Path Minimization

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6k: Solving New Differential Equation with Method of Undetermined Coefficients

In this series, I’m discussing how ideas from calculus and precalculus (with a touch of differential equations) can predict the precession in Mercury’s orbit and thus confirm Einstein’s theory of general relativity. The origins of this series came from a class project that I assigned to my Differential Equations students maybe 20 years ago.

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

where $u = \displaystyle \frac{1}{r}$ , $\displaystyle \frac{1}{\alpha} = \frac{GMm^2}{\ell^2}$ , $\delta = \displaystyle \frac{3GM}{c^2}$ , $G$ is the gravitational constant of the universe, $m$ is the mass of the planet, $M$ is the mass of the Sun, $\ell$ is the constant angular momentum of the planet, $c$ is the speed of light, and $P$ is the smallest distance of the planet from the Sun during its orbit (i.e., at perihelion).

In recent posts, we used the method of undetermined coefficients to show that the general solution of the differential equation is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} +\frac{\delta \epsilon}{\alpha^2} \theta \sin \theta- \frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta$ .

We now use the initial conditions to find the constants $c_1$ and $c_2$ . (We did this earlier when we solved the differential equation via variation of parameters, but we repeat the argument here for completeness.) From the initial condition $u(0) = \displaystyle \frac{1}{P} = \frac{1+\epsilon}{\alpha}$ , we obtain

$u(0) = \displaystyle c_1 \cos 0 + c_2 \sin 0 + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \cdot 0 \cdot \sin 0}{\alpha^2} -\frac{\delta \epsilon^2 \cos 0}{6\alpha^2}$

$\displaystyle \frac{1+\epsilon}{\alpha} = c_1 + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} -\frac{\delta \epsilon^2}{6\alpha^2}$

$\displaystyle \frac{\epsilon}{\alpha} = c_1 + \displaystyle \frac{3\delta +\delta \epsilon^2}{3\alpha^2}$ ,

so that

$c_1 = \displaystyle \frac{\epsilon}{\alpha} - \frac{\delta(3 + \epsilon^2)}{3\alpha^2}$ .

Next, we compute $u'(\theta)$ and use the initial condition $u'(0) = 0$ :

$u'(\theta) = \displaystyle -c_1 \sin \theta + c_2 \cos \theta + \frac{\delta \epsilon}{\alpha^2} (\sin \theta + \theta \cos \theta) + \frac{\delta \epsilon^2 \sin 2\theta}{3\alpha^2}$

$u'(0) = \displaystyle -c_1 \sin 0 + c_2 \cos 0 + \frac{\delta \epsilon}{\alpha^2} (\sin 0 + 0 \cos 0) + \frac{\delta \epsilon^2 \sin 0}{3\alpha^2}$

$0 = c_2$ .

Substituting these values for $c_1$ and $c_2$ , we finally arrive at the solution

$u(\theta) = \displaystyle \left(\frac{\epsilon}{\alpha} - \frac{\delta(3 + \epsilon^2)}{3\alpha^2} \right) \cos \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

$= \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{ \delta\epsilon}{\alpha^2} \theta \sin \theta - \frac{ \delta \epsilon^2}{6\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta$ .

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6j: Rationale for Method of Undetermined Coefficients VII

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

Let me summarize the partial results that we’ve found in the past few posts.

1. The general solution of the associated homogeneous differential equation

$u''(\theta) + u(\theta) = 0$

$u_0(\theta) = c_1 \cos \theta + c_2 \sin \theta$ .

2. One particular solution of the nonhomogeneous differentiatial equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$

$u_1(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

3. One particular solution of the nonhomogeneous differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{2\delta \epsilon \cos \theta}{\alpha^2}$

$u_2(\theta) = \displaystyle \frac{\delta \epsilon}{\alpha^2} \theta \sin \theta$ .

