Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers
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2.8 Exercises 701
16. (a) x (s, τ)=τ, y (s, τ)=
τ
2
2
+ aτ s + s, u (s, τ)=τ + as.
τ = x, s =(1+ax)
−1
y −
1
2
x
2
a, and hence,
u (x , y)=x + as =(1+ax)
−1
&
x + a
y +
1
2
x
2
'
, singular at x = −
1
a
.
(b) y =
u
2
2
+ f (u − x), 2y = u
2
+(u − x)
2
, u (0,y)=
√
y.
17. (a) Hint:
d(x+y +u)
2(x+y+u)
=
d(y −u)
−(y −u )
=
d(u−x)
−(u−x)
(x + y + u)(y − u)
2
= c
1
and (x + y + u)(u − x)
2
= c
2
.
(b) Hint:
dx
x
=
dy
−y
. Hence, xy = a.
dx
xu(u
2
+a)
=
du
x
4
.So,
dx
du
=
u
(
u
2
+a
)
x
3
giving x
4
= u
4
+2au
2
+ b
and, thus, x
4
− u
4
− 2u
2
xy = b.
(c)
dx
x+y
=
dy
x−y
=
dy
0
(exact equation). u = f
x
2
− 2xy − y
2
.
(d) f
x
2
− y
2
,u−
1
2
y
2
x
2
− y
2
=0.
(e) f
x
2
+ y
2
+ z
2
,ax+ by + cz
=0.
18. Hint:
dx
x
=
dy
y
=
dz
z
, and hence,
x
z
= c,
y
z
= d.
x
2
+ y
2
= a
2
and z = tan
−1
y
x
give
c
2
+ d
2
z
2
= a
2
and z = b tan
−1
d
c
.
c =
a
z
cos θ , d =
a
z
sin θ,andz = b tan
−1
(tan θ)=bθ.
Thus, the curves are xbθ= az cos θ and ybθ= az sin θ.
19. F
(
x + y + u, (x − 2y)
2
+3u
2
)
=0.Hint:
(dx−2dy)
9u
=
du
−3(x−2y)
.
(x − 2y)
2
+3u
2
=(x + y + u)
2
.
20. F
x
2
+ y, yu
=0,
x
2
+ y
4
= yu.
21. Hint: x − y + z = c
1
,
dz
−(x+y +z)
=
(dx+dy+dz)
8z
, and hence,
8z
2
+(x + y + z)
2
= c
2
. F
(
(x − y + z) , 8z
2
+(x + y + z)
2
)
=0.
c
2
1
+ c
2
=2a
2
,or(x − y + z)
2
+(x + y + z)
2
+8z
2
=2a
2
.
22. F
x
2
+ y
2
+ z
2
,y
2
− 2yz − z
2
=0.
(a) y
2
− 2yz − z
2
=0,twoplanesy =
1+
√
2
z.
(b) x
2
+2yz +2z
2
= 0, a quadric cone with vertex at the origin.
(c) x
2
− 2yz +2y
2
= 0, a quadric cone with vertex at the origin.
702 Answers and H ints to Selected Exercises
23. Use the Hint of 17(c).
dx
dt
= x + y,
dy
dt
= x − y,
d
2
x
dt
2
=2x.
dx
dt
2
=2x
2
+ c.Whenx =0=y,
dx
dt
=
√
2 x.
√
2 u =lnx + x
2
− 2xy +2y.
24. (a) a = f
x +
3
2
y
.
(b) x = at + c
1
, y = bt, u = c
2
e
ct
, c
2
= f (c
1
),
u (x , y)=f
x −
a
b
y
exp
cy
b
.
(c) u = f
x
1−y
(1 − y)
c
.
(d) x =
1
2
t
2
+ αst + s, y = t; u = y +
1
2
α (αy +1)
−1
2x − y
2
.
26. (a) Hint: (f
′
)
2
=1− (g
′
)
2
= λ
2
; f
′
(x)=λ and g
′
(y)=
√
1 − λ
2
.
f (x)=λx + c
1
and g (y)=y
√
1 − λ
2
+ c
2
.
