ma 201: partial differential equations lecture 4-6 · charpit’s method such a g should satisfy fp...
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MA 201: Partial Differential Equations
Lecture 4-6
MA201(2014):PDE
Method of Characteristicsfor
First Order Equations in two independent variables
MA201(2014):PDE
For a PDEF (x , y , z , p, q) = 0 (1)
the characteristic equations are
dx
dt= Fp (2)
dy
dt= Fq (3)
dz
dt= pFp + qFq (4)
dp
dt= −[Fx + pFz ] (5)
dq
dt= −[Fy + qFz ] (6)
A curve x = x(t), y = y(t), z = z(t) satisfying these equation is acharacteristic curve for (26).
MA201(2014):PDE
Example: Determine the characteristic curves of the equation
z = p2 − q2
and find the integral surface which passes through the parabola4z + x2 = 0, y = 0.
Solution: (a) Write the equation in the form F (x , y , z , p, q) = 0.
F (x , y , z , p, q) = z − p2 + q2 = 0. (7)
(b) Parametrize the given initial curve.
x0(s) = 2s, y0(s) = 0, z0(s) = −s2.
(c) Find initial values p0(s) and q0(s) from
F (x , y , z , p,q) = 0 anddz
ds= p
dx
ds+ q
dy
ds.
p0(s) = −s, q0(s) = ±√2s.
MA201(2014):PDE
(d) Write the characteristic equations.
Here F = z − p2 + q2. So, Fx = 0,Fy = 0,Fz = 1,Fp = −2p,Fq = 2q.Equations are
dx
dt= Fp = −2p (8)
dy
dt= Fq = 2q (9)
dz
dt= pFp + qFq = −2(p2 − q2) = −2z (10)
dp
dt= −[Fx + pFz ] = −p (11)
dq
dt= −[Fy + qFz ] = −q. (12)
(e) Solve the characteristic equations (ODE’s).
p = c1e−t , q = c2e
−t , x = 2c1e−t + c3, y = −2c2e
−t + c4, z = c5e−2t
(f) Find the constants satisfying the initial condition, i.e., at t = 0.
c1 = −s, c2 = ±√2s, c3 = 4s, c4 = ±2
√2s, c5 = −s2.
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(g) Write the characteristic curves passing through the initial curve.
x(s, t) = 2s(2− e−t) (13)
y(s, t) = ±[
2√2s(1− e−t)
]
(14)
z(s, t) = −s2e−2t (15)
For each value of s we get a characteristic curve.(h) Eliminate s and t to obtain the required surface.
4z + (x ±√2y)2 = 0.
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Compatible system of first-order equations
DefinitionIf every solution of the first-order pde
f (x , y , z , p, q) = 0 (16)
is also a solution of the pde
g(x , y , z , p, q) = 0 (17)
then the equations are said to be compatible.
TheoremA necessary condition that the two equations (16) and (17) arecompatible is
[f , g ] :=∂(f , g)
∂(x , p)+ p
∂(f , g)
∂(z , p)+∂(f , g)
∂(y , q)+ q
∂(f , g)
∂(z , q)= 0. (18)
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Proof. If
J ≡ ∂(f , g)
∂(p, q)6= 0, (19)
then one can solve (16) and (17) for p and q to obtain
p = φ(x , y , z), q = ψ(x , y , z). (20)
The equations (16) and (17) will be compatible only if the system (20) iscompletely integrable, i.e., the equation
φ dx + ψ dy − dz = 0 (21)
is integrable.
An equation of the formn
∑
i=1
Fidxi = 0 is called a Pfaffian
differential equation. A necessary condition that the Pfaffianequation is integrable is that ~X · curl ~X = 0, where~X = (F1,F2, . . . ,Fn).
Thus, (21) is integrable if φ(−ψz ) + ψ(φz )− (ψx − φy ) = 0, i.e.,
ψx + φψz = φy + ψφz . (22)
MA201(2014):PDE
Substituting from equations (20) into (16) and differentiating w.r.t. xand z , respectively, we get
fx + fpφx + fqψx = 0,
fz + fpφz + fqψz = 0,
from which it is deduced that
fx + φfz + fp(φx + φφz ) + fq(ψx + φψz ) = 0.
Similarly we can deduce from (17) that
gx + φgz + gp(φx + φφz ) + gq(ψx + φψz) = 0.
Solving these two equations, we find that
ψx + φψz =1
J
{
∂(f , g)
∂(x , p)+ φ
∂(f , g)
∂(z , p)
}
(23)
where J is as defined as in (19).
