Let us introduce the auxiliar variables x1(t) = y(t), and x2(t)= Ëy(t), with initial values x1(0) = A, and x2(0) = B. Then, we can rewrite (7) as the following system.
Solution to an integro-differential equation using Laplace transforms Let us consider the following equation t
Z
f (t − t0 )y(t ˙ 0 ) − by(t) = 0,
y¨(t) − a
(1)
0
with a, b ∈ R, and f : R+ → R a good function admiting a Laplace transform, i.e., so that there are two numbers M, p ∈ R so that |f (t)| ≤ M ept for t → ∞. We will represent Laplace transforms by capital letters; the transform of y(t) will be denoted by Y (s) and the transform of f (t) by F (s). Note that, given that the highest derivative in this Eq. (1) is of second order, the initial values y(0) and y(0) ˙ have to be known and given, otherwhise we will find a general solution with two undetermined constants. Applying Laplace transform directly in (1), and using the rule for the Laplace transform of the first and second derivatives, and of the convolution, we readily obtain 2 s Y (s) − sy(0) − y(0) ˙ − a F (s) sY (s) − y(0) − bY (s) = 0. (2) From (2) we can easily solve for Y (s): Y (s) =
As + B − aAF (s) , s2 − asF (s) − b
(3)
where we have adopted the notation A = y(0), B = y(0). ˙ From the previous result, the solution y(t) can be obtained by applying the inverse Laplace transfom y(t) =
1 2πi
Z
+i∞
dsets Y (s) ≡
−i∞
1 2πi
Z
+i∞
ds −i∞
ets (As + B − aAF (s)) , s2 − asF (s) − b
(4)
which can be evaluated using residue theory, or even advanced calculus programs, like Matlab or Mathematica. As an example, let us consider the simple case where f (t) is a simple exponential, f (t) = f0 e−kt . In that case we have f0 , k+s
(5)
As(k + s) + B(k + s) − aAf0 . s2 (k + s) − af0 s − b(k + s)
(6)
F (s) = and we have Y (s) =
The full problem then reduces to solve the integral in Eq. (4). To finish, note that in the case f (t) = f0 e−kt the solution can be found using another trick. Let us write the equation in this case: Z y¨(t) − af0
t
0
e−(t−t ) y(t ˙ 0 ) − by(t) = 0.
(7)
0
We still have the initial conditions y(0) = A, y(0) ˙ = b. Let us introduce the auxiliar variables x1 (t) = y(t), and x2 (t) = y(t), ˙ with initial values x1 (0) = A, and x2 (0) = B. Then, we can rewrite (7) as the following system x˙ 1 (t) = x2 (t), Z t 0 x˙ 2 (t) = af0 e−k(t−t ) x2 (t0 )dt0 + bx1 (t).
(8) (9)
0
To proceed further, we introduce a third variable, x3 (t) = x3 (0) = 0, and
Rt 0
0
e−k(t−t ) x2 (t0 )dt0 . Note that, by construction, we have
x˙ 3 (t) = −kx3 (t) + x2 (t).
(10)
2 Thus we are left with a linear system of three ordinary differential equations: x˙ 1 (t) = x2 (t), x˙ 2 (t) = af0 x3 (t) + bx1 (t), x˙ 3 (t) = −kx3 (t) + x2 (t),
(11) (12) (13)
with initial conditions x1 (0) = A, x2 (0) = B, and x3 (0) = 0, which is easy to solve. Our solution is simply y(t) = x1 (t). I hope this helps.
