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A Properties of linear differential equations
For Classical Mechanics A, Univ.Tokyo (2017)
Descriptions: Reply to the question associated with the ansatz for the damped harmonic
oscillator: why do we need to consider the linear combination?
[Apr. 29, 2017 ]1Damped harmonic oscillator

The equation of motion for the damped harmonic oscillator is given as()dd2
2m 2 + 2mγ + mω x(t) = 0.
dtdt

(1)

^ Then, the equation isAs in Feynman Ch. 25, let us denote the diﬀerential operator in this equation as L.
expressed
^L(x)
= 0.
(2)
In order to solve the diﬀerential equation, three approaches were attempted in the previous lectures: trigonometric functions, the McLaurin (power) series expansion, and the ansatz method.
The last one, the ansatz method, is the method that we studied most recently, and it assumes a possible
mathematical form of solutions with parameters to be determined. In the present case, the following ansatz
is known to work
x(t) = eλt ,(3)where λ is a complex parameter to be determined. Inserting this ansatz to the diﬀerential equation, we havea quadratic equation for λλ2 + 2γλ + ω2 = 0.
(4)With the quadratic formula, we obtain
λ=γ±√γ2 − ω2 .

Let us classify this solution by an integer k in the following way√λk = γ + (−1)k γ2 − ω2 ,(5)(6)

where k = 0, 1. Therefore, we have two possible solutions for the damped harmonic oscillator, which are
x0 (t) = eλ0 t

andx1 (t) = eλ1 t(7)Obviously,^ k) = 0L(x(8)

is satisfied for k = 0, 1.
In my previous lecture (on Wednesday, April 26, 2017), I said that the linear combination of the above twosolutions xk (t)x(t) = c0 x0 (t) + c1 x1 (t), ck being a constant for k = 0, 1(9)is regarded as the “general” solution of the damped harmonic oscillator. Then, there was a question raisedon this remark:1

“You said that the ansatz is assumed to have a single-term form of the exponential, but you also
said that the “general” solution is a linear combination of the two independent solutions, but it
consists of two terms. This result looks contradicting to what you assumed in the beginning,
doesn’t it? We should rather employ the independent solutions separately, shouldn’t we? Say,x(t) = x0 (t), or x(t) = x1 (t). What about this?”This question really hits the centre of the so-called “linear problem” of the diﬀerential equation, which isalso discussed in Feynman Ch. 25. Let us rebuild what is discussed in Feynman’s textbook.

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Linear problemndDiﬀerential operators dt
n satisfies the following property

dndndn(af(t)+bg(t))=af(t)+bg(t),dtndtndtn(10)where a, b are constants. This result means that the derivative of the linear combination of functions isequivalent to the linear combination of the derivatives of the individual functions. Therefore, if x0 (t) andx1 (t) are the two independent solutions of the equation of motion (1), then the linear combination of these
independent solutions
x(t) = c0 x0 (t) + c1 x1 (t)(11)
is another solution of Eq.(1). The proof of this remark relies on the “linear property” of Eq.(10):^ 0 ) = 0 and L(x^ 1 ) = 0. Then,Suppose L(x^ 0 x0 + c1 x1 ) = c0 L(x^ 0 ) + c1 L(x^ 1 ) = 0.
L(c(12)

The above equation is justified thanks to the linear form of the diﬀerential equation (1).
In this way, we have millions of solutions to satisfy the equation of motion (1), by changing ck . Suchindividual and special cases can be summarised into the general form (11). For example, x(t) = eλ0 t simply
corresponds to (c0 , c1 ) = (1, 0).
Furthermore, suppose that we can find more than three solutions to satisfy the equation of motion (1),^ 0 ) = 0, L(x^ 1 ) = 0, L(x^ 2 ) = 0, then the linear combination of these three is again the solution
say L(xof the equation of motion. However, the “degrees of freedom” of a diﬀerential equation of order n (i.e.,a diﬀerential equation containing a term dn /dtn ) is n, so that the equation of motion allows only two
independent solutions (the equation of motion is a second-order diﬀerential equation). When x0 and x1 aresuch independent solutions, then x2 can be expressed as a linear combination
x2 (t) = d0 x0 (t) + d1 x1 (t).

(13)

Hence,
x(t) = c0 x0 (t) + c1 x1 (t) + c2 x2 (t)= (c0 + d0 ) x0 (t) + (c1 + d1 ) x1 (t)= C0 x0 (t) + C1 x1 (t).