4. One particular solution of the nonhomogeneous differential equatio

$u''(\theta) + u(\theta) = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

$u_3(\theta) = \displaystyle -\frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta$ .

To solve the original differential equation, we will simply add these four solutions together:

$u(\theta) = u_0(\theta) + u_1(\theta) + u_2(\theta) + u_3(\theta)$

$= c_1 \cos \theta + c_2 \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} +\frac{\delta \epsilon}{\alpha^2} \theta \sin \theta- \frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta$ .

It’s a straightforward exercise to show that this new function satisfies the original differential equation:

$u''(\theta) + u(\theta) = u_0''(\theta) + u_1''(\theta) + u_2''(\theta) + u_3''(\theta) + u_0(\theta) + u_1(\theta) + u_2(\theta) + u_3(\theta)$

$= [u_0''(\theta) +u_0(\theta)]+ [u_1''(\theta) + u_1(\theta)]+[u_2''(\theta) + u_2(\theta)] + [u_3''(\theta) + u_3(\theta)]$

$= 0 + \displaystyle \left( \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} \right) + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$ ,

as required.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6i: Rationale for Method of Undetermined Coefficients VI

In the last few posts, I’ve used a standard technique from differential equations: to solve the $n$ th order homogeneous differential equation with constant coefficients

$a_n y^{(n)} + \dots + a_3 y''' + a_2 y'' + a_1 y' + a_0 y = 0$ ,

we first solve the characteristic equation

$a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 = 0$

using techniques from Precalculus. The form of the roots $r$ determines the solutions of the differential equation.

While this is a standard technique from differential equations, the perspective I’m taking in this series is scaffolding the techniques used to predict the precession in a planet’s orbit using only techniques from Calculus and Precalculus. So let me discuss why the above technique works, assuming that the characteristic equation does not have repeated roots. (The repeated roots case is a little more complicated but is not needed for the present series of posts.)

We begin by guessing that the above differential equation has a solution of the form $y = e^{rt}$ . Differentiating, we find $y' = re^{rt}$ , $y'' = r^2 e^{rt}$ , etc. Therefore, the differential equation becomes

$a_n r^n e^{rt} + \dots + a_3 r^3 e^{rt} + a_2 r^2 e^{rt} + a_1 r e^{rt} + a_0 e^{rt} = 0$

$e^{rt} \left(a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 \right) = 0$

$a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 = 0$

The last step does not “lose” any possible solutions for $r$ since $e^{rt}$ can never be equal to $0$ . Therefore, solving the differential equation reduces to finding the roots of this polynomial, which can be done using standard techniques from Precalculus.

For example, one of the differential equations that we’ve encountered is $y''+y=0$ . The characteristic equation is $r^2+1=0$ , which has roots $r=\pm i$ . Therefore, two solutions to the differential equation are $e^{it}$ and $e^{-it}$ , so that the general solution is

$y = c_1 e^{it} + c_2 e^{-it}$ .

To write this in a more conventional way, we use Euler’s formula $e^{ix} = \cos x + i \sin x$ , so that

$y = c_1 (\cos t + i \sin t) + c_2 (\cos (-t) + i \sin (-t))$

$= c_1 \cos t + i c_1 \sin t + c_2 \cos t - i c_2 \sin t$

$= (c_1 + c_2) \cos t + (ic_1 - ic_2) \sin t$

$= C_1 \cos t + C_2 \sin t$ .

Likewise, in the previous post, we encountered the fourth-order differential equation $y^{(4)}+5y''+4y = 0$ . To find the roots of the characteristic equation, we factor:

$r^4 + 5r^2 + 4r = 0$

$(r^2+1)(r^2+4) = 0$

$r^2 +1 = 0 \qquad \hbox{or} \qquad \hbox{or} r^2 + 4 = 0$

$r = \pm i \qquad \hbox{or} \qquad r = \pm 2i$ .

