Solution To Laplace Transform Differential Equation

The problem again:

Use Laplace transforms to solve the differential equation:

d ³ Y/ dt ³ -    d Y /dt   =   0 

Using the conditions:  Y(0) =  1,  Y’(0) =  0  and  Y" (0) = 1

Solution:  We Write:

£ { d ³ Y/ dt ³  } =  s ³   y(s)  -   Y(0) s ²    -    Y'(0) s  -  Y"(0)


=    s ³  y(s)  -   s ²    -   1   


£ {dY / dt}   =    s y(s)   -  Y(0)  - Y"(0)



=    s y(s) -  1   


Substitute the expression for each transform:

s ³  y(s)  -   s ²    - 1   =   0 


Collect like terms and transpose:

  (s ³  -     s)  y(s)   =      s ²    +  1  

 
Solve for y(s):

y(s)   =   ( s ²    +  1  ) /  (s ³  -     s)  =    ( s ²    +  1  ) /  s (  s ²    -  1   ) 

Use  of partial fractions yields:

A/ s   +    B/ (  s ²    -  1   )     =   s ²    +  1  

Then:


 A (  s ²    -  1   )   +  Bs   =   s ²    +  1  

Whence:

A s ²     -  A    +  B s    =   s ²   +   1


Yielding values (by equating coefficients):

A    = 1   

B   -     A   =  0 

So:  B  =    1



è


( s ²    +  1  ) /  s (  s ²    -  1   )  =

 1 / s     +   1 /  ( s ²    -  1   )


We then need to take the inverse Laplace transform : 

 £ ^-1  [1 / s     +   1 /  ( s ²    -  1   ) ]

=  

£ ^-1  [1 / s ]    + £ ^-1 [ 1 /  ( s ²    -  1   )]   


From a table of transforms (see bottom of previous post from  March 2nd):

We see:   1/s   is inverse transformed to  1

And similarly:  [ 1 /  ( s ²    -  1   )]  ->   cosh (t)

 
Then the solution to the DE is:  Y (t)   =   1   +    cosh (t)


Check soln.  to  d ³ Y/ dt ³ -    d Y /dt   =   0 :


d ³  (1+ cosh (t))/ dt ³   =   0 
  

d  (1   +    cosh (t))/dt   =   0 

Then:   0   -  0  =  0

So: 

d ³ Y/ dt ³ -    d Y /dt   =   0 

Revisiting Basic Laplace Transforms In Solving Differential Equations

Definition : Let  F be a function defied for t > 0. Then define a new function f by:

f(s) =   ò ^¥_o       exp(-st) F(t) dt

For all s such that the integral exists, then f is called the Laplace transform of F and is written as:

f =  £ {F} or f(s)  =    £ {F(t) }

Example:  Compute £ {t}   where F(t)  =   t

£ {t}  =   ò ^¥_o    exp(-st) t dt  =    lim _{R ® 0}       ò ^R_o     exp(-st) F(t) dt   

=     lim _R ® 0    [ - t/s  exp (-st)]  ^R_o        + 1/s   ò ^R_o     t exp(-st) dt

=    lim _R ® 0    - R/s  exp (-Rs)  +  (-1/ s 2      exp(-Rs)  + 1/ s 2  )

General properties of Laplace Transforms:

1)Let F1 and F2 both have Laplace transforms on some common interval. Let c1 and c2 be constants. Then:

£ {c1 F1 + c2 F2} =  c1 £ {F1}  + c2 £ {F2} 

Let F1 = 1, and F2 = cos t

Then:  £ {1 – cos t} =   £ {1} - £ {cos t} = 1/s  - s /1 +  s ² 

2)Let F be continuous for t > 0 and of exponential order exp (a t).  Assume F’ is piecewise continuous on every interval of the form [0, b], and 0 <  b <  ¥    .   Then,

£ {F’}exists and:

£ {F’(t) }  =  s  £ {F(t) }  -  F (0)

Solution example:

Solve:   dY / dt  +   2Y = cos t

Using Laplace transforms:

Then:

£ {Y’(t) }  + 2 £ {Y (t) }  =  £ {cos (t) } 

And:

£ {Y’(t) }  + 2 £ {Y (t) }  =    s / s ² +  1

Further:   £ {Y’(t) }   =  y(s)

s £ {Y’(t) }  -  Y(0)  + 2 £ {Y (t) }  =    s / s ² +  1

s y(s) + 1  + 2 y(s)   =  s / s ² +  1

y(s)  [s + 2}  =   s -   s ² +  1  / s ² +  1

  

Whence:  y(s)  =    -  s ² + s  -  1  / ( s ² +  2) ( s  +  2)

Separate using partial fractions:

As + B/ s ² +  1  +   C/ s + 2   =

(As + B)  (s + 2) + Cs² +  C/  ( s ² +  2) ( s  +  1)

