Download Solving Inhomogeneous ODEs: A Forced Harmonic Oscillator Example and more Study notes Calculus in PDF only on Docsity!
Inhomogeneous ODEs
Adrian Down
April 13, 2006
1 Linear homogeneous first order systems with
constant coefficients
1.1 Setup
Recall the situation considered last time of a system of first order differential equations with constant coefficients. We wrote the system in matrix notation,
x˙ = Ax
where the vector x contains the dependent variables,
x 1 (t) x 2 (t)
)T
, and A is the matrix of coefficients of the differential equations. In the example we considered last time, the coefficient matrix was
A =
1.2 Diagonalization
We assumed elementary solutions of the form
x = veλt
This led to the equation,
Av = λv
Hence the elementary solutions are found by diagonalizing A.
We also saw by taking the complex conjugate of the eigenvalue equation, that if v is an eigenvector with corresponding eigenvalue λ, then the complex conjugate v∗^ is also an eigenvector with corresponding eigenvalue λ∗ In the case that we considered previously, we found the solutions to be,
x 1 =
−ı
e(−+ı)t^ x 2 =
ı
e(−−ı)t
1.3 Real solutions
It is often more useful to express the complex solutions as purely real func- tions, especially in the case of physical problems. Notice that the real an imaginary parts of the solution are themselves solutions,
x˙ = Ax ⇒ (<˙x) = A(<x)
A similar argument holds for the imaginary part of the solution. The final solution will be a linear combination of these two solutions. In this particular example, the real and imaginary components of the solution are,
<x 1 =
e−t^ cos t e−t^ sin t
=x 1 =
e−t^ sin t −e−t^ cos t
In this case, we took the real an imaginary components of only the first solution. However, the components of the second solution give no new infor- mation. Considering the second solution,
x 2 =
ı
e(−−ı)t^ =
e−t(cos t − ı sin t) e−t(cos t + ı sin t)
Taking the real and imaginary components,
<x 2 =
e−t^ cos t e−t^ sin t
=x 2 =
−e−t^ sin t e−t^ cos t
The real component is the same as that obtained from x 1. The imaginary component is the negative of that obtained from x 1. This negative could be absorbed into the coefficient in a linear combination of solutions. Hence these two solutions give no new information relative to those obtained from x 1.
v = ˙y. With this definition, we can rewrite the differential equation and solve for ˙y,
v˙ + y = f (t) ⇒ v˙ = −y + f (t)
We can now formulate the system in matrix notation, ( y˙ v ˙
y v
f
Note. In the general case of the method demonstrated in this example, we would be solving the matrix equation
y˙ = Ay + f
where A is an n × n matrix, and y is an n × 1 matrix.
2.3 Solution
2.3.1 Homogeneous case
We could now solve the system by diagonalizing A and using the methods developed previously. However, it will be more instructive (and simpler) to use some mathematical intuition. In the homogeneous case, the differential equation becomes,
f = 0 ⇒ y = Ay
In the case of our particular example,
y˙ = v v˙ = −y
The trig functions satisfy these relations, so assume solutions of the form,
y 1 =
cos t − sin t
y 2 =
sin t cos t
These are two linearly independent solutions. We can form the matrix of homogeneous solutions,
S =
y 1 y 2
Note. This portion of the example is a particular case of the general situation x˙ = A(t)x in which A is a time-depend n × n matrix. The initial value problem
x(0) = ei =
← ith position
has a solution Si(t). We can then form the solution matrix,
S =
S 1 (t)... Sn(t)
By construction, the solution matrix satisfies the homogeneous differential equation,
S˙ =
S˙ 1... S˙n
AS 1... ASn
= A
S 1... Sn
= AS
2.3.2 Inhomogeneous case
We now want to solve the inhomogeneous differential equation,
y˙ = A(t)y + f (t)
We seek a solution of the form
y = S(t)c(t)
The matrix c is the set of coefficients of the homogeneous solutions contained in the matrix S. In the homogeneous case, c is a matrix of constants. In the inhomogeneous case, we allow c to be a function of time. To determine the solution y(t), we must determine the behavior of c. It is possible to show that matrices obey the product rule,
y˙ = S c˙ + Sc˙
y(t) = S(t)y 0 +
∫ (^) t
0
S(t)S−^1 (t′)f (t′)dt′
Note. The solution to the vector case has the same general form as that which we obtained in the scalar case,
y(t) = e−P^ (t)y 0 +
∫ (^) t
0
e−P^ (t)+P^ (t ′) f (t′)dt′
2.4 Particular example
We return now to the particular example that we were considering earlier. Forming the matrix of solutions,
S =
cos t sin t − sin t cos t
Recognize the solution matrix S as the matrix representation of a rotation through an angle −t. This realization allows us to easily form the inverse S−^1 and its product with S,
S(t) ︸︷︷︸ rot. by −t
S−^1 (t′) ︸ ︷︷ ︸ rot. by t′
= S(t − t′) =
cos(t − t′) sin(t − t′) − sin(t − t′) cos(t − t′)
Forming the product with f (t′),
S(t)S(t′)f (t′) = S(t − t′)
=
cos(t − t′) sin(t − t′) − sin(t − t′) cos(t − t′)
f (t′)
f (t′) sin(t − t′) f (t′) cos(t − t′)
Suppose we are given the initial conditions
y(0) = 0 y˙(0) = v(0) = 0
⇒ y(0) ≡ 0
This corresponds to the case that the particle is initially at rest.
The final form of the solution is,
y(t) =
∫ (^) t
0
f (t′) sin(t − t′)dt′
Imagine the case that the particle is initially at rest and then it receives a sharp impulse at t = t′, from a hammer blow, for example. The solution takes the form,
y(t) =
0 t ≤ t′ ∝ sin(t − t′) t > t′
An arbitrary forcing function can be considered as a superposition of many short impulses.
