APM3701 Assignment 1 (192480) Due 25 May 2026

APM3701
Assignment 1
Unique No: 192480
Due 25 May 2026

,Assignment 01 – Question 1(a)

We solve the PDE

∂2 𝑢
(𝑥, 𝑦, 𝑡) = 2𝑥𝑡
∂𝑥 ∂𝑦 ∂𝑡

with conditions

𝑦𝑡 2 𝑡 𝑦 ∂𝑢 𝑥𝑦 ∂2 𝑢
𝑢(1, 𝑦, 𝑡) = + + + 1, (𝑥, 𝑦, 0) = , (𝑥, 0, 𝑡) = 2𝑥𝑡 + 𝑥 − 𝑡.
2 2 4 ∂𝑥 2 ∂𝑥 ∂𝑡



Q: Solve the PDE with the given boundary and initial conditions.

A:

We start from

𝑢𝑥𝑦𝑡 = 2𝑥𝑡

Integrate with respect to 𝑡,

𝑢𝑥𝑦 = ∫ 2𝑥𝑡 𝑑𝑡 = 𝑥𝑡 2 + 𝑓1 (𝑥, 𝑦)

where 𝑓1 (𝑥, 𝑦) is an arbitrary function.

Then integrate with respect to 𝑦,

𝑢𝑥 = ∫ (𝑥𝑡 2 + 𝑓1 (𝑥, 𝑦)) 𝑑𝑦 = 𝑥𝑡 2 𝑦 + 𝐹(𝑥, 𝑦) + 𝑓2 (𝑥, 𝑡)

where 𝐹𝑦 = 𝑓1 and 𝑓2 is arbitrary.

Now integrate with respect to 𝑥,

𝑥2
𝑢 = ∫ (𝑥𝑡 2 𝑦 + 𝐹(𝑥, 𝑦) + 𝑓2 (𝑥, 𝑡)) 𝑑𝑥 ⟹ 𝑢 = 2
𝑦𝑡 2 + 𝐺(𝑥, 𝑦) + 𝐻(𝑥, 𝑡) + 𝑓3 (𝑦, 𝑡)

where 𝐺𝑥 = 𝐹 and 𝐻𝑥 = 𝑓2 .

Hence the general solution becomes

, 𝑥 2 𝑦𝑡 2
𝑢(𝑥, 𝑦, 𝑡) = + 𝐺(𝑥, 𝑦) + 𝐻(𝑥, 𝑡) + 𝑓3 (𝑦, 𝑡).
2



Use the condition

𝑥𝑦
𝑢𝑥 (𝑥, 𝑦, 0) = .
2

First compute 𝑢𝑥 ,

𝑢𝑥 = 𝑥𝑦𝑡 2 + 𝐺𝑥 (𝑥, 𝑦) + 𝐻𝑥 (𝑥, 𝑡).

At 𝑡 = 0 ,

𝑥𝑦
𝑢𝑥 (𝑥, 𝑦, 0) = 𝐺𝑥 (𝑥, 𝑦) + 𝐻𝑥 (𝑥, 0) = .
2

This implies

𝑥𝑦
𝐺𝑥 (𝑥, 𝑦) = , 𝐻 𝑥 (𝑥, 0) = 0.
2

Integrate

𝑥𝑦 𝑥2 𝑦
𝐺(𝑥, 𝑦) = ∫ 𝑑𝑥 = + 𝑔1 (𝑦).
2 4



Use the condition

𝑢𝑥𝑡 (𝑥, 0, 𝑡) = 2𝑥𝑡 + 𝑥 − 𝑡.

Compute

∂
𝑢𝑥𝑡 = (𝑥𝑦𝑡 2 + 𝐺𝑥 + 𝐻𝑥 ) = 2𝑥𝑦𝑡 + 𝐻 𝑥𝑡 .
∂𝑡

At 𝑦 = 0 ,

𝑢𝑥𝑡 (𝑥, 0, 𝑡) = 𝐻𝑥𝑡 (𝑥, 𝑡) = 2𝑥𝑡 + 𝑥 − 𝑡.