Lagrange and Charpit Methods for Solving First order PDEs - CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Method of Characteristics and Lagrange-Charpit
method
Consider the following quasilinear ?rst order equation.
a(x,y,u)u
x
+b(x,y,u)u
y
+c(x,y,u) = 0. (1)
Thefunction u(x,y) isour unknown,and a,band care C
1
functionsof their
arguments. Suppose we are given a function u(x,y) that satis?es the above
equation. Now, consider the curve x(t),y(t) satisfying
dx
dt
= a(x(t),y(t),u(x(t),y(t)),
dy
dt
= b(x(t),y(t),u(x(t),y(t)).
(2)
Along this curve, we have:
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= au
x
+bu
y
=-c(x(t),y(t),u(x(t),y(t)).
(3)
whereweused(2)inthesecondequalityand(1)inthelast. Thecurvetraced
by (x(t),y(t),u(x(t),y(t)) (or (x(t),y(t)))is called a characteristic curve.
The above observation suggests the following method of solving equation
(1). Suppose we are given a C
1
curve G given by (x
0
(s),y
0
(s)). On this
curve, we are given initial values u = u
0
(s). We suppose that the following
condition is satis?ed on this curve:
det

dx
0
/ds a
dy
0
/ds b

6= 0. (4)
Page 2


Method of Characteristics and Lagrange-Charpit
method
Consider the following quasilinear ?rst order equation.
a(x,y,u)u
x
+b(x,y,u)u
y
+c(x,y,u) = 0. (1)
Thefunction u(x,y) isour unknown,and a,band care C
1
functionsof their
arguments. Suppose we are given a function u(x,y) that satis?es the above
equation. Now, consider the curve x(t),y(t) satisfying
dx
dt
= a(x(t),y(t),u(x(t),y(t)),
dy
dt
= b(x(t),y(t),u(x(t),y(t)).
(2)
Along this curve, we have:
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= au
x
+bu
y
=-c(x(t),y(t),u(x(t),y(t)).
(3)
whereweused(2)inthesecondequalityand(1)inthelast. Thecurvetraced
by (x(t),y(t),u(x(t),y(t)) (or (x(t),y(t)))is called a characteristic curve.
The above observation suggests the following method of solving equation
(1). Suppose we are given a C
1
curve G given by (x
0
(s),y
0
(s)). On this
curve, we are given initial values u = u
0
(s). We suppose that the following
condition is satis?ed on this curve:
det

dx
0
/ds a
dy
0
/ds b

6= 0. (4)
This is called the noncharacteristic condition. Now, for each value of s, we
solve the following di?erential equation:
dx
dt
= a(x,y,u),
dy
dt
= b(x,y,u)
du
dt
=-c(x,y,u) (5)
with initial condition:
x(t = 0,s) = x
0
(s), y(t = 0,s) = y
0
(s), u(t = 0,s) = u
0
(s). (6)
We have now expressed x,y and u as a function of (t,s). Given the non-
characteristic condition (4), we can solve for (t,s) in terms of (x,y) (by the
inverse function theorem) near the curve G. We may then substitute this
into u(t,s) to ?nd the function u(x,y).
We can check that this indeed gives us the solution in the following way.
Note ?rst that u(x,y) is a C
1
function (by C
1
dependence of solutions of
ODEs to parameters and the inverse function theorem). Thus, using (5)
?
?t
(u(x(t,s),y(t,s))) = u
x
?x
?t
+u
y
?y
?t
= au
x
+bu
y
. (7)
On the other hand, again using (5),
?
?t
(u(x(t,s),y(t,s))) =-c, (8)
and therefore, (1) is satis?ed.
We would now like to generalize this construction to the fully nonliner sit-
uation. Consider:
F(x,y,u,p,q) = 0, p = u
x
, q = u
y
. (9)
We assume F is a C
1
function of its arguments. Let the C
2
function u(x,y)
solve this PDE. We want to ?nd a characteristic ODE system similar to (5).
To do so, we ?rst set:
dx
dt
=
?F
?p
= F
p
,
dy
dt
=
?F
?q
= F
q
. (10)
This is in analogy with (5). Indeed, applying this procedure to (1) results
in (5). Given this, we may ?nd the equations for u along the characteristic
lines as follows.
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= pF
p
+qF
q
(11)
Page 3


