Cauchy Problem for First Order PDEs - Partial Differential Equations, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Chapter2
First order PDE
2.1 How and Why First order PDE appear?
2.1.1 Physical origins
ConservationlawsformoneofthetwofundamentalpartsofanymathematicalmodelofContinuum
Mechanics. These models are PDEs. Discussion is beyond the scope of this course.
2.1.2 Mathematical origins
1. Two-parameter family of surfaces:
1
Let f : R
2
×A×B -?R be a smooth function.
Then
z =f(x,y,a,b), (2.1)
roughly speaking, represents a two-parameter family of surfaces inR
3
. Di?erentiating (2.1)
with respect to x and y yields the relations
z
x
=f
x
(x,y,a,b), (2.2a)
z
y
=f
y
(x,y,a,b). (2.2b)
Eliminating a and b from (2.1)-(2.2), we get a relation of the form
F(x,y,z,z
x
,z
y
)= 0.
This is a PDE for the unknown function of two independent variables.
Exercise 2.1 Let f(x,y,a,b)= (x-a)
2
+(y-b)
2
. Get a PDE by eliminating the parameters
a and b. (Answer: u
2
x
+u
2
y
= 4u.)
2. Unknown function of known functions:
(a) Unknown function of a single known function: Let u = f(g) where f is an
unknown function and g is a known function of two independent variables x and y.
Di?erentiating u = f(g) w.r.t. x and y yields the equations u
x
= f
'
(g)g
x
and u
y
=
f
'
(g)g
y
respectively. Eliminating the arbitraryfunction f from these two equations,we
obtain
g
y
u
x
-g
x
u
y
=0,
which is a ?rst order PDE for u.
1
Let y = f(x,c
1
,c
2
,··· ,cn) denote an n-parameter family of plane curves. By eliminating c
1
,··· ,cn we get an
n-th order ODE of the form F(x,y,y
'
,··· ,y
(n)
) = 0. If we consider a family of space curves, then we get systems
of ODE after eliminating parameters. Of course, all this we get only if we are able to eliminate the parameters!
Page 2


Chapter2
First order PDE
2.1 How and Why First order PDE appear?
2.1.1 Physical origins
ConservationlawsformoneofthetwofundamentalpartsofanymathematicalmodelofContinuum
Mechanics. These models are PDEs. Discussion is beyond the scope of this course.
2.1.2 Mathematical origins
1. Two-parameter family of surfaces:
1
Let f : R
2
×A×B -?R be a smooth function.
Then
z =f(x,y,a,b), (2.1)
roughly speaking, represents a two-parameter family of surfaces inR
3
. Di?erentiating (2.1)
with respect to x and y yields the relations
z
x
=f
x
(x,y,a,b), (2.2a)
z
y
=f
y
(x,y,a,b). (2.2b)
Eliminating a and b from (2.1)-(2.2), we get a relation of the form
F(x,y,z,z
x
,z
y
)= 0.
This is a PDE for the unknown function of two independent variables.
Exercise 2.1 Let f(x,y,a,b)= (x-a)
2
+(y-b)
2
. Get a PDE by eliminating the parameters
a and b. (Answer: u
2
x
+u
2
y
= 4u.)
2. Unknown function of known functions:
(a) Unknown function of a single known function: Let u = f(g) where f is an
unknown function and g is a known function of two independent variables x and y.
Di?erentiating u = f(g) w.r.t. x and y yields the equations u
x
= f
'
(g)g
x
and u
y
=
f
'
(g)g
y
respectively. Eliminating the arbitraryfunction f from these two equations,we
obtain
g
y
u
x
-g
x
u
y
=0,
which is a ?rst order PDE for u.
1
Let y = f(x,c
1
,c
2
,··· ,cn) denote an n-parameter family of plane curves. By eliminating c
1
,··· ,cn we get an
n-th order ODE of the form F(x,y,y
'
,··· ,y
(n)
) = 0. If we consider a family of space curves, then we get systems
of ODE after eliminating parameters. Of course, all this we get only if we are able to eliminate the parameters!
(b) Unknown function of two known functions: Let
u =f(x-ay)+g(x+ay). (2.3)
Denoting v(x,y) =x-ay and w(x,y) =x+ay, the above equation becomes
u =f(v)+g(w). (2.4)
where f,g are unknown functions and v,w are known functions.
Di?erentiating (2.4) with respect to x and y yields the relations
p =u
x
=f
'
(x-ay)+g
'
(x+ay), (2.5a)
q =u
y
=-af
'
(x-ay)+ag
'
(x+ay). (2.5b)
Eliminating f and g from (2.5a)-(2.5b) (after di?erentiating them w.r.t. y and x respec-
tively), we get a relation of the form
q
y
=a
2
p
x
In terms of u the above ?rst order PDE is the well-known Wave equation
u
yy
=a
2
u
xx
.
