Solutions to Partial Differential Equations (2)

 1)  Find a partial differential equation whose solution is:

z = (a + 2) x   +   ( a²  + 1) y  + b  

Soln.

Differentiate the function partially with respect to x and then y to obtain:

¶ z/ ¶ x = p  =  a + 2    

¶ z/ ¶ y = q   =  a²  + 1    

Eliminate a  from these eqns. to obtain:

q  =  (p - 2) ²  +  1

Then:

¶ z/ ¶ y =  (¶ z/¶ x -  2) ²  + 1  

2)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.

Soln.:     

For a cubic box: a = b = c

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

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

The independent solutions will be:

x= Ö(2/a)   sin ( n _xpx/a)] 

y= Ö(2/a)   sin ( n _y px/a)] 

z= Ö(2/a)   sin ( n _z px/a)] 

where:  n _x   = 1, 2, 3, 4.. etc.

Then the quantized energy is:

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

3)  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.

Soln.

u(x,t)  =  X(t)T(t)

¶ ² u/ ¶ t ²  =  X(x) (d ² T / d t ² )

¶ ⁴ u/ ¶ x⁴     = T(t)  d ⁴ X / dx⁴ 

XT"  -  c ²   X⁴ T  = 0

Separate variables to get;

X⁴ / X   =  -  T / c ²  T  =  k 

Let u(x,t) be defined over  0  <   x   <   ℓ  (t>0)

T(t)   =  c1 exp ( cr t)   +   exp (- cr t)   

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.

Soln.:

Use characteristic eqn.:  m ⁴  -  b⁴ =  0

Then:  (m ²  +  b² )   (m ²  -  b²)   =  0

Further factoring:

  ( m  +  ib )   (m   -  ib )  (m   +  b ) (m   -  b ) = 0

m  =  +  ib       m  =  +  b 

Then:  

X(x) =  C1 cos b x + C2 sin b x + C3 exp (b x) + C4 exp (-b x)

C1  = A,   C2 = B,  C3 =  ½ (C + D),  C4 = ½ (C - D)

This works since:  

X(x) =  A cos b x + B sin b x + ½ (C + D) exp (b x) 

+ ½ (C - D) exp (-b x)

=    A cos b x + B sin b x + C (e ^b x   + e ^-b x /2)  +

D (e ^b x   -   e ^-b x /2) 

=    A cos b x + B sin b x + C cosh b x + D sinh b x