Method of Characteristics and Lagrange-Charpit
method
Consider the following quasilinear ?rst order equation.
a(x,y,u)u
x
+b(x,y,u)u
y
+c(x,y,u) = 0. (1)
Thefunction u(x,y) isour unknown,and a,band care C
1
functionsof their
arguments. Suppose we are given a function u(x,y) that satis?es the above
equation. Now, consider the curve x(t),y(t) satisfying
dx
dt
= a(x(t),y(t),u(x(t),y(t)),
dy
dt
= b(x(t),y(t),u(x(t),y(t)).
(2)
Along this curve, we have:
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= au
x
+bu
y
=-c(x(t),y(t),u(x(t),y(t)).
(3)
whereweused(2)inthesecondequalityand(1)inthelast. Thecurvetraced
by (x(t),y(t),u(x(t),y(t)) (or (x(t),y(t)))is called a characteristic curve.
The above observation suggests the following method of solving equation
(1). Suppose we are given a C
1
curve G given by (x
0
(s),y
0
(s)). On this
curve, we are given initial values u = u
0
(s). We suppose that the following
condition is satis?ed on this curve:
det

dx
0
/ds a
dy
0
/ds b

6= 0. (4)
This is called the noncharacteristic condition. Now, for each value of s, we
solve the following di?erential equation:
dx
dt
= a(x,y,u),
dy
dt
= b(x,y,u)
du
dt
=-c(x,y,u) (5)
with initial condition:
x(t = 0,s) = x
0
(s), y(t = 0,s) = y
0
(s), u(t = 0,s) = u
0
(s). (6)
We have now expressed x,y and u as a function of (t,s). Given the non-
characteristic condition (4), we can solve for (t,s) in terms of (x,y) (by the
inverse function theorem) near the curve G. We may then substitute this
into u(t,s) to ?nd the function u(x,y).
We can check that this indeed gives us the solution in the following way.
Note ?rst that u(x,y) is a C
1
function (by C
1
dependence of solutions of
ODEs to parameters and the inverse function theorem). Thus, using (5)
?
?t
(u(x(t,s),y(t,s))) = u
x
?x
?t
+u
y
?y
?t
= au
x
+bu
y
. (7)
On the other hand, again using (5),
?
?t
(u(x(t,s),y(t,s))) =-c, (8)
and therefore, (1) is satis?ed.
We would now like to generalize this construction to the fully nonliner sit-
uation. Consider:
F(x,y,u,p,q) = 0, p = u
x
, q = u
y
. (9)
We assume F is a C
1
function of its arguments. Let the C
2
function u(x,y)
solve this PDE. We want to ?nd a characteristic ODE system similar to (5).
To do so, we ?rst set:
dx
dt
=
?F
?p
= F
p
,
dy
dt
=
?F
?q
= F
q
. (10)
This is in analogy with (5). Indeed, applying this procedure to (1) results
in (5). Given this, we may ?nd the equations for u along the characteristic
lines as follows.
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= pF
p
+qF
q
(11)
Wewouldnowliketouse(10)with(11)asasourcharacteristicODEsystem.
However, this is not possible since the ODE system, unlike (5) depends
not only on x,y and u but also on u
x
and u
y
. Therefore, we must derive
equations for p = u
x
and q = u
y
. We have:
dp
dt
=
d
dt
u
x
= u
xx
dx
dt
+u
xy
dy
dt
= u
xx
F
p
+u
xy
F
q
(12)
Let us take the derivative of (9) with respect to x:
F
x
+F
u
u
x
+F
p
u
xx
+F
q
u
xy
= 0. (13)
Therefore, (12) may be written as
dp
dt
=-F
x
-pF
u
. (14)
Likewise, we have:
dq
dt
=-F
y
-qF
u
(15)
WenowhaveaclosedODEsystem,whichwelistherebelowforconvenience:
dx
dt
= F
p
,
dy
dt
= F
q
,
du
dt
= pF
p
+qF
q
,
dp
dt
=-pF
u
-F
x
,
dq
dt
=-qF
u
-F
y
.
(16)
The above is called the Lagrange-Charpit system of ODEs.
This leads to the following method for solving (9). First, we are given a
non-characteristic curve G given by (x
0
(s),y
0
(s)) and values u = u
0
(s) on
this curve. In contrast to the quasilinear case (1), we need initial conditions
for p = p
0
(s) and q
0
(s) to solve (16). The initial conditions must satisfy the
PDE (9):
F(x
0
,y
0
,u
0
,p
0
,q
0
) = 0. (17)
Weneedanothercondition todetermine p
0
andq
0
. Toobtainthiscondition,
note that the solution u(x,y) to (9) must satisfy
d
ds
u(x
0
(s),y
0
(s)) = u
x
dx
0
ds
+u
y
dy
0
ds
. (18)
Therefore, we require that p
0
and q
0
also satisfy:
du
0
ds
= p
0
(s)
dx
0
ds
+q
0
(s)
dy
0
ds
. (19)
Page 4