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If we differentiate the given pair of equations w.r.t. y and z , respectively,we will obtain
φy + ψφz = − 1
J
{
∂(f , g)
∂(y , q)+ ψ
∂(f , g)
∂(z , q)
}
. (24)
so that, substituting from (23) and (24) into (22and replacing φ, ψ byp, q respectively, we see that the condition that the two
[f , g ] =∂(f , g)
∂(x , p)+ p
∂(f , g)
∂(z , p)+∂(f , g)
∂(y , q)+ q
∂(f , g)
∂(z , q)= 0. (25)
MA201(2014):PDE
Example: Show that the equations
xp = yq, z(xp + yq) = 2xy
are compatible and solve them.Solution: Here f = xp − yq, g = z(xp + yq) − 2xy , so that
∂(f , g)
∂(x , p)= 2xy ,
∂(f , g)
∂(z , p)= −x2p−xyq,
∂(f , g)
∂(y , q)= −2xy ,
∂(f , g)
∂(z , q)= xyp+y2q
which imply that [f , g ] = y2q2 − p2x2 = 0. Hence the given equationsare compatible.Solving the given equations for p and q we get p = y/z , q = x/z . Thus,dz = ∂z
∂x dx + ∂z∂y dy gives
z dz = y dx + x dy .
Hence the solutions arez2 = 2xy + c1.
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Charpit’s Method
For solvingf (x , y , z , p, q) = 0, (26)
findg(x , y , z , p, q, a) = 0 (27)
containing an arbitrary constant a such that
• The PDE (27) is compatible to (26).
• Equations (26) and (27) can be solved to give
p = p(x , y , z , a), q = q(x , y , z , a).
• The equation
dz = p(x , y , z , a)dx + q(x , y , z , a)dy (28)
is integrable.
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Charpit’s Method
Such a g should satisfy
fp∂g
∂x+ fq
∂g
∂y+ (pfp + qfq)
∂g
∂z− (fx + pfz)
∂g
∂p− (fy + qfz )
∂g
∂q= 0. (29)
which is equivalent to
dx
fp=
dy
fq=
dz
pfp + qfq=
dp
−(fx + pfz)=
dq
−(fy + qfz). (30)
These equations are known as Charpit’s equations.Once an integral g(x , y , z , p, q, a) of this kind has been found, theproblem reduces to solving for p, q and then integrating the equation
dz = p(x , y , z , a)dx + q(x , y , z , a)dy (31)
MA201(2014):PDE
Example. Find a complete integral of the equation
p2x + q2y = z . (32)
Solution: The Charpit’s equations are:
dx
2px=
dy
2qy=
dz
2(p2x + q2y=
dp
p − p2=
dq
q − q2
from which it follows that
p2dx + 2pxdp
p2x=
q2dy + 2qydq
q2y
⇒ p2x = aq2y , a is a constant. (33)
Solving (32) and (33) for p and q, we get
p =
{
az
(1 + a)x)
}1/2
, q =
{
z
(1 + a)y)
}1/2
So, equation (28) in this case becomes
(
1 + a
z
)1/2
dz =( a
x
)1/2dx +
(
1
y
)1/2
dy
⇒ {(1 + a)z}1/2 = (ax)1/2 + y1/2 + b
which is the required complete integral of (32).
MA201(2014):PDE
Special types of first-order PDEs
We discuss now some special types of first order PDEs.
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Equations involving only p and q
Consider an equationf (p, q) = 0. (34)
Charpit’s equations reduce to
dx
fp=
dy
fq=
dz
pfp + qfq=
dp
0=
dq
0(35)
Then, dp = dq = 0. Suppose we choose p = a, a constant. Thecorresponding value of q is obtained from
f (a, q) = 0 (36)
so that q = Q(a), a constant. Thus, we get a solution (completeintegral) of the equation as
z = ax + Q(a)y + b (37)
where a, b are parameters.Note: at times assuming q = a and solving f (p, a) = 0 for p may reducethe work considerably.
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Example: Find the complete integral of pq = 1.
Solution: The equation involves only p and q, and so from Charpit’sequation dp = dq = 0. Putting p = a we get q = 1/a. The completeintegral is, therefore,
z = ax +y
a+ b
i.e.,a2x + y − az = c , where a, c are arbitrary constants.
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Equations not involving the independent variable
Let the pde be of the form
f (z , p, q) = 0. (38)
Then the Charpit’s equations take the form
dx
fp=
dy
fq=
dz
pfp + qfq=
dp
−pfz=
dq
−qfz. (39)
The last equation givesdp
p=
dq
q, which leads to the relation
p = aq (40)
where a is a constant. Solving (38) and (40), we obtain expressions forp, q from which a complete integral follows immediately.
MA201(2014):PDE
Example: Find a complete integral of p2z2 + q2 = 1.
Solution: The equation does not involve the independent variables.From Charpit’s equation, we have p = aq. Therefore, we have
q2(1 + a2z2) = 1
⇒ q = (1 + a2z2)−1/2, p = a(1 + a2z2)−1/2
Hencedz = pdx + qdy
⇒ (1 + a2z2)1/2dz = adx + dy .
which leads to the complete integral
az(1 + a2z2)1/2 − log[az + (1 + a2z2)1/2] = 2a(ax + y + b).
