Week 02 : Method of C haracteristics
From now on we will study one by one classical techniques of obtaining solution formulas for PDEs. The
first one is the method of characteristics, which is particularly useful when solving first order equations.
1 . C lassification of first-order equations.
The general form of first-order PDE ( in R2 ) :
F( x , y , u , u x , u y ) = 0,
or in
R3 :
(1)
( x , y) ∈ D ⊂ R2 .
( 2)
F( x , y , z , u , u x , u y , u z ) = 0.
Often the following notation is used in 2D:
p = ux , q = u y
( 3)
F( x , y , u , p, q) = 0.
( 4)
thus the equation can be written as
The linear equations can be classificed into the following cases, from easier to more difficult:
1 . Linear:
2. Semi-linear:
3. Quasi-linear:
4. General case.
F( x , y , u , u x , u y ) = a( x , y) u x + b( x , y) u y + c( x , y) u − d( x , y) .
( 5)
F( x , y , u , u x , u y ) = a( x , y) u x + b( x , y) u y − c( x , y , u) .
( 6)
F( x , y , u , u x , u y ) = a( x , y , u) u x + b( x , y , u) u y − c( x , y , u) .
( 7)
2. Method of characteristics.
We try to find a method to solve the general first-order quasi-linear equation
( 8)
a( x , y , u) u x + b( x , y , u) u y = c( x , y , u) .
Let’ s start by thinking geometrically. Consider the 3 dimensional space with coordinates ( x , y , u) .
Assume that u = u( x , y) is a solution to the equation, it is clear that it represents a surface in the ( x , y, u)
space.
Now we explore the geometrical meaning of the equation. Observe that the equation can be written in
the form of an inner product
 


 


a( x , y , u)
a( x , y , u)
ux
ux
 b( x , y , u)  ⊥ u y  .
 b( x , y , u)  ·  u y  = 0
( 9)
c( x , y , u)
−1
c( x , y , u)
−1
Therefore all we need to do is to understand the relation between the vector
First introduce a new function Ψ : R
3
R through
Φ ( x , y , u) = u − u( x , y) .



ux
uy 
− 1
and the solution.
( 1 0)
Note that in the RHS of the above, the first u is a variable, the second u is a function. For example, suppose u( x , y) = x 2 + y 2 , then the corresponding Φ ( x , y, u) = u − x 2 + y 2 .
Now we easily see that




Φx
ux
 u y  = −  Φ y  = − ∇Φ .
(11)
Φu
−1
Recall
ux
 uy 
− 1
that geometrically, ∇Φ ( and also − ∇Φ ) is a normal vector of the surface Φ = 0. As a consequence
is perpendicular to the solution surface u = u( x , y) .
On the other hand, from the equation we know that
means
a
 b 
c
must be tangent to the surface u = u( x , y) .


a
 b 
c
is perpendicular to the vector



ux
uy 
− 1
which
Now we summarize.
We

 have shown that the equation is equivalent to the geometrical requirement
that the vector
curve


a
of  b  ,
c
a ( x , y , u)
 b ( x , y , u) 
c( x , y , u)

that is any
is tangent to the solution surface u = u( x , y) . As a consequence, any integral