Method of Characteristics and Lagrange-Charpit
method
Consider the following quasilinear ?rst order equation.
a(x,y,u)u
x
+b(x,y,u)u
y
+c(x,y,u) = 0. (1)
Thefunction u(x,y) isour unknown,and a,band care C
1
functionsof their
arguments. Suppose we are given a function u(x,y) that satis?es the above
equation. Now, consider the curve x(t),y(t) satisfying
dx
dt
= a(x(t),y(t),u(x(t),y(t)),
dy
dt
= b(x(t),y(t),u(x(t),y(t)).
(2)
Along this curve, we have:
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= au
x
+bu
y
=-c(x(t),y(t),u(x(t),y(t)).
(3)
whereweused(2)inthesecondequalityand(1)inthelast. Thecurvetraced
by (x(t),y(t),u(x(t),y(t)) (or (x(t),y(t)))is called a characteristic curve.
The above observation suggests the following method of solving equation
(1). Suppose we are given a C
1
curve G given by (x
0
(s),y
0
(s)). On this
curve, we are given initial values u = u
0
(s). We suppose that the following
condition is satis?ed on this curve:
det

dx
0
/ds a
dy
0
/ds b

6= 0. (4)
This is called the noncharacteristic condition. Now, for each value of s, we
solve the following di?erential equation:
dx
dt
= a(x,y,u),
dy
dt
= b(x,y,u)
du
dt
=-c(x,y,u) (5)
with initial condition:
x(t = 0,s) = x
0
(s), y(t = 0,s) = y
0
(s), u(t = 0,s) = u
0
(s). (6)
We have now expressed x,y and u as a function of (t,s). Given the non-
characteristic condition (4), we can solve for (t,s) in terms of (x,y) (by the
inverse function theorem) near the curve G. We may then substitute this
into u(t,s) to ?nd the function u(x,y).
We can check that this indeed gives us the solution in the following way.
Note ?rst that u(x,y) is a C
1
function (by C
1
dependence of solutions of
ODEs to parameters and the inverse function theorem). Thus, using (5)
?
?t
(u(x(t,s),y(t,s))) = u
x
?x
?t
+u
y
?y
?t
= au
x
+bu
y
. (7)
On the other hand, again using (5),
?
?t
(u(x(t,s),y(t,s))) =-c, (8)
and therefore, (1) is satis?ed.
We would now like to generalize this construction to the fully nonliner sit-
uation. Consider:
F(x,y,u,p,q) = 0, p = u
x
, q = u
y
. (9)
We assume F is a C
1
function of its arguments. Let the C
2
function u(x,y)
solve this PDE. We want to ?nd a characteristic ODE system similar to (5).
To do so, we ?rst set:
dx
dt
=
?F
?p
= F
p
,
dy
dt
=
?F
?q
= F
q
. (10)
This is in analogy with (5). Indeed, applying this procedure to (1) results
in (5). Given this, we may ?nd the equations for u along the characteristic
lines as follows.
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= pF
p
+qF
q
(11)
Wewouldnowliketouse(10)with(11)asasourcharacteristicODEsystem.
However, this is not possible since the ODE system, unlike (5) depends
