Solving Partial Differential Equations: Three Easy Examples & One (Somewhat) Hard

 This post will deal with the solutions of some partial differential equations starting with relatively simple ones.

Examples:

Ex. (1)   Solving a first order partial:

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

Let:   ¶  z/ ¶ x  =   p  =  (a  + 2)

Let:   ¶  z/ ¶ y    =  q   =    a ² +   1

So:   q   =   (p  - 2) ² +   1

Then  general solution can be written:

z   =   (p)  x   +  (q)  y   +   g

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

Ex. (2) Solving a 2nd order partial

¶ ² u/ ¶ y  ¶ x  =   x ³     -  y 

First, write in the form:

¶  y  (¶ u/ ¶ x)   =   x ³     -  y 

Next integrate partially with respect to y, holding x constant to obtain:

¶ u/ ¶ x =   x ³   y -  y ²  /2  +    j  (x)

Next integrate  the result  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.  We note here that while ordinary DEs have solutions which contain arbitrary  constants,  partial differential DE solutions have arbitrary functions.

Ex. (3) 2nd order  (homogenous with constant coefficients)

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

First  write the auxiliary equation:

m²  - a m – 6a = 0

Which  has roots:  m  = 3a,   m = -2a

The general solution will therefore be:

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

Ex. (4):  Application in three dimensions, using separation of variables.

A particle confined in a 3-dimensional Box

We want to find the applicable wave function solution ( y (x,y,z)), and the energy applicable

We can set out the rectangular regions (limits)  according to:

0 < x < a

0 < y < b

0 < z <  c

The partial differential  equation to solve then becomes:

¶² y/ ¶ x²  +   ¶² y/ ¶ y²  + ¶² y/ ¶ z²  + 2mE/  ħ² y = 0

This is a 2nd order partial differential equation easily solved by the separation of variables:

y  =   X(x) Y(y) Z(z)

Then: 

X’ =   ¶X/ ¶ x    Y’ =  ¶Y/ ¶ y  and Z’ =  ¶Z/ ¶ z   

This leads to the equation:

X”YZ  +  XY” Z   + XYZ” + 2mE/  ħ²  XYZ  = 0

The required normalization equation is then:

ò _o^a  ò ^b_o  ò ^c _o  ‖y‖ ²  dz dy dx = 1

Then, dividing the  equation by XYZ:

X”/ X + Y”/ Y  + Z”/Z + 2mE /  ħ²   = 0

We let:

X”/ X =  - a²,  Y”/ Y  =  -b²,    Z”/Z =  -g²

With a, b   and g constants.

Then, we have:

x²   +  a² x  = 0,   y²   +  b² y  = 0,  z²   +  g² z 

So the independent solutions will be:

x= Ö(2/a)   sin ( n _xpx/a)] 

y= Ö(2/b)   sin ( n _y px/b)] 

z= Ö(2/c)   sin ( n _z px/c)] 

where:  n _x   = 1, 2, 3, 3 etc.

Then: 

y  (x, y, z) = (8/ abc)^1/2 sin ( n _xpx/a) sin ( n _y px/b) sin ( n _zpx/c)

And further, to obtain the quantized energy, E:

n² _x p² /a²  -  n² _y p² /b²  -  n² _z p² /c²  +  2mE/  ħ²   = 0

And:

E =   p²  ħ² / 2m [n²_x  /a²  + n² _y  /b²  + n² _z  /c² )

If the box is a cube, so a = b = c:

E =   p²  ħ² / 2m a² (n²_x  + n² _y    + n² _z )

Problems for the Math Whiz:

1)  Solve:   ¶  z/ ¶ x   =   ax +  y

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

3)  Solve the following higher order partial DE:

¶ ⁴ z/ ¶ x⁴  -   6  [ ¶ ⁴ z/ ¶ x ³  ¶ y]  +  14 [¶ ⁴ z/ ¶ x²  ¶ y² ] -  

 16 [ ¶ ⁴ z/ ¶ x   ¶ y ³ ]     +   8  [ ¶ ⁴ z/ ¶ y ⁴ ]   =  0

3)  Consider a wire of length  ℓ :

]-------------- ℓ   ---------------------->[

for which the relevant wave equation is:

 ¶ ² u/ ¶ t ²  =   c ²   [¶ ² u/ ¶ x² ]

If the displacement satisfies:

u(0, t) = u(ℓ , t )  =   0     ( t >  0)

Suppose at time t= 0 the displacement is u(x,0)  =  f(x)

And:    ¶ u/ ¶ t ] t=  0   =  g(x)

Using the technique of separation of variables write two different general solutions for X( ℓ ) and state why one is unacceptable

Hints:  i)  u(0, t) =  X(0) T(t)   and

 (ii)  u(ℓ , t )  = X( ℓ ) T(t)