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Unformatted text preview: rabolic, or elliptic.
a)
2
(a) 1 u = 10, x x + sin(y 4 0
6. uxx a =01,xyb + 16uyy c =u16 ) b )u =ac. = 36 > 0. PDE is hyperbolic. (b) a = , b
Characteristic equations:
(a) Find1the = 10, c = 16 ) b2 4ac = 36 > 0. PDE is hyperbolic.
b)
characteristics.
p
p
b2 4ac
dy
b
dy
(b) Characteristic equations:b + b2 4ac
=
=
p2a
p2a
dx
dx
dy = b 8 b2 4ac
dy = b + b2 4ac
=2
=
dx
2a
dx
2a
Integrating:
=2
=8
y = 2 x + k1
y = 8 x + k2
Integrating:
y + 2 x = k1
y + 8 x = k2
y = 2 x + k1
y = 8 x + k2
y + ⇠y = k
(c) Let ⇠ = y + y x, 2x == +1 8x, then ⇠x = 2, 8x = 1, 2⌘x = 8, ⌘y = 1 and
2+ ⌘
yk
all second partial derivatives are zero. Noting that d = x, e = 0, f =
(c) Let ⇠ = y + 2x, ⌘ = 1 + 8x, then ⇠x = 2, ⇠y = 1, ⌘x = 8, ⌘y = 1 and
y
sin(y ), g = 0, and x = 6 (⌘ ⇠ ), y = 1 (4⇠ ⌘ ), we obtain g = 0 and
ˆ
3
all second partial derivatives are zero. Noting that d = x, e = 0, f =
1
B 0, 2(1)(2)(8)
+
sin(y ), g = = and x = 6 (⌘ 10 ), y = 1 (4⇠ ⌘ ),+ 2(16)(1)(1)= 0 and
⇠ ((2)(1) 3 (8)(1)) we obtain g = 36
ˆ
1
1
1
(⌘ ⇠ )(2) 0 ⇠
⌘
D = 2(1)(2)(8) 1=((2)(1) + (8)(1)) + 2(16)(1)(1) = 36
B
6
3
3
1
4
4
1 (⌘ ⇠ )(2) = 1 ⇠ 1 ⌘
D=
(⌘ ⇠ )(8) = ⇠
⌘
E= dx = 2a 2 dx = 8 2a Assignment #8 Solutions  Fall 2013
Page 5
Integrating:
y = 2 x + k1
y = 8 x + k2
y + 2 x = k1
y + 8 x = k2
c) Put the equation in its canonical form. AMATH 350 (c) Let ⇠ = y + 2x, ⌘ = y + 8x, then ⇠x = 2, ⇠y = 1, ⌘x = 8, ⌘y = 1 and
all second partial derivatives are zero. Noting that d = x, e = 0, f =
1
sin(y ), g = 0, and x = 6 (⌘ ⇠ ), y = 1 (4⇠ ⌘ ), we obtain g = 0 and
ˆ
3
B = 2(1)(2)(8) 10 ((2)(1) + (8)(1)) + 2(16)(1)(1) =
1
1
1
(⌘ ⇠ )(2) = ⇠
⌘
D=
6
3
3
4
4
1
(⌘ ⇠ )(8) = ⇠
⌘
E=
6
3
3
4
1
ˆ
f = sin(y (⇠ , ⌘ )) = sin( ⇠
⌘)
3
3 36 1
u3
u
Thus the PDE become...
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This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.
 Fall '12
 DavidHamsworth

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