2.2 Quasi-linear PDE
Consider the quasi-linear PDE given by
a(x,y,u)u
x
+b(x,y,u)u
y
=c(x,y,u), (2.6)
wherea,b,carecontinuouslydi?erentiablefunctionsonadomainO?R
3
. LetO
0
betheprojection
of O to the XY-plane.
De?nition 2.2 (Integral Surface) Let D ? O
0
and u : D -?R be a solution of the equation
(2.6). The surface S represented by z =u(x,y) is called an Integral Surface.
Remark 2.3
1. Any point on an integral surface S has the form (x,y,u(x,y)) for some (x,y)? D. Since S
is an integral surface, such an (x,y) is unique.
2. Note that any integral surface S is of the form z = u(x,y) for some solution u (de?ned on
its domain) of the equation (2.6). The projection of such an S to the XY-plane will be the
domain of u.
3. For the surface z = u(x,y), the normal at any point (x
0
,y
0
,u(x
0
,y
0
)) on S is given by
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1). We can write the PDE (2.6) in the form
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1)•(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) =0.
Thusthevector(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
)))belongstothetan-
gent space to S at the point (x
0
,y
0
,u(x
0
,y
0
)). By de?nition of tangent space, there exists
a curve ? : (-d,d) ? R
3
such that (i) ? lies on S, (ii) ?(0) = (x
0
,y
0
,u(x
0
,y
0
)), and
(iii) ?
'
(0) = (a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))). This motivates the
de?nition of a Characteristic curve that we are going to de?ne shortly.
Page 3


Chapter2
First order PDE
2.1 How and Why First order PDE appear?
2.1.1 Physical origins
ConservationlawsformoneofthetwofundamentalpartsofanymathematicalmodelofContinuum
Mechanics. These models are PDEs. Discussion is beyond the scope of this course.
2.1.2 Mathematical origins
1. Two-parameter family of surfaces:
1
Let f : R
2
×A×B -?R be a smooth function.
Then
z =f(x,y,a,b), (2.1)
roughly speaking, represents a two-parameter family of surfaces inR
3
. Di?erentiating (2.1)
with respect to x and y yields the relations
z
x
=f
x
(x,y,a,b), (2.2a)
z
y
=f
y
(x,y,a,b). (2.2b)
Eliminating a and b from (2.1)-(2.2), we get a relation of the form
F(x,y,z,z
x
,z
y
)= 0.
This is a PDE for the unknown function of two independent variables.
Exercise 2.1 Let f(x,y,a,b)= (x-a)
2
+(y-b)
2
. Get a PDE by eliminating the parameters
a and b. (Answer: u
2
x
+u
2
y
= 4u.)
2. Unknown function of known functions:
(a) Unknown function of a single known function: Let u = f(g) where f is an
unknown function and g is a known function of two independent variables x and y.
Di?erentiating u = f(g) w.r.t. x and y yields the equations u
x
= f
'
(g)g
x
and u
y
=
f
'
(g)g
y
respectively. Eliminating the arbitraryfunction f from these two equations,we
obtain
g
y
u
x
-g
x
u
y
=0,
which is a ?rst order PDE for u.
1
Let y = f(x,c
1
,c
2
,··· ,cn) denote an n-parameter family of plane curves. By eliminating c
1
,··· ,cn we get an
n-th order ODE of the form F(x,y,y
'
,··· ,y
(n)
) = 0. If we consider a family of space curves, then we get systems
of ODE after eliminating parameters. Of course, all this we get only if we are able to eliminate the parameters!
(b) Unknown function of two known functions: Let
u =f(x-ay)+g(x+ay). (2.3)
Denoting v(x,y) =x-ay and w(x,y) =x+ay, the above equation becomes
u =f(v)+g(w). (2.4)
where f,g are unknown functions and v,w are known functions.
Di?erentiating (2.4) with respect to x and y yields the relations
p =u
x
=f
'
(x-ay)+g
'
(x+ay), (2.5a)
q =u
y
=-af
'
(x-ay)+ag
'
(x+ay). (2.5b)
Eliminating f and g from (2.5a)-(2.5b) (after di?erentiating them w.r.t. y and x respec-
tively), we get a relation of the form
q
y
=a
2
p
x
In terms of u the above ?rst order PDE is the well-known Wave equation
u
yy
=a
2
u
xx
.
2.2 Quasi-linear PDE
Consider the quasi-linear PDE given by
a(x,y,u)u
x
+b(x,y,u)u
y
=c(x,y,u), (2.6)
wherea,b,carecontinuouslydi?erentiablefunctionsonadomainO?R
3
. LetO
0
betheprojection
of O to the XY-plane.
