Solving More Difficult Partial Differential Equations (Part 1)

Partial differential equations in a more complicated  form generally are distinguished by more variables, more dimensions and more boundary conditions.  They can often be solved by the variables separable method which  occupy perhaps 5 percent of the total constellation of such equations. However, others often require Fourier series solutions which I examine in Part 2.   In this post we examine three examples of PDEs using the variables separable method and I also provide a couple of challenging problems for more ambitious math mavens at the end. 

Example 1: Consider a wire of length  ℓ :

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

For which the relevant wave equation is:

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

And 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 provide two different general solutions for X( ℓ )   and X(x).

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

 (ii)  u(ℓ , t )  = X( ℓ ) T(t)

 

Solution:  By separation of variables we have: 

u(x, t) =   X(x) T(t) 

Now let:  dT/ dt = T ',    d ² T/ dt ²   = T"

dX/ dt =  X '        d ²X/  dt ²   = X "

So that:  ¶ ² u/ ¶ t ²  =  X(t) T "(t)

And:   ¶ ² u/ ¶ t ²  = X" (x) T (t)

The wave equation can then be written:

XT "   =   c ²   X"  T

Or:    T "/ c ²  T   =    X" / X

We set both sides equal to a constant K (separation constant):

 T "/ c ²  T   = K  =      X" / X

Two equations result:

i) X"   -   K X = 0

ii) T "  -   c ² KT    =   0

 

Look at two cases:

i) K  = 0

Then X" - KX = 0  becomes X" = 0

For which: X(x) = c1 X + c2

And: X(0) = 0   Þ  c2 = 0  so X(x) = c1 X

However,  X(x) = 0 ⇒  c1 = 0  and X( ℓ ) = 0

Which is unacceptable, so K cannot be zero. So next look at K > 0

ii)  K  > 0

Then:  Let K  =    r ²

 So that:  X"(x) -  r ² X  = 0

Has solution:  c1 exp (r x)  +  c2 exp (-r x) 

After eliminating c2  (as fn. of  c1)  

 X(x) =  c1 [exp ( r x)   -   exp (-r x) ]   ⇒

X(ℓ) =  c1 [exp ( r ℓ)   -   exp (-r ℓ) ]   =   2 c1 sin n r ℓ 

Example 2:

 Provide two general solutions for the transverse mode (H _z  =  0 ) of an EM wave starting with the Maxwell equation:

Ñ ² E   =  me  ¶ ² E  / ¶ t ²  

Solution:

In Cartesian coordinates, the H, E   z-components partials can be written (with the angular frequency  w  , magnetic permeability  m  and the permittivity  e ):

i)     ¶ ² E _z  / ¶ x²  +   ¶ ² E _z  / ¶ y ²    +     g ² E _z  =    mew²  E _z 

   ii)  ¶ ² H _z  / ¶ x²  +   ¶ ² H _z  / ¶ y ²    +     g ² H _z  =    mew²  H _z 

The transverse mode (H _z  =  0 )  implies:

¶ ² E _z  / ¶ x²  +   ¶ ² E _z  / ¶ y ²    +     g ² E _z  =    - w²  me E _z 

We assume:  E _z   (x, y, z)  =   E _z   (x, y) exp (- g z)

At time t = 0

Using a variables separable approach we first try the solution:

E _z   (x, y, z)  =   X(x) Y(y) exp (- g z)

Þ  Y d ² X / d x²  +  X d ² Y / d y ²  +  g ² XY    =  - w²  me XY  

Then divide by XY:

1/ X (d ² X / d x² )  +  1/Y (d ² Y / d y ²  )  +   g ²  +   w²  me    

And let  h ² =  g ²  +   w²  me    

Then rewrite equation as:

1/ X (d ² X / d x² )  +  1/Y (d ² Y / d y ² )  + h ² =  0  

Þ 1/ X  ( d ² X / d x² )  +  h ² =  A ² =    1/Y (d ² Y / d y ² )      

X   = C1 cos Bx  +  C2 sin Bx

Where: B ² = A ² + h ² 

Similarly for Y:  d ² Y / d y ² +  A ² Y =   o     

Where:  Y =  C3 cos Ay  +  C4 sin Ay

The constants C1, C2, C3 and C4 are determined from boundary conditions.

Example 3:

 Consider a particle in a rectangular box, e.g.

Find three independent solutions and the quantized 

energy given the partial differential equation below with wave function y :

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

And: 0 < x < a,   0 < y  < b,   0 <  z  <   c

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

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

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, 4.. 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 )

Suggested Problems:

1)  Use separation of variables to solve the partial differential equation:

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

For a particle in a cubic box.  Include the quantized energy for the particle.

2)  The partial differential equation for a deflected beam is given as:

¶ ² u/ ¶ t ²  +   c ²   [¶ ⁴ u/ ¶ x⁴ ] =   0

And:   c ²   =    EI/ r A

Where EI is the flexural rigidity, r   is the density of the wood, and A is the area.

a)   If u(x,t) is defined such that: t > 0   and

0  <   x   <   ℓ

Use separation of variables to arrive at an initial general (but not optimal) solution and write the two ordinary differential equations arising from the approach.

b) Apply the characteristic equati0n (i.e. in order 4)  to obtain improved general solutions X(x) and T(t)  and also suggest values for the constants arising: C1, C2, C3 and C4.