Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers
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2.7 Method of Separation of Variables 51
ln u (ξ, η)=−η +lnf (ξ) ,
where f (ξ) is an arbitrary function of ξ only.
Equivalently,
u (ξ ,η)=f (ξ) e
−η
.
In terms of the original variables x and y , the general solution of equa-
tion (2.6.8) is
u (x , y)=f (x + y) e
−y
, (2.6.10)
where f is an arbitrary function.
The characteristic equations of (2.6.9) are
dx
y
=
dy
1
=
du
x
.
It follows from the first two equations that ξ (x, y)=x−
y
2
2
= c
1
; we choose
η (x, y)=y = c
2
. Consequently, u
x
= u
ξ
and u
y
= −yu
ξ
+ u
η
and hence,
equation (2.6.9) reduces to
u
η
= ξ +
1
2
η
2
.
Integrating this equation gives the general solution
u (ξ ,η)=ξη +
1
6
η
3
+ f (ξ) ,
where f is an arbitrary function.
Thus, the general solution of (2.6.9) in terms of x and y is
u (x , y)=xy −
1
3
y
3
+ f
x −
y
2
2
.
2.7 Method of Separation of Variables
During the last two centuries several methods have been developed f or solv-
ing partial differential equations. Among these, a technique known as the
method of separation of variables is perhaps the oldest systematic method
for solving partial differential equations. Its essential feature is to transform
the partial differential equations by a set of ordinary differential equations.
The required solution of the partial differential equations is then exposed as
a product u (x, y)=X (x) Y (y) =0,orasasumu (x, y)=X (x)+Y (y),
where X (x)andY (y) are functions of x and y, respectively. Many signifi-
cant problems in partial differential equations can be solved by the method
52 2 First-Order, Quasi-Linear Equations and Method of Characteristics
of separation of variables. This method has been considerably r efined and
generalized over the last two centuries and is one of the classical techniques
of applied mathematics, mathematical physics and engineering science.
Usually, the fi rst-order partial differential equation can be solved by sep-
aration of variables without the need for Fourier series. The main purpose
of this section is to illustrate the method by examples.
Example 2.7.1. Solve the initial-value problem
u
x
+2u
y
=0,u(0,y)=4e
−2y
. (2.7.1ab)
We seek a separable solution u (x, y)=X (x) Y (y) = 0 and substitute
into the equation to obtain
X
′
(x) Y (y)+2X (x) Y
′
(y)=0.
This can also be expressed in the form
X
′
(x)
2X (x)
= −
Y
′
(y)
Y (y)
. (2.7.2)
Since the left-hand side of this equation is a function of x only and the
right-hand is a function of y only, it follows that (2.7.2) can be true if both
sides are equal to the same constant value λ which is called an arbitrary
separation constant. Consequently, (2.7.2) gives two ordinary differential
equations
X
′
(x) − 2λX (x)=0,Y
′
(y)+λY (y)=0. (2.7.3)
These equations have solutions given, respectively, by
X (x)=Ae
2λx
and Y (y)=Be
−λy
, (2.7.4)
where A and B are arbitrary i ntegrating constants.
Consequently, the general solution is given by
u (x , y)=AB exp (2λx − λy)=C exp (2λx − λy) , (2.7.5)
where C = AB is an arbitrary constant.
Using the condition (2.7.1b), we find
4 e
−2y
= u (0,y)=Ce
−λy
,
and hence, we deduce that C = 4 and λ = 2. Therefore, the final solution
is
u (x , y)=4exp(4x − 2y) . (2.7.6)
2.7 Method of Separation of Variables 53
Example 2.7.2. Solve the equation
y
2
u
2
x
+ x
2
u
2
y
=(xyu)
2
. (2.7.7)
We assume u (x, y)=f (x) g (y) = 0 is a separable solution of (2.7.7),
and substitute into the equation. Consequently, we obtain
y
2
{f
′
(x) g (y)}
2
+ x
2
{f (x) g
′
(y)}
2
= x
2
y
2
{f (x) g (y)}
2
,
or, equivalently,
1
x
2
f
′
(x)
f (x)
2
+
1
y
2
g
′
(y)
g (y)
2
=1,
or
1
x
2
f
′
(x)
f (x)
2
=1−
1
y
2
g
′
(y)
g (y)
2
= λ
2
,
where λ
2
is a separation constant. Thus,
1
x
f
′
(x)
f (x)
= λ and
g
′
(y)
yg(y)
=
1 − λ
2
. (2.7.8)
Solving these ordinary differential equations, we find
f (x)=A exp
λ
2
x
2
and g (y)=B exp
1
2
y
1 − λ
2
,
where A and B are arbitrary con stant. Thus, the general solution is
u (x , y)=C exp
λ
2
x
2
+
1
2
y
2
1 − λ
2
, (2.7.9)
where C = AB is an arbitrary constant.