De?nition 2.2 (Integral Surface) Let D ? O
0
and u : D -?R be a solution of the equation
(2.6). The surface S represented by z =u(x,y) is called an Integral Surface.
Remark 2.3
1. Any point on an integral surface S has the form (x,y,u(x,y)) for some (x,y)? D. Since S
is an integral surface, such an (x,y) is unique.
2. Note that any integral surface S is of the form z = u(x,y) for some solution u (de?ned on
its domain) of the equation (2.6). The projection of such an S to the XY-plane will be the
domain of u.
3. For the surface z = u(x,y), the normal at any point (x
0
,y
0
,u(x
0
,y
0
)) on S is given by
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1). We can write the PDE (2.6) in the form
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1)•(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) =0.
Thusthevector(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
)))belongstothetan-
gent space to S at the point (x
0
,y
0
,u(x
0
,y
0
)). By de?nition of tangent space, there exists
a curve ? : (-d,d) ? R
3
such that (i) ? lies on S, (ii) ?(0) = (x
0
,y
0
,u(x
0
,y
0
)), and
(iii) ?
'
(0) = (a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))). This motivates the
de?nition of a Characteristic curve that we are going to de?ne shortly.
De?nition 2.4 (Characteristic Vector Field) The vector ?eld (a(x,y,z), b(x,y,z), c(x,y,z))
is called the Characteristic Vector Field of the equation (2.6). The direction of the vector
(a(x,y,z), b(x,y,z), c(x,y,z)) is called the Characteristic Direction at (x,y,z)? O.
De?nition 2.5 (Characteristic Curve) A curve inR
3
which is tangential to the characteristic
direction at each of its points is called a Characteristic Curve (i.e., the tangent to the curve is
parallel to the characteristic direction at every point on the curve).
De?nition 2.6 (Base Characteristics) TheprojectionsofCharacteristicCurvestotheXY-
plane are called Base characteristics.
Remark 2.7
1. The characteristic curves of the equation (2.6) are solutions of the following autonomous
system of ODEs:
dx
dt
= a(x,y,z)
dy
dt
= b(x,y,z)
dz
dt
= c(x,y,z)
Note that the above system is autonomous and we are interested not on the solutions but
only the trace of the solutions in the XYZ-space as t varies in the interval on which solutions
tothe above system exist. Thus the parameter tis somewhat arti?cial (see [18]) and replacing
it with any other parameter along a characteristic curve will amount to replacing a,b,c by
proportional quantities.
2. Since a,b,c are assumed to be continuously di?erentiable on O, through any point of O
there exists a unique characteristic curve of the equation (2.6) (Prove this!). Hence distinct
characteristic curves do not intersect (why?), however their projections to the XY-plane
might intersect(Does it not contradict the existence and uniqueness theorem? Explain).
3. In the case of semi-linear PDE, neither distinct characteristic curves nor their projections
on XY-plane intersect. However, if at least one of the functions a,b is such that existence
and uniqueness theorem for solutions to IVPs cannot be applied, then it may happen that
characteristic curves and their projections to XY-plane may intersect. For example, see
Exercise 2.23.
4. The characteristic curves form a two-parameter family. However the solutions to the char-
acteristic system of ODEs form a three-parameter family.
Exercise 2.8 Justify the above remarks.
Theorem 2.9 Let S : z = u(x,y) be a surface in R
3
. Then the following statements are equiva-
lent.
1. The surface S is an integral surface of equation (2.6).
2. The surface S is a union of characteristic curves of the equation (2.6).
Proof :
Proof of (1)=? (2): Let S : z = u(x,y) be an integral surface of equation (2.6). That is, there is a
domain D?R
2
and a function u: D-?R such that
a(x,y,u(x,y))u
x
(x,y)+b(x,y,u(x,y))u
y
(x,y) =c(x,y,u(x,y)) for all (x,y)? D. (2.7)
Page 4


Chapter2
First order PDE
2.1 How and Why First order PDE appear?
2.1.1 Physical origins
ConservationlawsformoneofthetwofundamentalpartsofanymathematicalmodelofContinuum
Mechanics. These models are PDEs. Discussion is beyond the scope of this course.
2.1.2 Mathematical origins
1. Two-parameter family of surfaces:
1
Let f : R
2
×A×B -?R be a smooth function.
Then
z =f(x,y,a,b), (2.1)
roughly speaking, represents a two-parameter family of surfaces inR
3
. Di?erentiating (2.1)
with respect to x and y yields the relations
z
x
=f
x
(x,y,a,b), (2.2a)
z
y
=f
y
(x,y,a,b). (2.2b)
Eliminating a and b from (2.1)-(2.2), we get a relation of the form
F(x,y,z,z
x
,z
y
)= 0.
