APM3701
Assignment 1
Unique No: 192480
Due 25 May 2026
,Assignment 1
Question 1(a)
Q
Solve the PDE:
∂3 𝑢
= 2𝑥𝑡
∂𝑥 ∂𝑦 ∂𝑡
Subject to:
𝑦𝑡 2 𝑡 𝑦 ∂𝑢 𝑥𝑦 ∂ 2 𝑢
𝑢(1, 𝑦, 𝑡) = +2+4+1 (𝑥, 𝑦, 0) = (𝑥, 0, 𝑡) = 2𝑥𝑡 + 𝑥 − 𝑡
2 ∂𝑥 2 ∂𝑥 ∂𝑡
A
Given:
𝑢𝑥𝑦𝑡 = 2𝑥𝑡
Integrate successively.
Integrate w.r.t. 𝑦
𝑢𝑥𝑡 = ∫ 2𝑥𝑡 𝑑𝑦 = 2𝑥𝑡𝑦 + 𝑓 1 (𝑥, 𝑡)
Integrate w.r.t. 𝑡
𝑢𝑥 = ∫ (2𝑥𝑡𝑦 + 𝑓1 (𝑥, 𝑡))𝑑𝑡 𝑢𝑥 = 𝑥𝑦𝑡 2 + ∫ 𝑓1 (𝑥, 𝑡)𝑑𝑡 + 𝑓2 (𝑥, 𝑦)
, Let:
𝐹(𝑥, 𝑡) = ∫ 𝑓1 (𝑥, 𝑡)𝑑𝑡
Then
𝑢𝑥 = 𝑥𝑦𝑡 2 + 𝐹(𝑥, 𝑡) + 𝑓2 (𝑥, 𝑦)
Integrate w.r.t. 𝑥
𝑥 2 𝑦𝑡 2
𝑢 = ∫ (𝑥𝑦𝑡 2 + 𝐹(𝑥, 𝑡) + 𝑓2 (𝑥, 𝑦))𝑑𝑥 𝑢 = 2
+ ∫ 𝐹(𝑥, 𝑡)𝑑𝑥 + ∫ 𝑓2 (𝑥, 𝑦)𝑑𝑥 + 𝑓3 (𝑦, 𝑡)
Let:
𝐺(𝑥, 𝑡) = ∫ 𝐹(𝑥, 𝑡)𝑑𝑥 𝐻(𝑥, 𝑦) = ∫ 𝑓2 (𝑥, 𝑦)𝑑𝑥
Thus
𝑥 2 𝑦𝑡 2
𝑢= + 𝐺(𝑥, 𝑡) + 𝐻(𝑥, 𝑦) + 𝑓3 (𝑦, 𝑡)
2
Use boundary conditions
Condition 1
𝑥𝑦
𝑢𝑥 (𝑥, 𝑦, 0) =
2
Differentiate general solution:
𝑢𝑥 = 𝑥𝑦𝑡 2 + 𝐺𝑥 + 𝐻𝑥
At 𝑡 = 0 :
𝑥𝑦
0 + 𝐺𝑥 (𝑥, 0) + 𝐻𝑥 (𝑥, 𝑦) =
2
Assignment 1
Unique No: 192480
Due 25 May 2026
,Assignment 1
Question 1(a)
Q
Solve the PDE:
∂3 𝑢
= 2𝑥𝑡
∂𝑥 ∂𝑦 ∂𝑡
Subject to:
𝑦𝑡 2 𝑡 𝑦 ∂𝑢 𝑥𝑦 ∂ 2 𝑢
𝑢(1, 𝑦, 𝑡) = +2+4+1 (𝑥, 𝑦, 0) = (𝑥, 0, 𝑡) = 2𝑥𝑡 + 𝑥 − 𝑡
2 ∂𝑥 2 ∂𝑥 ∂𝑡
A
Given:
𝑢𝑥𝑦𝑡 = 2𝑥𝑡
Integrate successively.
Integrate w.r.t. 𝑦
𝑢𝑥𝑡 = ∫ 2𝑥𝑡 𝑑𝑦 = 2𝑥𝑡𝑦 + 𝑓 1 (𝑥, 𝑡)
Integrate w.r.t. 𝑡
𝑢𝑥 = ∫ (2𝑥𝑡𝑦 + 𝑓1 (𝑥, 𝑡))𝑑𝑡 𝑢𝑥 = 𝑥𝑦𝑡 2 + ∫ 𝑓1 (𝑥, 𝑡)𝑑𝑡 + 𝑓2 (𝑥, 𝑦)
, Let:
𝐹(𝑥, 𝑡) = ∫ 𝑓1 (𝑥, 𝑡)𝑑𝑡
Then
𝑢𝑥 = 𝑥𝑦𝑡 2 + 𝐹(𝑥, 𝑡) + 𝑓2 (𝑥, 𝑦)
Integrate w.r.t. 𝑥
𝑥 2 𝑦𝑡 2
𝑢 = ∫ (𝑥𝑦𝑡 2 + 𝐹(𝑥, 𝑡) + 𝑓2 (𝑥, 𝑦))𝑑𝑥 𝑢 = 2
+ ∫ 𝐹(𝑥, 𝑡)𝑑𝑥 + ∫ 𝑓2 (𝑥, 𝑦)𝑑𝑥 + 𝑓3 (𝑦, 𝑡)
Let:
𝐺(𝑥, 𝑡) = ∫ 𝐹(𝑥, 𝑡)𝑑𝑥 𝐻(𝑥, 𝑦) = ∫ 𝑓2 (𝑥, 𝑦)𝑑𝑥
Thus
𝑥 2 𝑦𝑡 2
𝑢= + 𝐺(𝑥, 𝑡) + 𝐻(𝑥, 𝑦) + 𝑓3 (𝑦, 𝑡)
2
Use boundary conditions
Condition 1
𝑥𝑦
𝑢𝑥 (𝑥, 𝑦, 0) =
2
Differentiate general solution:
𝑢𝑥 = 𝑥𝑦𝑡 2 + 𝐺𝑥 + 𝐻𝑥
At 𝑡 = 0 :
𝑥𝑦
0 + 𝐺𝑥 (𝑥, 0) + 𝐻𝑥 (𝑥, 𝑦) =
2