Hence, u (x, y)=λx + y
√
1 − λ
2
+ c.
(b) Hint: (f
′
)
2
+(g
′
)
2
= f (x)+g (y)or(f
′
)
2
−f (x)=g (y)−(g
′
)
2
= λ.
Hence, (f
′
)
2
= f (x)+λ and g
′
=
g (y) − λ.
Or,
df
√
f+λ
= dx and
dg
√
g −λ
= dy.
f (x)+λ =
x+c
1
2
2
and g (y) − λ =
y +c
2
2
2
.
u (x , y)=
x+c
1
2
2
+
y +c
2
2
2
.
(c) Hint: (f
′
)
2
+ x
2
= −g
′
(y)=λ
2
.
Or f
′
(x)=
√
λ
2
− x
2
,and g (y)=−λ
2
y + c
2
.
Putting x = λ sin θ, we obtain
f (x)=
1
2
λ
2
sin
−1
x
λ
+
x
2
√
λ
2
− x
2
+ c
1
,
u (x , y)=
1
2
λ
2
sin
−1
x
λ
+
x
2
√
λ
2
− x
2
− λ
2
y +(c
1
+ c
2
).
(d) Hint: x
2
(f
′
)
2
= λ
2
and 1 − y
2
(g
′
)
2
= λ
2
.
Or, f (x)=λ ln x + c
1
and g (y)=
√
1 − λ
2
ln y + c
2
.
27. (a) Hint: v =lnu gives v
x
=
1
u
· u
x
,andv
y
=
1
u
· u
y
.
x
2
u
x
u
2
+ y
2
u
y
u
2
=1.
Or, x
2
v
2
x
+ y
2
v
2
y
= 1 gives x
2
(f
′
)
2
+ y
2
(g
′
)
2
=1.
2.8 Exercises 703
x
2
{f
′
(x)}
2
=1− y
2
(g
′
)
2
= λ
2
.
Or, f (x)=λ ln x + c
1
and g (y)=
√
1 − λ
2
(ln y)+c
2
.
Thus, v (x, y)=λ ln x +
√
1 − λ
2
(ln y)+lnc,(c
1
+ c
2
=lnc).
u (x , y)=cx
λ
y
√
1−λ
2
.
(b) Hint: v = u
2
and v (x, y)=f (x)+g (y)maynotwork.
Try u = u (s), s = λxy,sothatu
x
= u
′
(y)·(λy)andu
y
= u
′
(s)·(λx).
Consequently, 2λ
2
1
u
du
ds
2
=1.Or,
1
u
du
ds
=
1
√
2
1
λ
.
Hence, u (s)=c
1
exp
s
λ
√
2
. u (x, y)=c
1
exp
xy
√
2
.
28. Hint: v
x
=
1
2
u
x
√
u
, v
y
=
1
2
u
y
√
u
.Thisgivesx
4
(f
′
)
2
+ y
2
(g
′
)
2
=1.
Or, x
4
(f
′
)
2
=1− y
2
(g
′
)
2
= λ
2
.
Or, x
4
(f
′
)
2
= λ
2
and y
2
(g
′
)
2
=1− λ
2
.
Hence, f (x)=−
λ
x
+ c
1
and g (y)=
√
1 − λ
2
ln y + c
2
u (x , y)=
−
λ
x
+
√
1 − λ
2
ln y + c
2
.
29. Hint: v
x
=
u
x
u
, v
y
=
u
y
u
.
v
2
x
x
2
+
v
2
y
y
2
=1,andv = f (x)+g (y).
Or,
(
f
′
)
2
x
2
=1−
1
y
2
(g
′
)
2
= λ
2
.
f
′
(x)=λx,andg
′
(y)=
√
1 − λ
2
y.
