Solving More Difficult Partial Differential Equations (Pt. 2)

  In line with the topic of difficult partial differential equations, we consider the problem of the electrical potential distribution for an insulated cylinder with cross-section shown below:

A partial differential equation can be applied starting with two particular forms of what is called the Poisson equation.    For the case with position vector r  such as shown for the cross section shown, we write:

Ñ ²  j  =   f(r)

For the case  f(r) = 0 we obtain the Laplace equation:

Ñ ²  j  =  0  

Based on the proceeding the equation must be satisfied in a charge free region 

j (r)=  0 

The field E can easily be found from the electrostatic potential j which must satisfy the Laplace equation written in cylindrical coordinates.   

From symmetry considerations  j must be independent of z so that: j = j (r, q)  and:

1/r  ¶/ ¶r (r  ¶ j /¶ r)  +   1/ r ²  (¶  ² j /¶ q² )  =  0 

 The following boundary conditions are imposed on j:

j (r, q)  =   +V  (0  <  q <  p )

j (r, q)  =   -V  (- p  <   q  < 0)

These basics allow us to proceed with the solution which can be applied to analogous PDEs. In this case it is understood that  j must be periodic everywhere inside the cylinder (r < a).  But it is expected that this condition will break down at r = a,   q = 0 or p  (where the insulation is) since E is expected to be infinite at these points.  The method of separation of variables can be used again such that:

j (r, q)  = R(r) q(q)

Then:

1/r  d/dr[r (dR/dr) Q ]+   1/r ²  R(r) d ² Q/d q ²  =  0

Þ     

1/r  [dR/dr Q + r d ²R/dr ²  Q ]+  R/r ²  d ² Q/d q ²  =  0

 

Þ        r  dR/dr + r ² d ²R/dr ²  = -R/q   d ² Q/d q ² 

Þ     r ²  /R {d ² R/dr² } +  r/R  dR/dr = - 1/q d ² Q /d q ² 

We next let l  be the separation constant, yielding.

1) r ²   d ²R/dr ²  + r  dR/dr + lR =  0

2) d ² Q /d q ² =  lQ

Leading to:

(3) 1/r  d/dr[r (dR/dr) Q ]+   l/r ²  R =  0

Since  Q  is periodic in q  then  Q must be also which immediately implies that: l = - m ²

(m = 0, 1, 2, 3,....)

And Q is then a linear combination of sin m q, and  cos m q 

The radial equation (3) becomes:

 d ²R/dr ²  +  1/R  dR/dr  -  m ² /r ²  R =  0

which is just the Euler equation with the solutions r ^m  and r ^-m

for m > 0,  C = const.  and log r (for m = 0)  

Since  j (r, q)  must have a gradient at r = 0,  the solutions   

  r ^-m and log r are not admissible.  According to the applicable general principles we can form a linear (infinite) superposition of solutions, i.e.

j (r, q) =   

A _o  +   å^¥ _{m = 1}   r ^-m (A _m  cos mq   +    B _m  sin mq ) = f(q)

 

To satisfy the boundary conditions we set:

 j (r, q) = f(q)  =  {+V  (0  <  q <  p )

                              { -V  (- p  <   q  < 0)  

So the constants A _m   and B _m  are related to the Fourier coefficients of f(q).  Then A _o  = A _m  (all m) owing to the asymmetry of  under the interchange  q -> - q.  Note that this property could have been inferred for j (r, q)   as well.  Being aware of these facts can considerably simplify solutions in more complicated PDE problems.  In this case direct calculation leads to the result:

B _m  = 2V/p a _m   ò ^p  ₀  sin mq dq   = {0  (m even)

                                                                 { 4V/ a _m p m     m odd

So that: j (r, q)  = 4V/p  å^¥ _{m = 1,3,5...}   1/m (r/a) ^m sin mq

Note: It's not difficult to verify that the above function satisfies the conditions of the problem.   I.e. if r < r₁  , where  r₁ is any positive integer less than a, the series can be differentiated term by term twice with respect to both r and q  .  The factor (r/a) will lead to uniform convergence in the 2x differentiated series. Thus the 'difficult' PDE is easily verified. Try it and see!

   Challenge Problem:

 A partial differential equation:    

¶  u/ ¶ t   =   c ²   [¶ ² u/ ¶ x² ]

with u = u(x, t)

Such that 0 <  x <  ℓ   and t > 0

is used when considering a perfectly insulated rod of length (with no lateral heat loss) We include the initial and end conditions for the rod:  u(0, t) = u(ℓ  , t)    

And: u (x, 0) = f(x)

Where u represents the temperature of the rod as a function of position x.  Obtain the solutions in terms of Fourier series for:

1) u( x, 0) = f(x)

and:

2) u(x, t) =  u(ℓ, t) 

Hint: Use a variables separable method such that:

u(x, t) = X(x)  T (t)

So that:  ¶ ² u/ ¶ x²  =  X''  T

And:   X T' =  =    c ²  X''  T