(14)

The linear property of the equation of motion can be broken down, for instance, if we replace the linear
frictional force −2γv(t) with the quadratic friction −2γ ′ v(t)2 , or if we replace the harmonic term −ω2 x(t)()
with an anharmonic force −ω2 x(t) + ϵx(t)3 . Such equations of motion, or diﬀerential equations that
contain “non-linear terms” are called “non-linear equations”, in which the superposition principle, i.e., Eqs.
(11) and (12), cannot be applied.
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In fundamental theories of physics, linear equations are everywhere. For instance, the Maxwell equations in
the classical electromagnetism, which is to be lectured in the A-semester, are linear diﬀerential equations.
The Schrödinger equation in Quantum mechanics is also a linear equation. Therefore, the superposition
principle can be applied to the electrodynamics and quantum mechanics, which plays a significant role toformulate the theories. Students are strongly recommended to confirm that the Maxwell and Schrödinger
equations satisfy the linear property. (For instance, suppose Φ0 and Φ1 are two independent wave functionsof the Schrödinger equation. Then, you should examine whether a linear combination c0 Φ0 + c1 Φ1 satisfies
the same Schrödinger equation or not.) The concept of the superposition principle leads to an applicationof the Fourier series expansion and/or the Fourier transform to these linear problems. For instance, theelectromagnetic field (or the vector potential) can be decomposed in accordance with the Fourier transform.

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Practical problems if sticking with single-term solution

Let us now discuss what is not written in Feynman’s textbook. Suppose that the single-term solution, x0 (t)or x1 (t), is employed rather than the general solution of the linear combination form. Then, what is wrong
with it?
Firstly, we write down a solution, for instance, asx(t) = eλ0 t .

(15)Then, what is wrong with this? For the sake of simplicity, we consider the case γ = 0 in the below. Thenthe “solution” turns to bex(t) = eiωt .
(16)
As explained in the lecture, the position of a particle of interest needs to be described with a real dynamicalvariable. The above quantity is a complex variable, not a real one. You cannot measure a position ofparticles with complex numbers.
By utilising a freedom to multiply a complex number C = a + ib to the solution, we might be able totransform the single-term solution to a real variable. Here, a and b are constant real numbers.
Ceiωt = (a + ib) (cos(ωt) + i sin(ωt))= {a cos(ωt) − b sin(ωt)} + i {b cos(ωt) + a sin(ωt)}(17)

To make the imaginary part vanish, we requireb cos(ωt) + a sin(ωt) = 0,(18)

but we know that these trigonometric functions are linearly independent to each other. The above conditionis thus satisfied only when a = b = 0, which means C = 0. Surely, x(t) = 0 is the trivial solution, but does
not carry any physical information with regard to the dynamics in Eq.(1). We have to conclude that thechoice of a single-term solution ends up in an absurd result.
Next, we should examine whether or not the linear combination of Eq.(11) satisfies the physical requirement.
To demand x(t) to be a real variable, we should havex(t) = x(t)∗ .

(19)

In other words, the complex conjugate of the position variable must be equal to the original. With thesuperposition, we havec0 eiωt + c1 e−iωt = c∗0 e−iωt + c∗1 eiωt ,(20)

3or(c0 − c∗1 )eiωt + (c1 − c∗0 )e−iωt = 0.

(21)

Exponential functions eix and e−ix are linearly independent since their Wronskian
eixe−ix
ix
ie−ie−ix= −i − i = −2i(̸= 0).

(22)

is evaluated to be non-vanishing. Therefore, we can conclude thatc∗0 = c1 ,(23)

orx(t) = c0 eiωt + c∗0 e−iωt= (c0 + c∗0 ) cos(ωt) + i (c0 − c∗0 ) sin(ωt)= 2Re(c0 ) cos(ωt) − 2Im(c0 ) sin(ωt)(24)

If we express C0 = 2Re(c0 ), C1 = −2Im(c0 ) (Ck are both real constants), then x(t) is nothing but a linear
combination of the trigonometric functions, as we already saw in the first lecture (on April 5, 2017).
This result also reminds us of a mathematical nature of a complex number z = x + iy, which carries
information of two “eﬀective” degrees of freedom (x, y).
In summary, only through the linear combination, Eq.(11), one can construct a consistent solution withthe previously obtained solutions, which are expressed by means of the trigonometric functions and theMcLaurin series expansion.

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