This is a PDE for the unknown function of two independent variables.
Exercise 2.1 Let f(x,y,a,b)= (x-a)
2
+(y-b)
2
. Get a PDE by eliminating the parameters
a and b. (Answer: u
2
x
+u
2
y
= 4u.)
2. Unknown function of known functions:
(a) Unknown function of a single known function: Let u = f(g) where f is an
unknown function and g is a known function of two independent variables x and y.
Di?erentiating u = f(g) w.r.t. x and y yields the equations u
x
= f
'
(g)g
x
and u
y
=
f
'
(g)g
y
respectively. Eliminating the arbitraryfunction f from these two equations,we
obtain
g
y
u
x
-g
x
u
y
=0,
which is a ?rst order PDE for u.
1
Let y = f(x,c
1
,c
2
,··· ,cn) denote an n-parameter family of plane curves. By eliminating c
1
,··· ,cn we get an
n-th order ODE of the form F(x,y,y
'
,··· ,y
(n)
) = 0. If we consider a family of space curves, then we get systems
of ODE after eliminating parameters. Of course, all this we get only if we are able to eliminate the parameters!
(b) Unknown function of two known functions: Let
u =f(x-ay)+g(x+ay). (2.3)
Denoting v(x,y) =x-ay and w(x,y) =x+ay, the above equation becomes
u =f(v)+g(w). (2.4)
where f,g are unknown functions and v,w are known functions.
Di?erentiating (2.4) with respect to x and y yields the relations
p =u
x
=f
'
(x-ay)+g
'
(x+ay), (2.5a)
q =u
y
=-af
'
(x-ay)+ag
'
(x+ay). (2.5b)
Eliminating f and g from (2.5a)-(2.5b) (after di?erentiating them w.r.t. y and x respec-
tively), we get a relation of the form
q
y
=a
2
p
x
In terms of u the above ?rst order PDE is the well-known Wave equation
u
yy
=a
2
u
xx
.
2.2 Quasi-linear PDE
Consider the quasi-linear PDE given by
a(x,y,u)u
x
+b(x,y,u)u
y
=c(x,y,u), (2.6)
wherea,b,carecontinuouslydi?erentiablefunctionsonadomainO?R
3
. LetO
0
betheprojection
of O to the XY-plane.
De?nition 2.2 (Integral Surface) Let D ? O
0
and u : D -?R be a solution of the equation
(2.6). The surface S represented by z =u(x,y) is called an Integral Surface.
Remark 2.3
1. Any point on an integral surface S has the form (x,y,u(x,y)) for some (x,y)? D. Since S
is an integral surface, such an (x,y) is unique.
2. Note that any integral surface S is of the form z = u(x,y) for some solution u (de?ned on
its domain) of the equation (2.6). The projection of such an S to the XY-plane will be the
domain of u.
3. For the surface z = u(x,y), the normal at any point (x
0
,y
0
,u(x
0
,y
0
)) on S is given by
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1). We can write the PDE (2.6) in the form
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1)•(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) =0.
Thusthevector(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
)))belongstothetan-
gent space to S at the point (x
0
,y
0
,u(x
0
,y
0
)). By de?nition of tangent space, there exists
a curve ? : (-d,d) ? R
3
such that (i) ? lies on S, (ii) ?(0) = (x
0
,y
0
,u(x
0
,y
0
)), and
(iii) ?
'
(0) = (a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))). This motivates the
de?nition of a Characteristic curve that we are going to de?ne shortly.
De?nition 2.4 (Characteristic Vector Field) The vector ?eld (a(x,y,z), b(x,y,z), c(x,y,z))
is called the Characteristic Vector Field of the equation (2.6). The direction of the vector
(a(x,y,z), b(x,y,z), c(x,y,z)) is called the Characteristic Direction at (x,y,z)? O.
De?nition 2.5 (Characteristic Curve) A curve inR
3
which is tangential to the characteristic
direction at each of its points is called a Characteristic Curve (i.e., the tangent to the curve is
parallel to the characteristic direction at every point on the curve).
De?nition 2.6 (Base Characteristics) TheprojectionsofCharacteristicCurvestotheXY-
plane are called Base characteristics.
Remark 2.7
1. The characteristic curves of the equation (2.6) are solutions of the following autonomous
system of ODEs:
dx
dt
= a(x,y,z)
dy
dt
= b(x,y,z)
dz
dt
= c(x,y,z)
Note that the above system is autonomous and we are interested not on the solutions but
only the trace of the solutions in the XYZ-space as t varies in the interval on which solutions
tothe above system exist. Thus the parameter tis somewhat arti?cial (see [18]) and replacing
it with any other parameter along a characteristic curve will amount to replacing a,b,c by
proportional quantities.
