## Tuesday, June 21, 2022

### Revisiting Partial Differential Equations (1)

Partial differential equations occupy a wide swath of physics and mathematical physics areas, from quantum mechanics, to plasma physics to the Lagrangian and Hamiltonian formulations of planetary motion and non-holonomic systems to magneto-hydrodynamics.

Perhaps the simplest form of such an equation is the 1st order partial:

z/  y    =   (  z/  x   -   2) 2 +   1

For which we can use substitution and let:

z/  x  =   p  =  (a  + 2)

And:

z/  y    =  q   =    a 2 +   1

Then  general solution can be written:

z   =   (p)  x   +  (q)  y   +   g

Or:  z  =   (a  +  2) x   +   (a 2 +   1) y  +  f

A simple example of a 2nd order partial would be:

2 u/  y  x  =   x 3     -  y

Which we can rewrite in the form:

y  ( u/  x)   =   x 3     -  y

Then integrate partially with respect to y and holding x constant to get:

u/  x =   x 3   y -  y  /2  +    j(x)

Next, integrate partially with respect to x, holding y  constant to obtain the solution in the form:

u  =   x 4   y /4   -    x 2  y/  2  +  f(x)   +  g(y)

Where  f(x)  =    ò j(x) dx

is  an arbitrary function of x and g is an arbitrary function  of y.

I point out at this point that while ordinary differential equations when integrated yield results with arbitrary constants, PDEs yield results with arbitrary functions.

Homogeneous linear partial differential equations of 2nd order are also relatively tame to solve, and have the general form:

a( 2 u/  x2 ) +  b (  2 u/  x   y )  +  c ( 2 u/  y2 ) = 0      (Original eqn.)

Where a, b, c are constants.  And solutions will be of the form: f (y + mx)

Where  f is an arbitrary function and m is a constant.

Then rewrite.:

2 u/  x=   m2 f" (y + mx)

2 u/  x   y  =   m f" (y + mx)

2 u/  y2   =    f" (y + mx)

Substituting these into the original eqn.:

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

Or:   f" (y + mx)  [am2  +   bm  +  c]  = 0

Then f(y + mx) is a solution of the original equation  if m satisfies the quadratic:

am2  +   bm  +  c  = 0

Note here that four conditions attach to the possible solutions:

i) a ≠  0  and roots of quadratic are distinct

ii)  a ≠  0  and roots of quadratic are equal

iii) a = 0,  b    0

iv) a = 0,  b = 0,   c    0

Another 2nd order homogeneous example but with the constant coefficient a:

2 z/  x2  -   a  [  2 z/  x   y]   -  6a 2 [  2 z/  y2] =   0

To solve, first write the auxiliary equation, i.e. bearing the quadratic:

m2  - a m – 6a = 0

This will be found to have the roots: m  = 3a,   m = -2a

The general solution will therefore be:

z  =   j(y  +   3  ax)   +  x  y  (y -  2 ax)

Suggested Problems For The Math Whiz:

1) Solve:     z/  x   =   ax +  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?

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

What would the solutions be?

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

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