{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ma2aPracLectures25-26

# Ma2aPracLectures25-26 - Lecture 25 The utility of the...

This preview shows pages 1–4. Sign up to view the full content.

Lecture 25 The utility of the Laplace transform in solving ODE’s Recall that if f is a function on [0 , ) which is piecewise continuous and is “ a -nice,” i.e., | f ( t ) | ≤ ce at , for large enough t , for a positive constant c , then its Laplace transform L ( f ( t )) = F ( s ) = 0 f ( t ) e st dt ( = lim A →∞ A 0 f ( t ) e st dt ) is well defined. (In fact, F ( s ) makes sense for any complex number s as long as Re ( s ) > a .) Note: L (1) = 0 e st dt = lim A →∞ A 0 e st dt = lim A →∞ ( e st s ) A 0 = lim A →∞ ( e sA s + 1 s ) = 1 s , since e sA 0 as A → ∞ , for s > 0 . An oft-used function is the Heaviside function u c attached to any c > 0: u c ( t ) = { 1 , if t c 0 , if 0 t < c The Laplace transform of t n can be calculated by using induction and integration by parts (using e st dt = 1 s d ( e st )): L ( t n ) = 0 t n e st dt = lim A →∞ 1 s ( t n e st A 0 n A 0 t n 1 e st dt ) = n s L ( t n 1 ) . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
f ( t ) F ( s ) = L ( f ( t )) 1 1 s (for s > 0) e at 1 s a ( s > a ) u c ( t ) e - at s (for s > 0) e at cos( bt ) s a ( s a ) 2 + b 2 (for s > 0) e at sin( bt ) b ( s a ) 2 + b 2 ( s > 0) t n ( n 0) n ! s n +1 ( n > 0) Important properties of L 1) L is linear, i.e., L ( af + bg ) = a L ( f ) + b L ( g ), for all constants a, b 2) L ( f ( t )) = 1 c F ( s c ), if c > 0 and F ( s ) = L ( f ( t )) 3) L ( u c ( t ) f ( t c )) = e cs L ( f ( t )) (check this!) 4) If f is continuous (not just piecewise continuous) and F ( s ) = 0 for all s > M , for some M > 0, then f ( t ) = 0 for all t . 5) Suppose f is n -times differentiable with f ( n ) being piecewise continuous and a -nice, then L ( f ( n ) ( t )) = s n L ( f ( t )) s n 1 f (0) s n 2 f (0) . . . f ( n 1) (0) . Consequence of 4) and 1): Given continuous functions f, g on [0 , ), if F ( s ) = G ( s ) for all s > c for some c > 0, then f ( t ) = g ( t ), for all t . 2
Reason : By 1), L ( f ( t ) g ( t )) = L ( f ( t )) − L ( g ( t )) = F ( s ) G ( s ) , which is 0 by hypothesis. So by 4), f ( t ) g ( t ) = 0 , t.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

Ma2aPracLectures25-26 - Lecture 25 The utility of the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online