Using the condition u (x, 0) = 3 exp
x
2
/4
, we can determine both C
and λ in (2.7.9). It turns out that C = 3 and λ =(1/2), and the solution
becomes
u (x , y)=3exp
1
4
x
2
+ y
2
√
3
. (2.7.10)
Example 2.7.3. Use the separation of variables u (x, y)=f (x)+g (y)to
solve the equation
u
2
x
+ u
2
y
=1. (2.7.11)
Obviously,
54 2 First-Order, Quasi-Linear Equations and Method of Characteristics
{f
′
(x)}
2
=1−{g
′
(y)}
2
= λ
2
,
where λ
2
is a separation constant. Thus, we obtain
f
′
(x)=λ and g
′
(y)=
1 − λ
2
.
Solving these ordinary differential equations, we find
f (x)=λx + A and g (y)=y
1 − λ
2
+ B,
where A and B ar e constants of integration. Finally, the solution of (2.7.11)
is given by
u (x , y)=λx + y
1 − λ
2
+ C, (2.7.12)
where C = A + B is an arbitr ary constant.
Example 2.7.4. Use u (x, y)=f (x)+g (y) to solve the equation
u
2
x
+ u
y
+ x
2
=0. (2.7.13)
Obviously, equation (2.7.13) has the separable form
{f
′
(x)}
2
+ x
2
= −g
′
(y)=λ
2
,
where λ
2
is a separation constant.
Consequently,
f
′
(x)=
λ
2
− x
2
and g
′
(y)=−λ
2
.
These can be integrated to obtain
f (x)=
λ
2
− x
2
dx + A
= λ
2
cos
2
θdθ+ A, (x = λ sin θ)
=
1
2
λ
2
sin
−1
x
λ
+
x
λ
1 −
x
2
λ
2
+ A
and
g (y)=−λ
2
y + B.
Finally, the general solution is given by
u (x , y)=
1
2
λ
2
sin
−1
x
λ
+
x
2
λ
2
− x
2
− λ
2
y + C, (2.7.14)
where C = A + B is an arbitr ary constant.
2.8 Exercises 55
Example 2.7.5. Use v =lnu and v = f (x)+g (y) to solve the equation
x
2
u
2
x
+ y
2
u
2
y
= u
2
. (2.7.15)
In view of v =lnu, v
x
=
1
u
u
x
and v
y
=
1
u
u
y
, and hence, equation (2.7.15)
becomes
x
2
v
2
x
+ y
2
v
2
y
=1. (2.7.16)
Substitute v (x, y)=f (x)+g (y) into (2.7.16) to obtain
x
2
{f
′
(x)}
2
+ y
2
{g
′
(y)}
2
=1
or
x
2
{f
′
(x)}
2
=1− y
2
{g
′
(y)}
2
= λ
2
,
where λ
2
is a separation constant. Thus, we obtain
f
′
(x)=
λ
x
and g
′
(y)=
1
y
1 − λ
2
.
Integrating these equations gives
f (x)=λ ln x + A and g (y)=
1 − λ
2
ln y + B,
where A and B are integrating constants. Therefore, the general solution
of (2.7.16) is given by
v (x, y)=λ ln x +
1 − λ
2
ln y +lnC
=ln
x
λ
· y
√
1−λ
2
· C
, (2.7.17)
where ln C = A + B. The final solution is
u (x , y)=e
v
= Cx
λ
· y
√
1−λ
2
, (2.7.18)
where C is an integrating constant.
2.8 Exercises
1. (a) Show that the family of right circular cones whose axes coincide
with the z-axis
x
2
+ y
2
=(z − c)
2
tan
2
α
satisfies the first-order, partial differential equation
56 2 First-Order, Quasi-Linear Equations and Method of Characteristics
yp − xq =0.
(b) Show that all the surfaces of revolution, z = f
x
2
+ y
2
with the
z-axis as th e axis of symmetry, where f is an arbitrary function, satisfy
the partial differential equation
yp − xq =0.
(c) Show that the two-parameter family of curves u −ax −by −ab =0
satisfies the nonlinear equation
xp + yq + pq = u.
