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Formation of partial differential equations:

There are two methods to form a partial differential equation.

By elimination of arbitrary constants.

By elimination of arbitrary functions.

 

Formation of partial differential equations by elimination of arbitrary constants:

Let f(x, y, z, a, b) = 0 be an equation which contains two arbitrary constants ‘a’ and ‘b’.

Therefore partially differentiate the equation w.r.t. x and y to eliminate the constant ‘a’ and ‘b’

 

Note: we use following notations in PDE

p = ∂z/∂x     q = ∂z/∂y     r = ∂2z/∂x2        s = ∂2z/∂x∂y     t = ∂2z/∂y2       

Types of solution:

(a) A solution in which the number of arbitrary constants is equal to number of independent variables is called complete integral or complete solution.

(b) In complete integral if we give particular values to the arbitrary constants we get particular integral.

(c) Singular integral: let f(x, y, z, p, q) = 0 be a partial differential equation whose complete integral is    Φ(x, y, z, a, b) = 0 ……………..  (1)

Differentiating (1) partially w.r.t a and b and then equate to zero, we get

                                   ∂Φ/∂a = 0……………. (2)

                                   ∂Φ/∂b = 0 ……………. (3)

Eliminate a and b by using (1), (2) and (3).

The eliminate of a and b is called singular integral.

 

Types of first order non-linear partial differential equations:

Type 1: F (p, q) = 0

Type 2: z = px +qy + f (p, q) [Caliraut’s form]

Type 3: F (z, p, q) = 0

Type 4: F1 (x, p) = F2(y, q)

Type 5: F (xmp, ynq) = 0 and F (z, xmp, ynq) = 0

Type 6: f (zmp, zmq) = 0 and F1 (x, zmp) = F2(y, zmq)

 

Lagrange’s linear equations:

The equation of the form

    Pp + Qq = R …………. (1)

Is known as Lagrange’s equation, where P, Q and R are functions of x, y and z. To solve this equation it is enough to solve the subsidiary equations.

 dx/P = dy/Q = dz/R ……………(2)

If the solution of the subsidiary equation is of the form u(x, y) = c1 and v(x, y) = c2 then the solution of the given Lagrange’s equation is Φ(u, v) = 0.

To solve the subsidiary equations we have two methods:

     Method of Grouping:             
                Consider the subsidiary equation dx/P = dy/Q = dz/R. take any two members say first two or last two  or first and last members. Now consider the first two members dx/P = dy/Q. If  P and Q contain z(other than x and y) try to eliminate it. Now direct integration gives u(x, y) = c1. Similarly take another two members dy/Q = dz/R. If  Q and R contain x(other than y and z) try to eliminate it. Now direct integration gives v(y, z) = c2 . Therefore solution of the given Lagrange’s equation is Φ(u, v) = 0.

 

          Method of  multiplier’s           
                 Choose any three multipliers l, m, n may be constants or function of x, y and z such that in

                     dx/P = dy/Q = dz/R = (ldx + mdy + ndz) / (lP + mQ + nR)

  the expression lP + mQ + nR = 0

Hence    ldx + mdy + ndz = 0  

[ since each of the above ratios equal to a constant

         dx/P = dy/Q = dz/R = (ldx + mdy + ndz) / (lP + mQ + nR) = k (say)

          (ldx + mdy + ndz) =  k (lP + mQ + nR)

If  lP + mQ + nR = 0 then  ldx + mdy + ndz = 0]

Now direct integration gives u(x, y, z) = c1.

similarly choose another set of multipliers l′, m′, n′

dx/P = dy/Q = dz/R = (l′dx +m′dy + n′dz) / (l′P + m′Q + n′R)

  the expression l′P + m′Q + n′R = 0

therefore l′dx +m′dy + n′dz  = 0 (as explained earlier)

Now direct integration gives v(x, y, z) = c2.

Therefore solution of the given Lagrange’s equation is Φ(u, v) = 0.

 

Homogeneous Linear partial differential equations:

F (x, y) = [a0 Dn + a1 Dn-1 D′ + a2 Dn-2 D′2 + ……. + anD′n]z   where D = ∂/∂x and D′ = ∂/∂y

Solution of Homogeneous Linear partial differential equations:

      The Complementary function.                                          
    The particular integral.
 

To find the Complementary function (C.F.):

The complementary function is the solution of the equation

                        a0 Dn + a1 Dn-1 D′ + a2 Dn-2 D′2 + ……. + anD′n = 0.

In this equation , put D = m and D′ = 1 then we get an equation, which is called auxiliary equation

Hence the auxiliary equation is

a0 mn + a1 mn-1  + a2 mn-2 + ……. + an   = 0.

Let the root of this equation be m1, m2, m3,……….. mn.

Case 1: If the roots are real or imaginary and different say m1 ≠ m2  m3 ≠……..  ≠ mn. then the C.F. is

                          Z = f1 (y + m1x) + f2 (y + m2x) +……… + fn (y + mnx)

Case 2: If any two roots are equal, say m1 = m2 = m, and others are different then the C.F. is

                          Z = f1 (y + mx) + xf2 (y + mx) + f3 (y + m3x) +……… + fn (y + mnx)

  Case 3: If three roots are equal, say m1 = m2 = m3 = m,  then the C.F. is

                          Z = f1 (y + mx) + xf2 (y + mx) + x2f3 (y + mx) +……… + fn (y + mnx) .

 

 To find the Particular Integral:

Rule1: If F(x, y) = eax+by  then

           P.I.  = 1/ Φ (D, D′). eax+by 

                   = 1/ Φ (a, b). eax+by   provided Φ (a, b) ≠ 0 [Replace D by a and D′ by b]

           If   Φ (a, b) = 0 refer rule 4.

 

Rule2: If F(x, y) = sin (mx + ny) or cos (mx + ny) then

           P.I.  = 1/ Φ (D, D′).  sin (mx + ny) or cos (mx + ny)

        Replace D2 by –m2, D′2 by –n2 and DD′ by –mn in provided the denominator

is not equal to zero. If the denominator is zero refer rule 4.

  Rule3: If F(x, y) = xm yn

          P.I.  = 1/ Φ (D, D′). xm yn

                        =  [Φ (D, D′)]-1 xm yn

Expand [Φ (D, D′)]-1    by using binomial theorem and then operate on xm yn

Note: 1/D denotes integration w.r.t x, 1/ D′ denotes integration w.r.t y.

 

Rule4: If F(x, y) is any other function, resolve Φ (D, D′) in to linear factor say (D – m1 D′) (D – m2 D′) etc. then the

P.I. = 1/(D – m1 D′) (D – m2 D′)  F(x, y)

 Note: 1/(D – mD′)  F(x, y)  =  ∫ F(x, c-mx) dx, where y = c-mx.

 

Note: If the denominator is zero in rule (1) and (2) then apply Rule (4)

 








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