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Linear ODEs with Constant Coefficients
Adrian Down
April 11, 2006
1 General method
1.1 Review
1.1.1 Differential equation
Last time, we found solutions to the differential equation,
y′^ + p(x)y = f (x) (1)
where f (x) is a known function and y(x 0 ) is a given boundary condition.
1.1.2 Homogeneous solution
In the case that f ≡ 0, the ODE (1) is homogeneous. The solutions involve an anti-derivative P (x) of p(x),
y = y(x 0 )e−P^ (x)
1.1.3 Inhomogeneous solution
In the case that f ′^6 = 0, the ODE (1) is inhomogeneous. We found solutions by multiplying both sides of (1) by eP^ (x). The left side could then be written as a perfect differential. We integrated both sides of the equation, using the Fundamental Theorem of Calculus to simplify the integration of the differential. Utilizing the given boundary conditions,
y(x) = y(x 0 )e−P^ (x)^ +
∫ (^) x
x 0
eP^ (t)−p(x)f (t)dt
The anti-derivative P (x) was chosen such that P (x 0 ) = 0.
1.2 Example: charged particle in electromagnetic fields
1.2.1 Setup
Consider the case of a charged particle moving in a magnetic and electric field applied simultaneously. Take the fields to be isotropic^1 functions of time. Take the electric field E(t) to be in the ˆx direction and the magnetic field B(t) to be in the −ˆz direction.
1.2.2 Differential equations
The charged particle experiences a force due to the magnetic field propor- tional to the cross product of velocity of the charge and the B field. The magnetic field causes the charged particle to rotate in the clockwise direction. The force equations involve the acceleration of the particle, which is the rate of change of the velocity. The Cartesian components of the velocity satisfy a set of first order differential equations resulting from the force equations,
x˙ = u(t)ˆx + v(t)ˆy u ˙ = −B(t)v + E(t) (2) v˙ = B(t)u (3)
Note. • The dot denotes differentiation with respect to time.
- The units are chosen so that constants, such as the mass of the particle, do not appear in the differential equations.
1.2.3 Complex solution
As in the case of the spiral example considered previously, we seek a complex solution of the form,
z(t) = u(t) + ıv(t) (4)
Substitute (2) and (3) into (4),
z˙ = ˙u + ı v˙ = −B(t)v + E(t) + ıB(t)u = ıB(t)(u + ıv) + E(t) = ıB(t)z + E(t) (5) (^1) Isotropic means independent of position.
The left side of (5) can be integrated using the Fundamental Theorem of calculus.
e−ıΘ(t)z(t) − e^0 z 0 =
∫ (^) t
0
e−ıθ(s)E(s)ds
z(t) = z 0 eıΘ(t)^ +
∫ (^) t
0
eı(Θ(t)−Θ(s))E(s)ds (9)
Note. The name of the dummy variable of integration on the right has been changed to avoid confusion with the actual independent variable t.
The first term on the right of (9) is a solution to the homogeneous IVP. The integral on the right of (9) is a particular solution to the IVP. In (9) we have an explicit integral formula for the solution z(t) to the differential equation given any analytic field functions E(t) and B(t). It is rare when con- sidering physical problems to be able to construct explicit analytic solutions such as these.
1.2.6 Explicit fields
Consider the case of a cyclotron or similar device, in which the fields are,
B(t) = Ω = constant E(t) = cos Ωt
The boundary condition is z 0 = 0, indicating that the particle is at rest at z = 0. Substituting these fields into (9),
z(t) =
∫ (^) t
0
eıΩ(t−s)^ cos(Ωs)ds
It will be more convenient to write the cosine in terms of complex exponen- tials,
z(t) =
∫ (^) t
0
eıΩ(t−s)^ ·
eıΩs^ + e−ıΩs
ds
The integral can be simplified. t is independent of the integration variable s, and can be factored out of the integral. Distributing the e−ıΩs^ term,
z(t) =
eıΩt 2
∫ (^) t
0
1 − e−^2 ıΩs
ds
eıΩt 2
t −
2 ıΩ
e−^2 ıΩt^ − 1
teıΩt 2
sin Ωt
Plotting the solution in the complex plane, z(t) is a spiral outward from the origin. The spiral is not an equal-angle spiral, since the particle is speed- ing up as it moves away from the origin. This spiral behavior was used in early physical devices to accelerate charged particles to high speeds.
2 Examples of exponential solutions
2.1 Elastic rod
2.1.1 Setup
Consider the case of some elastic rod described by a plane curve y = y(x). The rod is connected to the the horizontal by springs. The rod could repre- sent the surface of a mattress, for example. In the absence of a disturbance, the rod would be horizontal. If the rod is disturbed, the variation propagates along the rod for some distance. This behavior is similar to the flexible rod example considered when we studied splines, although in this case, the elastic rod is not confined at intermediate points.
2.1.2 Differential equation
There is an energy associated with the position of the elastic rod due to the potential energy of the springs. The physical solution for the vertical position of the rod y(x) is that which minimizes the energy. Energy minimization is equivalent to satisfying the differential equation,
y(4)^ + y = 0 ∀x > 0
2.1.4 Physical solution
To make the solutions more intelligible, convert the complex solutions to real solutions. The real and imaginary component of each solution are separately solutions,
y 1 = e √x (^2) cos √x 2 y 2 = e √x (^2) sin √x 2 y 3 = e−^ √x (^2) cos √x 2 y 4 = e−^ √x (^2) sin √x 2 The general solution is a linear combination of all of these solutions. To determine the coefficients of the linear combination, employ the boundary conditions. The requirement that y → 0 as x → ∞ implies that the general linear combination cannot contain y 1 or y 2 , as these functions approach ∞ as x → ∞. We also have the requirement that y(0) = 0. This excludes y 3 , which is equal to 1 at x = 0. The final boundary condition is that y′(0) = θ. This boundary condition determines the coefficient of y 4 in the final solution,
y = Ce−^ √x (^2) sin √x 2
⇒ y′^ = C
e−^ √x (^2) sin √x 2
e−^ √x (^2) cos √x 2
y′(0) = y 0 =
C
−e^0 sin 0 + e^0 cos 0
C
⇒ C =
2 y 0 y(x) =
2 y 0 · e−^ √x (^2) sin √x 2
2.2 Spiral example
2.2.1 Motivation
Before, we solved this problem using some special techniques. We return to this problem to show how it can be solved efficiently using no special
techniques. The example also illustrates the use of diagonalization in solving ODEs.
2.2.2 Problem setup
Recall the set of differential equations,
x˙ = −x − y y ˙ = −x + y
We made a column vector of the form,
x(t) =
x(t) y(t)
The differential equation can be written as a matrix equation ˙x = Ax, where the coefficient matrix is,
A =
2.2.3 Exponential solutions
Assume exponential solutions of the form
x = veλt
where v is a constant vector and λ is a constant to be determined. Substitute into the ODE system ˙x = Ax. Canceling the exponential,
Av = λv
This is the form of the algebraic eigenvalue problem. The eigenvalues and eigenvectors can be obtained in the usual way,
0 = det (A − λ 1 1) = det
− − λ − 1 1 − − λ
= ( + λ)^2 + 1 = 0
