Second order equations and superposition

An initial position no longer determines a trajectory once the equation involves acceleration, and everything built for first order equations has to be redone.

Mechanics forces the change. Newton's second law relates force to the second derivative of position, so the equation for a mass on a spring is $m x'' = - k x$ , and knowing where the mass is says nothing about where it goes next: it might be moving either way at any speed. Two numbers are needed, a position and a velocity, and that count is the visible signature of second order.

This lesson builds the structure that makes second order linear equations tractable. It produces no solutions at all, which is deliberate: the structure says how many solutions to look for and what to do once one is found, and the next lesson does the finding.

The operator and its linearity

Write the general second order linear equation in standard form, having divided by the coefficient of $y''$ :

y'' + p (t) y' + q (t) y = g (t)

It is convenient to name the left side. Define $L [y] = y'' + p y' + q y$ , a rule that takes a twice differentiable function and returns another function. The equation is then $L [y] = g$ , homogeneous when $g = 0$ and otherwise driven by the forcing $g$ .

The one property that matters is that $L$ is linear:

L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}]

which follows immediately from the derivative of a sum being the sum of derivatives, and from constants passing through derivatives. Everything in this lesson is a consequence of that identity, and nothing in it survives if the equation contains $y^{2}$ or $\sin y$ .

The first consequence is the principle of superposition: if $y_{1}$ and $y_{2}$ both solve $L [y] = 0$ , then so does $c_{1} y_{1} + c_{2} y_{2}$ for any constants, since $L$ of that combination is $c_{1} \cdot 0 + c_{2} \cdot 0$ . Solutions of the homogeneous equation can be added and scaled freely, which in the language of the linear algebra course makes them a vector space. The question that then decides everything is what its dimension is.

Existence and uniqueness, and what it buys

The relevant theorem is stronger than Picard's, precisely because linearity removes the possibility of blow-up. If $p$ , $q$ and $g$ are continuous on an open interval $I$ containing $t_{0}$ , then for any numbers $y_{0}$ and $v_{0}$ there is exactly one solution of $L [y] = g$ on the whole of $I$ with $y (t_{0}) = y_{0}$ and $y' (t_{0}) = v_{0}$ .

Three points. The interval is the whole of $I$ , not some unknown piece of it. Two numbers are required, matching the order. And uniqueness has an immediately useful corollary: a solution whose value and derivative both vanish at one point is identically zero, because the zero function is one solution with that data and there is only one.

The solution space has dimension exactly two

Take any two solutions $y_{1}$ , $y_{2}$ of the homogeneous equation, and ask when every solution can be written as $c_{1} y_{1} + c_{2} y_{2}$ .

Let $y$ be any solution, with $y (t_{0}) = y_{0}$ and $y' (t_{0}) = v_{0}$ . We want constants making the combination match those two numbers:

c_{1} y_{1} (t_{0}) + c_{2} y_{2} (t_{0}) = y_{0}, c_{1} y_{1}' (t_{0}) + c_{2} y_{2}' (t_{0}) = v_{0}

Two linear equations in two unknowns. They have a solution for every right hand side exactly when the determinant of the coefficients is not zero, and that determinant is

W (t_{0}) = y_{1} (t_{0}) y_{2}' (t_{0}) - y_{1}' (t_{0}) y_{2} (t_{0})

called the Wronskian. If $W (t_{0}) \neq 0$ the constants exist, the combination $c_{1} y_{1} + c_{2} y_{2}$ agrees with $y$ in value and derivative at $t_{0}$ , and uniqueness forces the two functions to be the same everywhere. So every solution is a combination of the two, the pair is a fundamental set, and $c_{1} y_{1} + c_{2} y_{2}$ deserves the name general solution.

The dimension is therefore two, no more and no less: two because two constants are needed to meet arbitrary initial data, and no more because those two suffice. This is why a second order equation is solved once two independent solutions are found, and why finding a third is a sign of an algebra error rather than a discovery.

Abel's identity sharpens the test. Differentiating $W = y_{1} y_{2}' - y_{1}' y_{2}$ and using the equation to replace both second derivatives gives $W' = - p (t) W$ , a separable equation, so $W = C e^{- \int p d t}$ . The exponential is never zero, so either $C = 0$ and the Wronskian vanishes identically, or it is never zero at all. Checking independence at one convenient point, usually $t = 0$ , settles it for the whole interval.

Example. Show that $y_{1} = \cos 2 t$ and $y_{2} = \sin 2 t$ form a fundamental set for $y'' + 4 y = 0$ .

Both are solutions: differentiating twice multiplies each by $- 4$ , and the equation holds. The Wronskian is

W = \cos 2 t \cdot 2 \cos 2 t - (- 2 \sin 2 t) \cdot \sin 2 t = 2 \cos^{2} 2 t + 2 \sin^{2} 2 t = 2

which is never zero, so the pair is independent and $y = c_{1} \cos 2 t + c_{2} \sin 2 t$ is the general solution. Note that $p = 0$ here, so Abel's identity predicts a constant Wronskian, which is what appeared.

Now you. Show that $y_{1} = e^{3 t}$ and $y_{2} = e^{- 2 t}$ form a fundamental set for $y'' - y' - 6 y = 0$ , and compute the Wronskian.

Answer

Substituting $e^{3 t}$ gives $9 - 3 - 6 = 0$ and $e^{- 2 t}$ gives $4 + 2 - 6 = 0$ , so both solve it. The Wronskian is $e^{3 t} (- 2 e^{- 2 t}) - 3 e^{3 t} e^{- 2 t} = - 5 e^{t}$ , never zero. Abel's identity agrees: $p = - 1$ , so $W$ is a constant times $e^{t}$ .