Or, f (x)=
λ
2
x
2
+ c
1
,andg (y)=
1
2
y
2
√
1 − λ
2
+ c
2
.
v (x, y)=
λ
2
x
2
+
y
2
2
√
1 − λ
2
+ c =lnu.
u (x , y)=c exp
λ
2
x
2
+
y
2
2
√
1 − λ
2
,c
1
+ c
2
=lnc.
e
x
2
= u (x, 0) = ce
λ
2
x
2
, which gives c = 1 and λ =2.
30. (a) Hint: ξ = x − y, η = y; u (x, y)=e
y
f (x − y),
(b) ξ = x, η = y −
x
2
2
, u
ξ
= η +
1
2
ξ
2
, u = ξη +
1
6
ξ
3
+ f (η).
u (x , y)=xy −
1
3
x
3
+ f
y −
x
2
2
.
(c) ξ = y exp
−x
2
, η = y,ande
2u
f (ξ)=η, e
2u
f
ye
−x
2
= y,
(d)
dx
1
=
dy
−y
=
du
1+u
, ξ = ye
x
, η = y.
Thus, (1 + u) f (ξ)=
1
η
. Or, (1 + u) f (ye
x
)=y
−1
.
31. (c) u (x, y)=α exp
βx −
a
b
βy
.
704 Answers and H ints to Selected Exercises
32. (a) v (x, t)=x + ct, u (x, t)=
(
6x+3ct
2
+5ct
3
)
6(1+2t)
.
(b) v (x, t)=x + ct, u (x, t)=
(
6x+3ct
2
+4ct
3
)
6(1+2t)
.
33. (a) v (x, t)=e
x+at
, u −
1
a
e
at
= c
1
,and
u −
1
a
e
at
= f
x − ut +
t
a
e
at
−
1
a
2
e
at
.
u (x , t) = (1 + t)
−1
&
(x − ut)+
1
a
+
t
a
−
1
a
2
e
at
+
1
a
2
−
1
a
'
.
(b) v = x − ct, u (x, t)=
(
6x−3ct
2
+4ct
3
)
6(1−2t)
.
34.
dt
1
=
dy
−x
=
du
u
, t +lnx = c
1
,andxu = c
2
. g (xu, t +lnx)=0.
Or, u =
1
x
h (t +lnx). u (x, t)=e
t
ln (xe
t
),
where g and h are arbitrary function s.
3.9 Exercises
11. Hint: Differentiate the first equation with respect to t to obtain ρ
tt
+
ρ
0
divu
t
= 0. Take gradient of the last equation to get ∇ρ = −
ρ
0
/c
2
0
u
t
.
We next combin e these two equations to obtain ρ
tt
= c
2
0
∇
2
ρ.Ap-
plication of ∇
2
to p − p
0
= c
2
0
(ρ − ρ
0
)leadsto∇
2
p = c
2
0
∇
2
ρ.Also
p
tt
= c
2
0
ρ
tt
= c
4
0
∇
2
ρ = c
2
0
∇
2
p.
Using u = ∇φ in the first equation gives ρ
t
+ ρ
0
∇
2
φ = 0, and dif-
ferenting the last equation with respect to t yields ρ
t
= −
ρ
0
/c
2
0
φ
tt
.
Combining these two equations produces the wave equation for φ. Fi-
nally, we take gradient of the first and the last equations to obtain
∇ρ
t
+ ρ
0
∇
2
u = 0 and ∇ρ = −
ρ
0
/c
2
0
u
t
that leads to the wave equa-
tion for u
t
.
14. (a) Differentiate the first equation with respect to t and the second
equation with respect to x. Then eliminate V
xt
and V
x
to obtain the
desired telegraph equation.
(e) (i)
∂
2
∂x
2
(I, V )=
1
c
2
∂
2
∂t
2
(I, V ) ,c
2
=
1
LC
.
3.9 Exercises 705
(ii)
∂
∂t
(I, V )=κ
∂
2
∂x
2
(I, V ) ,κ=
1
RC
.
