ma 201: partial differential equations lecture - 5 · charpit’s method it is a general method for...
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Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
MA 201: Partial Differential EquationsLecture - 5
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Definition (Compatible systems of first-order PDEs)A system of two first-order PDEs
f (x , y , u, p, q) = 0 (1)
andg(x , y , u, p, q) = 0 (2)
are said to be compatible if they have a common solution.
TheoremEquations (1) and (2) are compatible on a domain D if
(i) J = ∂(f ,g)∂(p,q) =
∣
∣
∣
∣
fp fqgp gq
∣
∣
∣
∣
6= 0 on D.
(ii) p and q can be explicitly solved from (1) and (2) as p = φ(x , y , u)and q = ψ(x , y , u). Further, the equation
du = φ(x , y , u)dx + ψ(x , y , u)dy
is integrable.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
TheoremA necessary and sufficient condition for the integrability of the equationdu = φ(x , y , u)dx + ψ(x , y , u)dy is
[f , g ] ≡ ∂(f , g)
∂(x , p)+∂(f , g)
∂(y , q)+ p
∂(f , g)
∂(u, p)+ q
∂(f , g)
∂(u, q)= 0. (3)
In other words, equations (1) and (2) are compatible iff (3) holds.
ExampleShow that the equations
xp − yq = 0, xup + yuq = 2xy
are compatible and solve them.
Solution. Take f ≡ xp − yq = 0, g ≡ u(xp + yq)− 2xy = 0. Then
fx = p, fy = −q, fu = 0, fp = x , fq = −y ,
gx = up − 2y , gy = uq − 2x , gu = xp + yq, gp = ux , gq = uy .
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Compute
J ≡ ∂(f , g)
∂(p, q)=
∣
∣
∣
∣
fp fqgp gq
∣
∣
∣
∣
=
∣
∣
∣
∣
x −yux uy
∣
∣
∣
∣
= uxy + uxy = 2uxy 6= 0
for x 6= 0, y 6= 0, u 6= 0. Further,
∂(f , g)
∂(x , p)=
∣
∣
∣
∣
fx fpgx gp
∣
∣
∣
∣
=
∣
∣
∣
∣
p xup − 2y ux
∣
∣
∣
∣
= uxp − x(up − 2y) = 2xy
∂(f , g)
∂(u, p)=
∣
∣
∣
∣
fu fpgu gp
∣
∣
∣
∣
=
∣
∣
∣
∣
0 xxp + yq ux
∣
∣
∣
∣
= 0− x(xp + yq) = −x2p − xyq
∂(f , g)
∂(y , q)=
∣
∣
∣
∣
fy fqgy gq
∣
∣
∣
∣
=
∣
∣
∣
∣
−q −yuq − 2x uy
∣
∣
∣
∣
= −quy + y(uq − 2x) = −2xy
∂(f , g)
∂(u, q)=
∣
∣
∣
∣
fu fqgu gq
∣
∣
∣
∣
=
∣
∣
∣
∣
0 −yxp + yq zy
∣
∣
∣
∣
= y(xp + yq) = y2q + xyp.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
It is an easy exercise to verify that
[f , g ] ≡ ∂(f , g)
∂(x , p)+∂(f , g)
∂(y , q)+ p
∂(f , g)
∂(u, p)+ q
∂(f , g)
∂(u, q)
= 2xy − x2p2 − xypq − 2xy + y2q2 + xypq
= y2q2 − x2p2
= 0.
So the equations are compatible.• Next step is to determine p and q from the two equationsxp− yq = 0, u(xp+ yq) = 2xy . Using these two equations, we have
uxp + uyq − 2xy = 0 =⇒ xp + yq =2xy
u
=⇒ 2xp =2xy
u=⇒ p =
y
u= φ(x , y , u).
and
xp − yq = 0 =⇒ q =xp
y=
xy
yu
=⇒ q =x
u= ψ(x , y , u).
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Substituting p and q in du = pdx + qdy , we get
udu = ydx + xdy = d(xy),
and hence integrating, we obtain
u2 = 2xy + k ,
where k is a constant.NOTE:
For the compatibility of f (x , y , u, p, q) = 0 and g(x , y , u, p, q) = 0, it isnot necessary that every solution of f (x , y , u, p, q) = 0 be a solution ofg(x , y , u, p, q) = 0 or vice-versa. For instance, the equations
f ≡ xp − yq − x = 0 (4)
g ≡ x2p + q − xu = 0 (5)
are compatible. They have common solutions u = x + c(1 + xy), where cis an arbitrary constant. Note that u = x(y + 1) is a solution of (4) butnot of (5).
