Download HW 1 solutions. - Department of Mathematics

HOMEWORK 1
SHUANGLIN SHAO
1. P5. #1.
Proof. Equations 1, 2, 4, 7, 8 are linear equations. Equations 3, 5, 6 are
nonlinear equations.
2. P5. #2.
Proof. (a). linear.
L(au + bv) = (au + bv)x + x(au + bv)y
= aux + bvx + xauy + xbvy
= a(ux + xuy ) + b(vx + xvy )
= aL(u) + bL(v).
This proves that L is linear.
(b). nonlinear.
L(au + bv) = (au + bv)x + (au + bv)(au + bv)y
= aux + bvx + (au + bv)(auy + bvy )
= aux + bvx + a2 uuy + b2 vvy + abuvy + bavuy .
On the other hand,
aL(u) + bL(v) = a(ux + uuy ) + b(vx + vvy )
= aux + bvx + auuy + bvvy .
If a = 2, b = 0, these two identities above not equal; otherwise
uuy = 0, implies that (u2 )y = 0, i.e., u2 (x, y) = u2 (x, 0).
The last condition put a constraint on u, which is not known. So the linearity
fails.
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(c). nonlinear.
L(au + bv) = (au + bv)x + ((au + bv)y )2
= aux + bvx + (auy + bvy )2
= aux + bvx + a2 u2y + b2 vy2 + 2abuy vy .
On the other hand,
aL(u) + bL(v) = a(ux + u2y ) + b(vx + vy2 )
= aux + bvx + au2y + bvy2 .
Similarly as in (b), if a = 2, b = 0, if L is linear, u2y = 0. So
uy = 0.
This is not known. So linearity fails.
(d). nonlinear.
L(au + bv) = (au + bv)x + (au + bv)y + 1
= a(ux + uy ) + b(vx + vy ) + 1.
On the other hand
aL(u) + bL(v) = a(ux + uy + 1) + b(vx + vy + 1)
= a(ux + uy ) + b(vx + vy ) + a + b.
So if a = b = 1, if L is linear, we have 1 = 2. So linearity fails. Another way
to see this is that L does not map zero functions to zero.
(e). linear. The proof is similar to that in (a). So we omit it.
3. P5. # 3.
Proof. (a). It is a linear inhomogeneous equation. Define
L(u) = ut − uxx , g = −1.
(b). It is a linear homogeneous equation. Define
L(u) = ut − uxx + xu.
(c). nonlinear. The proof is similar to that in (b) in #2; so we omit it.
(d). It is a linear inhomogeneous equation. Define
L(u) = ut − uxx , g = −x2 .
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(e). It is a linear homogeneous equation. Define
L(u) = iut − uxx +
1
u.
x
(f ). nonlinear. If the operator is defined,
uy
ux
L(u) = p
+q
,
1 + u2x
1 + u2y
then
L(2u) 6= 2L(u).
Otherwise
2ux
1
1

