Partial Differential Equation - Notes

Lecture notes for Partial Differential equation.
Student an go through theory and solve in built exercise..

1. Partial Dierential Equations
Introduction
Partial Dierential Equations(PDE) arise when the functions involved or
depend on two or more independent variables. Physical and Engineering
problems like solid and uid mechanics, heat transfer, vibrations, electro-
magnetic theory and other areas lead to PDE.
Partial Dierential Equations are those which involve partial
derivatives with respect to two or more independent variables.
E.g.:
@
2
u
@x
2
+
@
2
u
@y
2
= 0 Two-dimensional Laplace equation ..(1)
@
2
u
@t
2
=c
2
@
2
u
@x
2
One-dimensional Wave equation ..(2)
@
2
u
@x
2
+
@
2
u
@y
2
+
@
2
u
@z
2
= 0 Three-dimensional Laplace equation ..(3)
@u
@t
=c
2
@
2
u
@x
2
One-dimensional heat equation ..(4)
@z
@x
+
@z
@y
= 1 ..(5)
TheORDER of a PDE is the order of the highest derivatives in the
equation.
NOTE: We restrict our study to PDE's involving one dependent vari-
able z and only two independent variables x and y.
1

2. Partial Dierential Equations
NOTATIONS:
@z
@x
=p,
@z
@y
=q,
@
2
z
@x
2
=r,
@
2
z
@x@y
=sand
@
2
z
@y
2
=t
Ex. 1Show thatu=sin9t sin(
x
4
) is a solution of a one dimensional wave
equation.
Sol.Hereu=sin9t sin(
x
4
).
@
2
u
@t
2
=c
2
@
2
u
@x
2
.. (1) is one dimensional wave equation.
@u
@t
=
@u
2
@t
2
=
@u
@x
=
@u
2
@x
2
=
Substitute above values in eq. (1), we get
2 Dept. of Mathematics, AITS - Rajkot

3. Partial Dierential Equations
c=
)Forc= ,u=sin9t sin(
x
4
) is a solution of a wave equation.
Ex. 2Verify that the functionu=e
x
cosyis the solution of the Laplace
equation
@
2
u
@x
2
+
@
2
u
@y
2
= 0.
Sol.
Dept. of Mathematics, AITS - Rajkot 3

4. Partial Dierential Equations
Ex. 3Verify that the functionu=x
3
+ 3xt
2
is the solution of the wave
equation
@
2
u
@t
2
=c
2
@
2
u
@x
2
for a suitable value ofc.
Sol.
4 Dept. of Mathematics, AITS - Rajkot

5. Partial Dierential Equations
Exercise 1.1
Q.1Verify the following functions are the solutions of the Laplace's equation
@
2
u
@x
2
+
@
2
u
@y
2
= 0.
(a)u=log(x
2
+y
2
)
(b)u=sinxsinhy
(c)u=tan
1

y
x

Q.2Verify the following functions are the solutions of the wave equation
@
2
u
@t
2
=c
2
@
2
u
@x
2
for a suitable value ofc.
(a)u=x
3
+ 3xt
2
(b)u=cosctsinx
Q.3Verify the following functions are the solutions of the heat equation
@u
@t
=c
2
@
2
u
@x
2
for a suitable value ofc.
(a)u=e
2t
cosx
Dept. of Mathematics, AITS - Rajkot 5

6. Partial Dierential Equations
Formation of PDE
Partial Dierential equations can be formed either by the elimination of
(1) arbitrary constants present in the functional relation present in the rela-
tion between the variable
(2) arbitrary functions of these variables.
By elimination of arbitrary constants
Consider the functionf(x; y; z; a; b) = 0 ..(i)
where a and b are two independent arbitrary constants. To eliminate two
constants, we require two more equations.
Dierentiating eq. (i) partially with respect toxandyin turn, we obtain
@f
@x
+
@f
@z
@z
@x
= 0)
@f
@x
+p
@f
@z
= 0.. (ii)
@f
@y
+
@f
@z
@z
@y
= 0)
@f
@y
+q
@f
@z
= 0.. (iii)
Eliminatingaandbfrom the set of equations (i), (ii) and (iii) , we get a
PDE of the rst order of the formF(x; y; z; p; q) = 0. .. (iv)
Ex. 1Eliminate the constantsaandbformz= (x+a)(y+b).
Sol.Given thatz= (x+a)(y+b) ... (1)
Dierentiating partially eq. (1) with respect toxandyrespectively,
we get
@z
@x
=p= ..(2)
@z
@y
=q= ..(3)
Eliminatingaandbfrom equation (1), (2) and (3), we get
6 Dept. of Mathematics, AITS - Rajkot