(iii)
∂
2
∂t
2
+2k
∂
∂t
+ k
2
(I, V )=c
2
∂
2
∂x
2
(I, V ).
17. (a) The two-dimensional unsteady Euler equations are
du
dt
= −
1
ρ
∂p
∂x
,
dv
dt
= −
1
ρ
∂p
∂y
,
where
d
dt
=
∂
∂t
+ u ·∇=
∂
∂t
+ u
∂
∂x
+ v
∂
∂y
,andu =(u, v).
(b) For two-dimensional steady flow, the Euler equations are
uu
x
+ vu
y
= −
1
ρ
p
x
, uv
x
+ vv
y
= −
1
ρ
p
y
.
Using
dp
dρ
= c
2
, these equations become
uu
x
+ vu
y
= −c
2
(ρ
x
/ρ), uv
x
+ vv
y
= −c
2
(ρ
y
/ρ).
Multiply the first equation by u and the second by v and add to obtain
u
2
u
x
+ uv (u
y
+ v
x
)+v
2
v
y
= −
c
2
ρ
(uρ
x
+ vρ
y
).
Using the continuity equation (ρu)
x
+(ρv)
y
= 0, the right hand of this
equation becomes c
2
(u
x
+ v
y
). Hence is the desired equation.
(c) Using u = ∇φ =(φ
x
,φ
y
), the result follows.
(d) Substitute ρ
x
and ρ
y
from 17(b) into the continuity equation
uρ
x
+ vρ
y
+ ρ (u
x
+ v
y
)=0toobtain
c
2
− u
2
φ
xx
− 2uvφ
xy
+
c
2
− v
2
φ
yy
=0.
Also
du = u
x
dx + u
y
dy = −φ
xx
dx − φ
xy
dy,
dv = v
x
dx + v
y
dy = −φ
xy
dx − φ
yy
dy.
Denoting D for the coefficient determinant of the above equations for
φ
xx
, φ
xy
and φ
yy
gives the solutions
φ
xx
= −
D
1
D
,φ
xy
=
D
2
D
,φ
yy
=
−D
3
D
.
D = 0 gives a quadratic equation for the slope of the characteristic C,
that is,
c
2
− u
2
dy
dx
2
+2uv
dy
dx
+
c
2
− v
2
=0.
Thus, directions are r eal and distinct provided
706 Answers and H ints to Selected Exercises
4u
2
v
2
− 4
c
2
− u
2
c
2
− v
2
> 0, or
u
2
+ v
2
>c
2
.
D
2
= 0 gives −
dy
dx
=
(
c
2
−v
2
)
(c
2
−u
2
)
dv
du
.
Substitute into the quadratic equation to obtain
c
2
− v
2
dv
du
2
− 2uv
dv
du
+
c
2
− u
2
=0.
Note that when D
1
= D
2
= D
3
= D = 0, any one of the second order
φ derivatives can be discontinuous.
18. (a) Hint: Use ∇×
∂u
∂t
=
∂ω
∂t
, ∇×(u ·∇u) u ·∇ω −ω∇u, where we have
used ∇·u = 0 and ∇·ω =0.Since∇×∇f = 0 for any scalar function
f, these lead to the vorticity equation in this simplified model.
(b) The rate of change of vorticity as we follow the fluid is given by the
term ω ·∇u.
(c) u = i u (x, y)+j v (x, y)andω = ω (x, y) k and hence,
ω ·∇u = ω (x, y)
∂
∂z
[i u (x, y)+j v (x, y)] = 0. This gives the result.
20. We differentiate the first equation partially wit h respect to t to find
E
tt
= c curl H
t
. We then substitute H
t
from the second equation to ob-
tain E
tt
= −c
2
curl (curl E). Using the vector identity curl (curl E)=
grad (div E)−∇
2
E with div E = 0 gives the desired equation. A similar
procedure shows that H satisfies the same equation.