Therefore, four solutions of this differential equation are $e^{it}$ , $e^{-it}$ , $e^{2it}$ , and $e^{-2it}$ , so that the general solution is

$y = c_1 e^{it} + c_2 e^{-it} + c_3 e^{2it} + c_4 e{-2it}$ .

Using Euler’s formula as before, this can be rewritten as

$y = C_1 \cos t + C_2 \sin t + C_3 \cos 2t + C_4 \sin 2t$ .

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6h: Rationale for Method of Undetermined Coefficients V

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In the two previous posts, we derived the method of undetermined coefficients for the simplified differential equations

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

and

$u''(\theta) + u(\theta) = \displaystyle \frac{2\delta \epsilon \cos \theta}{\alpha^2}$ .

In this post, we consider the simplified differential equation if the right-hand side has only the fifth term,

$u''(\theta) + u(\theta) = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$ .

Let $v(\theta) = \displaystyle \frac{\delta \epsilon^2 }{2\alpha^2} \cos 2\theta$ . Then $v$ satisfies the new differential equation $v'' + 4v = 0$ . Also, $v = u'' + u$ . Substituting, we find

$(u''+u)'' + 4(u''+u) = 0$

$u^{(4)} + u'' + 4u'' + 4u = 0$

$u^{(4)} + 5u'' + 4u = 0$

The characteristic equation of this new differential equation is

$r^4 + 5r^2 + 4 = 0$

$(r^2 + 1)(r^2 + 4) = 0$

$r^2 + 1 = 0 \qquad \hbox{or} \qquad r^2 + 4 = 0$

$r = \pm i \qquad \hbox{or} \qquad r = \pm 2i$

Therefore, the general solution of the new differential equation is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + c_3 \cos 2\theta + c_4 \sin 2\theta$ .

The constants $c_3$ and $c_4$ can be found by substituting back into the original differential equation:

$u''(\theta) + u(\theta) = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

$-c_1 \cos \theta - c_2 \sin \theta - 4c_3 \cos 2\theta - 4c_4 \sin 2\theta + c_1 \cos \theta + c_2 \sin \theta + c_3 \cos 2\theta + c_4 \sin 2\theta = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

$- 3c_3 \cos 2\theta - 3c_4 \sin 2\theta = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

Matching coefficients, we see that $c_3 = \displaystyle -\frac{\delta \epsilon^2}{6\alpha^2}$ and $c_4 = 0$ . Therefore, the solution of the simplified differential equation is

$u(\theta) = c_1 \theta + c_2 \theta \displaystyle -\frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta$ .

In particular, setting $c_1 = 0$ and $c_2 = 0$ , we see that

$u(\theta) = \displaystyle -\frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta$

is a particular solution to the simplified differential equation.

In the next post, we put together the solutions of these three simplified differential equations to solve the original differential equation,

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6g: Rationale for Method of Undetermined Coefficients IV

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In this post, we will use the guesses

$u(\theta) = f(\theta) \cos \theta \qquad \hbox{or} u(\theta) = f(\theta) \sin \theta$

that arose from the technique/trick of reduction of order, where $f(\theta)$ is some unknown function, to find the general solution of the differential equation

$u^{(4)} + 2u'' + u = 0$ .

To do this, we will need to use the Product Rule for higher-order derivatives that was derived in the previous post:

$(fg)'' = f'' g + 2 f' g' + f g''$

and

$(fg)^{(4)} = f^{(4)} g + 4 f''' g' + 6 f'' g'' + 4f' g''' + f g^{(4)}$ .

In these formulas, Pascal’s triangle makes a somewhat surprising appearance; indeed, this pattern can be proven with mathematical induction.

We begin with $u(\theta) = f(\theta) \cos \theta$ . If $g(\theta) = \cos \theta$ , then

$g'(\theta) = - \sin \theta$ ,

$g''(\theta) = -\cos \theta$ ,

$g'''(\theta) = \sin \theta$ ,

$g^{(4)}(\theta) = \cos \theta$ .

