dt dx2 dy , ди u(0, у, 1) dx (1, y, t) = 0, и(х, 0, г) %3D ди - (х, 1, г) %3D 0, dx и(х, у, 0) %3D 2 sin(%3 лх) sin(%3 лу). %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Solve the problem for a heat equation on a square plate of length 1, with no flux boundary condition in the right
and top sides:
Pu
ди
= k
dt
Pu
+
dx2
0 < y < 1, t> 0,
0 < x < 1,
ду
u(0, у, t) —
ди
-(1, у, t) 3D 0,
dx
ди
- (х, 1, t) %3D 0,
и(х, 0, t) %—
dx
u(x, y, 0) = 2 sin(Grx) sin(Ty).
Transcribed Image Text:Solve the problem for a heat equation on a square plate of length 1, with no flux boundary condition in the right and top sides: Pu ди = k dt Pu + dx2 0 < y < 1, t> 0, 0 < x < 1, ду u(0, у, t) — ди -(1, у, t) 3D 0, dx ди - (х, 1, t) %3D 0, и(х, 0, t) %— dx u(x, y, 0) = 2 sin(Grx) sin(Ty).
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