2. Since a,b,c are assumed to be continuously di?erentiable on O, through any point of O
there exists a unique characteristic curve of the equation (2.6) (Prove this!). Hence distinct
characteristic curves do not intersect (why?), however their projections to the XY-plane
might intersect(Does it not contradict the existence and uniqueness theorem? Explain).
3. In the case of semi-linear PDE, neither distinct characteristic curves nor their projections
on XY-plane intersect. However, if at least one of the functions a,b is such that existence
and uniqueness theorem for solutions to IVPs cannot be applied, then it may happen that
characteristic curves and their projections to XY-plane may intersect. For example, see
Exercise 2.23.
4. The characteristic curves form a two-parameter family. However the solutions to the char-
acteristic system of ODEs form a three-parameter family.
Exercise 2.8 Justify the above remarks.
Theorem 2.9 Let S : z = u(x,y) be a surface in R
3
. Then the following statements are equiva-
lent.
1. The surface S is an integral surface of equation (2.6).
2. The surface S is a union of characteristic curves of the equation (2.6).
Proof :
Proof of (1)=? (2): Let S : z = u(x,y) be an integral surface of equation (2.6). That is, there is a
domain D?R
2
and a function u: D-?R such that
a(x,y,u(x,y))u
x
(x,y)+b(x,y,u(x,y))u
y
(x,y) =c(x,y,u(x,y)) for all (x,y)? D. (2.7)
The statement (2) of the theorem is equivalent to
S =
[
? is a characteristic curve
?.
Thus, to prove that S is a union of characteristic curves, it is su?cient to prove that the charac-
teristic curve ?
p
lies entirely
1
on S for every p ? S (why?). Let p = (x
0
,y
0
,z
0
) be an arbitrary
point on the surface S. Through p, there exists a unique characteristic curve ?
p
and we want to
prove that ?
p
lies entirely
2
on S. Suppose that ?
p
is given by
x =x(t), y =y(t), z =z(t), t?I and P =(x
0
,y
0
,z
0
)= (x(t
0
),y(t
0
),z(t
0
)) for somet
0
? I.
Without loss of generality assume that (x(t),y(t)) ? D for all t ? I; if not we replace I by an
interval I
'
for which this holds. To prove that ?
p
lies entirely on S, we will prove
z(t)=u(x(t),y(t))
3
for all t?I.
Thus we are led to consider the following function which is de?ned on I:
V(t) =z(t)-u(x(t),y(t)) for all t? I.
We need to show that V is the zero function. Note that V(t
0
)= 0 as P ? S. Let us compute the
derivative of V.
V
'
(t) = z
'
(t)-u
x
(x(t),y(t))
dx
dt
-u
y
(x(t),y(t))
dy
dt
= c(x(t),y(t),z(t))-u
x
(x(t),y(t))a(x(t),y(t),z(t))-u
y
(x(t),y(t))b(x(t),y(t),z(t))
= c(x(t),y(t),V(t)+u(x(t),y(t)))-u
x
(x(t),y(t))a(x(t),y(t),V(t)+u(x(t),y(t)))
-u
y
(x(t),y(t))b(x(t),y(t),V(t)+u(x(t),y(t))).
Thus the function V : I -?R is a solution (why?) of the ODE
U
'
=f(t,U), (2.8)
where
f(t,U) = c(x(t),y(t),U +u(x(t),y(t)))-u
x
(x(t),y(t))a(x(t),y(t),U +u(x(t),y(t)))
-u
y
(x(t),y(t))b(x(t),y(t),U +u(x(t),y(t)))
The RHS of (2.8) is a locally Lipschitz function w.r.t. U since a,b,c,u are continuously di?eren-
tiable functions on D if u is assumed to be continuously di?erentiable. Further note that U(t)=0
is a solution of the ODE (2.8) (why?). Also U(t
0
)= 0 in view of (2.7) as z =u(x,y) is an integral
surface. Hence by uniqueness of solutions of Initial value problems to ODEs, we conclude that
V = 0.
Proof of (2)=? (1): Letthe surfaceS : z =u(x,y)be aunionofcharacteristiccurvesofthe equation
(2.6). We want to show that S is an integral surface. In other words, we want to show that u
solves the equation (2.6). Let p
0
= (x
0
,y
0
,u(x
0
,y
0
)) be any point on the surface S. We want to
show
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1)•(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) =0. (2.9)
Since S is a union of characteristics, there is a characteristic passing through p
0
that lies
entirely on S. Since (u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1) is the normal direction to S at p
0
and
(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) is the direction of tangent to ? at p
0
,
we get (2.9). This ?nishes the proof.