SO:

(C  + A) s²   +  (2A + B) s  + 2B + C  =  =    -  s ² + s  -  1  

From which we see by inspection:

A + C = -1,   2A +  B  = 1,  2B  + C  = -1

Add:

-2A – 2C  = 2

 2A   + B = 1

-----------------

B – 2C   =   3

Add:

B  - 2C   =  3

4B  + 2C =  -2

----------------

5B        =  1     Therefore:  B = 1/5  

2A + B = 1 and 2A =   1 - 1/5   =   4/5

A =   ½ (4/5)   =   2/5  so:   C = -1 – 2B = -1 – 2(1/5) = -7/5

The inverse transform is therefore:

£ ^-1 {y(s)}  =  2 cos t/ 5 -  sin t/5 – 7/5 exp (-2t)  = Y(1)

Problem for Math Mavens:

Solve , using the Laplace transform:

d ³ Y/ dt ³ -    d Y /dt   =   0 

Using the conditions:  Y(0) =  1,  Y’(0) =  0  and  Y’ (0) = 1

----------------------------------

Some  common Laplace transforms:

More on Laurent Series

Note to readers: I reiterate once more that this set of blog posts isn’t intended to be comprehensive nor should they be seen as replacing the hard work of diligently working problems on your own using available text books. They are merely intended to spur on curiosity and perhaps offer a few different perspectives from what one might have seen already.

Included here are the coefficients in typical Laurent series which are generally obtained by other means (i.e. than appealing directly to integral representations). Some examples I have included below might suffice to expose this.

 

Ex. (1):  You have the Maclaurin series:

 

exp (z) = å^¥ _{n = 0}   (z) ⁿ  / n!  = 1 + z + z² / 2!  + z³ / 3!  +…..

 

 

Now, replace z by 1/z in the expansion:

 

 

exp( 1/z) = å^¥ _{n = 0}   1  / n!   z ⁿ  = 1 + 1/ 1! z   + 1/ 2!  z² 

 +  1/ 3! z³   +…..

 

for: 0   <   ÷ z ÷    <     ¥ 

 

Which then becomes a Laurent series expansion.

 

To establish this on inspection note that no positive powers of z appear, only negative, i.e. then 1/z is 

z ^{- 1}   

 So, in effect we can say that the coefficients of the positive powers are zero.  It’s also important to note here that the coefficient of : 

 

1/ 1! z     = 1/ z

 

is unity, so according to Laurent’s theorem we designate that the coefficient:

 

c _n  =  1/2 pi  ò_C  exp( 1/z)  dz

 

where C is any positively oriented simple closed contour around the origin.  Now, since from our prior  Dec. 12, Introducing Laurent Series) examination c _n  =  1 then:
 

1  =1/2 pi  ò_C  exp( 1/z)  dz   and:  ò_C  exp( 1/z)  dz    =  2 pi 

Ex. (2):  Consider the function:

f(z) = 1/ (z – i) ²

 

This is already in the form of a Laurent series , where z ₀  =   i.  That is,

 

å^¥ _{n = -¥}   c _n (z – i) ⁿ     (For:  0   <   ÷ z - i ÷    <     ¥  )

Where in this case,  c _{- 2}  =  1  and all other coefficients are zero.  Then from the previous theorems, relations we’ve explored (blog post of December 12, ‘Introduction to Laurent Series’, sub-header: ‘More intricacies and singularities’):

 

c _n  =  1/2 pi  ò_C   dz/ ( z – i) ^{n +3}      n = 0, +1, +2, +3 …..

 

 

where C denotes any positively oriented circle ÷ z - i ÷   = R  about the point: z ₀  =   i

 

 

Then it follows from this:

 

 

a)    ò_C   dz/ ( z – i) ^{n +3}      =  0   (when n ¹   2)

 

b) ò_C   dz/ ( z – i) ^{n +3}      =  1    (when n=    2)

 

Problems for the Math Maven:

 

1)     The function: f(z) =  -1/ (z – 1) (z – 2)
 

a) Rewrite it using the partial fraction form

 

b) Identify the two singular points.

 

 

c) If  the function is analytic in the domains:

 

÷ z ÷    <   1;   1     < ÷ z ÷     <      2   and:   2    <  ÷ z ÷    <     ¥,  

 

 

Draw a sketch showing the different domains.

 

 

2) Find the Laurent series that represents the function:

 

 

f(z) =  z²  sin (1 / z² )

 

In the domain:  : 0   <  ÷ z ÷   <     ¥ 

 

 

Hint:  Recall the series for sin(z) = z -  z³ / 3!  +  z⁵ / 5!  …..-  

   (-1) ^{n- 1} z ^{2n -1}/  (2n -1)!  +   ……   (For ÷ z ÷   <    ¥  )