X( s)
 Y( s) 
U( s)
satisfying
dX
= a( X , Y , U)
( 1 2)
ds
dY
= b( X , Y , U)
( 1 3)
ds
dU
= c( X , Y , U)
( 1 4)
ds
must be contained in one of the solution surfaces. Conversely, any surface “woven” by such integral curves
is a solution surface.
The above understanding leads to the following “method of characteristics” due to Lagrange.
Theorem 1 . The general solution of a first- order, quasi- linear PDE
a( x , y , u) u x + b( x , y , u) u y = c( x , y , u)
( 1 5)
F( φ , ψ) = 0,
( 1 6)
satisfies
where F is an arb itrary function of φ( x , y, u) and ψ( x , y , u) , and any intersection of the level sets of φ
and ψ is a solution of the characteristic equations
dx d y du
=
=
.
a
b
c
( 1 7)
Remark 2. As we will see soon, φ , ψ are obtained through solving the characteristic equations. And each
F gives a solution to the original equation.
Remark 3. The curves mentioned above are called the families of characteristic curves of the equation.
Remark 4. As we will see, the main technique in getting φ and ψ is
a±c a c
= = .
b±d b d
a c
=
b d
( 1 8)
In the following, we will show how to apply this method. We start with the simplest case.
3. Solving linear first-order equations.
3. 1 . Equations with constant coefficients.
We start from the simplest case, where a , b , c and d are just constants.
Example 5. ( § 2. 8, 3( b) ) Find the general solution of the equation
a u x + b u y = 0;
a , b are constant.
( 1 9)
Solution. The characteristic equations are
dx d y du
=
=
.
b
0
a
( 20)
What we need are two functions φ( x , y , u) and ψ( x , y , u) such that d φ = 0, dψ = 0 along the characteristics.
Obviously we can take φ = u. For ψ, notice that
thus we can take
d( a y − b x) = a d y − b dx = 0,
( 21 )
ψ = a y − b x.
( 22)
F( a y − b x , u) = 0
( 23)
u = f( a y − b x) .
( 24)
As a consequence, the solution satisfies
for any function F. This means
for an arbitrary function f.
Example 6. ( C auchy problem, § 2. 8, 5( a) ) Oftentimes, the value of the solution along some specific
curve in the plane is prescribed. For example, solve
u( x , 0) = sin x.
3 u x + 2 u y = 0,
( 25)
Solution. From the above example we know that the general solution takes the form
( 26)
u = f( 2 x − 3 y) .
Now substituting this into the initial condition, we obtain
x
f( x) = sin .
2
f( 2 x) = sin x
Therefore
u( x , y) = sin
( 27)
2x−3y
.
2
Example 7. ( c, d 0) What happens when the c, d are not 0? The method still works. We write
the characteristic equations
dx dy
du
=
=
.
a
b
d− cu
We have
dx dy
=
d( b x − a y) = 0,
a
b
−1
−1
dx
du
du
=
= a− 1 d − a− 1 c u
u = C e − a c x + c− 1 d
d e a c x u − c− 1 d = 0
a
d− cu
dx
As a consequence
φ = b x − a y,
and
ψ = ea
− 1
cx
u − c− 1 d
F( φ , ψ) = 0
down
( 28)
( 29)
( 30)
( 31 )
( 32)
gives
u = c− 1 d + e − a
− 1
cx
f( b x − a y) .
Remark 8. In the above, one may be tempted to conclude
− 1
d u − C e− a cx = 0
−1
( 33)
( 34)
and try to use u − C e − a c x as ψ. This is obviously wrong as d somehow disappeared. One should keep
in mind that neither φ or ψ can involve arbitrary constants. It is their values that are arbitrary constants.
Example 9. Some times people use the following “method of characteristics”:
First solve
dy
=b
ds
dx
= a,
ds
then solve
us + c u = d
( 35)
( 36)
to obtain the solution u in the form u = u( s , t) . Finally represent s , t by x , y and obtain the solution. This
is equivalent to our method but considerably more complicated to use. Anyone who does not believe this
should try using this method to the following examples with non-constant coefficients.
Remark 1 0. One may wonder, how to find out φ and ψ efficiently? Unfortunately there may not be any
short-cut. One way to systematically find φ and ψ is the following. The characteristic equations consists
of three equations. Pick any two of them. If you can find general solutions, then you have φ and ψ. But
this fails when any coefficient involves all other variables.
3. 2. Equations with non-constant coefficients.
Example 1 1 . ( § 2. 8, 3( h) ) Find the general solution of
y u y − x ux = 1 .
( 37)
dx
d y du
=
=
.
−x
y
1
( 38)
dy
dx
=
y
−x
( 39)
Solution. The characteristic equations are
Using
we obtain
y dx + x dy = 0
d( x y) = 0.
( 40)
Thus
On the other hand, from
