Solutions To Partial Differential Equation Problems

 1)  Solve:   ¶  z/ ¶ x   =   ax +  y

Solution:  Since 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/ ¶ x ³  ¶ y]  +  14 [¶ ⁴ z/ ¶ x²  ¶ y² ] -  

 16 [ ¶ ⁴ z/ ¶ x   ¶ 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  =   j _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 :

]-------------- 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(ℓ , 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( ℓ ) and state why one is unacceptable

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

 (ii)  u(ℓ , t )  = X( ℓ ) T(t)

---------

Solution:  By separation of variables we have: 

u(x, t) =   X(x) T(t)

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

Or:    T "/ c ²  T   =    X" / X

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

 T "/ c ²  T   = K  =      X" / X

Two equations result:

i) X"   -   K X = 0

ii) T "  -   c ² KT    =   0

We know (hints):

a) u(0, t) =  X(0) T(t) =  0 

b)  u(ℓ , t )  = X( ℓ ) 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( ℓ ) = 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(ℓ) =  c1 [exp ( r ℓ)   -   exp (-r ℓ) ]   =   2 c1 sin n r ℓ