7. Partial Dierential Equations
is the required partial dierential equation.
Ex. 2Find the dierential equation of the set of all spheres whose centres
lie on the z-axis.
Sol.The equation of the spheres with centres on the z-axis is
..(1)
Dierentiating partially eq. (1) with respect toxandyrespectively,
we get
Eliminating arbitrary constant from (2) and (3), we get
is the required partial dierential equation.
Dept. of Mathematics, AITS - Rajkot 7

8. Partial Dierential Equations
By elimination of arbitrary functions
Consider a relation between x, y and z of the type(u; v) = 0 ..(v)
where u and v are known functions of x, y and z andis an arbitrary
function of u and v.
Dierentiating equ. (v) partially with respect to x and y, respectively,
we get
@
@u

@u
@x
+
@u
@z
p

+
@
@v

@v
@x
+
@v
@z
p

= 0.. (vi)
@
@u

@u
@y
+
@u
@z
q

+
@
@v

@v
@y
+
@v
@z
q

= 0.. (vii)
Eliminating
@
@u
and
@
@v
from the eq. (vi) and (vii), we get the
equations
@(u;v)
@(y;z)
p+
@(u;v)
@(z;x)
q=
@(u;v)
@(x;y)
.. (viii)
which is a PDE of the type (iv).
since the power of p and q are both unity it is also linear equation,
whereas eq. (iv) need not be linear.
Ex. 1Eliminate the arbitrary function from the equation
z=xy+f(x
2
+y
2
).
Sol.Letx
2
+y
2
=u.
)z=xy+f(u) .. (1)
dierentiating eq. (1) with respect to x and y respectively, we get
8 Dept. of Mathematics, AITS - Rajkot

9. Partial Dierential Equations
p= ..(2)
q= ..(3)
eliminating from (2) and (3), we get
Ex. 2Eliminate arbitrary function from the equation
f(x+y+z; x
2
+y
2
+z
2
) = 0.
Sol.Letx+y+z=u,x
2
+y
2
+z
2
=vthen
f(u; v) = 0 .. (1)
Dierentiating (1) with respect to x and y, we get
@f
@u

@u
@x
+
@u
@z
p

+
@f
@v

@v
@x
+
@v
@z
p

= 0 .. (2)
@f
@u

@u
@y
+
@u
@z
q

+
@f
@v

@v
@y
+
@v
@z
q

= 0 .. (3)
Now
@u
@x
=
@v
@x
=
@u
@y
=
@v
@y
=
@u
@z
=
@v
@z
=
Dept. of Mathematics, AITS - Rajkot 9

10. Partial Dierential Equations
substituting above values in eq. (2) and (3), we get
eliminating and from above eq. (4) and (5), we get
is the required partial dierential equation of the rst order.
10 Dept. of Mathematics, AITS - Rajkot

11. Partial Dierential Equations
Exercise 1.2
Q.1Form the partial dierential equationby eliminating the arbitrary
constantsfrom the following:
1.z= (x
2
+a)(y
2
+b)
2. (xa)
2
+ (yb)
2
+z
2
= 1
Q.2Form the partial dierential equationsby eliminating the arbitrary
functionsfrom the following:
1.z=f(x
2
y
2
)
2.z=x+y+f(xy)
3.f(xy+z
2
; x+y+z) = 0
4.f(x
2
+y
2
+z
2
; xyz) = 0
Dept. of Mathematics, AITS - Rajkot 11

12. Partial Dierential Equations
Integrals of Partial Dierential Equation
Asolutionorintegralof a partial dierential equation is a relation
between the dependent and the independent variables that satises
the dierential equation.
A solution which contains a number of arbitrary constants equal to
the independent variables, is called acomplete integral.
A solution obtained by giving particular values to the arbitrary
constants in a complete integral is called aparticular integral.
Singular integral
LetF(x; y; z; p; q) = 0 .. (1)
be the partial dierential equation whose complete integral is
f(x; y; z; a; b) = 0 .. (2)
eliminating a, b between eq. (2) and
@f
@a
= 0,
@f
@b
= 0
if it exists, is called asingular integral.
General Integral
In eq. (2), if we assumeb=(a), then (2) becomes
f[x; y; z; a; (a)] = 0 .. (3)
dierentiating (2) partially with respect to a,
@f
@a
+
@f
@b

0
(a) = 0 .. (4)
eliminating a between theses two equations (3) and (4), if it exists, is
called thegeneral integralof eq. (1).
12 Dept. of Mathematics, AITS - Rajkot