21. When Hooke’s law is used to the rod of variable cross section, the
tension at point P is given by T
P
= λA(x) u
x
,whereλ is a constant. A
longitudinal vibration would displace each cross sectional element of the
rod along the x-axis of the rod. An element PQ of length δx and mass
m = ρA(x) δx will be displaced to P
′
Q
′
with length (δx + δu)withthe
same mass m. The acceleration of the element P
′
Q
′
is u
tt
so that the
difference of the tensions at P
′
and Q
′
must be equal to the product
mu
tt
. Hence, mu
tt
= T
Q
′
− T
P
′
=
∂
∂t
T
P
′
δx =
∂
∂x
(λA(x) u
x
) δx.
This gives the equation.
4.6 Exercises 707
4.6 Exercises
1. (a) x<0, hyperboli c;
u
ξη
=
1
4
ξ−η
4
4
−
1
2
1
ξ−η
(u
ξ
− u
η
),
x = 0, parabolic, the given equation is then in canonical form;
x>0, elliptic and the canonical form is
u
αα
+ u
ββ
=
1
β
u
β
+
β
4
16
.
(b) y = 0, parabolic; y = 0, elliptic, and hence,
u
αα
+ u
ββ
= u
α
+ e
α
.
(d) Parabolic everywhere and hence,
u
ηη
=
2ξ
η
2
u
ξ
+
1
η
2
e
ξ/η
.
(f) Elliptic everywhere for finite values of x and y,then
u
αα
+ u
ββ
= u −
1
α
u
α
−
1
β
u
β
.
(g) Parabolic everywhere
u
ηη
=
1
1−e
2(η−ξ)
sin
−1
e
η−ξ
− u
ξ
!
.
(h) B
2
−4AC = y −4x. Equation is hyperbolic if y>4x, parabolic if
y =4x and elliptic if y<4x.
(i) y = 0, parabolic; and y = 0, hyp erbolic,
u
ξη
=
(1+ξ −ln η)
η
u
ξ
+ u
η
+
1
η
u.
2. (i) u (x, y)=f (y/x)+g (y/x) e
−y
,
(ii) Hint: ϕ = ru and check the solution by substitution.
u (r, t)=(1/r) f (r + ct)+(1/r) g (r − ct);
(iii) A =4,B = 12, C = 9. Hence, B
2
−4AC = 0. Parabol ic at every
point in (x, y)-plane.
dy
dx
=
3
2
or y =
3
2
x+c ⇒ 2y −3x = c
1
, ξ =2y −3x,
η = y. The canonical form is u
ηη
− u =1⇒ u (ξ, η)=f (ξ) cosh η +
g (ξ)sinhη −1. Or, u (x, y)=f (2y − 3x) cosh y +g (2y − 3x)sinhy −1.
(iv) Hyperbolic at all points in the (x, y)-plane. ξ = y − 2x, η = y + x.
Thus, u
ξη
+ u
η
= ξ, u (ξ, η)=η (ξ − 1) + f (ξ) e
−ξ
+ g (ξ).
708 Answers and H ints to Selected Exercises
u (x , y)=(x + y)(y − 2x − 1) + f (x + y)exp(2x − y)+g (y − 2x).
(v) Hyperbolic. ξ = y, η = y−3x. ξu
ξη
+u
η
=0,u (ξ, η)=
1
ξ
f (η)+g (ξ).
u (x , y)=
1
y
fu(y − 3x)+g (y).
(vi) A =1,B =0,C =1,B
2
− 4AC = −4 < 0. So, this equation
is elliptic.
dy
dx
=+
i or
dx
dy
=
+ i or ξ = x + iy = c
1
and
η = x − iy = c
2
.
The general solution is u = φ (ξ)+ψ (η)=φ (x + iy)+ψ (x − iy).
(vii) u = φ (x +2iy)+ψ (x − 2iy).
(viii) B
2
− 4AC = 0. Equation is parabolic. The general solution is
given by (4.3.16), where λ =
B
2A
= −1 and hence, the general solution
becomes u = φ (y + x)+yψ(y + x).
