Solutions to Partial Differential Equations Revisited (Pt. 1)

1)  Solve:   ¶  z/ ¶ x   =   ax +  y

Soln.:

Since the differentiation is with respect to x only, we can treat y as a constant and separate the variables thus:

dz =   (ax + y) dx

On integration we obtain the general solution:

z   =    1/2  a x ²     +   xy   +   y  (y) 

2) For the 2nd order PDE: 

am² f" (y + mx)  +   b m f" (y + mx) + cf" (y + mx)  =  0

Let the distinct roots be m1 and m2.  What would the solutions be?

Soln.  Given distinct roots of m1 and m2, the solutions will be: 

f(y + m1 x)   and g(y  +  m2 x)

Where f and g are arbitrary functions of their respective arguments.  Then we can write the solution as the linear combination:

f(y + m1 x)   + g(y  +  m2 x)

  3)  Same PDE as (2) but now let there be a double root, e.g. m _o.

 What would the solutions be? 

Soln.

We can write the solution for the double root as: 

f(y + m _o x)   where f is again an arbitrary function of its argument. Further, in this case the solution: xg (y + m _o x)   also applies where g is an arbitrary function.   Then the solution can be expressed as the linear combination:

f(y + m _o x)   +  xg (y + m _o x)  

4) Apply condition (iii) to the same equation what would the solution be?

Soln.: 

Condition (iii) specifies:  a = 0,  b  ≠  0   Then the relevant quadratic equation reduces to bm + c = 0. And hence will have only one root.  Denote this single root by m1 then the solution will be f(y + m1 x)  where f is an arbitrary function of its argument.  Further, it is also valid to write the full solution as:  f(y + m1 x)  +   g(x)

Where g is an arbitrary function of x only.

5) Solve:  ¶ ² u/ ¶ x²  -  4  [ ¶ ² u / ¶ x  ¶ y]  + 4[ ¶ ² u/ ¶ y ² ] = 0

Solution:  The auxiliary (quadratic) equation corresponding to the differential equation is:

m²  - 4 m  + 4 = 0

Factoring yields:   (m - 2)  (m - 2)  =  0  

This discloses a double (e.g. repeat) root of m  = 2

So the general solution is written in terms of two arbitrary functions, 

each with (y  + 2x):

Thus:   u   =   f( y + 2x)  +  xg(y + 2x)