The Laplace transform is perhaps one of the most useful devices for solving differential equations. In this post I look at some of the basics underlying its use, especially in solving DEs.
Definition : Let F be a function defined for t > 0. Then define a new function f by:
f(s) = ò ¥o exp(-st) t dt
f(s) = ò ¥o exp(-st) 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 Ro exp(-st) F(t) dt
= lim R ® 0 [ - t/s exp (-st)] Ro + 1/s ò Ro 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)
Problem example:
Solve: dY / dt + 2Y = cos t
Using Laplace transforms:
Write:
£ {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)
Table of common Laplace transforms:
1) Solve:
d 3 Y/ dt 3 - d Y /dt = 0
Using the Laplace transform and the conditions:
Y(0) = 1, Y’(0) = 0 and Y’ (0) = 1
2) Solve:
d2 y/
dt 2 + 4y
= 3 sin t
Using the appropriate Laplace transform, given:
y= F(t), F(0)= 1, F'(t) = 0
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