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Charpit’s method
It is a general method for finding the general solution of a nonlinear PDEof first-order of the form
f (x , y , u, p, q) = 0. (6)
Basic Idea: To introduce another partial differential equation of the firstorder
g(x , y , u, p, q, a) = 0 (7)
which contains an arbitrary constant a and is such that(i) equations (6) and (7) can be solved for p and q to obtain
p = p(x , y , u, a), q = q(x , y , u, a).
(ii) the equation
du = p(x , y , u, a)dx + q(x , y , u, a)dy (8)
is integrable.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
When such a function g is found, the solution
F (x , y , u, a, b) = 0
of (8) containing two arbitrary constants a and b will be the solution of(6).The compatibility of equations (6) and (7) yields
[f , g ] ≡ ∂(f , g)
∂(x , p)+∂(f , g)
∂(y , q)+ p
∂(f , g)
∂(u, p)+ q
∂(f , g)
∂(u, q)= 0.
Expanding it, we are led to the following linear PDE in g(x , y , u, p, q):
fp∂g
∂x+ fq
∂g
∂y+ (pfp + qfq)
∂g
∂u− (fx + pfu)
∂g
∂p− (fy + qfu)
∂g
∂q= 0. (9)
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Now solve (9) to determine g by finding the integrals of the followingauxiliary equations:
dx
fp=
dy
fq=
du
pfp + qfq=
dp
−(fx + pfu)=
dq
−(fy + qfu)(10)
These equations are known as Charpit’s equations. Once an integralg(x , y , u, p, q, a) of this kind has been found, the problem reduces tosolving for p and q, and then integrating equation (8).Remarks.
• For finding integrals, all of Charpit’s equations (10) need not beused.
• p or q must occur in the solution obtained from (10).
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
ExampleFind a general solution of
p2x + q2y = u. (11)
Solution. To find a general solution, we proceed as follows:
• Step 1: (Computing fx , fy , fu, fp , fq).Set f ≡ p2x + q2y − u = 0. Then
fx = p2, fy = q2, fu = −1, fp = 2px , fq = 2qy ,
and hence,
pfp + qfq = 2p2x + 2q2y , −(fx + pfu) = −p2 + p,
−(fy + qfu) = −q2 + q.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
• Step 2: (Writing Charpit’s equations and finding a solutiong(x , y , u, p, q, a)).The Charpit’s equations (or auxiliary) equations are:
dx
fp=
dy
fq=
du
pfp + qfq=
dp
−(fx + pfu)=
dq
−(fy + qfu)
=⇒ dx
2px=
dy
2qy=
du
2(p2x + q2y)=
dp
−p2 + p=
dq
−q2 + q
From which it follows that
p2dx + 2pxdp
2p3x + 2p2x − 2p3x=
q2dy + 2qydq
2q3y + 2q2y − 2q3y
=⇒ p2dx + 2pxdp
p2x=
q2dy + 2qydq
q2y
On integrating, we obtain
log(p2x) = log(q2y) + log a
=⇒ p2x = aq2y , (12)
where a is an arbitrary constant.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
• Step 3: (Solving for p and q).Using (11) and (12), we find that
p2x + q2y = u, p2x = aq2y
=⇒ (aq2y) + q2y = u =⇒ q2y(1 + a) = u
=⇒ q2 =u
(1 + a)y=⇒ q =
[
u
(1 + a)y
]1/2
.
and
p2 = aq2y
x= a
u
(1 + a)y
y
x=
au
(1 + a)x
=⇒ p =
[
au
(1 + a)x
]1/2
.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
• Step 4: (Writing du = p(x , y , u, a)dx + q(x , y , u, a)dy and findingits solution).Writing
du =
[
au
(1 + a)x
]1/2
dx +
[
u
(1 + a)y
]1/2
dy
=⇒(
1 + a
u
)1/2
du =( a
x
)1/2
dx +
(
1
y
)1/2
dy .
Integrate to have
[(1 + a)u]1/2
= (ax)1/2 + (y)1/2 + b
the general solution of equation (11).
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Some Special Types of First-Order PDEs
• Equations involving only p and qIf the equation is of the form
f (p, q) = 0, (13)
then Charpit’s equations take the form
dx
fp=
dy
fq=
du
pfp + qfq=
dp
0=
dq
0
the last two are actually equivalent todp
dt= 0,
dq
dt= 0 and hence
an immediate solution is given by p = a, where a is an arbitraryconstant. Substituting p = a in (13), we obtain a relation
q = Q(a).