Substituting into the fourth-order differential equation, we find the differential equation becomes

$(f \cos \theta)^{(4)} + 2 (f \cos \theta)'' + f \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 6 f'' \cos \theta + 4 f' \sin \theta + f \cos \theta + 2 f'' \cos \theta - 4 f' \sin \theta - 2 f \cos \theta + f \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 6 f'' \cos \theta + 2 f'' \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 4 f'' \cos \theta = 0$

The important observation is that the terms containing $f$ and $f'$ cancelled each other. This new differential equation doesn’t look like much of an improvement over the original fourth-order differential equation, but we can make a key observation: if $f'' = 0$ , then differentiating twice more trivially yields $f''' = 0$ and $f^{(4)} = 0$ . Said another way: if $f'' = 0$ , then $u(\theta) = f(\theta) \cos \theta$ will be a solution of the original differential equation.

Integrating twice, we can find $f$ :

$f''(\theta) = 0$

$f'(\theta) = c_1$

$f(\theta) = c_1 \theta + c_2$ .

Therefore, a solution of the original differential equation will be

$u(\theta) = c_1 \theta \cos \theta + c_2 \cos \theta$ .

We now repeat the logic for $u(\theta) = f(\theta) \sin \theta$ :

$(f \sin \theta)^{(4)} + 2 (f \sin \theta)'' + f \sin \theta = 0$

$f^{(4)} \sin \theta + 4 f''' \cos \theta - 6 f'' \sin \theta - 4 f' \cos\theta + f \sin \theta + 2 f'' \sin \theta + 4 f' \cos \theta - 2 f \sin \theta + f \sin \theta = 0$

$f^{(4)} \sin\theta + 4 f''' \cos \theta - 6 f'' \sin \theta + 2 f'' \sin \theta = 0$

$f^{(4)} \sin\theta - 4 f''' \cos\theta - 4 f'' \sin\theta = 0$ .

Once again, a solution of this new differential equation will be $f(\theta) = c_3 \theta + c_4$ , so that $f'' = f''' = f^{(4)} = 0$ . Therefore, another solution of the original differential equation will be

$u(\theta) = c_3 \theta \sin \theta + c_4 \sin \theta$ .

Adding these provides the general solution of the differential equation:

$u(\theta) = c_1 \theta \cos \theta + c_2 \cos \theta + c_3 \theta \sin \theta + c_4 \sin \theta$ .

Except for the order of the constants, this matches the solution that was presented earlier by using techniques taught in a proper course in differential equations.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6f: Rationale for Method of Undetermined Coefficients III

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In the previous post, I used a standard technique from differential equations to find the general solution of

$u^{(4)} + 2u'' + u = 0$ .

to be

$u(theta) = c_1 \cos \theta + c_2 \sin \theta + c_3 \theta \cos \theta + c_4 \theta \sin \theta$ .

However, as much as possible in this series, I want to take the perspective of a talented calculus student who has not yet taken differential equations — so that the conclusion above is far from obvious. How could this be reasonable coaxed out of such a student?

To begin, we observe that the characteristic equation is

$r^4 + 2r^2 + 1 = 0$ ,

$(r^2 + 1)^2 = 0$ .

Clearly this has the same roots as the simpler equation $r^2 + 1 = 0$ , which corresponds to the second-order differential equation $u'' + u = 0$ . We’ve already seen that $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ are solutions of this differential equation; perhaps they might also be solutions of the more complicated differential equation also? The answer, of course, is yes:

$u_1^{(4)} + 2 u_1'' + u_1 = \cos \theta - 2 \cos \theta + \cos \theta = 0$

and

$u_2^{(4)} + 2u_2'' + u_2 = \sin \theta - 2 \sin \theta + \sin \theta = 0$ .

