Calculus
Classification of 2nd Order PDE
Classification of Second Order Homogenous Linear Equation” A second order
linear homogeneous PDE of the form
      ∂ 2φ ∂ 2φ     ∂ 2φ     ∂φ     ∂φ
A ∂x 2 + B ∂x∂y + C ∂y 2 + D ∂x + E ∂y + Fφ (x, y) = 0 ------ (1)
Where A, B, C, D, E and F are either functions of x and y only or constants is called
(i) a parabolic equation, if B2 − 4AC = 0
(ii) an elliptic equation, if B2 − 4AC < 0
(iii) a hyperbolic equation, if B2 − 4AC > 0
For Example
(1) consider the one dimensional heat equation
         ∂u      ∂ 2u
               2
         ∂t = c ∂x 2
                       ∂ 2u     ∂u
         ⇒ c ∂x 2 2
                              − ∂t = 0
         Comparing it with (1), we have
         A = c2, B = 0 and C = 0
              ∴ B2 – 4AC = 02 − 4 × c2 × 0 = 0
         ∴ One dimensional heat equation is parabolic
         Similarly, it can be easily observed that
(2)      one dimensional wave equation
         ∂2y            2
                              ∂2y        2
             ∂t
              2 = c ∂x 2 is hyperbolic (B − 4AC > 0) and (3). The laplace equation
         ∂ 2u            ∂ 2u
         ∂x 2
                      + ∂y 2 = 0 is elliptic (B2 − 4AC < 0)
Method of Separation of Variables: Consider a PDE involving a dependent variable u
and two independent variables x and y. In the method of separation of variables, we
find a solution of the PDE in the form of a product of a function of x and a function of y
                                                                         Calculus
i.e. we write u(x, y) = X(x). Y(y) ----- (1)
       ∂u      ∂               ∂u       ∂
then   ∂x
            = ∂x (xy) = x1y;   ∂y
                                    =   ∂y
                                             (xy) = xy1
∂ 2u        ∂ 2u    1 1 ∂ u
                          2
     = x y, ∂x∂y = x y , ∂y2 = xy11
        11
                                               and so on
∂x 2
         dX       dY       d2 X      d2 Y
        1      1        11        11
Here x = dx ; y = dy ; x =      ;y =
                           dx 2      dy 2
Substituting these in the given PDE, separating x and its derivatives from y and its
derivatives, finding solutions for x and y and substituting them in (1), we get the
solution of the given PDE. This is best explained through the examples given