!
p
−p
1 + u2x
1 + 4u2x

1
1
 = 0.
+ 2uy  q
−q
2
2
1 + uy
1 + 4uy
This equation puts a constraint on u, which is not known.
(g). It is a linear homogeneous equation. Define
L(u) = ux + ey uy .
(h). It is a linear inhomogeneous equation. If we define
√
L(u) = ut + uxxxx + 1 + u,
then L(0) = 1 6= 0. So linearity fails.
4. P5. # 4.
Proof. Let u and v solve the linear homogeneous equation L(u) = g. Then
we have
L(u) = g,
L(v) = g.
So we have
L(u − v) = 0.
So the difference u − v solves the equation L(u) = 0.
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5. P5. #10.
000
00
000
00
Proof. Define L(u) = u − 3u + 4u. Then the equation u − 3u + 4u = 0
reduces to the linear homogeneous equation
L(u) = 0.
Let V be the solution space of L(u) = 0. Let u, v ∈ V. For any a, b ∈ R,
then
au + bv ∈ V
because L(au + bv) = aL(u) + bL(v) = 0. So V is a vector space over R.
From the theory of ODE, V is spanned by the basis
{e−x , e2x , xe2x }.
6. P5. #11.
Proof. To the equation, uuxy = ux uy ,
LHS = f (x)g(y) (f (x)g(y))xy = f (x)fx (x)g(y)gy (y).
On the other hand,
RHS = ux uy = (f (x)g(y))x (f (x)g(y))y = f (x)fx (x)g(y)gy (y).
Thus u(x, y) = f (x)g(y) is a solution.
7. P9. #1.
Proof. By formula (2) in Section 1.2,
u(x, t) = f (2x − 3t),
where f is a function of one variable. Given the condition, we have
u(x, 0) = f (2x) = sin x.
So f (x) = sin x3 . Thus
u(x, t) = sin
2x − 3t
3
= sin(x − t).
2
2
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8. P9. #2.
Proof. Let v = uy . Then 3v + vx = 0. We multiply both sides by e3x ,
vx e3x + 3e3x v = 0.
By using the chain rule of differentiation,
(ve3x )x = 0.
Thus by the fundamental theorem of calculus,
v(x, y)e3x − v(0, y) = 0.
Set g(y) = v(0, y). Then
v(x, y) = e−3x g(y).
To solve for u,
uy = e−3x g(y).
Again by the fundamental theorem of calculus,
Z y
−3x
g(t)dt.
u(x, y) − u(x, 0) = e
0
Let f (x) = u(x, 0) and h(y) =
Ry
0
g(t)dt. We have
u(x, y) = f (x) + e−3x h(y),
where f and h are both functions of one variable.
9. P10. # 3.
Proof. We follow the method in solving Equation (4) in section 1.2. The
equation is
1
ux +
uy = 0.
1 + x2
1
So the directional derivative of u along the direction (1, 1+x
2 ) is zero. We
look for the equations of these characteristic curves y = y(x) in the plane
1
such that their directions are (1, 1+x
2 ). Their equations are
dy(x)
1
=
.
dx
1 + x2
Thus integrating from 0 to x, we obtain
Z x
1
y(x) − y(0) =
dt = tan−1 x.
2
1
+
t
0
Hence
y(x) = tan−1 x + y(0);
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that is to say, y − tan−1 x = C, where C = y(0). On each of these curves
u(x, y) is a constant because
uy
d
u(x, tan−1 x + C) = ux +
= 0.
dx
1 + x2
Thus u(x, tan−1 x + C) = u(0, tan−1 0 + C) = u(0, C) is independent of x.
Therefore there exists a function of one variable f such that
u(x, y) = f (y − tan−1 x).
10. P10. #5.
Proof. We still use the method in solving Equation (4) in section 1.2. The
equation is
y
ux + uy = 0.
x
So the directional derivative of u along the direction (1, xy ) is zero. We look
for the equations of these characteristic curves y = y(x) in the plane such
that their directions are (1, xy ). Their equations are
dy(x)
y
= .
dx
x
To solve this equation,
y 0 (x)
y(x)
= x1 . Thus
d(ln y(x))
1
= .
dx
x
Integrating this equation from 1 to x, we see that
ln y(x) = ln x + ln y(1);
that is to say, ln y(x) = ln y(1)x, where y(x) = Cx, where C = y(1). On
each of these curves u(x, y) is a constant because
d
y
u(x, Cx) = ux + Cuy = ux + uy = 0.
dx
x
Thus u(x, Cx) = u(1, C) is independent of x. Therefore there exists a function of one variable f such that
u(x, y) = f (y/x).
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11. P10. # 8.
Proof. The equation is
(1)
aux + buy + cu = 0.
Thus
uy
ux
+b
= −c.
u
u
Let v = ln u. The equation above is
a
avx + bvy + c = 0.
Thus without loss of generality, we assume that a 6= 0. Thus
c
c
a(v + x)x + b(v + x)y = 0.
a
a
Therefore we have
c
v + x = f (bx − ay),
a
where f is a function of one variable. Since u = ev ,
c
u(x, y) = e− a x+f (bx−ay) .
(2)
Another way is to use the coordinate method. From the equation a uux +b
−c. Let v = ln u. Then
avx + bvy = −c.
0
We change variables: x = bx − ay and y 0 = ax + by. Then
vx = bvx0 + avy0 ,
vy = −avx0 + bvy0 .
Thus
−c = avx + bvy = (a2 + b2 )vy0 .
Hence we have
vy 0 = −
c
.
a2 + b2
Integrating from 0 to y 0 , we have
v(x0 , y 0 ) − v(x0 , 0) = −
cy 0
.
a2 + b2
Therefore
−
u(x0 , y 0 ) = u(x0 , 0)e
cy 0
a2 +b2
,
i.e.,
(3)
u(x, y) = f (bx − ay)e
−
c
(ax+by)
a2 +b2
.
Remark. Equation (2) and (3) is in the same form. Writing
c
−
e− a x = e
c
bc
(ax+by)−
(bx−ay)
a2 +b2
a(a2 +b2 )
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.
uy
u
=
Then we write
− ac x+f (bx−ay)
e
−
=e
c
(ax+by)
a2 +b2
bc
−
2 +b2 ) (bx−ay) f (bx−ay)
a(a
e
,
e
where the second factor is a function of bx − ay.
12. P10. # 9.
Proof. Let v = u − x. Then the original equation is
vx + vy = 0.
Thus we see that
v(x, y) = f (x − y),
where f is a function of one variable. Therefore v(x, y) = x + f (x − y). 13. P10. # 10.
Proof. Let x0 = −x + y and y 0 = x + y. Then
ux = −ux0 + uy0 ,
uy = ux0 + uy0 ,
y 0 − x0
,
2
x0 + y 0
y=
.
2
Thus the equation that ux + uy + u = ex+2y becomes
x=
1 x0 3 0
1
uy0 + u = e 2 + 2 y .
2
2
y0
Multiplying both sides by e 2 ,
0 y
d
1 x0 +2y0
u
2
e
=
e2
.
dy 0
2
This implies that
y0
1 x0
e 2 u(x0 , y 0 ) − u(x0 , 0) = e 2
2
Hence
Z
0
y0
1 x0
0
e2t dy 0 = e 2 (e2y − 1).
4
y0
1 x0 −y0
0
u(x0 , y 0 ) = e 2 (e2y − 1) − e− 2 u(x0 , 0).
4
In terms of (x, y)-variables,
x+y
1 x+2y
u(x, y) =
e
− e−x − e− 2 u(y − x, 0).
4
8
One can verify that u(x, y) satisfies that ux +uy +u = ex+2y . But if imposing
the condition that u(x, 0) = 0, we obtain a contradiction. Indeed, since
u(x, 0) = 0 for any x,
u(y − x, 0) = 0.
Thus
1
(4)
u(x, y) = (ex+2y − e−x ).
4
This is also a solution to the equation. Let y = 0 in (4). Thus
1
u(x, 0) = (ex − e−x ),
4
which is generally not zero. Therefore there is no solution for this equation
with u(x, 0) = 0.
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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