The far trickier part is finding the two additional solutions. To find these, we use a standard trick/technique called reduction of order. In this technique, we guess that any additional solutions much have the form of either

$u(\theta) = f(\theta) \cos \theta \qquad \hbox{or} \qquad u(\theta) = f(\theta) \sin \theta$ ,

where $f(\theta)$ is some unknown function that we’re multiplying by the solutions we already have. We then substitute this into the differential equation $u^{(4)} + 2u'' + u = 0$ to form a new differential equation for the unknown $f$ , which we can (hopefully) solve.

Doing this will require multiple applications of the Product Rule for differentiation. We already know that

$(fg)' = f' g + f g'$ .

We now differentiate again, using the Product Rule, to find $(fg)''$ :

$(fg)'' = ( [fg]')' = (f'g)' + (fg')'$

$= f''g + f' g' + f' g' + f g''$

$= f'' g + 2 f' g' + f g''$ .

We now differential twice more to find $(fg)^{(4)}$ :

$(fg)''' = ( [fg]'')' = (f''g)' + 2(f'g')' + (fg'')'$

$= f'''g + f'' g' + 2f'' g' + 2f' g'' + f' g'' + f g'''$

$= f''' g + 3 f'' g' + 3 f' g'' + f g'''$ .

A good student may be able to guess the pattern for the next derivative:

$(fg)^{(4)} = ( [fg]''')' = (f'''g)' + 3(f''g')' +3(f'g'')' + (fg''')'$

$= f^{(4)}g + f''' g' + 3f''' g' + 3f'' g'' + 3f'' g'' + 3f'g''' + f' g''' + f g^{(4)}$

$= f^{(4)} g + 4 f''' g' + 6 f'' g'' + 4f' g''' + f g^{(4)}$ .

In this way, Pascal’s triangle makes a somewhat surprising appearance; indeed, this pattern can be proven with mathematical induction.

In the next post, we’ll apply this to the solution of the fourth-order differential equation.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6e: Rationale for Method of Undetermined Coefficients II

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In the previous post, we derived the method of undetermined coefficients for the simplified differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

In this post, we consider the simplified differential equation if the right-hand side has only the fourth term,

$u''(\theta) + u(\theta) = \displaystyle \frac{2\delta \epsilon }{\alpha^2}\cos \theta$ .

Let $v(\theta) = \displaystyle \frac{2\delta \epsilon }{\alpha^2}\cos \theta$ . Then $v$ satisfies the new differential equation $v'' + v = 0$ . Since $u'' + u = v$ , we may substitute:

$(u''+u)'' + (u'' + u) = 0$

$u^{(4)} + u'' + u'' + u = 0$

$u^{(4)} + 2u'' + u = 0$ .

The characteristic equation of this homogeneous differential equation is $r^4 + 2r^2 + 1 = 0$ , or $(r^2+1)^2 = 0$ . Therefore, $r = i$ and $r = -i$ are both double roots of this quartic equation. Therefore, the general solution for $u$ is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + c_3 \theta \cos \theta + c_4 \theta \sin \theta$ .

Substituting into the original differential equation will allow for the computation of $c_3$ and $c_4$ :

$u''(\theta) + u(\theta) = -c_1 \cos \theta - c_2 \sin \theta - 2c_3 \sin \theta - c_3 \theta \cos \theta + 2c_4 \cos \theta - c_4 \theta \sin \theta$

$+ c_1 \cos \theta + c_2 \sin \theta + c_3 \theta \cos \theta + c_4 \theta \sin \theta$

$\displaystyle \frac{2\delta \epsilon }{\alpha^2}\cos \theta = - 2c_3 \sin \theta+ 2c_4 \cos \theta$

Matching coefficients, we see that $c_3 = 0$ and $c_4 = \displaystyle \frac{\delta \epsilon }{\alpha^2}$ . Therefore,

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + \displaystyle \frac{\delta \epsilon }{\alpha^2} \theta \sin \theta$

is the general solution of the simplified differential equation. Setting $c_1 = c_2 = 0$ , we find that

$u(\theta) = \displaystyle \frac{\delta \epsilon }{\alpha^2} \theta \sin \theta$

is one particular solution of this simplified differential equation. Not surprisingly, this matches the result is the method of undetermined coefficients had been blindly followed.