Fitting the initial conditions

With a fundamental set in hand, solving an initial value problem is two linear equations in two unknowns, every time.

Example. The general solution of $y'' - y' - 6 y = 0$ is $y = c_{1} e^{3 t} + c_{2} e^{- 2 t}$ . Solve it with $y (0) = 1$ and $y' (0) = 8$ .

Setting $t = 0$ : $c_{1} + c_{2} = 1$ . Differentiating, $y' = 3 c_{1} e^{3 t} - 2 c_{2} e^{- 2 t}$ , so $3 c_{1} - 2 c_{2} = 8$ . Substituting $c_{2} = 1 - c_{1}$ into the second gives $5 c_{1} = 10$ , so $c_{1} = 2$ and $c_{2} = - 1$ . The solution is $y = 2 e^{3 t} - e^{- 2 t}$ .

Now you. Same equation, with $y (0) = 4$ and $y' (0) = - 3$ .

Answer

$c_{1} + c_{2} = 4$ and $3 c_{1} - 2 c_{2} = - 3$ . Substituting gives $5 c_{1} = 5$ , so $c_{1} = 1$ and $c_{2} = 3$ , and $y = e^{3 t} + 3 e^{- 2 t}$ .

Reduction of order

Suppose one solution $y_{1}$ is known, by guesswork or from the physics. There is a systematic way to manufacture the second, due to d'Alembert: look for a solution of the form $y = v (t) y_{1} (t)$ , with $v$ a function rather than a constant.

Substituting is a short calculation. With $y = v y_{1}$ , $y' = v' y_{1} + v y_{1}'$ and $y'' = v'' y_{1} + 2 v' y_{1}' + v y_{1}''$ , so

L [y] = v'' y_{1} + v' (2 y_{1}' + p y_{1}) + v (y_{1}'' + p y_{1}' + q y_{1})

The last bracket is $L [y_{1}]$ , which is zero because $y_{1}$ solves the equation. That is the point of the method: the term in $v$ disappears, leaving an equation containing only $v''$ and $v'$ , which is a first order linear equation in the unknown $v'$ and therefore already solvable by the integrating factor.

Example. Verify that $y_{1} = t^{2}$ solves $t^{2} y'' - 3 t y' + 4 y = 0$ for $t > 0$ , and find a second independent solution.

Substituting $y_{1} = t^{2}$ gives $2 t^{2} - 6 t^{2} + 4 t^{2} = 0$ , so it is a solution. Put $y = v t^{2}$ . Then $y' = v' t^{2} + 2 t v$ and $y'' = v'' t^{2} + 4 t v' + 2 v$ , and the equation becomes

t^{4} v'' + 4 t^{3} v' + 2 t^{2} v - 3 t^{3} v' - 6 t^{2} v + 4 t^{2} v = t^{4} v'' + t^{3} v' = 0

The terms in $v$ cancel, as promised. Dividing by $t^{4}$ leaves $v'' + v' / t = 0$ , so $v' = C / t$ and $v = C \ln t$ . Taking $C = 1$ , the second solution is $y_{2} = t^{2} \ln t$ , and the general solution is $y = c_{1} t^{2} + c_{2} t^{2} \ln t$ . Their Wronskian is $t^{3}$ , nonzero for $t > 0$ .

Now you. Verify that $y_{1} = e^{2 t}$ solves $y'' - 4 y' + 4 y = 0$ , and find a second independent solution by reduction of order.

Answer

Substituting gives $4 - 8 + 4 = 0$ . With $y = v e^{2 t}$ , $y' = (v' + 2 v) e^{2 t}$ and $y'' = (v'' + 4 v' + 4 v) e^{2 t}$ , so the equation becomes $(v'' + 4 v' + 4 v - 4 v' - 8 v + 4 v) e^{2 t} = v'' e^{2 t} = 0$ . Hence $v'' = 0$ , $v = A + B t$ , and the new solution is $y_{2} = t e^{2 t}$ .

That last result is worth holding on to. An equation whose characteristic algebra will turn out to have a repeated root produces a second solution with an extra factor of $t$ , and reduction of order is where that factor comes from rather than being pulled out of a hat.

Adding forcing

Now let $g$ be present. Suppose $y_{p}$ is any single solution of $L [y] = g$ , called a particular solution, and let $y$ be any other. Then by linearity

L [y - y_{p}] = L [y] - L [y_{p}] = g - g = 0

so the difference solves the homogeneous equation. Therefore every solution of the driven equation is

y = y_{p} + c_{1} y_{1} + c_{2} y_{2}

one particular response plus the general homogeneous solution, which is called the complementary function. This is the same split that appeared for first order equations as steady state plus transient, and it is now a theorem rather than an observation about one formula.

The practical consequences are two. The initial conditions must be applied to the whole expression, never to the complementary function alone, which is the single commonest error in this subject. And superposition extends to forcing: if $L [y_{1}] = g_{1}$ and $L [y_{2}] = g_{2}$ , then $L [y_{1} + y_{2}] = g_{1} + g_{2}$ , so a complicated forcing can be split into pieces, each handled separately and the results added. A circuit driven by a battery and a sinusoid can be analysed twice and the answers summed, which is the whole basis of frequency domain engineering.

What is still missing

Nothing so far produces a single solution. The theory says that two independent ones exist, that the Wronskian recognises them, that a third would be redundant, and that reduction of order gets the second from the first. It does not say how to get the first.

For general $p (t)$ and $q (t)$ there is no method, and even innocuous cases have no elementary solutions: $y'' + t y = 0$ , the Airy equation, needs power series and defines new functions. But one case covers most of physics and all of elementary circuit theory, the case of constant coefficients, and there a single guess turns the differential equation into a quadratic. That is the next lesson.