Monday, September 28, 2020

Solutions To Partial Differential Equation Problems

 1)  Solve:     z/  x   =   ax +  y


SolutionSince the differentiation is with respect to x only, we can treat y as a constant and separate the variables thus:

dz =   (ax + y) dx

On integration we obtain the general solution:

z   =    1/2  a x 2     +   xy   +   y  (y) 


2) Solve:   2 u/  x2  -  4  [  2 u /  x   y]  + 4[  2 u/  y 2 ] = 0

Solution:  The auxiliary (quadratic) equation corresponding to the differential equation is:

m2  - 4 m  + 4 = 0

Factoring yields:   (m - 2)  (m - 2)  =  0  

This discloses a double (e.g. repeat) root of m  = 2

So the general solution is written in terms of two arbitrary functions, 
each with (y  + 2x):

Thus:   u   =   f( y + 2x)  +  xg(y + 2x) 


3)  Solve the following higher order partial DE:

 4 z/  x4  -   6  [  4 z/  3   y]  +  14 [ 4 z/  x2   y2 ] -  

 16 [  4 z/     y 3 ]     +   8  [  4 z/  y 4 ]   =  0


Solution:  Again we write out the auxiliary (quadratic) equation corresponding to the DE:

m4  - 6  m3 + 14 m2  - 16 m  + 8  = 0

Factoring yields:  

(m - 2) (m - 2) (m - (1 +i))(m - (1 - i))

so  m  =  2, 2,  1 + i, 1 - i    


Hence, the general solution (noting the repeat root of m = 2) is:

z  =   o (y  +   2 x)   +  x  y o (y +  2 x)  +  j1  (y  +  x  +  ix)   +   j1 (y   + x  - ix) 

+  i[ y1  (y   + x  +  ix)  -   y1  (y   + x  -  ix)]


3)  Consider a wire of length  ℓ :

]--------------    ---------------------->[

for which the relevant wave equation is:

  2 u/  t 2  =   c 2   [ 2 u/  x2 ]

If the displacement satisfies:

u(0, t) = u( , t )  =   0     ( t >  0)

Suppose at time t= 0 the displacement is u(x,0)  =  f(x)

And:     u/  t t=  0   =  g(x)

Using the technique of separation of variables write two different general solutions for X(  ) and state why one is unacceptable

Hints:  i)  u(0, t) =  X(0) T(t)   and

 (ii)  u( , t )  = X(  ) T(t)
---------

Solution:  By separation of variables we have: 

u(x, t) =   X(x) T(t)

Now let:  dT/ dt =  T ',    d Tdt 2   = T"

dX/ dt =  X '        d 2X/  dt 2   = X "

So that:   2 u/  t 2  =  X(t) T "(t)

And:

 2 u/  t 2  = X" (x) T (t)

The wave equation can then be written:

XT "   =   2   X"  T

Or:    T "2  T   =    X" / X

We set both sides equal to a constant K (separation constant):

 T "/ 2  T   = K  =      X" / X

Two equations result:

i) X"   -   K X = 0

ii) T "  -   2 KT    =   0

We know (hints):

a) u(0, t) =  X(0) T(t) =  0 

b)  u( , t )  = X(  ) T(t)  = 0 


Look at two cases:

i) K  = 0

Then X" - KX = 0  becomes X" = 0

For which: X(x) = c1 X + c2

And: X(0) = 0   è   c2 = 0  so X(x) = c1 X

However,  X(x) = 0 ⇒  c1 = 0  and X(  ) = 0

Which is unacceptable, so K cannot be zero. So next look at K > 0


ii)  K  > 0

Then:  Let K  =    r 2

So that:  X"(x) -  r 2 X  = 0

Has solution:  c1 exp (r x)  +  c2 exp (-r x) 

After eliminating c2  (as fn. of  c1)  

 X(x) =  c1 [exp ( r x)   -   exp (-r x) ]   

X() =  c1 [exp ( r )   -   exp (-r ) ]   =   2 c1 sin n r ℓ 

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