we obtain
du = d log y
( 41 )
d y du
=
1
y
( 42)
d( u − log y) = 0.
As a consequence we can take
( 43)
ψ = u − log y.
( 44)
F( x y , u − log y) = 0
( 45)
u = log y + f( x y) .
( 46)
Putting these together we obtain
which gives
φ = x y;
Example 1 2. ( § 2. 8, 5( c) ) Find the solution of the following Cauchy problem:
x ux + y u y = 2 x y ,
with u = 2 on y = x 2 .
( 47)
Solution. We need to first find the general solution, then using the value on y = x 2 to determine the arbitrary function involved.
−
Find the general solution
The characteristic equations are
Using
dx d y
du
=
=
.
x
y
2xy
( 48)
dx d y
=
y
x
( 49)
we obtain
d
On the other hand, we have
y
=0
x
can take φ =
y
.
x
( 50)
d( u − x y) = 0
du = 2 x dy = 2 y dx = x dy + y dx = d( x y)
therefore
( 52)
ψ = u − x y.
−
The general solution satisfies
y
F , u− xy =0
x
Determine the solution.
We have u = 2 along y = x 2 , that is
( 51 )
u=xy+ f
y
.
x
( 53)
u x , x 2 = 2.
( 54)
Using the formula for the general solution, we have
x 3 + f( x) = 2
As a consequence
f( x) = 2 − x 3 .
( 55)
y 3
.
x
( 56)
u( x , y) = x y + 2 −
4. Solving semi-linear first-order equations.
Example 1 3. ( § 2. 8, 3( g) ) Find the general solution of the following equation:
y 2 u x − x y u y = x ( u − 2 y) .
( 57)
dy
du
dx
=
=
.
− x y x ( u − 2 y)
y2
( 58)
dy
dx
=
y2
− xy
( 59)
Solution. The characteristic equations are
From
we have
On the other hand, we have
dy
du
=
− x y x ( u − 2 y)
d x2 + y2 = 0
du
u
=− +2
dy
y
( 60)
φ = x2 + y2 .
d( u − y)
u− y
=−
y
dy
d( ( u − y) y) = 0.
( 61 )
Thus we take
ψ = y ( u − y) .
Now
F x 2 + y 2 , y ( u − y) = 0
gives
Example 1 4. ( § 2. 8, 5( g) ) Solve
u = y + y− 1 f x2 + y2 .
x ux + y u y = u + 1
with u( x , y) = x 2 on y = x 2 .
( 62)
( 63)
( 64)
( 65)
Solution. The characteritic equations are
dx dy
du
=
=
x
y
u+1
( 66)
which easily lead to
y
,
x
φ=
Thus
u=xf
Now the Cauchy data implies
ψ=
u+1
.
x
( 67)
y
− 1.
x
( 68)
x f( x) − 1 = u x , x 2 = x 2
thus
( 70)
f( x) = x + x − 1 .
As a consequence
u( x , y) = x
( 69)
y
y −1
x2
− 1 = y+
+x
− 1.
x
x
y
( 71 )
5. Solving quasi-linear first-order equations.
Example 1 5. ( § 2. 8, 3( f) ) Find the general solution of
( y + u) u x + y u y = x − y.
( 72)
dy
du
dx
=
=
.
y
x− y
y+u
( 73)
Solution. The characteristic equations are
From this we have
d x 2 − ( y + u) 2 = 0.
d( y + u)
dx
=
y+u
x
Thus we can take
φ = x 2 + ( y + u) 2 .
( 74)
( 75)
On the other hand, we have
2
d u 2 − ( x − y) = 0.
d( x − y)
du
=
x− y
u
As a consequence, the solution is given by
F x 2 − ( y + u) 2 , u 2 − ( x − y) 2 = 0.
Remark 1 6. Note that, in the above we can also use
d y d( x + u)
x+u
=
d
= const.
x+u
y
y
( 76)
( 77)
( 78)
Thus the formula may not be unique.
Example 1 7. ( § 2. 8, 5( h) ) solve
u u x − u u y = u 2 + ( x + y)
2
with u = 1 on y = 0.
( 79)
Solution. The characteristic equation are
We have
Then we have
dy
du
dx
=
=
.
− u u 2 + ( x + y) 2
u
dx
dy
=
u
−u
d( x + y) = 0
dy
du
=
− u u2 + φ2
φ = x + y.
ψ = e2 y u2 + φ2 .
( 80)
( 81 )
( 82)
Now from the Cauchy data we have
Therefore effectively F has to be
and the solution satisfies
which leads to
F φ , 1 + φ 2 ≡ 0.
F( φ , ψ) = ψ − φ 2 + 1
( 83)
( 84)
e 2 y u 2 + ( x + y) 2 − 1 + ( x + y) 2 = 0
u= ±
hn
1 + ( x + y)
2
o
e − 2 y − ( x + y)
i
2 1 /2
( 85)
( 86)
.
6. Equations with more than two variables.
The method of characteristics can be applied to higher dimensional problems with no difficulty in principle – it indeed becomes more difficult in practice!
Example 1 8. ( § 2. 8, 8( d) ) Solve the following equation
y z u x − x z u y + x y x 2 + y 2 u z = 0.
( 87)
Solution. The characteristic equations are
dy
dz
du
dx
=
=
=
.
0
y z − x z x y ( x2 + y2 )
( 88)
This time we need three invariants, let’ s denote them by φ , ψ and η.
Clearly one can take
( 89)
φ = u.
For the second invariant, we observe
dx
dy
=
yz − xz
dx
dy
=
y
−x
d x2 + y2 = 0
dy
dz
=
− xz xyφ
dy
dz
=
− z yφ
d z 2 + y2 φ = 0
The last invariant can be obtained through
Therefore the solutions are obtained by setting
F u , x2 + y2 , z 2 + y2 x2 + y2
which gives
u = f x 2 + y2 , z 2 + y2 x2 + y2
for arbitrary
f. 1
ψ = x2 + y2 .
η = z 2 + y 2 φ.
=0
1 . K eep in mind that this f will be determined once some C auchy data is given.
( 90)
( 91 )
( 92)
( 93)