1
Many books claim this statement. This is false. Note that u is given to us. We cannot assume anything about
the domain on which u is de?ned.
2
In fact, that part of ?p lies on S for which the corresponding base characterisitcs reside in D, the domain on
which u is de?ned.
3
We dont have the right to write this equation if (x(t),y(t)) does not belong to D for all t ? I.
Page 5


Chapter2
First order PDE
2.1 How and Why First order PDE appear?
2.1.1 Physical origins
ConservationlawsformoneofthetwofundamentalpartsofanymathematicalmodelofContinuum
Mechanics. These models are PDEs. Discussion is beyond the scope of this course.
2.1.2 Mathematical origins
1. Two-parameter family of surfaces:
1
Let f : R
2
×A×B -?R be a smooth function.
Then
z =f(x,y,a,b), (2.1)
roughly speaking, represents a two-parameter family of surfaces inR
3
. Di?erentiating (2.1)
with respect to x and y yields the relations
z
x
=f
x
(x,y,a,b), (2.2a)
z
y
=f
y
(x,y,a,b). (2.2b)
Eliminating a and b from (2.1)-(2.2), we get a relation of the form
F(x,y,z,z
x
,z
y
)= 0.
This is a PDE for the unknown function of two independent variables.
Exercise 2.1 Let f(x,y,a,b)= (x-a)
2
+(y-b)
2
. Get a PDE by eliminating the parameters
a and b. (Answer: u
2
x
+u
2
y
= 4u.)
2. Unknown function of known functions:
(a) Unknown function of a single known function: Let u = f(g) where f is an
unknown function and g is a known function of two independent variables x and y.
Di?erentiating u = f(g) w.r.t. x and y yields the equations u
x
= f
'
(g)g
x
and u
y
=
f
'
(g)g
y
respectively. Eliminating the arbitraryfunction f from these two equations,we
obtain
g
y
u
x
-g
x
u
y
=0,
which is a ?rst order PDE for u.
1
Let y = f(x,c
1
,c
2
,··· ,cn) denote an n-parameter family of plane curves. By eliminating c
1
,··· ,cn we get an
n-th order ODE of the form F(x,y,y
'
,··· ,y
(n)
) = 0. If we consider a family of space curves, then we get systems
of ODE after eliminating parameters. Of course, all this we get only if we are able to eliminate the parameters!
(b) Unknown function of two known functions: Let
u =f(x-ay)+g(x+ay). (2.3)
Denoting v(x,y) =x-ay and w(x,y) =x+ay, the above equation becomes
u =f(v)+g(w). (2.4)
where f,g are unknown functions and v,w are known functions.
Di?erentiating (2.4) with respect to x and y yields the relations
p =u
x
=f
'
(x-ay)+g
'
(x+ay), (2.5a)
q =u
y
=-af
'
(x-ay)+ag
'
(x+ay). (2.5b)
Eliminating f and g from (2.5a)-(2.5b) (after di?erentiating them w.r.t. y and x respec-
tively), we get a relation of the form
q
y
=a
2
p
x
In terms of u the above ?rst order PDE is the well-known Wave equation
u
yy
=a
2
u
xx
.
2.2 Quasi-linear PDE
Consider the quasi-linear PDE given by
a(x,y,u)u
x
+b(x,y,u)u
y
=c(x,y,u), (2.6)
wherea,b,carecontinuouslydi?erentiablefunctionsonadomainO?R
3
. LetO
0
betheprojection
of O to the XY-plane.
De?nition 2.2 (Integral Surface) Let D ? O
0
and u : D -?R be a solution of the equation
(2.6). The surface S represented by z =u(x,y) is called an Integral Surface.
Remark 2.3
1. Any point on an integral surface S has the form (x,y,u(x,y)) for some (x,y)? D. Since S
is an integral surface, such an (x,y) is unique.
2. Note that any integral surface S is of the form z = u(x,y) for some solution u (de?ned on
its domain) of the equation (2.6). The projection of such an S to the XY-plane will be the
domain of u.
3. For the surface z = u(x,y), the normal at any point (x
0
,y
0
,u(x
0
,y
0
)) on S is given by
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1). We can write the PDE (2.6) in the form
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1)•(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) =0.
Thusthevector(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
)))belongstothetan-
gent space to S at the point (x
0
,y
0
,u(x
0
,y
0
)). By de?nition of tangent space, there exists
a curve ? : (-d,d) ? R
3
such that (i) ? lies on S, (ii) ?(0) = (x
0
,y
0
,u(x
0
,y
0
)), and
(iii) ?
'
(0) = (a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))). This motivates the
de?nition of a Characteristic curve that we are going to de?ne shortly.
