February 2, 2020
APM 346
Justin Ko
Classifying PDEs
Problem 1. Consider first order equations and determine if they are linear homogeneous, linear
inhomogeneous, or nonlinear; for nonlinear equations, indicate if they are also semilinear, or quasilinear:
ut + xux − u = 0,
2
(1)
ut + ux − u = 0,
(2)
ut + uux + x = 0.
(3)
Solution 1.
(1) Since the coefficients in front of ut , ux , and u are functions of x and t only, the equation is linear.
There is also no term that depends only on x or t, so it is homogeneous. To prove that the equation
is linear, notice that
L[au + bv] = (au + bv)t + x(au + bv)x − (au + bv)
= a(ut + xux − u) + b(vt + xvx − v)
= aL[u] + bL[v].
(2) There is a u2 term, so the function is nonlinear. However, the coefficients of the highest order
terms are functions of x and t, so the function is semilinear. To prove that the operator is non-linear,
we show that the scaling property fails,
L[2x] = (2x)t + (2x)x − (2x)2 = 2 − 4x2 6= 2 − 2x2 = 2L[x].
(3) There is a uux term, so the function is nonlinear. However, the coefficients of the highest order
terms are functions of x, t and u, so the function is quasilinear.
Solving Basic PDEs
PDEs with only one term
Problem 2. Find the general solutions to the following equations:
uxxy = 0,
(1)
uxyz = sin(x) + sin(y) sin(z).
(2)
Solution 2.
(1) We integrate out each of the partial derivatives and introduce an integration constant in each
step,
uxxy = 0
⇒ uxx = f (x)
⇒ ux = f˜(x) + g(y)
⇒
f˜x = f
˜
˜
u = f˜(x) + xg(y) + h(y) f˜xx = f
˜
where f˜ is a twice differentiable function.
Page 1 of 5
February 2, 2020
APM 346
Justin Ko
(2) We integrate out each of the partial derivatives and introduce an integration constant in each
step,
uxyz = sin(x) + sin(y) sin(z)
⇒ uxy = z sin(x) − sin(y) cos(z) + f (x, y)
⇒ ux = yz sin(x) + cos(y) cos(z) + f˜(x, y) + g(x, z)
⇒
f˜y = f
˜
˜
u = −yz cos(x) + x cos(y) cos(z) + f˜(x, y) + g̃(x, z) + h(y, z) g̃x = g, f˜yx = f ,
˜
where f˜ is differentiable in each of its coordinates, and g̃ is differentiable in its first coordinate.
Semilinear First Order PDEs
Problem 3. Find the general solutions to the following equations
ut − 4ux + u = 0,
−2ux + 4uy = e
x+3y
− 5u.
(1)
(2)
Solution 3.
(1) We have the system of equations
dx
du
dt
=
=
.
1
−4
−u
Characteristic Curve: We start by solving the equation involving the first and second term,
dt
dx
dx
=
⇒
= −4 ⇒ C = x + 4t.
1
−4
dt
General Solution: We now solve the equation involving the first and third term,
du
du
dt
=
⇒
= −u.
1
−u
dt
This is a separable ODE, which has solution
log |u| = −t + f (C) ⇒ u = ±ef (C) e−t
Since f (C) is an arbitrary function, we might can redefine ±ef (C) =: g(C). Since C = x + 4t, we have
our general solution is
u(t, x) = g(x + 4t)e−t .
(2) We have the system of equations
dx
dy
du
=
= x+3y
.
−2
4
e
− 5u
Characteristic Curve: We start by solving the equation involving the first and second term,
dx
dy
dy
=
⇒
= −2 ⇒ C = y + 2x.
−2
4
dx
General Solution: We now solve the equation involving the first and second term,
dx
du
du
1
du 5
1
= x+3y
⇒
= − (ex+3y − 5u) ⇒
− u = − ex+3y .
−2
e
− 5u
dx
2
dx 2
2
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February 2, 2020
APM 346
Justin Ko
There is a y variable appearing in this ODE that we must eliminate it. Since y = C − 2x, we need to
solve
du 5
1
− u = − e−5x+3C .
dx 2
2
5
This is a linear ODE, which can be solved using an integrating factor of the form φ(x) = e− 2 x , which
gives us
Z
15
5
5
5
1 −5x+3C 1
1
1 5 2e− 2 x+3C
e−5x+3C e− 2 x dx = − e 2 x
+ f (C) ⇒ u =
e
− f (C)e 2 x .
u = e2x −
2
2
−15
15
2
Since C = y + 2x, if we set g(z) = − 12 f (z) then we get the general solution
u(x, y) =
5
1 x+3y
e
+ g(y + 2x)e 2 x .
15
Problem 4. Solve the initial value problem
2xyux + (x2 + y 2 )uy = 0
with u(x, y) = exp(x/(x − y)) on {x + y = 1}.