2. Find the partial differential equation arising from each of the following
surfaces:
(a) z = x + y + f (x y), (b) z = f (x − y),
(c) z = xy + f
x
2
+ y
2
,(d)2z =(αx + y)
2
+ β.
3. Find the general solution of each of the following equations:
(a) u
x
=0, (b)au
x
+ bu
y
=0;a, b, are constant,
(c) u
x
+ yu
y
=0, (d)
1+x
2
u
x
+ u
y
=0,
(e) 2xy u
x
+
x
2
+ y
2
u
y
=0, (f) (y + u) u
x
+ yu
y
= x − y,
(g) y
2
u
x
− xy u
y
= x (u − 2y), (h) yu
y
− xu
x
=1,
(i) y
2
up + u
2
xq = −xy
2
,(j)(y − xu) p +(x + yu) q = x
2
+ y
2
.
4. Find the general solution of the equation
u
x
+2xy
2
u
y
=0.
5. Find the solution of the following Cauchy problems:
(a) 3u
x
+2u
y
=0,withu (x, 0) = sin x,
(b) yu
x
+ xu
y
=0,withu (0,y)=exp
−y
2
,
(c) xu
x
+ yu
y
=2xy,withu =2ony = x
2
,
(d) u
x
+ xu
y
=0,withu (0,y)=siny,
(e) yu
x
+ xu
y
= xy, x ≥ 0, y ≥ 0, with u (0,y)=exp
−y
2
for y>0, and u (x, 0) = exp
−x
2
for x>0,
2.8 Exercises 57
(f) u
x
+ xu
y
=
y −
1
2
x
2
2
,withu (0,y)=exp(y),
(g) xu
x
+ yu
y
= u +1,with u (x, y)=x
2
on y = x
2
,
(h) uu
x
− uu
y
= u
2
+(x + y)
2
,withu =1ony =0,
(i) xu
x
+(x + y) u
y
= u +1,with u (x, y)=x
2
on y =0.
(j)
√
xu
x
+ uu
y
+ u
2
=0, u (x, 0) = 1, 0 <x<∞.
(k) ux
2
u
x
+ e
−y
u
y
+ u
2
=0, u (x , 0) = 1, 0 <x<∞.
6. Solve the initial-value problem
u
t
+ uu
x
=0
with the initial curve
x =
1
2
τ
2
,t= τ,u = τ.
7. Find the solution of the Cauchy problem
2xy u
x
+
x
2
+ y
2
u
y
=0, with u =exp
x
x − y
on x + y =1.
8. Solve the following equations:
(a) xu
x
+ yu
y
+ zu
z
=0,
(b) x
2
u
x
+ y
2
u
y
+ z (x + y) u
z
=0,
(c) x (y − z) u
x
+ y (z − x) u
y
+ z (x − y) u
z
=0,
(d) yz u
x
− xz u
y
+ xy
x
2
+ y
2
u
z
=0,
(e) x
y
2
− z
2
u
x
+ y
z
2
− y
2
u
y
+ z
x
2
− y
2
u
z
=0.
9. Solve the equation
u
x
+ xu
y
= y
with the Cauchy data
(a) u (0,y)=y
2
, (b) u (1,y)=2y.
58 2 First-Order, Quasi-Linear Equations and Method of Characteristics
10. Show that u
1
= e
x
and u
2
= e
−y
are solutions of the nonlinear equation
(u
x
+ u
y
)
2
− u
2
=0
but that their sum (e
x
+ e
−y
) is not a solution of the equation.
11. Solve the Cauchy problem
(y + u) u
x
+ yu
y
=(x − y) , with u =1+x on y =1.
12. Find the integral surfaces of the equation uu
x
+ u
y
=1foreachofthe
following initial data:
(a) x (s, 0) = s, y (s, 0) = 2s, u (s, 0) = s,
(b) x (s, 0) = s
2
, y (s, 0) = 2s, u (s, 0) = s,
(c) x (s, 0) =
s
2
2
, y (s, 0) = s, u (s, 0) = s.
Draw characteristics in each case.
13. Show that the solution of the equation
yu
x
− xu
y
=0
containing the curve x
2
+ y
2
= a
2
, u = y, does not exist.