(xi) B
2
− 4AC =25> 0. Hyperbolic. The general solution is
u = φ
y −
3
2
x
+ ψ
y −
1
6
x
.
3. (a) ξ =(y − x)+i
√
2 x, η =(y − x) − i
√
2 x, α = y − x, β =
√
2 x,
u
αα
+ u
ββ
= −
1
2
u
α
− 2
√
2 u
β
−
1
2
u +
1
2
exp
β/
√
2
.
(b) ξ = y + x, η = y; u
ηη
= −
3
2
u.
(c) ξ = y − x, η = y − 4x; u
ξη
=
7
9
(u
ξ
+ u
η
) −
1
9
sin [(ξ − η) /3].
(d) ξ = y + ix, η = y − ix.Thus,α = y, β = x.
The given equation is already in canonical form.
(e) ξ = x, η = x − (y/2); u
ξη
=18u
ξ
+17u
η
− 4.
(f) ξ = y +(x/6), η = y; u
ξη
=6u − 6η
2
.
(g) ξ = x, η = y; the given equation is already in canonical form.
(h) ξ = x, η = y; the given equation is already in canonical form.
(i) Hyperbolic in the (x, y) plane except the axes x = y =0.ξ = xy, η =
(y/x); y =
√
ξη, x =
ξ/η; u
ξη
=
1
2
1+
1
2
%
η
ξ
u
η
−
1
4
1
√
ξη
−
1
4ξη
−
1
2
.
(j) Elliptic when y>0;
dy
dx
=+
i
√
y, α =2
√
y and β = −x;
u
αα
+ u
ββ
= α
2
u
β
. Parabolic when y =0; u
xx
+
1
2
u
y
=0.
4.6 Exercises 709
Hyperbolic when y<0; ξ = x − 2
√
−y, η = −x − 2
√
−y.
The canonical form is u
ξη
=
1
16
(ξ + η)
2
(u
η
− u
ξ
).
(k) Parabolic,
dy
dx
=(xy)
−1
. Integrating gives
1
2
y
2
=lnx+ln ξ,where
ξ is an integrating constant. Hence, ξ =
1
x
exp
1
2
y
2
, η = x.
u
xx
= x
−4
e
y
2
u
ξξ
− 2x
−2
exp
1
2
y
2
u
ξη
+ u
ηη
+2x
−2
exp
1
2
y
2
u
ξ
,
u
xy
= −yx
−3
exp
y
2
u
ξξ
+ yx
−1
exp
1
2
y
2
u
ξη
− yx
−2
exp
1
2
y
2
u
ξ
,
u
yy
=
y
2
x
−2
e
y
2
u
ξξ
+
y
2
x
−2
exp
1
2
y
2
u
ξ
. u
ηη
+
ξ/η
2
u
ξ
=0.
(l) Elliptic if y>0, ξ = x +2i
√
y, η = x − 2i
√
y,
α =
1
2
(ξ + η)=x, β =
1
2i
(ξ − η)=2
√
y; u
αα
+ u
ββ
=
1
β
u
β
.
Hyperbolic y<0, ξ = x +2i
√
y, η = x − 2i
√
y, ξ − η =4i
√
y;
u
ξη
+
1
2
u
ξ
−u
η
ξ−η
= 0. The equations of the characteristic curves are
dy
dx
=+
i
√
y that gives 2
√
y =+i (x − c), or y = +
1
4
(x − c)
2
,
where c is an integrating constant. Two branches of parabolas with
positive or negative slopes.
4. (i) u (x, y)=f (x + cy)+g (x − cy); (ii) u (x, y)=f (x + iy)+f (x − iy);
(iii) Use z = x + i y. Hence, u (x , y)=(x − iy) f
1
(x + iy)+f
2
(x + iy)
+(x + iy)+f
3
(x − iy)+f
4
(x − iy)
(iv) u (x, y)=f (y + x)+g (y +2x); (v) u (x, y)=f (y)+g (y − x);
(vi) u (x, y)=(−y/128) (y − x)(y − 9x)+f (y − 9x)+g (y − x).
