = K0 T + α L T + α ∞u
Assignment 2
DUE : 7 SEPTEMBER 2026
, ASSIGNMENT 02
Chapters 1–3 — Applied Partial Differential Equations (Haberman, 4th Edition)
Unique Number: 196817
Closing Date: 07 September 2026
Question 1 — Exercise 1.4.1(h) [7 Marks]
Problem (PB, p. 19): Equilibrium temperature of a rod, Q = 0, with
du du
( 0 ) − [ u ( 0 ) −T ) = 0 ,− K0 ( L ) = α [ u ( L ) −u ∞ )
dx dx
Solution
2
d u
With Q = 0at equilibrium, 2
= 0 (Sec. 1.3), so
dx
u ( x ) = c1 + c2 x .
Left BC: u ′ ( 0 ) = c2, u ( 0 ) = c1, so
c 2 − ( c 1 −T ) = 0 ⇒c 2 = c1 −T ( ⋆)
Right BC: u ′ ( L ) = c2, u ( L ) = c1 + c2 L, so
− K0 c 2 = α ( c 1 + c2 L−u ∞ )
Substituting ( ⋆):
− K0 ( c 1 −T ) = α c1 + α L( c 1 −T ) − α u∞
c 1 [ α ( 1+ L ) + K0 ) = K0 T + α L T + α ∞u
