Solutions to Difficult Partial Differential Equations (Pt. 2)

 We are given the 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 

Solution:

From the given initial conditions we can write the two equations:

T'  -   c ²  k  = 0   and:  X'' - k X = 0

From the given initial conditions we can write:

u(0, t) = X(0) T(t) = 0  Þ   X(0) = 0

u(ℓ, t)   =  X(ℓ) T(t)  = 0  Þ   X(ℓ) = 0

But, we cannot have X, T and k identically 0, so we choose: 

k  =   - r ²

   And:  X(x)  =  A cos ( r x)  +  B sin  ( r x)

But, X(x) = 0 Þ   A =  (0)

And:  X(x)  =   B sin  ( r x)     B ≠  0

Further:   X(ℓ) = 0   Þ  B sin  ( r ℓ) = 0     (B ≠  0)

Now let:  r =    n p /  ℓ   (For n = 1, 2.....)    And take B = 1, then:

X _n (x)    =   sin (n p x  / ℓ )    Where  X _n (x)  denotes many solns.

Going now to the 2nd equation: 

(With  k  =   - r ² ):

T'  +   c ²  r ² T = 0

Þ   T(t) = a exp (- c ²  r ² t )   

r ² =    n ² p ²  /  ℓ ²    =  l_n²    (For n = 1, 2.....) 

Then:  T_n(t) = a exp (- c ²  l_n²  t )   

And:  

u_n(x, t) =  X _n (x) T_n(t) = 

å^¥ _n=1  a n sin (nx p / ℓ )  exp (- c ²  l_n²  t )   

But (from condition 1):   

u( x, 0) = f(x) =  å^¥ _n=1  a n sin (nx p / ℓ ) 

Where:   _a n   =  2/ℓ  ò ℓ ₀   f(x) sin (np / ℓ ) x dx

( n = 1, 2, 3.....)