Then, integrating the expression
du = adx + Q(a)dy
we obtainu = ax + Q(a)y + b, (14)
where b is a constant. Thus, (14) is a general solution of (13).IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Note: Instead of taking dpdt
= 0, we can take dqdt
= 0 ⇒ q = a. In someproblems, taking dq = 0 the amount of computation involved may bereduced considerably.
ExampleFind a general solution of the equation pq = 1.
Solution. If p = a then pq = 1 ⇒ q = 1a. In this case, Q(a) = 1/a.
From (14), we obtain a general solution as
u = ax +y
a+ b
=⇒ a2x + y − au = b,
where a and b are arbitrary constants.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
• Equations not involving the independent variables
For equation of the type
f (u, p, q) = 0, (15)
Charpit’s equation becomes
dx
fp=
dy
fq=
du
pfp + qfq=
dp
−pfu=
dq
−qfu.
From the last two relations, we have
dp
−pfu=
dq
−qfu=⇒ dp
p=
dq
q
=⇒ p = aq, (16)
where a is an arbitrary constant. Solving (15) and (16) for p and q,we obtain
q = Q(a, u) =⇒ p = aQ(a, u).
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Now
du = pdx + qdy
=⇒ du = aQ(a, u)dx + Q(a, u)dy
=⇒ du = Q(a, u) [adx + dy ] .
It gives general solution as∫
du
Q(a, u)= ax + y + b, (17)
where b is an arbitrary constant.
ExampleFind a general solution of the PDE p2u2 + q2 = 1.
Solution. Putting p = aq in the given PDE, we obtain
a2q2u2 + q2 = 1
=⇒ q2(1 + a2u2) = 1
=⇒ q = (1 + a2u2)−1/2.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
Now,
p2 = (1− q2)/u2 =
(
1− 1
(1 + a2u2)
)(
1
u2
)
=⇒ p2 =a2
1 + a2u2
=⇒ p = a(1 + a2u2)−1/2.
Substituting p and q in du = pdx + qdy , we obtain
du = a(1 + a2u2)−1/2dx + (1 + a2u2)−1/2dy
=⇒ (1 + a2u2)1/2du = adx + dy
=⇒ 1
2a
{
au(1 + a2u2)1/2 − log[au + (1 + a2u2)1/2]}
= ax + y + b,
which is the general solution of the given PDE.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
• Separable equations
A first-order PDE is separable if it can be written in the form
f (x , p) = g(y , q). (18)
For this type of equation, Charpit’s equations become
dx
fp=
dy
−gq=
du
pfp − qgq=
dp
−fx=
dq
gy.
From the last two relations, we obtain an ODE
dp
−fx=
dx
fp=⇒ dp
dx+
fxfp
= 0 (19)
which may be solved to yield p as a function of x and an arbitraryconstant a. Writing (19) in the form fpdp + fxdx = 0, we see thatits solution is f (x , p) = a. Similarly, we get g(y , q) = a. Determinep and q from the equation
f (x , p) = a, g(y , q) = a
and then use the relation du = pdx + qdy to determine a completeintegral.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
ExampleFind a general solution of p2y(1 + x2) = qx2.
Solution. First we write the given PDE in the form
p2(1 + x2)
x2=
q
y(separable equation)
It follows that
p2(1 + x2)
x2= a2 =⇒ p =
ax√1 + x2
,
where a is an arbitrary constant. Similarly,
q
y= a2 =⇒ q = a2y .
Now, the relation du = pdx + qdy yields
du =ax√1 + x2
dx + a2ydy =⇒ u = a√
1 + x2 +a2y2
2+ b,
where a and b are arbitrary constants, a general solutionfor the givenPDE.
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
• Clairaut’s equation
A first-order PDE is said to be in Clairaut form if it can be written as
u = px + qy + f (p, q). (20)
Charpit’s equations take the form
dx
x + fp=
dy
y + fq=
du
px + qy + pfp + qfq=
dp
0=
dq
0.
Now, equivalently considering dpdt
= 0 =⇒ p = a, where a is anarbitrary constant.dqdt
= 0 =⇒ q = b, where b is an arbitrary constant.Substituting the values of p and q in (20), we obtain the requiredgeneral solution
u = ax + by + f (a, b).
IIT Guwahati MA201(2016):PDE
Compatible Systems and Charpit’s MethodCharpit’s Method
Some Special Types of First-Order PDEs
ExampleFind a general solution of (p + q)(u − xp − yq) = 1.
Solution. The given PDE can be put in the form
u = xp + yq +1
p + q, (21)
which is of Clairaut’s type. Putting p = a and q = b in (21), a generalsolution is given by
u = ax + by +1
a+ b,
where a and b are arbitrary constants.
IIT Guwahati MA201(2016):PDE