14. Solve the following Cauchy problems:
(a) x
2
u
x
− y
2
u
y
=0,u → e
x
as y →∞,
(b) yu
x
+ xu
y
=0,u =sinx on x
2
+ y
2
=1,
(c) −xu
x
+ yu
y
=1for0<x<y, u =2x on y =3x,
(d) 2xu
x
+(x +1)u
y
= y for x>0, u =2y on x =1,
(e) xu
x
− 2yu
y
= x
2
+ y
2
for x>0, y>0, u = x
2
on y =1,
(f) u
x
+2u
y
=1+u, u (x, y)=sinx on y =3x +1,
(g) u
x
+3u
y
= u, u (x, y)=cosx on y = x,
(h) u
x
+2xu
y
=2xu, u (x, 0) = x
2
,
(i) uu
x
+ u
y
= u, u (x, 0) = 2x,1≤ x ≤ 2,
(j) u
x
+ u
y
= u
2
, u (x, 0) = tanh x.
2.8 Exercises 59
Show that the solu tion of (j) is unbounded on the critical curve
y tanh (x − y)=1.
15. Find the solution surface of the equation
u
2
− y
2
u
x
+ xy u
y
+ xu =0, with u = y = x, x > 0.
16. (a) Solve the Cauchy problem
u
x
+ uu
y
=1,u(0,y)=ay,
where a is a constant.
(b) Find the solution of the equation in (a) with the data
x (s, 0) = 2s, y (s, 0) = s
2
,u
0,s
2
= s.
17. Solve the following equations:
(a) (y + u) u
x
+(x + u) u
y
= x + y,
(b) xu
u
2
+ xy
u
x
− yu
u
2
+ xy
u
y
= x
4
,
(c) (x + y) u
x
+(x − y) u
y
=0,
(d) yu
x
+ xu
y
= xy
x
2
− y
2
,
(e) (cy − bz) z
x
+(az − cx) z
y
= bx − ay.
18. Solve the equation
xz
x
+ yz
y
= z,
and find the curves which satisfy the associated characteristic equations
and intersect the helix x + y
2
= a
2
, z = b tan
−1
y
x
.
19. Obtain the family of curves which represent the general solution of the
partial differential equation
(2x − 4y +3u) u
x
+(x − 2y − 3u) u
y
= −3(x − 2y) .
Determine the particular member of the family which contains the line
u = x and y =0.
20. Find the solution of the equation
yu
x
− 2xy u
y
=2xu
with the condition u (0,y)=y
3
.
60 2 First-Order, Quasi-Linear Equations and Method of Characteristics
21. Obtain the general solution of the equation
(x + y +5z) p +4zq +(x + y + z)=0 (p = z
x
,q= z
y
) ,
and find the particular solution which passes through the circle
z =0,x
2
+ y
2
= a
2
.
22. Obtain the general solution of the equation
z
2
− 2yz − y
2
p + x (y + z) q = x (y − z)(p = z
x
,q= z
y
) .
Find the integral surfaces of this equation passing throu gh
(a) the x-axis, (b) the y-axis, and (c) the z-axis.
23. Solve the Cauchy problem
(x + y) u
x
+(x − y) u
y
=1,u(1,y)=
1
√
2
.
24. Solve the following Cauchy problems:
(a) 3 u
x
+2u
y
=0,u (x, 0) = f (x),
(b) au
x
+ bu
y
= cu, u (x, 0) = f (x), where a, b, c are constants,
(c) xu
x
+ yu
y
= cu, u (x, 0) = f (x),
(d) uu
x
+ u
y
=1,u (s, 0) = αs, x (s, 0) = s, y (s, 0) = 0.
25. Apply the method of separation of variables u (x, y)=f (x) g (y)to
solve the following equations:
(a) u
x
+ u = u
y
, u (x, 0) = 4e
−3x
,(b)u
x
u
y
= u
2
,
(c) u
x
+2u
y
=0, u (0,y)=3e
−2y
,(d)y
2
u
2
x
+ x
2
u
2
y
=(xyu)
2
,
(e) x
2
u
xy
+9y
2
u =0, u (x, 0) = exp
1
x
,(f)yu
x
− xu
y
=0,
(g) u
t
= c
2
(u
xx
+ u
yy
), (h) u
xx
+ u
yy
=0.
26. Use a separable solution u (x, y)=f (x)+g (y) to solve the following
equations:
(a) u
2
x
+ u
2
y
=1, (b)u
2
x
+ u
2
y
= u,
(c) u
2
x
+ u
y
+ x
2
=0, (d)x
2
u
2
x
+ y
2
u
2
y
=1,
(e) yu
x
+ xu
y
=0,u(0,y)=y
2
.