Solving Difficult Partial Differential Equations (Pt. 3) Extension Of Problem Conditions

 We continue with previous problem but change the conditions to have a very long rod such that:   -oo  <   ℓ   <   + oo   (i.e. no end conditions).

Here:

u (x, 0) = f(x)  (  -oo  <   ℓ   <   + oo )

Assume again: 

u(x, t) = X(x)  T (t)     But u(x,t)  ≠  0

T' /  c ²   T    =    X''/X  = const.

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

T(t) = a exp (- c ²  r ² t ) 

Then:  XT =

A cos ( r x)  +  B sin  (r x) exp (- c ²  r ² t )      set a = 1

This leads us to write;

u(x, t) = ò ^¥_o     [A cos ( r x)  +  B sin  (r x)] exp (- c ²  r ² t ) dr

The preceding allows the rod to be continuous instead of discrete (i.e.  n p /ℓ)

Analogously:

u (x, 0) = f(x) = ò ^¥_o   [A cos ( r x)  +  B sin  (r x)] dr

Where:

A (r) = 1/p  ò ^¥_-¥ f(w) cos ( r w)  dw

B (r) = 1/p  ò ^¥_-¥ f(w) sin ( r w)  dw

Writing solution in more compact form:

u(x, t) = 1/p [ò ^w  _o  [ò ^¥_-¥ f(w) cos ( r w) dw cos (rx)  +  

ò ^¥_-¥ f(w) sin ( r w) dw sin (rx)] exp (- c ²  r ² t ) dr

note that each term in the integral involves f(w), which can be factored out.

Use trig identity:

cos ( r w) cos (rx) + sin ( r w) sin (rx)  = cos r (x - w)

Then:

u(x, t) = 1/p [ò ^w  _o  [ò ^¥_-¥ f(w) cos r (x - w)exp (- c ²  r ² t )  dw dr

or:

u(x, t) = 1/p [ò ^w  _o  [ò ^¥_-¥ f(w) cos r (x - w)exp (- c ²  r ² t dr) dw 

This can be written in definite integral form, using:

s = c r Ö t        b = (x - w)/ 2c Ö t 

And using the definite integral:

ò ^¥_o     [exp (- s ² )  cos 2bs ds =   Ö p/ 2  {exp (- b ² ) }

But:  2bs  =   [2 (x - w)/ 2c Ö t ] c r Ö t

So the final solution can be written:

u(x, t) = 1/2c Öp t  ò ^¥_-¥ f(w) exp (- x - w) ²/ 4 c ² t   dw