not only on x,y and u but also on u
x
and u
y
. Therefore, we must derive
equations for p = u
x
and q = u
y
. We have:
dp
dt
=
d
dt
u
x
= u
xx
dx
dt
+u
xy
dy
dt
= u
xx
F
p
+u
xy
F
q
(12)
Let us take the derivative of (9) with respect to x:
F
x
+F
u
u
x
+F
p
u
xx
+F
q
u
xy
= 0. (13)
Therefore, (12) may be written as
dp
dt
=-F
x
-pF
u
. (14)
Likewise, we have:
dq
dt
=-F
y
-qF
u
(15)
WenowhaveaclosedODEsystem,whichwelistherebelowforconvenience:
dx
dt
= F
p
,
dy
dt
= F
q
,
du
dt
= pF
p
+qF
q
,
dp
dt
=-pF
u
-F
x
,
dq
dt
=-qF
u
-F
y
.
(16)
The above is called the Lagrange-Charpit system of ODEs.
This leads to the following method for solving (9). First, we are given a
non-characteristic curve G given by (x
0
(s),y
0
(s)) and values u = u
0
(s) on
this curve. In contrast to the quasilinear case (1), we need initial conditions
for p = p
0
(s) and q
0
(s) to solve (16). The initial conditions must satisfy the
PDE (9):
F(x
0
,y
0
,u
0
,p
0
,q
0
) = 0. (17)
Weneedanothercondition todetermine p
0
andq
0
. Toobtainthiscondition,
note that the solution u(x,y) to (9) must satisfy
d
ds
u(x
0
(s),y
0
(s)) = u
x
dx
0
ds
+u
y
dy
0
ds
. (18)
Therefore, we require that p
0
and q
0
also satisfy:
du
0
ds
= p
0
(s)
dx
0
ds
+q
0
(s)
dy
0
ds
. (19)
Equations (17) and (19) may be solved for each s to obtain the initial func-
tions p
0
(s) and q
0
(s). One di?erence between (9) and (1) is that there could
be multiple solutions p
0
and q
0
. Once the functions p
0
and q
0
are chosen,
one requires that the following non-characteristic condition is satis?ed:
det

x
'
0
(s) F
p
y
'
0
(s) F
q

6= 0. (20)
We may now solve (16) for each s with initial conditions given on the non-
characteristic curve. This procedure produces the functions:
x(t,s), y(t,s), u(t,s), p(t,s), q(t,s). (21)
We may solve for t and s in terms of x and y (thanks to (20) and the inverse
function theorem), andsubstitutethis into theexpression u(t,s) to obtain u
as a funciton x and y. We thus obtain a solution u(x,y) in a neighborhood
U of G.
We now show that this procedureindeed producesan equation that satis?es
(9). We ?rst show that
G(t,s)= F(x(t,s),y(t,s),u(t,s),p(t,s),q(t,s)) = 0. (22)
Note ?rst that G(0,s) =0 by design (see (17)). We see that
?G
?t
= F
x
dx
dt
+F
y
dy
dt
+F
u
du
dt
+F
p
dp
dt
+F
q
dq
dt
= F
x
F
p
+F
y
F
q
+F
u
(pF
p
+qF
q
)-F
p
(F
x
+pF
u
)-F
q
(F
y
+qF
u
)
= 0.
(23)
Therefore, G(t,s) = 0. Next, we must check that:
p(x,y) =
?u
?x
(x,y), q(x,y) =
?u
?y
(x,y). (24)
To check this, we show that the function
H(t,s)=
?u
?s
-p(t,s)
?x
?s
-q(t,s)
?y
?s
= 0. (25)
Note that H(0,s) = 0 given (19). Let us compute the derivative of H with
Page 5


