## Tuesday, July 5, 2022

### 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 ħy = 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

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

Dividing the  equation by XYZ:

X”/ X + Y”/ Y  + Z”/Z + 2mE /  ħ  = 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 px/a)]

z= Ö(2/a)   sin ( n px/a)]

where:  n   = 1, 2, 3, 4.. etc.

Then the quantized energy is:

E =   p2  ħ/ 2m a2 (n2 + n2  + n2 )

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

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

And:   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   <

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) (2 T d t 2 )

4 u/  x4     = T(t)  4 X dx4

XT"  -  c 2   X4 T  = 0

Separate variables to get;

X4 / X   =  -  T / 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.:  4  -  b4 =  0

Then:  (2  +  b2   (2  -  b2)   =  0

Further factoring:

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

=  +  ib        =  +

Then:

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

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

This works since:

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

½ (C - D) exp (-x)

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

D (e b x   -   e -b x /2)

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