De?nition 2.4 (Characteristic Vector Field) The vector ?eld (a(x,y,z), b(x,y,z), c(x,y,z))
is called the Characteristic Vector Field of the equation (2.6). The direction of the vector
(a(x,y,z), b(x,y,z), c(x,y,z)) is called the Characteristic Direction at (x,y,z)? O.
De?nition 2.5 (Characteristic Curve) A curve inR
3
which is tangential to the characteristic
direction at each of its points is called a Characteristic Curve (i.e., the tangent to the curve is
parallel to the characteristic direction at every point on the curve).
De?nition 2.6 (Base Characteristics) TheprojectionsofCharacteristicCurvestotheXY-
plane are called Base characteristics.
Remark 2.7
1. The characteristic curves of the equation (2.6) are solutions of the following autonomous
system of ODEs:
dx
dt
= a(x,y,z)
dy
dt
= b(x,y,z)
dz
dt
= c(x,y,z)
Note that the above system is autonomous and we are interested not on the solutions but
only the trace of the solutions in the XYZ-space as t varies in the interval on which solutions
tothe above system exist. Thus the parameter tis somewhat arti?cial (see [18]) and replacing
it with any other parameter along a characteristic curve will amount to replacing a,b,c by
proportional quantities.
2. Since a,b,c are assumed to be continuously di?erentiable on O, through any point of O
there exists a unique characteristic curve of the equation (2.6) (Prove this!). Hence distinct
characteristic curves do not intersect (why?), however their projections to the XY-plane
might intersect(Does it not contradict the existence and uniqueness theorem? Explain).
3. In the case of semi-linear PDE, neither distinct characteristic curves nor their projections
on XY-plane intersect. However, if at least one of the functions a,b is such that existence
and uniqueness theorem for solutions to IVPs cannot be applied, then it may happen that
characteristic curves and their projections to XY-plane may intersect. For example, see
Exercise 2.23.
4. The characteristic curves form a two-parameter family. However the solutions to the char-
acteristic system of ODEs form a three-parameter family.
Exercise 2.8 Justify the above remarks.
Theorem 2.9 Let S : z = u(x,y) be a surface in R
3
. Then the following statements are equiva-
lent.
1. The surface S is an integral surface of equation (2.6).
2. The surface S is a union of characteristic curves of the equation (2.6).
Proof :
Proof of (1)=? (2): Let S : z = u(x,y) be an integral surface of equation (2.6). That is, there is a
domain D?R
2
and a function u: D-?R such that
a(x,y,u(x,y))u
x
(x,y)+b(x,y,u(x,y))u
y
(x,y) =c(x,y,u(x,y)) for all (x,y)? D. (2.7)
The statement (2) of the theorem is equivalent to
S =
[
? is a characteristic curve
?.
Thus, to prove that S is a union of characteristic curves, it is su?cient to prove that the charac-
teristic curve ?
p
lies entirely
1
on S for every p ? S (why?). Let p = (x
0
,y
0
,z
0
) be an arbitrary
point on the surface S. Through p, there exists a unique characteristic curve ?
p
and we want to
prove that ?
p
lies entirely
2
on S. Suppose that ?
p
is given by
x =x(t), y =y(t), z =z(t), t?I and P =(x
0
,y
0
,z
0
)= (x(t
0
),y(t
0
),z(t
0
)) for somet
0
? I.
Without loss of generality assume that (x(t),y(t)) ? D for all t ? I; if not we replace I by an
interval I
'
for which this holds. To prove that ?
p
lies entirely on S, we will prove
z(t)=u(x(t),y(t))
3
for all t?I.
Thus we are led to consider the following function which is de?ned on I:
V(t) =z(t)-u(x(t),y(t)) for all t? I.
We need to show that V is the zero function. Note that V(t
0
)= 0 as P ? S. Let us compute the
derivative of V.
V
'
(t) = z
'
(t)-u
x
(x(t),y(t))
dx
dt
-u
y
(x(t),y(t))
dy
dt
= c(x(t),y(t),z(t))-u
x
(x(t),y(t))a(x(t),y(t),z(t))-u
y
(x(t),y(t))b(x(t),y(t),z(t))
= c(x(t),y(t),V(t)+u(x(t),y(t)))-u
x
(x(t),y(t))a(x(t),y(t),V(t)+u(x(t),y(t)))
-u
y
(x(t),y(t))b(x(t),y(t),V(t)+u(x(t),y(t))).