Solution 4. We have the system of equations
dy
du
dx
= 2
=
.
2
2xy
(x + y )
0
Characteristic Curves: We start by solving the equation involving the first and second term,
dx
dy
dy
1 x 1 y
= 2
⇒
= · + · .
2
2xy
(x + y )
dx
2 y
2 x
This is a Homogenous ODE, which can be solved using the change of variables w =
dy
dw
dx = x dx + w, so under this change of variables we have
y
x.
We have
1
1
dw
1
1
1 − w2
dw
+ w = · w−1 + · w ⇒ x
= · w−1 − · w =
.
dx
2
2
dx
2
2
2w
This is a separable equation, so
x
2w
x2 − y 2
1
.
dw = dx ⇒ − ln(1 − w2 ) = ln x + D ⇒ e−D = x(1 − w2 ) =
2
1−w
x
x
If we set C = e−D , then C =
x2 −y 2
x
is our characteristic curve.
General Solution: We now solve the equation involving the first and third term,
dx
du
du
=
⇒
= 0 ⇒ u = f (C).
2xy
0
dx
Since C =
x2 −y 2
x ,
we have our general solution is
u(x, y) = f
x2 − y 2 .
x
Particular Solution: We now use the initial value to solve for f . Since u(x, y) = exp(x/(x − y)) when
x + y = 1, we have
x2 − y 2 (x − y)(x + y) x − y
x
e x−y = u(x, y) x+y=1 = f
=f
=f
.
x
x
x
x+y=1
x+y=1
If we set z =
x−y
x ,
1
then the above implies e z = f (z), so our particular solution is of the form
x
u(x, y) = e x2 −y2 .
Page 3 of 5
February 2, 2020
APM 346
Justin Ko
Semilinear First Order PDEs in Higher Dimensions
Problem 5. Find the general solution to the equation
ux + 3uy − 2uz = u.
Solution 5. We have the system of equations,
dx
dy
dz
du
=
=
=
.
1
3
−2
u
Characteristic Curves Part 1: We start by solving the equation involving the first and second term,
dx
dy
=
=⇒ C = 3x − y.
1
3
Characteristic Curves Part 2: We now solve the equation involving the first and third term,
dz
dx
=
=⇒ D = 2x + z.
1
−2
General Solution: We now solve the equation involving the first and fourth term,
du
dx
=
=⇒ x = log |u| + f (C, D) =⇒ u(x, y, z) = f (C, D)ex = f (3x − y, 2x + z)ex ,
1
u
for some function f of two variables.
Problem 6. Find the general solution to the equation
ut + yux + xuy = 0.
Find the particular solution when u(0, x, y) = f (x, y).
Solution 6. We have the system of equations,
dx
dy
du
dt
=
=
=
.
1
y
x
0
Characteristic Curves Part 1: We start by solving the equation involving the second and third term,
dx
dy
=
=⇒ x2 = y 2 + C =⇒ C = x2 − y 2 .
y
x
Characteristic
√ Curves Part 2: We now solve the equation involving the first and second term using
the fact y = x2 − C,
p
dt
dx
dx
(x + y)
=
=√
=⇒ t = log | x2 − C + x| + D = log |y + x| + D =⇒ D =
.
2
1
y
et
x −C
General Solution: We now solve the equation involving the first and fourth term,
du
(x + y) dt
=
=⇒ u(t, x, y) = g(C, D) = g x2 − y 2 ,
.
1
0
et
Particular Solution: Plugging in our initial conditions, we have
u(0, x, y) = g(x2 − y 2 , x + y) = f (x, y).
Page 4 of 5
February 2, 2020
APM 346
Justin Ko
We set u = x2 − y 2 and v = x + y. Our goal is to write x and y as some functions of u and v. We see
that
u
u = x2 − y 2 = (x − y)(x + y) = (x − y)v =⇒ x − y = .
v
Since x + y = v and x − y = uv , we can add and subtract our answers to conclude
x=
1
u
v+
2
v
so
u(0, x, y) = g(u, v) = f
y=
1
2
1
u
v−
,
2
v
v+
u 1
u ,
v−
.
v 2
v
Therefore, our particular solution is given by
(x + y) u(t, x, y) = g x2 − y 2 ,
et
1 x + y x2 − y 2 1 x + y x2 − y 2 =f
+ (x+y) ,
− (x+y)
2
et
2
et
et
et
1 x + y x − y 1 x + y x − y =f
+ −t ,
− −t
.
2
et
e
2
et
e
Remark: We were a bit sloppy with the constants and domains of our functions above. The constants
C and D changed each line and writing x2 − y 2 = C implicitly in terms of y depends on the value of
C. We should check our general solution to ensure that it is a solution to our PDE by differentiating.
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