As we’ll see in a future post, the presence of this $\theta \sin \theta$ term is what predicts the precession of a planet’s orbit under general relativity.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6d: Rationale for Method of Undetermined Coefficeints I

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

We now take the perspective of a student who is taking a first-semester course in differential equations. There are two standard techniques for solving a second-order non-homogeneous differential equations with constant coefficients. One of these is the method of constant coefficients. To use this technique, we first expand the right-hand side of the differential equation and then apply a power-reduction trigonometric identity:

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos^2 \theta}{\alpha^2}$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2}{\alpha^2} \frac{1 + \cos 2\theta}{2}$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

This is now in the form for using the method of undetermined coefficients. However, in this series, I’d like to take some time to explain why this technique actually works. To begin, we look at a simplified differential equation using only the first three terms on the right-hand side:

$u''(\theta) + u(\theta) = \displaystyle\frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

Let $v(\theta) =\displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ . Since $v$ is a constant, this function satisfies the simple differential equation $v' = 0$ . Since $u''+u=v$ , we can substitute:

$(u'' + u)' = 0$

$u''' + u' = 0$

(We could have more easily said, “Take the derivative of both sides,” but we’ll be using a more complicated form of this technique in future posts.) The characteristic equation of this differential equation is $r^3 + r = 0$ . Factoring, we obtain $r(r^2 + 1) = 0$ , so that the three roots are $r = 0$ and $r = \pm i$ . Therefore, the general solution of this differential equation is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + c_3$ .

Notice that this matches the outcome of blindly using the method of undetermined coefficients without conceptually understanding why this technique works.

The constants $c_1$ and $c_2$ are determined by the initial conditions. To find $c_3$ , we observe

$u''(\theta) +u(\theta) = -c_1 \cos \theta - c_2 \sin \theta +c_1 \cos \theta + c_2 \sin \theta + c_3$

$\displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} = c_3$ .

Therefore, the general solution of this simplified differential equation is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

Furthermore, setting $c_1 = c_2 = 0$ , we see that

$u(\theta) = \displaystyle\frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$

is a particular solution to the differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

In the next couple of posts, we find the particular solutions associated with the other terms on the right-hand side.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6c: Solving New Differential Equation with Variation of Parameters

Under general relativity, the motion of a planet around the Sun —in polar coordinates $(r,\theta)$ , with the Sun at the origin — satisfies the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

We now take the perspective of a student who is taking a first-semester course in differential equations. There are two standard techniques for solving a second-order non-homogeneous differential equations with constant coefficients. One of these is the method of variation of parameters. First, we solve the associated homogeneous differential equation

$u''(\theta) + u(\theta) = 0$ .

The characteristic equation of this differential equation is $r^2 + 1 = 0$ , which clearly has the two imaginary roots $r = \pm i$ . Therefore, two linearly independent solutions of the associated homogeneous equation are $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ .

(As an aside, this is one answer to the common question, “What are complex numbers good for?” The answer is naturally above the heads of Algebra II students when they first encounter the mysterious number $i$ , but complex numbers provide a way of solving the differential equations that model multiple problems in statics and dynamics.)