Thus the function V : I -?R is a solution (why?) of the ODE
U
'
=f(t,U), (2.8)
where
f(t,U) = c(x(t),y(t),U +u(x(t),y(t)))-u
x
(x(t),y(t))a(x(t),y(t),U +u(x(t),y(t)))
-u
y
(x(t),y(t))b(x(t),y(t),U +u(x(t),y(t)))
The RHS of (2.8) is a locally Lipschitz function w.r.t. U since a,b,c,u are continuously di?eren-
tiable functions on D if u is assumed to be continuously di?erentiable. Further note that U(t)=0
is a solution of the ODE (2.8) (why?). Also U(t
0
)= 0 in view of (2.7) as z =u(x,y) is an integral
surface. Hence by uniqueness of solutions of Initial value problems to ODEs, we conclude that
V = 0.
Proof of (2)=? (1): Letthe surfaceS : z =u(x,y)be aunionofcharacteristiccurvesofthe equation
(2.6). We want to show that S is an integral surface. In other words, we want to show that u
solves the equation (2.6). Let p
0
= (x
0
,y
0
,u(x
0
,y
0
)) be any point on the surface S. We want to
show
(u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1)•(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) =0. (2.9)
Since S is a union of characteristics, there is a characteristic passing through p
0
that lies
entirely on S. Since (u
x
(x
0
,y
0
),u
y
(x
0
,y
0
),-1) is the normal direction to S at p
0
and
(a(x
0
,y
0
,u(x
0
,y
0
)),b(x
0
,y
0
,u(x
0
,y
0
)),c(x
0
,y
0
,u(x
0
,y
0
))) is the direction of tangent to ? at p
0
,
we get (2.9). This ?nishes the proof.
1
Many books claim this statement. This is false. Note that u is given to us. We cannot assume anything about
the domain on which u is de?ned.
2
In fact, that part of ?p lies on S for which the corresponding base characterisitcs reside in D, the domain on
which u is de?ned.
3
We dont have the right to write this equation if (x(t),y(t)) does not belong to D for all t ? I.
Remark 2.10 (On the domain D) In the quasi-linear equation (2.6), recall that a,b,c are de-
?ned on O?R
3
and were assumed to be continuously di?erentiable. Let O
0
be the projection of
O into the XY-plane. That is,
O
0
=

(x,y)?R
2
: (x,y,z)? O for some z?R
	
.
Then z = u(x,y) is an intgral surface on some domain D ? O
0
. Note that characteristic curve
lives in O, by virtue of being a soluton of the system of ODE where the vector ?eld is de?ned on
O. Projection of characteristic curves to XY-plane live inside O
0
.
Example 2.11 Ifu:D-?RbeasolutiontotheQuasi-linearPDE (2.6), then soisv :D
1
-?R
where v is de?ned by v(x,y) = u(x,y). The integral surfaces z = u(x,y) and z = v(x,y) are
di?erent since u and v are di?erent as functions. But both the integral surfaces coincide on D
1
.
Thus interesection of two integral surfaces could be another integral surface.
The following corollary follows immediately from Theorem 2.9
Corollary 2.12 Let S
1
and S
2
be two integral surfaces such that p?S
1
nS
2
. Then some part of
the characteristic passing through p lies on both S
1
and S
2
1
.
Corollary 2.13 If two integral surfaces intersect without touching and the intersection is a curve
?, then ? is a characteristic curve.
Proof :
To prove that ? is a characteristic curve, we have to prove that at any point P on the curve ?,
the tangent has the characteristic direction. At P the tangent to the curve ? lies in the tangent
planestoS
1
aswellasS
2
andalsothecharacteristicdirection(a(P),b(P),c(P)). Sincethetangent
planes do not coincide (why?), the only direction commonto both S
1
and S
2
is (a(P),b(P),c(P)).
Hence tangent to the curve ? at P is proportional to the characterisitc direction at P. Since P is
an arbitrary point on ?, it follows that ? is a characteristic curve.
Exercise 2.14 Suppose ? is a curve that lies on two integral surfaces. Can we use the above proof
to conclude that ? is a characteristic curve? If yes, give a proof. If not, explain where the above
proof fails.
2.2.1 Cauchy Problem for Quasi-linear PDE
Cauchy Problem
To ?nd an integral surface z = u(x,y) of the quasi-linear PDE (2.6), containing a given space
curve G whose parametric equations are
x =f(s), y =g(s), z =h(s), s? I, (2.10)
wheref,g,hareassumedtobecontinuouslydi?erentiableontheintervalI andh(s)=u(f(s),g(s))
for s?I.
Initial Value Problem
Initial value problem for the quasi-linearPDE (2.6) is a special Cauchy problemfor (2.6), wherein
theinitialcurveGliesintheZX-planeandtheyvariablehasaninterpretationofthetime-variable.
That is, G has the following parametric form:
x=f(s), y = 0, z =h(s), s?I, (2.11)
1
This is stated more confusingly and also wrongly as “If two integral surfaces intersect at a point p, then they
intersect along the entire characteristic curve through p.” Exercise: Which part of this statement is confusing and
which part is wrong?