APM3701 Assignment 1 semester 1 2026 (Answer Guide) 192480 – Due 25 May 2026

APM3701 Assignment 1 semester 1 2026 (Answer Guide)
192480 – Due 25 May 2026

APM3701 – Assignment 01 (2026)
Question 1



Question 1(a)

Given problem
Solve the following initial–boundary value problem:

∂3u
(x, y, t) = 2xt, x, y, t ∈ R
∂x∂y∂t

Subject to:


yt2 + t + y + 1
u(1, y, t) =
2 2 4

∂u xy
(x, y, 0) =
∂x 2
∂2u
(x, 0, t) = 2xt + x − t
∂x∂t

,Step 1: Integrate the PDE with respect to t
Starting from:

∂3u
= 2xt
∂x∂y∂t
Integrating with respect to t:


∂2u
= xt2 + f1(x, y)
∂x∂y


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




Step 2: Integrate with respect to y
∂u
= xyt2 + ∫ f1(x, y) dy + f2(x, t)
∂x
Let:




∫ f1(x, y) dy = g(x, y)

Thus:

∂u
= xyt2 + g(x, y) + f2(x, t)
∂x



∂u
Step 3: Apply condition ∂x (x, y, 0) = xy2
Substitute t = 0:
∂u xy
(x, y, 0) = g(x, y) + f2(x, 0) =
∂x 2
Choose:


So:

, x
g(
x, ,
y) f
2
= (
x
,
0
)
=
0
2