According to the method of variation of parameters, the general solution of the original nonhomogeneous differential equation

$u''(\theta) + u(\theta) = g(\theta)$

$u(\theta) = f_1(\theta) u_1(\theta) + f_2(\theta) u_2(\theta)$ ,

where

$f_1(\theta) = -\displaystyle \int \frac{u_2(\theta) g(\theta)}{W(\theta)} d\theta$ ,

$f_2(\theta) = \displaystyle \int \frac{u_1(\theta) g(\theta)}{W(\theta)} d\theta$ ,

and $W(\theta)$ is the Wronskian of $u_1(\theta)$ and $u_2(\theta)$ , defined by the determinant

$W(\theta) = \displaystyle \begin{vmatrix} u_1(\theta) & u_2(\theta) \\ u_1'(\theta) & u_2'(\theta) \end{vmatrix} = u_1(\theta) u_2'(\theta) - u_1'(\theta) u_2(\theta)$ .

Well, that’s a mouthful.

The only good news is that $W$ is easy to compute. Since $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ , we have

$W(\theta) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1$

from the usual Pythagorean trigonometric identity. Therefore, the denominators in the integrals for $f_1(\theta)$ and $f_2(\theta)$ essentially disappear.

Unfortunately, computing $f_1(\theta)$ and $f_2(\theta)$ , using

$g(\theta) = \displaystyle \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2$ ,

is a beast, requiring the creative use of multiple trigonometric identities. We begin with $f_1(\theta)$ , using the substitution $t = \cos \theta$ :

$f_1(\theta) = -\displaystyle \int \left[ \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2 \right] \sin \theta \, d\theta$

$= \displaystyle \int \left[ \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon t}{\alpha} \right)^2 \right] \, dt$

$= \displaystyle \frac{t}{\alpha} + \frac{\alpha\delta}{3\epsilon} \left( \frac{1 + \epsilon t}{\alpha} \right)^3 + a$

$= \displaystyle \frac{\cos \theta}{\alpha} + \frac{\alpha\delta}{3\epsilon} \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^3 + a$ ,

where we use $+a$ for the constant of integration instead of the usual $+C$ . Second,

$f_2(\theta) = \displaystyle \int \left[ \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2 \right] \cos\theta \, d\theta$ .

Unfortunately, this is not easily simplified with a substitution, so we have to expand the integrand:

$f_2(\theta) = \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta \cos \theta}{\alpha^2} + \frac{2 \delta \epsilon \cos^2 \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos^3 \theta}{\alpha^2} \right] \, d\theta$

$= \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta \cos \theta}{\alpha^2} + \frac{\delta \epsilon (1 + \cos 2 \theta)}{\alpha^2} + \frac{\delta \epsilon^2 \cos \theta \cos^2 \theta}{\alpha^2} \right] \, d\theta$

$= \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta \cos \theta}{\alpha^2} + \frac{\delta \epsilon}{\alpha^2} + \frac{\delta \epsilon \cos 2 \theta}{\alpha^2}+ \frac{\delta \epsilon^2 \cos \theta (1- \sin^2 \theta)}{\alpha^2} \right] \, d\theta$

$= \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta (1 + \epsilon^2) \cos \theta}{\alpha^2} + \frac{\delta \epsilon}{\alpha^2} + \frac{\delta \epsilon \cos 2 \theta}{\alpha^2} - \frac{\delta \epsilon^2 \cos \theta \sin^2 \theta}{\alpha^2} \right] \, d\theta$

$= \displaystyle \frac{\sin\theta}{\alpha} + \frac{\delta (1 + \epsilon^2) \sin\theta}{\alpha^2} + \frac{\delta \epsilon \theta}{\alpha^2} + \frac{\delta \epsilon \sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^3 \theta}{3\alpha^2} + b$ ,

using $+b$ for the constant of integration. Therefore, by variation of parameters, the general solution of the nonhomogeneous differential equation is

$u(\theta) = f_1(\theta) u_1(\theta) + f_2(\theta) u_2(\theta)$

$= \displaystyle \left( \frac{\cos \theta}{\alpha} + \frac{\delta}{3\alpha^2\epsilon} \left( 1 + \epsilon \cos \theta \right)^3 + a \right) \cos \theta$