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Separable equations
We say that a first-order pde is separable, if it can be written in the form
f (x , p) = g(y , q) (41)
For such equations, Charpit’s equations become
dx
fp=
dy
−gq=
dz
pfp − qgq=
dp
−fx=
dq
−gy. (42)
We havedx
fp=
dp
−fx, i.e., df = fxdx + fpdp = 0, i.e., f (x , p) = a, a
constant. Similarly, g(y , q) = a. Now, solve them for p and q to getp = r(x , a), q = s(y , a). Solve
dz = p dx + q dy , i.e., dz = r(x , a)dx + s(y , a)dy
to get a complete integral.
MA201(2014):PDE
Example. Determine the complete integral of
p2y(1 + x2) = qx2.
Solution: Given thatp2y(1 + x2) = qx2.
⇒ p2(1 + x2)
x2=
q
y,
so that p =ax√1 + x2
, q = a2y .
Hence a complete integral is
z = a√
1 + x2 +1
2a2y2 + b,
where a and b are constants.
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Clairaut’s Equation
A first-order PDE is said to be of Clairaut type if it can be written in theform
z = px + qy + g(p, q) (43)
The corresponding Charpit’s equations are
dx
x + gp=
dy
y + gq=
dz
px + qy + pgp + qgq=
dp
0=
dq
0
⇒ p = a, q = b.
Using these values in (43), we get the complete integral
z = ax + by + g(a, b) (44)
as is readily verified by direct differentiation.
MA201(2014):PDE
Example Find a complete integral of
(p + q)(z − xp − yq) = 1.
Solution: The given equation can be written in the form
z = xp + yq +1
p + q,
i.e., the equation is a Clairaut’s equation. The complete integral is
z = ax + by +1
a+ b.
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Solution/Integral Surface of First Order PDE
Passing through a Given Curve
Refer: Sneddon, Chapter 2, Section 12
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Suppose a complete integral (a two parameter family of surfaces)
f (x , y , z , a, b) = 0 (45)
is known for the PDEF (x , y , z , p, q) = 0. (46)
Can we use the complete integral for finding an integral surface passingthrough a given curve C given by
x = x(t), y = y(t), z = z(t) (47)
We expect such a surface can be
1 A particular case of the complete integral f (x , y , z , a, b) = 0obtained by giving a or b particular values,
2 A particular case of the general integral which is the envelope of aone parameter subsystem of f (x , y , z , a, b) = 0,
3 The envelope of the two parameter system f (x , y , z , a, b) = 0.
Most frequently one finds a surface of type (2).
MA201(2014):PDE
Suppose E is an integral surface of type (2) containing C . Then E is theenvelope of a one parameter subfamily each of which touches C at apoint. A one-parameter subfamily of (45) is obtained by puttingb = φ(a).For each t, the corresponding point on C is on a member of (45), and so
f (x(t), y(t), z(t), a, b) = 0. (48)
Because the point touches the surface, it implies that
∂
∂t
(
f (x(t), y(t), z(t), a, b))
= 0. (49)
This implies, (48) has two equal roots for t. Eliminating t from (48) and(49) we get relation
ψ(a, b). (50)
Each solution b = φ(a) of (50) gives a one parameter subfamily of (45)whose envelope gives a required integral surface.
MA201(2014):PDE
Find a complete integral of the PDE (p2 + q2)x = pz and deduce thatthe solution which passes through the curve x = 0, z2 = 4y .
Step 1: Find solution of (p2 + q2)x = pz using Charpit’s Method.Using Charpit’s method, we find the complete integral of the given PDEas
z2 = a2x2 + (ay + b)2 (51)
where a and b are real parameters.
Step 2: Find one Parameter Subsystem of (51) passing through givencurve,
Step 3: Find Envelope of One Parameter Subsystem of (51)
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Jacobi’s Method
Refer: Sneddon, Chapter 2, Section 13
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Let u(x , y , z) = 0 be a solution of F (x , y , z , p, q) = 0.Then ux + puz = 0 gives p = −ux/uz . Similarly, q = −uy/uz . Puttingthese in the equation, we get a new equation
f (x , y , z , ux , uy , uz) = 0.
An auxiliary equation is obtained as follows:
dx
fux=
dy
fuy=
dz
fuz=
du1
−fx=
duy
−fy=
duz
−fz
Solve these equations to obtain ux , uy , uz and get a complete solutionsfrom
du = uxdx + uydy + uzdz
involving three parameters a, b, c . For each value of one of theparameters, we get a complete solution.
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Example - Jacobi’s Method
Using Jacobi’s method, find a complete integral of the equation
p2x + q2y = z . (52)
Step 1: Convert (52) into the form f (x , y , z , ux , uy , uz) = 0
Step 2: Solving the PDE f (x , y , z , ux , uy , uz) = 0 by Jacobi’s method.
Step 3: Getting the Solution of (52) from the solution of f = 0.
MA201(2014):PDE