PARTIAL DIFFERENTIAL EQUATIONS

1. FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS

1.1 Elimination of Arbitrary Constants

Let the given equation be f(x,y,z,a,b) = 0 (1)

1.1.1 METHOD

1. Differentiate Equation (1) partially with respect to ‘a’, (2)

2. Differentiate Equation (1) partially with respect to ‘B’, (3)

3. Eliminate ‘a’ & ‘b’ from Eqns. (1), (2) & (3), we get F(x,y,z,p,q) = 0

1.1.2 PROBLEMS

Eliminate arbitrary constants from the following equations and form partial differential equations

1) z = (x2 + a2) (y2 + b2)

2)

3)

4)

5)

1.2 Elimination of Arbitrary Functions

1.2.1 Type I f(u,v) = 0 or u = f(x,y,z), where u and v are functions of x&y.

1.2.1.2 METHOD

Let f(u, v) = 0 be given (1)

1) Compute

2) Obtain , ,

3) The solution is given by P p + Q q = R.

1.2.1.2 PROBLEMS

Eliminate arbitrary functions from the following equations and form partial differential equations

1) z = x + y =f(xy)2) z = xy + f(x2 +y2 + z2)

3)

4)

5)

1.2.2 TYPE II z =F(x, y, z, f, φ)(1)

1.2.2.1 METHOD

1) Differentiate Eqn.(1) partially with respect to x, p = F1(x,y,z,f,f’,φ,φ’) = 0 (2)2) Differentiate Eqn.(1) partially with respect to y, q = F2(x,y,z,f,f’,φ,φ’) = 0 (3)3) Differentiate Eqn.(2) partially with respect to x, r = F3(x,y,z,f,f’,φ,φ’, f”,φ”) = 0 (4)4) Differentiate Eqn.(2) partially with respect to y, s = F4(x,y,z,f,f’,φ,φ’,f”,φ”) = 0 (5)5) Differentiate Eqn.(3) partially with respect to y, t = F5(x,y,z,f,f’,φ,φ’,f”,φ”) = 0 (6)6) Eliminate f & g from Eqns.(1) to (6), we get G(x,y,z,p,q,r,s,t) = 0

1.2.2.2 PROBLEMS

Form p.d.e by eliminating arbitrary functionsfrom the following equations.

1) z = f(y) + φ(x + y + z)

2)

3) z = f(x2 + y) + φ(x2 – y)4) z = f(x + y + z) + φ(x – y)5) z = f(ax + by) + g(cx + dy)

2. SOLUTIONS OF FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

A general first order homogeneous differential equation is of the form F(x,y,z,p,q) = 0

2.1 STANDARD I F(p, q ) = 0

2.1.1 METHOD

1) Assume the trial solution of F(p, q) = 0 as z = ax + by + c (1) where F(a, b) = 0

2) Express a = φ(b) or b = φ(a)3) Substitute a or b in Eqn(1), the Complete solution is z = φ(b)x + by + c or (2)

z = ax + φ(a)y + c.(3)

4) Put c = f(b) Eqn (2) becomes z = φ(b)x + by + f(b) (4) 5) To get general solution differentiate Eqn.(4) w.r.to ‘b’ and eliminate the constant b.

2.1.2 PROBLEMS

Solve the following equations

1)2) p + q = pq3) pq = a4) p2 + q2 = npq5) p + pq = 1

2.2 STANDARD II z = px + qy + f(p, q) = 0 (1)

2.2.1 Method

1) Put p =a & q = b, we get the complete integral as z = ax + by + f(a, b) = 0. (2)

2) Differentiate Eqn.(2) w.r.to a & b, we get (3) (4)

3) Express Eqn. (3) as x = F(a, b) (5)4) Express Eqn. (4) as y = G(a, b) (6)

5) Eliminate a& b from Eqns.(3) to (6), we get singular solution.

2.2.2 PROBLEMS

Solve the following equations

1) z = px + qy + pq2) z = px + qy + p2q2

3) z = px + qy + p2 – q2

4) z = px + qy + p2q + pq2

5)

2.3 STANDARD III f(z, p, q) = 0 (1)

2.3.1 METHOD

1) Assume the trial solution as z = f(u).