13. Partial Dierential Equations
Solutions of PDE by the method of Direct
Integration
This method is applicable to those problems, where direct integration is
possible.
Ex. 1Solve
@
2
z
@x@y
=x
3
+y
3
Sol.Given that
@
@x

@z
@y

=x
3
+y
3
Integrating w.r.t tokeeping constant, we get
)
@z
@y
=
Now integrating w.r.t. tokeeping as constant
)z=
Dept. of Mathematics, AITS - Rajkot 13

14. Partial Dierential Equations
Ex. 2Solve
@
2
u
@x@t
=e
t
cosx
Sol.Given that
@
@x

@u
@t

=e
t
cosx
Integrating w.r.t tokeeping constant, we get
Now integrating w.r.t. tokeeping as constant
14 Dept. of Mathematics, AITS - Rajkot

15. Partial Dierential Equations
Ex. 3Solve
@
3
z
@x
2
@y
=cos(2x+ 3y)
Sol.Given that
@
2
@x
2

@z
@y

=cos(2x+ 3y)
Integrating w.r.t tokeeping constant, we get
)
@
@x

@z
@y

=
Integrating w.r.t. tokeeping as constant
Finally integrating w.r.t. tokeeping as constant
Dept. of Mathematics, AITS - Rajkot 15

16. Partial Dierential Equations
Lagrange's Equation
We have
@(u; v)
@(y; z)
p+
@(u; v)
@(z; x)
q=
@(u; v)
@(x; y)
This can be expressed in the formPp+Qq=R, where P, Q and R are
functions of x, y and z. This partial dierential equation is known as
Lagrange's equation.
Method to solvePp+Qq=R
In order to solve the equationPp+Qq=R
1 Form the subsidiary (auxiliary ) equation
dx
P
=
dy
Q
=
dz
R
2 Solve these subsidiary equations by the method of grouping or by the
method of multiples or both to get two independent solutionsu=c1
andv=c2.
3 Then(u; v) = 0 oru=f(v) orv=f(u) is the general solution of
the equationPp+Qq=R.
Ex. 1Solve
y
2
z
x
@z
@x
+xz
@z
@y
=y
2
.
Sol.The given equation can be written as ( in for of p and q)
y
2
zp+x
2
zq=xy
2
comparing withPp+Qq=R, we get
P= ,Q= andR=
The subsidiary equations are
dx
P
=
dy
Q
=
dz
R
Taking rst two, we have
16 Dept. of Mathematics, AITS - Rajkot

17. Partial Dierential Equations
Now taking rst and third, we have
Ex. 2Find the general solution of the dierential equation
x
2
p+y
2
q= (x+y)z
Sol.Comparing withPp+Qq=R, we get
P= ,Q= andR=
The subsidiary equations are
dx
P
=
dy
Q
=
dz
R
Dept. of Mathematics, AITS - Rajkot 17

18. Partial Dierential Equations
Taking rst two, we have
Ex. 3Solve (y+z)p+ (z+x)q=x+y
Ex. 4Solvepzqz=z
2
+ (x+y)
2
Ex. 5Solvex
2
(yz)p+y
2
(zx) =z
2
(xy)
Ex. 6Solve (z
2
2yzy
2
)p+ (xy+zx)q=xyzx
Exercise
1.xp+yq=z
2.xpyq=xy
3.(1x)p+ (2y)q= 3z
4.zxpzyq=y
2
x
2
5.(yz)p+ (zx)q=xy
6.p2q= 2xe
y
+ 1
7.x(y
2
z
2
)p+y(z
2
x
2
)q=z(x
2
y
2
)
18 Dept. of Mathematics, AITS - Rajkot

19. Partial Dierential Equations
Special type of Nonlinear PDE of the rst
order
A PDE which involves rst order derivatives p and q with degree more than
one and the products of p and q is called anon-linear PDE of the rst
order.
There are four standard forms of these equations.
1. Equations involving only p and q
2. Equations not involving the independent variables
3. Separable equations
4. Clairaut's form
Standard Form 1. Equations involving only p and q
Such equations are of the formf(p,q)=0
z=ax+by+c... (1)
is a solution of the equationf(p; q) = 0
providedf(a; b) = 0 (means putp=aandq=b) ... (2)
solving (2) for b,b=F(a)
Hence the complete integral isz=ax+F(a)y+c
where a and c are arbitrary constants.
Ex. 1Solvepq=p+q
Sol.Here the give DE involves on p and q (so standard form 1)
The solution is
putp= andq=
therefore,
b=
Dept. of Mathematics, AITS - Rajkot 19