$+ \displaystyle \left(\frac{\sin\theta}{\alpha} + \frac{\delta (1 + \epsilon^2) \sin\theta}{\alpha^2} + \frac{\delta \epsilon \theta}{\alpha^2} + \frac{\delta \epsilon \sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^3 \theta}{3\alpha^2} + b \right) \sin \theta$

$= a\cos \theta + b \sin \theta + \displaystyle \frac{\cos^2 \theta + \sin^2 \theta}{\alpha} + \frac{\delta \cos \theta}{3\alpha^2\epsilon} + \frac{\delta \cos^2 \theta}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2}+ \frac{\delta\epsilon^2 \cos^4 \theta}{3\alpha^2}$

$+ \displaystyle \frac{\delta \sin^2 \theta}{\alpha^2} + \frac{\delta \epsilon^2 \sin^2\theta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta\sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^4 \theta}{3\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon} + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta (\cos^2 \theta+\sin^2 \theta)}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2}+ \frac{\delta\epsilon^2 \cos^4 \theta}{3\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2 \sin^2\theta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta\sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^4 \theta}{3\alpha^2}$

$= a\cos \theta+ \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon} + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta\sin 2 \theta}{2\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2 \sin^2\theta}{\alpha^2} + \frac{\delta \epsilon^2 (\cos^4 \theta -\sin^4 \theta)}{3\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon} + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta \cdot 2 \sin \theta \cos \theta}{2\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2 (1-\cos 2\theta)}{2\alpha^2} + \frac{\delta \epsilon^2 (\cos^2 \theta +\sin^2\theta)(\cos^2 \theta - \sin^2 \theta)}{3\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2}{2\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2} + \frac{\delta \epsilon^2 \cos 2 \theta}{3\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon}+ b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

$\displaystyle + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2} + \frac{\delta \epsilon \sin^2 \theta\cos \theta}{\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon}+ b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

$\displaystyle + \frac{\delta \epsilon \cos \theta (\cos^2 \theta + \sin^2 \theta)}{\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon}+ b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2} + \frac{\delta \epsilon \cos \theta}{\alpha^2}$

$= \displaystyle \left(a + \displaystyle \frac{\delta}{3\alpha^2\epsilon} + \frac{\delta \epsilon}{\alpha^2} \right) \cos \theta + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

$= \displaystyle A \cos \theta + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$ ,

where $A$ is another arbitrary constant.

Next, we use the initial conditions to find the constants $A$ and $b$ . From the initial condition $u(0) = \displaystyle \frac{1}{P} = \frac{1+\epsilon}{\alpha}$ , we obtain

$u(0) = \displaystyle A \cos 0 + b \sin 0 + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \cdot 0 \cdot \sin 0}{\alpha^2} -\frac{\delta \epsilon^2 \cos 0}{6\alpha^2}$

$\displaystyle \frac{1+\epsilon}{\alpha} = A + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} -\frac{\delta \epsilon^2}{6\alpha^2}$

$\displaystyle \frac{\epsilon}{\alpha} = A + \displaystyle \frac{3\delta +\delta \epsilon^2}{3\alpha^2}$ ,

so that

$A = \displaystyle \frac{\epsilon}{\alpha} - \frac{\delta(3 + \epsilon^2)}{3\alpha^2}$ .

Next, we compute $u'(\theta)$ and use the initial condition $u'(0) = 0$ :

$u'(\theta) = \displaystyle -A \sin \theta + b \cos \theta + \frac{\delta \epsilon}{\alpha^2} (\sin \theta + \theta \cos \theta) + \frac{\delta \epsilon^2 \sin 2\theta}{3\alpha^2}$

$u'(0) = \displaystyle -A \sin 0 + b \cos 0 + \frac{\delta \epsilon}{\alpha^2} (\sin 0 + 0 \cos 0) + \frac{\delta \epsilon^2 \sin 0}{3\alpha^2}$

$0 = b$ .

Substituting these values for $A$ and $b$ , we finally arrive at the solution