January 9th, 2017, 09:27 AM  #1 
Member Joined: Nov 2014 From: SG Posts: 55 Thanks: 1  Laplace transform property Could anyone explain to me why d/dt = s when initial conditions are zero ? If possible , please show me an example why when initial condition are zero , d/dt =s. Thank you so much for your help Last edited by KaiL; January 9th, 2017 at 09:33 AM. 
January 9th, 2017, 11:48 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,746 Thanks: 891 
$\mathscr{L}\{f^{(n)}\} = s^n \mathscr{L}\{f\}  \displaystyle{\sum_{k=1}^n}~ s^{nk}f^{(k1)}(0)$ so all the $f^{(nk)}(0),~k=1,n$ must equal zero. 
January 9th, 2017, 06:32 PM  #3  
Member Joined: Nov 2014 From: SG Posts: 55 Thanks: 1  Quote:
Thank you for your reply but I still don't really get it. $f^{(k1)}(0)=0, k=1$ since initial conditions are zero > I understand this part. Since k = 1 and subbing it to this equation $\mathscr{L}\{f^{(n)}\} = s^n \mathscr{L}\{f\}  \displaystyle{\sum_{k=1}^n}~$$s^{(n1)}$ $f^{(n1)}(0)$ Then I am not sure why is n equal to 0 ? and why d/dt = s ?  
January 9th, 2017, 09:04 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,746 Thanks: 891  Quote:
Consider the transform of $f^\prime(t)$ $\mathscr{L}\{f^\prime(t)\} = \displaystyle{\int_0^\infty}~f^\prime(t) e^{s t}~dt$ $u = e^{s t},~du = se^{st}$ $v = f^\prime(t)~dt,~v = f(t)$ $\left . e^{s t}f(t) \right_0^\infty + \displaystyle{\int_0^\infty}~f(t)s e^{s t}~dt =$ $f(0) + s F(s)$ You can use induction to prove the general formula. Last edited by skipjack; January 10th, 2017 at 03:47 AM.  
January 9th, 2017, 11:30 PM  #5  
Member Joined: Nov 2014 From: SG Posts: 55 Thanks: 1  Quote:
$\mathscr{L}\{f^\prime(t)\} = f(0) + s F(s)$ Given that initial conditions are zero, $f(0) = 0 $ So $\mathscr{L}\{f^\prime(t)\} = 0 + s F(s)$ $\mathscr{L}\{f^\prime(t)\} = s F(s)$ $\mathscr{L}\{f(t)\} = F(s)$ $\mathscr{L}\{f^\prime(t)\} = s \mathscr{L}\{f(t)\} $ However, I still can't see the reason why $d/dt = s ?$ So sorry for asking this question again. Last edited by skipjack; January 10th, 2017 at 03:47 AM.  
January 10th, 2017, 12:37 AM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 1,746 Thanks: 891  Quote:
$\dfrac{d}{dt} f(t) \overset{\mathscr{L}}{\Leftrightarrow} s F(s)$ Last edited by skipjack; January 10th, 2017 at 03:44 AM.  

Tags 
laplace, property, transform 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Laplace Transform pde  pass92  Differential Equations  0  November 28th, 2015 04:51 AM 
Relatioship between Laplace and Inverse Laplace Transform from tables  szz  Differential Equations  1  November 2nd, 2014 04:18 AM 
LAPLACE TRANSFORM  capea  Complex Analysis  2  August 23rd, 2013 11:43 AM 
Laplace tranform and inverse of Laplace transform  Deiota  Calculus  1  April 28th, 2013 11:28 AM 
Laplace transform.  gelatine1  Calculus  5  November 2nd, 2012 01:17 PM 