Question
Solution of β2π = 0 in a square domain (0 < x < 1 and 0 < y < 1) with boundary conditions: T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1,y) = 1 + y is
Options :
T(x, y) = x β xy + y
T(x, y) = x + y
T(x, y) = βx + y
T(x, y) = x + xy + y
Answer :
T(x, y) = x + y
Solution :
T(x, 0) = x β option (c) is not correct.
T(0, y) = y β all options satisfied.
T(x, 1) = 1 + x β only option (b) is satisfied.
T(1, y) = 1 + y β only option (b) is satisfied.
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