2) Put u = x + ay, so that in Eqn. (1)

3) Then Eqn. (1) becomes = 0

4) Separate du on one side and g(z)dz on other side5) Integrate on both sides we get ux + b = F(z)6) The complete solution is x + ay + b = F(z)

2.3.2 PROBLEMS

Solve the following equations

1) p(1 + q) = qz2

2) q(1 + p2) = pz3) z2 = 1 + p2 + q2

4) 16(p2z + q2) = 255) z = p2 +q2 6) p2z2 + q2 = p2q7) q2 = z2p2(1 – p2)8) p2 + q = z

2.4 STANDARD IV f(x,y,p,q) = 0 (1)2.4.1 METHOD

1) Separate x & p on one side and y & q on other side, so that F(x, p) = G(y, q). (2)2) Let f1(x, p) = f2(y, q) = a (3)3) Rewrite the above equation as p = f1(x, a) and q = f2(y, a)

4) Then the complete integral is

2.4.2 PROBLEMS

Solve the following equations

1) p2y(1 + x2) = qx2

2) p – x2 = q + y2

3) p + q = sin x+ sin y

4) yp = 2yx + log q

5) p2 + qx + y

2.5 EQUATIONS REDUCIBLE TO STANDARS FORMS

2.5.1 TYPE I f(xmp, ynq) = 0(1) f(z, xmp, ynq) = 0 (2)

2.5.2 METHOD

CASE I

If m ≠ 1 & n ≠ 1

1) Put X = x1-m, Y = y1-n .

2) Then xmp = P(1 – m) and ynp = Q(1- n), where

3) Then Eqn. (1) becomes STANDARD I problem and Eqn. II becomes STANDARD III problem

CASE II

If m = n = 1.

1) Put X = log x & Y = log y

2) Then P = xp & Q = yq

3) Now Eqn.(1) reduces to STANDARD I problem and Eqn. (2) reduces to STANDARD II problem.

2.5.3 PROBLEMS

Solve the following equations

1) x2p + y2q = 0 2) xp +yq = 0 3) x2p + y2q = z

4) xp +yq = z 5) x4 p2 +y2 zq = 2z2 6) p2 + x2y2q2 = x2z2

7) x2p2 + y2q2 = z2 8) z2(p2x2 +q2) = 1

2.5.3 TYPE II f(zmp, zmq) = 0 (1) f(x, y, zmp, zmq) = 0 (2)

CASE I m ≠ 1

1) Put Z = zm+1

2) Then

3) Now Eqn.(1) reduces to STANDARD I problem and Eqn.(2) reduces to STANDARD IV problem.

CASE II If m = -1

1) Put X = log z

2) Then

3) Now Eqn.(1) reduces to STANDARD I problem and Eqn.(2) reduces to STANDARD IV problem.

2.5.4 PROBLEMS

Solve the following equations

1) z2(p2 + q2) = x2 + y2 2) (p2 + q2) = z2( x2 + y2)

3) z(p2 - q2) = x2 - y2 4) (zp + x)2 + (zq +y)2 = 1

5) 4zq2 = y + 2zp –x

2.6 LAGRANGE’S LINEAR EQUATION

An equation of the form pP + qQ = R, where P, Q, R are functions of x,y&z is called Lagrange’s equation.

2.6.1 Type I Grouping Method

2.6.1.1 METHOD

1) Form the auxiliary equation

2) Take any two equations which do not contain the third variable.

3) Let the solutions be a = u(x, y, z) & b = v(x, y, z)

4) The general solution is φ(u, v) = 0

PROBLEMS

Solve the following

1) px + qy = z 2) p – q = log(x + y) 3) p yz + q xz = xy

4) p x2 + q y2 = z2 5) p y2z + q x2z = xy2 6)

2.6.2 Type II Multiplier Method

2.6.2.1 METHOD

1) Form the auxiliary equation

2) Choose multipliers l, m, n & l’, m’, n’ and write the above equation as

.

3) Choose mulitipliers so that , so that ldx + mdy + ndz = 0 or derivative of the denominator is in the numerator, so that on integrating we get v = log( Dr).

4) Obtain u = a & v= b, the general solution is φ(u, v ) = 0.

Note: Usually the multipliers are in the following pairs:

1, 1, 1; x, y, z; 1/x, 1/y, 1/z or the pairs with sign changed.