20. Partial Dierential Equations
Hence the complete solution is
Exercise
Solve the following equations
1.pq= 1
2.p=q
2
3.p
2
+q
2
= 4
4.pq+p+q= 0
20 Dept. of Mathematics, AITS - Rajkot

21. Partial Dierential Equations
Standard Form 2. Equations not involving the
independent variables
Such equations are of the formf(z,p,q)=0
Considerf(z; p; q) = 0 .. (1)
sincezis a function ofxandy
dz=
@z
@x
dx+
@z
@y
dy=pdx+qdy
letq=ap
then eq (1) becomesf(z; p; ap) = 0 .. (2)
solving (2) for p i.e.p=(z; a)
)dz=(z; a)dx+a (z; a)dy
)
dz
(z; a)
=dx+ady
integrating, we get
Rdz
(z; a)
=x+ay+bwhich is a complete integral.
Ex. 1Solve:p
2
z
2
+q
2
= 1
Sol.Given equation involves only p, q and z (standard form 2)
Therefore puttingq=apin given equation, we get
Nowdz=pdx+qdy, substituting above values in this equation and
integrate
Dept. of Mathematics, AITS - Rajkot 21

22. Partial Dierential Equations
Ex. 2Solve:p(1 +q) =qz
Ex. 3Solve:z=p
2
+q
2
Exercise
1q(p
2
z+q
2
) = 4
2pq=z
2
3p+q=z
4zpq=p+q
5z
2
= 1 +p
2
+q
2
22 Dept. of Mathematics, AITS - Rajkot

23. Partial Dierential Equations
Standard Form 3. Separable equations
Such equations are of the formf1(x,p) = f2(y,q)
A rst order PDE isseperableif it can be written in the form
f1(x; p) =f2(y; q)
We assume each side equal to an arbitrary constant a.
solvingf1(x; p) =a)p=1(x; a)
solvingf2(y; q) =a)q=2(y; a)
)z=
R
1(x; a)dx+
R
2(y; a)dywhich is complete integral
Ex. 1Solve:px
2
=q+y
2
Sol.Writing equation in the form
px
2
=q+y
2
=a
)p=a+x
2
andq=ay
2
Nowdz=pdx+qdy
dz= dx+ dy
Integrating both sides, we get
z=
Ex. 2Solve:p
2
+q
2
=x
2
+y
2
Sol.Writing equation in the form
p
2
x
2
=y
2
q
2
Letp
2
x
2
=aandy
2
q
2
=a
Hence,p
2
=x
2
+a,q
2
=y
2
a
Dept. of Mathematics, AITS - Rajkot 23

24. Partial Dierential Equations
)p= andq=
Nowdz=pdx+qdy
dz=
Exercise
1.qp+xy= 0
2.p+q=x+y
3.P
2
+q
2
=x+y
4.pq=xy
5.py+qx=pq
6.p+q=sinx+siny
24 Dept. of Mathematics, AITS - Rajkot

25. Partial Dierential Equations
Standard Form 4. Clairaut's form
A rst-order PDE is said to be of Clairaut type if it can be written in the
form,
z=px+qy+f(p; q)
substitutep=aandq=binf(p; q)
The solution of the the equation isz=ax+by+f(a; b)
Ex. 1Solve:z=px+qy+
p
1 +p
2
+q
2
Sol.The complete integral is obtained
z=px+qy+
p
1 +a
2
+b
2
Ex. 2Solve: (p+q)(zxpyq) = 1.
Sol.
Exercise
Solve the following equations.
1.z=px+qy+p
2
q
2
2.z=px+qy+ 2
p
pq
3.z=px+qy+
q
p
p
4. (1x)p+ (2y)q= 3z
5.
z
pq
=
x
q
+
y
p
+
p
pq
Dept. of Mathematics, AITS - Rajkot 25

26. Partial Dierential Equations
Classication of the PDE of second order
Consider the equation of the second order as
A
@
2
u
@x
2
+B
@
2
u
@x@y
+C
@
2
u
@
2
y
+f

x; y; u;
@u
@x
;
@u
@y

= 0 (1)
where A is positive
Now the equation (1) is
(i)ellipticifB
2
4AC <0
(ii)hyperbolicifB
2
4AC >0
(iii)parabolicifB
2
4AC= 0.
Exercise
Classify the following dierential equations.
1 2
@
2
u
@t
2
+ 4
@
2
u
@x@t
+ 3
@
2
u
@x
2
= 0
2 4
@
2
u
@t
2
9
@
2
u
@x@t
+ 5
@
2
u
@x
2
= 0
3
@
2
u
@t
2
+ 4
@
2
u
@x@t
+ 4
@
2
u
@x
2
= 0
26 Dept. of Mathematics, AITS - Rajkot