Partial Differential Equations

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Partial Differential Equations
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OUTLINE

• Formation of pde by eliminating the arbitrary
• constants
• Formation of pde by eliminating the arbitrary
• functions
• Solutions to first order first degree pde of
  the type
• P p Q q R

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• Charpits method
• Solution by Separation of Variables method
• Derivation of One Dimensional Wave equation
• Derivation of One Dimensional Heat
  equation
• Numerical solutions to,
• One Dimensional Wave
  equation
• One Dimensional Heat
  equation
• Two Dimensional Laplacian equation

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Formation of pde by eliminating the arbitrary
constants
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.(i)
(1)
Differentiating (i) partially with respect to x
and y,
Substituting these values of 1/a2 and 1/b2 in
(i), we get
2 z x p y q
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(2) z (x2 a) (y2 b)
Differentiating the given relation partially
w. r. t. x and y,
Or p q 4xyz

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(3) (x-a)2 (y-b)2 z2 k2 (i)
Differentiating (i) partially w. r. t. x and y,
Substituting for (x- a) and (y- b) from these in
(i), we get


This is the required partial differential
equation.
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(4) z ax by cxy ...(i)
Differentiating (i) partially w.r.t. x y, we get
It is not possible to eliminate a,b,c from
relations (i)-(iii). Partially differentiating
(ii),
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Using this in (ii) and (iii)
Substituting for a, b, c in (i), we get


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(5)
Differentiating partially w.r.t. x,
Differentiating this partially w.r.t. x, we get

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Is the required p. d. e..
Note Differentiating the given equation
partially w.r.t. y twice we get

As another required partial differential equation.
P.D.E. obtained by elimination of arbitrary
constants need not be not unique
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Formation of p d e by eliminating the arbitrary
functions
(1) z f(x2 y2)

Differentiating z partially w. r.t. x and y,
p /q x / y or y p x q0 as the pde
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Formation of p d e by eliminating the arbitrary
functions
(1) z f(x2 y2)

Differentiating z partially w. r.t. x and y,
p /q x / y or y p x q0 as the pde
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Formation of p d e by eliminating the arbitrary
functions
(1) z f(x2 y2)

Differentiating z partially w. r.t. x and y,
p /q x / y or y p x q0 as the pde
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(2) z f ( x ct ) g (x -ct)
Differentiating z partially with respect to x and
t,
Thus the pde is
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(3) x y z f(x2 y2 z2)
Differentiating partially w.r.t. x and y
required pde
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(4) z f ( xy / z ).
Differentiating partially w.r.t. x and y
or xp yq
required pde
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(5) z y2 2 f(1/x logy)
Hence
ans
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(6) Z xF(y) y ?(x)
Substituting for
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LAGRANGES FIRST ORDER FIRST DEGREE PDE PpQqR
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(1)Solve yzp zxq xy.
subsidiary equations are
From the first two and the last two terms, we
get, respectively,
and
Integrating we get x2 - y2 a, y2 z2
b.
F(x2-y2, y2 z2) 0 ans
Hence, a general solution is
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(2) Solve y2p - xyq x(z-2y)
From the first two ratios we get x2 y2 a

from the last ratios two we get
ordinary linear differential equation hence yz
y2 b
solution is F( x2 y2, yz y2) 0

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(3) Solve z(xp yq) y2 x2

on integration, yields xy a
xdx ydy zdz 0 x2 y2 z2 b
Hence, a general solution of the given equation
F(xy,x2y2z2)0
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(4)
(4)
x dx y dy z dz 0 (i)
x2 y2 z2 a
Integrating (i) we get
yz dx zx dy xy dz 0 (ii)

Dividing (ii) throughout by xyz and then
integrating, we get xyz b
F( x2 y2 z2, xyz ) 0 ans
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(5)
(x2z)p (4zx y)q 2x2 y


Using multipliers 2x, -1, -1 we obtain 2x dx dy
dz 0 which on integration gives
x2 y z a .(ii)
Using multipliers y, x, -2z in (i), we obtain y
dx x dy 2z dz 0 which on integration yields


xy z2 b
.(iii)
Hence, a general solution of the given equation is
F(x2 y z, xy z2) 0.