Math 447/547 Partial Differential Equations Homework 2
Section 1.2
2. Solve
3uy + uxy = 0.
Let v = uy , so the equation becomes
3v + vx = 0.
Multiply by e3x to get
e3x [3v + vx ] = 0 =
∂ 3x
e v.
∂x
Thus
e3x v(x, y) = f (y),
uy (x, y) = f (y)e−3x ,
and finally
u(x, y) = F (y)e−3x + g(x),
where
′
F (y) = f (y),
or F (y) =
Z
y
f (s) ds.
3. Solve
(1 + x2 )ux + uy = 0.
The characteristic curves satisfy
dy
1
=
.
dx
1 + x2
Integration then gives
y(x) = tan−1 (x) + c,
or
c = y − tan−1 (x).
Solutions have the form
u(x, y) = f (x) = f (y − tan−1 (x)).
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5. Solve the equation
xux + yuy = 0.
The characteristic curves satisfy
y
dy
= ,
dx
x
or
1
1
dy = dx.
y
x
Integration then gives
log y = logx + c,
or
y = ec x,
y/x = K = ec .
Solutions have the form
u(x, y) = f (y/x).
8. Solve
aux + buy + cu = 0.
There are at least two ways to solve this problem. One approach is to
multiply by an integrating factor
eax/c [aux + buy + cu] = 0.
Letting v = ecx/a u the equation then becomes
avx + bvy = 0,
and this can be solved by the method of characteristics.
A second approach starts with the substitution from page 7,
s = ax + by,
t = bx − ay.
The equation becomes
(a2 + b2 )us (s, t) + cu(s, t) = 0.
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Rewrite this as
us (s, t) + αu(s, t) = 0,
α=
c
.
a2 + b2
Now use the integrating factor eαs , getting
eαs [us (s, t) + αu(s, t)] = 0 =
∂ αs
[e u(s, t)].
∂s
This gives
[eαs u(s, t)] = f (t),
and so
u(x, y) = exp(−
u(s, t) = e−αs f (t)
c
(ax + by))f (bx − ay).
a2 + b2
If the first approach is used you will probably obtain a solution formula
with a different exponential factor. The ’missing’ factor is just absorbed
into the f (bx − ay) term, although this may not be obvious.
9. Solve the equation
ux + uy = 1.
This is a nonhomogeneous linear equation, so our first job is to find a
single solution up (x, y). We can take up (x, y) = x. The general solution to
the nonhomogeneous equation is up plus any solution of the homogeneous
equation
ux + uy = 0.
These have the form
uh (x, y) = f (y − x),
so the solutions of the original equation are
u(x, y) = x + f (y − x).
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