Method of Characteristics and Lagrange-Charpit
method
Consider the following quasilinear ?rst order equation.
a(x,y,u)u
x
+b(x,y,u)u
y
+c(x,y,u) = 0. (1)
Thefunction u(x,y) isour unknown,and a,band care C
1
functionsof their
arguments. Suppose we are given a function u(x,y) that satis?es the above
equation. Now, consider the curve x(t),y(t) satisfying
dx
dt
= a(x(t),y(t),u(x(t),y(t)),
dy
dt
= b(x(t),y(t),u(x(t),y(t)).
(2)
Along this curve, we have:
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= au
x
+bu
y
=-c(x(t),y(t),u(x(t),y(t)).
(3)
whereweused(2)inthesecondequalityand(1)inthelast. Thecurvetraced
by (x(t),y(t),u(x(t),y(t)) (or (x(t),y(t)))is called a characteristic curve.
The above observation suggests the following method of solving equation
(1). Suppose we are given a C
1
curve G given by (x
0
(s),y
0
(s)). On this
curve, we are given initial values u = u
0
(s). We suppose that the following
condition is satis?ed on this curve:
det

dx
0
/ds a
dy
0
/ds b

6= 0. (4)
This is called the noncharacteristic condition. Now, for each value of s, we
solve the following di?erential equation:
dx
dt
= a(x,y,u),
dy
dt
= b(x,y,u)
du
dt
=-c(x,y,u) (5)
with initial condition:
x(t = 0,s) = x
0
(s), y(t = 0,s) = y
0
(s), u(t = 0,s) = u
0
(s). (6)
We have now expressed x,y and u as a function of (t,s). Given the non-
characteristic condition (4), we can solve for (t,s) in terms of (x,y) (by the
inverse function theorem) near the curve G. We may then substitute this
into u(t,s) to ?nd the function u(x,y).
We can check that this indeed gives us the solution in the following way.
Note ?rst that u(x,y) is a C
1
function (by C
1
dependence of solutions of
ODEs to parameters and the inverse function theorem). Thus, using (5)
?
?t
(u(x(t,s),y(t,s))) = u
x
?x
?t
+u
y
?y
?t
= au
x
+bu
y
. (7)
On the other hand, again using (5),
?
?t
(u(x(t,s),y(t,s))) =-c, (8)
and therefore, (1) is satis?ed.
We would now like to generalize this construction to the fully nonliner sit-
uation. Consider:
F(x,y,u,p,q) = 0, p = u
x
, q = u
y
. (9)
We assume F is a C
1
function of its arguments. Let the C
2
function u(x,y)
solve this PDE. We want to ?nd a characteristic ODE system similar to (5).
To do so, we ?rst set:
dx
dt
=
?F
?p
= F
p
,
dy
dt
=
?F
?q
= F
q
. (10)
This is in analogy with (5). Indeed, applying this procedure to (1) results
in (5). Given this, we may ?nd the equations for u along the characteristic
lines as follows.
d
dt
u(x(t),y(t)) = u
x
dx
dt
+u
y
dy
dt
= pF
p
+qF
q
(11)
Wewouldnowliketouse(10)with(11)asasourcharacteristicODEsystem.
However, this is not possible since the ODE system, unlike (5) depends
not only on x,y and u but also on u
x
and u
y
. Therefore, we must derive
equations for p = u
x
and q = u
y
. We have:
dp
dt
=
d
dt
u
x
= u
xx
dx
dt
+u
xy
dy
dt
= u
xx
F
p
+u
xy
F
q
(12)
Let us take the derivative of (9) with respect to x:
F
x
+F
u
u
x
+F
p
u
xx
+F
q
u
xy
= 0. (13)
Therefore, (12) may be written as
dp
dt
=-F
x
-pF
u
. (14)
Likewise, we have:
dq
dt
=-F
y
-qF
u
(15)
WenowhaveaclosedODEsystem,whichwelistherebelowforconvenience:
dx
dt
= F
p
,
dy
dt
= F
q
,
du
dt
= pF
p
+qF
q
,
dp
dt
=-pF
u
-F
x
,
dq
dt
=-qF
u
-F
y
.
(16)
The above is called the Lagrange-Charpit system of ODEs.
This leads to the following method for solving (9). First, we are given a
non-characteristic curve G given by (x
0
(s),y
0
(s)) and values u = u
0
(s) on
this curve. In contrast to the quasilinear case (1), we need initial conditions
for p = p
0
(s) and q
0
(s) to solve (16). The initial conditions must satisfy the
PDE (9):
F(x
0
,y
0
,u
0
,p
0
,q
0
) = 0. (17)
Weneedanothercondition todetermine p
0
andq
0
. Toobtainthiscondition,
note that the solution u(x,y) to (9) must satisfy
d
ds
u(x
0
(s),y
0
(s)) = u
x
dx
0
ds
+u
y
dy
0
ds
. (18)
Therefore, we require that p
0
and q
0
also satisfy:
du
0
ds
= p
0
(s)
dx
0
ds
+q
0
(s)
dy
0
ds
. (19)
Equations (17) and (19) may be solved for each s to obtain the initial func-
tions p
0
(s) and q
0
(s). One di?erence between (9) and (1) is that there could
be multiple solutions p
0
and q
0
. Once the functions p
0
and q
0
are chosen,
one requires that the following non-characteristic condition is satis?ed:
det