2.6.2.2 PROBLEMS

Solve the following equations

1) (y2 + z2)p –xy q + xz = 0

2) (x2- y2 – z2) p + 2xy q = 2 xz

3) z(x – y) = px2 – q y2

4) x(y – z) p + y(z – x) q = z( x – y)

5) (x2 – yz)p + (y2 – xz) q = (z2 –xy)

6) x(y2 – z2) p + y(z2 – x2) q = z(x2 – y2)

7) (y + z) p + (z + x) q = x + y

8) (x + y)(y + 2z) q –(x + y)(x + 2z) p = z(y – x)

9) (y + xz) p – q(yz + x) = y2 – x2

10) x( y2 + z ) p – y(x2 + z) q = (x2 – y2)z

11) x2(y – z) p + y2(z – x) q = z2( x – y)

3. SOLUTION OF HIGHER ORDER DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

A general higher order partial differential equation with constant coefficient is of the form.

a0 Dn + a1 Dn-1 D’ + a2 Dn-2 D’2+….+an D’n)z = F(x, y) (1)

or f(D, D’) = F(x, y)

3.1 METHOD

1) Compute Complementary Function (C.F) using 3.1.1.

2) Obtain Particular Integral (P.I) using 3.1.2.

3) The general solution is z = C.F + P.I

3.1.1 Method of finding C.F

1) Replace D by m and D’ by1 in f(D, D’) = 0 (1)

2) Form auxiliary equation f(m) = 0, (2) which is a polynomial of degree m.

3) Let the roots of Eqn.(3) be m1, m2, m3,…..mn.

4.1) Let all the roots m1, m2, m3,…..mn are real and distinct, then the

C.F = φ1(y + m1x) + φ2(y + m2x) + …. φn(y + mnx)

4.2) Let m1 = m2 = m, m3, m4, …… are real and distinct, then the

C.F = φ1(y + mx) + x φ2(y + mx) +φ3(y + m3x) + …. φn(y + mnx)

Method of Finding P.I

Case 1 F(x, y ) = eax + by

3.1.2.1 METHOD

1) Replace D by a and D’ by b, if φ(a, b) ≠ 0.

2) If φ(a, b) = 0, then

3.1.2.2 PROBLEM

Solve the following equations

1. (D2 + 5D D’ + 6 D’2)z = e2x + y

2. (D3 – 3 D D’2 + 2D’3) = e2x – y + ex + y

3. (D2 –D D’ – 6 D’2)z = e3x + y

4. (D’3 -3D2 D’ + 3 D D’2 + 1)z = ex + y

5. (D2 + 7 D’ D + 14 ) z = e2x - y

Case 2. F(x, y ) = xr ys

3.1.2.3 METHOD

1) Make the denominator as (1 ± φ(D, D’)).

2) Rewrite P.I as P.I = (1 ± φ(D, D’))-1xr ys

3) Expand (1 ± φ(D, D’))-1 as a series of powers of D & D’.

4) Operate with xr ys.

3.1.2.4 PROBLEMS

Solve the following equations

1) ( D’2 + D D’ - 12 D’2)z = x2y

2) (D2 – 4 D D’ + 4 D’2) = x2  + y2

3) (6D2 + D D’ – D’3) z = xy

4) (D2 + 5 D D’ + 6 D’2) z = xy2

5) (2D2 + 5 D D’ + 6D’3) z = x + y

Case 3. F(x, y) = sin (ax +by) or cos(ax + by)

3.1.2.5 METHOD

1. Rewrite the denominator as φ(D2, DD’,D’2).

2. Replace D2 by –a2, D’2 by –b2 and DD’ by –ab, if φ(-a2, -ab, -b2) ≠ 0

3. If φ(-a2, -ab, -b2) = 0, then Denominator has a factor of the form

In this case

Or

3.1.2.6 PROBLEMS

Solve the following equations

1. (D3 – 4 D2 D’ + 4 D D’2) z = sin (3x – 4y)

2. (D3 – 7 D D’2 – 6 D’3) z = cos (2x –y)

3. (4D2 – 9 D’2 ) z = sin ( 3x – 2y)

4.

5. (D2 + 2 D D’ + 2D’2) z = cosx

CASE 4 F(x, y ) = eax + by φ(x, y)

3.1.2.7 METHOD

1. Replace D by D +a and D’ by D’ + b.

2. Proceed as in Case 2 or Case 3.

3.1.2.8 PROBLEMS

1.

2. (D2

–D D’ – D’2

) z = (y -1) ex

3. (D2

+ DD’ + 6 D’2

) z = y cos x

4. ( D2

+ 2 D D’ + D’)2

z = ( 2x + 3y)ey

5. (D2

+ D’2

) z = (xy + ex

)2

CASE 5

3.1.2.4. METHOD

1. Replace y by y – mx in f(x, y) and integrate w.r.to x, treating y as constant.

2. In the resulting integral replace y by y + mx.