18MAB201T-U1-Week1.pdf

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18MAB201T-Transforms and Boundary Value

Problems

Prepared by

Dr. S. ATHITHAN

Assistant Professor

Department of of Mathematics

Faculty of Engineering and Technology

SRM INSTITUTE OF SCIENCE AND TECHNOLOGY

Kattankulathur-603203, Kancheepuram District.

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PARTIAL DIFFERENTIAL EQUATIONS

TOPICS:

? Formation of partial differential equation by eliminating arbitrary constants

? Formation of partial differential equation by eliminating arbitrary functions

? Formation of partial differential equation by eliminating arbitrary functions of the form

φ(u, v) = 0.

? Solution of standard types of first order equations

? Reducible to standard type

? Lagrange’s linear equation: Method of grouping, method of multipliers

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1

Formation of PDE’s

Example: 1. Form a partial differential equation (pde) by eliminating arbitrary constants from

(x−a)2+ (y−b)2 +z2−1 = 0.

Hints/Solution: Let (x−a)2 + (y − b)2 +z2 −1 = 0 − − − − − − − − > (1)

Differentiating (1) partially w.r.t. xandy, we get

2(x−a) + 2z∂z ∂x = 0

=⇒ x−a = −pz− − − − − −− > (2)

and

2(y−b) + 2z∂z ∂y = 0

=⇒ y−b = −qz− − − − − −− > (3)

Substituting (2) and (3) in (1), we get the required pde as(p2+q2+ 1)z2 = 1

Example: 2. Form a partial differential equation (pde) by eliminating arbitrary constants from

z =ax+by +a2 +b2.

Hints/Solution: Letz = ax+by +a2 +b2 − − − − − − − − > (1)Differentiating (1) partially w.r.t.xandy, we get

∂z

∂x =p =a

and

∂z

∂y =q = b

Substituting in (1), we get the required pde asz = px+qy+p2 +q2

Example: 3. Form a partial differential equation (pde) by eliminating arbitrary constants from

z =ax+by +ab.

Hints/Solution:Letz =ax+by +ab− − − −− > (1)

Differentiating (1) partially w.r.t. xandy, we get ∂z

∂x = p = aand ∂z

∂y = q = b.

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Example: 4. Form a partial differential equation (pde) by eliminating arbitrary function from

z =xy +f(x2 +y2 +z2).

Hints/Solution:

Givenz =xy +f(x2 +y2 +z2)− − − − − − − − >(1)

Differentiating (1) partially w.r.t. xandy, we get

∂z

∂x = p= y+f

0

(x2+y2 +z2)(2x+ 2zp)− − − − − −− > (2)

and

∂z

∂y =q =x+f

0

(x2 +y2 +z2)(2zq+ 2y)− − − − − −− > (3)

=⇒ f0 = q−x

2zq+ 2y − − − − − −− > (4)

Substituting (4) in (2), we get the required pde asp = q−x

2zq + 2y(2x+ 2zp) +y

Example: 5. Form a partial differential equation (pde) by eliminating arbitrary function from

z =f(x2 +y2 +z2).

Hints/Solution:

Givenz =f(x2 +y2 +z2)− − − − − − − − >(1)

Differentiating (1) partially w.r.t. xandy, we get

∂z

∂x = p= f

0

(x2+y2 +z2)(2x+ 2zp)− − − − − −− > (2)

and

∂z

∂y =q = f

0

(x2+y2 +z2)(2zq + 2y)− − − − − −− > (3)

=⇒ f0 = q

2zq + 2y − − − − − −− > (4)

Substituting (4) in (2) we get,p= q

2zq + 2y(2x+ 2zp)

Simplifying we get the required pde aspy = qx

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Givenz =xf(x+y) +g(x+y)− − − −−> (1)

Differentiating (1) partially w.r.t. xandy, we get

∂z

∂x =p =xf

0

(x+y) +f(x+y) +g0(x+y)− − − −− >(2)

and

∂z

∂y =q =xf

0

(x+y) +g0(x+y)− − − −− > (3)

Now, differentiating (2) partially w.r.t.xandy, we get

∂2_z

∂x2 =r =xf

00

(x+y) + 2f0(x+y) +g00(x+y)− − − −− > (4)

∂2_z

∂x∂y = s=xf

00

(x+y) +f0(x+y) +g00(x+y)− − − −− > (5)

Differentiating (3) partially w.r.t. y, we get

∂2_z

∂y2 = t =xf

00

(x+y) +g00(x+y)− − − −− > (6)

Substituting (6) in (5) we get,f0 =s−t− − − −− >(7)

Substituting (7) in (4) we get,r = t+ 2(s−t)− − − −−> (8)

Simplifying we get the required pde as r = 2s−t

Example: 7. Form a partial differential equation (pde) by eliminating arbitrary function from

f(xy+z2, x+y+z) = 0.

Hints/Solution:

Givenf(xy+z2, x+y+z) = 0− − − −−> (1)

Letu = xy+z2 andv = x+y+zWe eliminatef by evaluation the determinant below:

∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y

= 0 =⇒

y+ 2zp 1 +p

x+ 2zq 1 +q

= 0

Simplifying we get the required pde as (2z −x)p+q(y−2x) = x−y

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Differentiating (1) partially w.r.t. xandy, we get

y+ 2zp =f0(x+y+z)·(1 +p)− − − −−> (2)

and

x+ 2zq = f0(x+y+z)·(1 +q)− − − −−> (3)

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2

Exercise/Practice/Assignment Problems

1. Form the differential equation of all spheres whose centres lie on the z-axis. Ans:

xq =yp

2. Form a partial differential equation (pde) by eliminating arbitrary constants from (a) z = (x2 +a)(y2 +b). Ans: pq = 4xyz

(b) z = (x−a)2 + (b−y)2 + 5. Ans:4z =p2+q2 + 20

(c) x

2

a2 +

y2

b2 +

z2

c2 = 1. Ans:−zp+xp 2

+zxr = 0

(d) z =ax+by +cxy. Ans:z = (p−sy)x+ (q−sx)y+sxy

3. Form a partial differential equation (pde) by eliminating arbitrary function/s from (a) z =φ(x2+y2 −z2). Ans:px(y2+z2)−qy(z2+x2) = z(x2 −y2)

(b) xy+yz+zx=f

_z

x+y

. Ans:z(x−y) = p(x+y)(x+ 2z)−q(x+

y)(y+ 2z)

(c) z =x2f(y) +y2g(x). Ans: xyr= 2[px+qy−2z]

(d) z =xf(y/x) +yφ(x). Ans:x ∂ 3_z

∂x∂y2 +y ∂3_z

∂y3 + ∂2_z

∂y2 = 0

(e) z =f(x3 + 2y) +φ(x3−2y). Ans:r = 2p

x +

9tx4

4 (f) z =x2f(y) +y2φ(x). Ans:xyr = 2[px+qy −2z] (g) g(z2−xy,x

z) = 0. Ans: zx=px

2 _{−q(xy −2z}2₎

(h) h(x2+y2+z2, xyz) = 0. Ans:px(z2−y2) +qy(x2−z2) = z(y2−x2) (i) f(y

x, x

2 _+y2 _+z2_{) = 0. Ans:xzp+yzq+x}2 _+y2 _{= 0}

PRACTICE MORE PROBLEMS ON SOME OF THE REFERENCE BOOKS.

Acknowledgement:

Some of the portions of this material are derived from various sources. I thank the authors for those books and related materials.