5. (i) v
ξη
= −(1/16) v, (ii) v
ξη
= (84/625) v.
7. (ii) Use α =
3y
2
, β = −x
3
/2.
8. x = r cos θ, y = r sin θ; r =
x
2
+ y
2
, θ = tan
−1
y
x
.
∂u
∂x
=
∂u
∂r
·
∂r
∂x
+
∂u
∂θ
·
∂θ
∂x
= u
r
·
x
r
+ u
θ
·
−
y
r
2
.
u
xx
=(u
x
)
x
=(u
x
)
r
·
∂r
∂x
+(u
x
)
θ
∂θ
∂x
=
x
r
u
r
−
y
r
2
u
θ
r
=
x
r
+
x
r
u
r
−
y
r
2
u
θ
−
y
r
2
=
x
r
u
rr
−
x
r
2
u
r
+
1
r
u
r
∂x
∂r
x
r
−
y
r
2
u
rθ
−
2y
r
3
u
θ
+
1
r
2
∂y
∂r
u
θ
x
r
+
x
r
u
rθ
+
1
r
u
r
·
∂x
∂θ
−
y
r
2
+
y
r
2
u
θθ
+
1
r
2
u
θ
·
∂y
∂θ
y
r
2
710 Answers and H ints to Selected Exercises
=
x
2
r
2
u
rr
−
2xy
r
3
u
rθ
+
y
2
r
4
u
θθ
+
y
2
r
3
u
r
+
2xy
r
4
u
θ
.
Similarly,
u
yy
=
y
2
r
2
u
rr
+
2xy
r
3
u
rθ
+
x
2
r
4
u
θθ
+
x
2
r
3
u
r
−
2xy
r
4
u
θ
.
Adding gives the result: ∇
2
u = u
xx
+ u
yy
= u
rr
+
1
r
u
r
+
1
r
2
u
θθ
=0.
9. (c) Use Exercise 8.
10. (a) u
x
= u
ξ
ξ
x
+ u
η
η
x
= au
ξ
+ cu
η
=
a
∂
∂ξ
+ c
∂
∂η
u,
u
y
= u
ξ
ξ
y
+ u
η
η
y
= bu
ξ
+ du
η
=
b
∂
∂ξ
+ d
∂
∂η
u.
u
xx
=(u
x
)
x
=
a
∂
∂ξ
+ c
∂
∂η
a
∂
∂ξ
+ c
∂
∂η
u
=
a
2
u
ξξ
+2ac u
ξη
+ c
2
u
ηη
.
u
yy
=(u
y
)
y
=
b
∂
∂ξ
+ d
∂
∂η
b
∂
∂ξ
+ d
∂
∂η
u
= b
2
u
ξξ
+2bd u
ξη
+ d
2
u
ηη
.
u
xy
=(u
y
)
x
=
a
∂
∂ξ
+ c
∂
∂η
b
∂
∂ξ
+ d
∂
∂η
u
= ab u
ξξ
+(ad + bc) u
ξη
+ cd u
ηη
.
Consequently,
0=Au
xx
+2Bu
xy
+ Cu
yy
=
Aa
2
+2Bab + Cb
2
u
ξξ
+2 [ac A +(ad + bc) B + bd C] u
ξη
+
Ac
2
+2Bcd + Cd
2
u
ηη
.
Choose arbitrary constants a, b, c and d such that a = c = 1 and such
that b and d are the two roots of the equation
Cλ
2
+2Bλ + A =0, and
λ =
−B +
√
D
C
= b, d, D = B
2
− AC.
Thus, the transformed equation with a = c = 1 is given by
[A +(b + d) B + bd C] u
ξη
=0.
Or,
2
C
AC − B
2
u
ξη
=0.
If B
2
−AC > 0, the equation is hyperbolic, and the equation u
ξη
=0is
in the canonical form. The general solution of this canonical equation