x
'
0
(s) F
p
y
'
0
(s) F
q

6= 0. (20)
We may now solve (16) for each s with initial conditions given on the non-
characteristic curve. This procedure produces the functions:
x(t,s), y(t,s), u(t,s), p(t,s), q(t,s). (21)
We may solve for t and s in terms of x and y (thanks to (20) and the inverse
function theorem), andsubstitutethis into theexpression u(t,s) to obtain u
as a funciton x and y. We thus obtain a solution u(x,y) in a neighborhood
U of G.
We now show that this procedureindeed producesan equation that satis?es
(9). We ?rst show that
G(t,s)= F(x(t,s),y(t,s),u(t,s),p(t,s),q(t,s)) = 0. (22)
Note ?rst that G(0,s) =0 by design (see (17)). We see that
?G
?t
= F
x
dx
dt
+F
y
dy
dt
+F
u
du
dt
+F
p
dp
dt
+F
q
dq
dt
= F
x
F
p
+F
y
F
q
+F
u
(pF
p
+qF
q
)-F
p
(F
x
+pF
u
)-F
q
(F
y
+qF
u
)
= 0.
(23)
Therefore, G(t,s) = 0. Next, we must check that:
p(x,y) =
?u
?x
(x,y), q(x,y) =
?u
?y
(x,y). (24)
To check this, we show that the function
H(t,s)=
?u
?s
-p(t,s)
?x
?s
-q(t,s)
?y
?s
= 0. (25)
Note that H(0,s) = 0 given (19). Let us compute the derivative of H with
respect to t.
?H
?t
=
?
2
u
?t?s
-
?p
?t
?x
?s
-p
?
2
x
?t?s
-
?q
?t
?y
?s
-q
?
2
y
?t?s
=
?
?s
(pF
p
+qF
q
)+(F
x
+pF
u
)
?x
?s
-p
?
?s
F
p
+(F
y
+qF
u
)
?y
?s
-q
?
?s
F
q
= F
p
?p
?s
+F
q
?q
?s
+(F
x
+pF
u
)
?x
?s
+(F
y
+qF
u
)
?y
?s
(26)
Recall that G(t,s) = 0 for all (t,s). If we take the derivative of this with
respect to s, we obtain:
?G
?s
= F
x
?x
?s
+F
y
?y
?s
+F
u
?u
?s
+F
p
?p
?s
+F
q
?q
?s
= 0. (27)
Using this fact in the last line of (26), we obtain:
?H
?t
=-F
u

?u
?s
-p
?x
?s
-q
?y
?s

=-F
u
H. (28)
This is an ODE for each value of s. The solution to the above ODE with
initial condition H(0,s) = 0 is H(t,s) = 0. We now show that H(t,s) = 0
implies (24). To see this, note that
?u
?t
= pF
p
+qF
q
= p
?x
?t
+q
?y
?t
. (29)
Thus, the above together with (25) shows that

?u/?t
?u/?s

= J

p
q

, J =

?x/?t ?y/?t
?x/?s ?y/?s

(30)
Viewing u as a function of x and y, we have
J

?u/?x
?u/?y

=

?u/?t
?u/?s

(31)
Using (30) and (31) and the fact that J is invertible (thanks to the inverse
function theorem and (20)) we obtain the equality (24).