then - islamic university of gaza
Post on 07-Apr-2022
12 Views
Preview:
TRANSCRIPT
1
.
PDE
2
Then
3
4
5
T equation is
And
6
7
π’π’π‘π‘π‘π‘ β ππ2π’π’π₯π₯π₯π₯ = π»π»(π₯π₯), 0 < π₯π₯ < πΏπΏ, π‘π‘ > 0.
π’π’(0, π‘π‘) = 0, π’π’(πΏπΏ, π‘π‘) = 0, (3)
π’π’π‘π‘(π₯π₯, 0) = 0, π’π’(π₯π₯, 0) = 0.
8
The new equation is π£π£π‘π‘π‘π‘ = ππ2π£π£π₯π₯π₯π₯ is homogeneous .
Applying the conditions π’π’(0, π‘π‘) = π’π’(πΏπΏ, π‘π‘) we have
Now the problem is π£π£π‘π‘π‘π‘ = ππ2π£π£π₯π₯π₯π₯ , 0 < π₯π₯ < πΏπΏ, π‘π‘ > 0.
9
10
11
12
13
Then ππ(π₯π₯) = ππ1βππ0πΏπΏ
π₯π₯ + ππ0 .
14
15
ππβ²β² β ππππ = 0, ππβ²β² + ππππ = 0.
and continue to find X and Y.
16
17
Excercises:
Ex 1.
Ex2.
Ex 3.
Ex 4.
18
Ex 5.
Exercise 7.9 page 265 Q(1,2,3,4,11,12,13)
top related