Factorization method for solving differential equations

Description of the method

The factorization method is a method that can be used to solve certain kinds of differential equations. The idea behind the method is to start with a differential equation:

$F(x,y,y^{\prime },\dots ,y^{(k)})=0$

and try to factor the expression $F$ as a product of two expressions, say $G$ and $H$. The differential equation now becomes:

$G(x,y,y^{\prime },\dots ,y^{(k)})H(x,y,y^{\prime },\dots ,y^{(k)})=0$

We now consider the two equations:

$G(x,y,y^{\prime },\dots ,y^{(k)})=0$

$H(x,y,y^{\prime },\dots ,y^{(k)})=0$

It is clear that any functional (respectively, relational) solution for either of the equations is a functional (respectively, relational) solution for $F$. Conversely, any functional (respectively, relational) solution for $F$ must, at every point in the domain, satisfy one of the differential equations. However, we need to worry about the possibility of a functional solution for $F$ that is a solution for $G$ on part of its domain and a solution for $H$ on part of its domain. It is quite rare for these "mixed" solutions to occur, but it does make sense to check for them.

The general solution for $F$ can thus be expressed as:

(General solution for $G$) $\bigcup$ (General solution for $H$) $\bigcup$ (Any solution for $F$ that is a solution for $G$ on part of the domain and a solution for $H$ on part of the domain)

Applicability

The factorization method is limited to situations where the expression can be factorized. Further, even if factorization is possible, there is no guarantee that the factor differential equations can be solved. Thus, the method has limited applicability.

However, there are some situations where this method can be used reliably:

• Using the quadratic formula, it can help convert second-degree differential equations to first-degree differential equations. In other words, if the differential equation is quadratic in the highest order derivative, we could factor it as a product of differential equations that are linear in the highest order derivative using the quadratic formula. The quadratic formula involves radicals, so this may come at the expense of introducing radicals into the differential equation.
• In particular, any first-order second-degree autonomous differential equation can be factored to give two first-order first-degree autonomous differential equations, which we have a general method for solving.

Counting degrees of freedom

The total number of degrees of freedom (i.e., freely varying parameters in the general solution) for $F$ is generally taken to be the maximum of the number of degrees of freedom for each of the solution families. In particular, assuming there are no mixed solutions, the number of degrees of freedom for $F$ is the maximum of the number of degrees of freedom for $G$ and the number of degrees of freedom for $H$.

The reason we take the maximum instead of the sum is that the freely varying parameters for the separate solution families are not interacting with each other. Geometrically, this is related to the idea that when we take the union of two one-dimensional things (lines or curves) we still get a one-dimensional thing (line or curve).

Examples

Example without mixed solutions

Consider the differential equation:

$(y^{\prime }{)}^2-3y{y}^{\prime }+2y^2=0$

This can be factored to give:

$(y^{\prime }-y)(y^{\prime }-2y)=0$

Thus, the differential equations we need to solve individually are:

$y^{\prime }=y,y^{\prime }=2y$

The solutions are respectively:

$y=C_1{e}^x,y=C_2{e}^{2x},C_1,C_2\in \mathbb{R}$

The general solution includes both solution families. Let's now consider the possibility of a mixed solution, i.e., a function $y=f(x)$ such that $y^{\prime }=y$ for part of the domain and $y^{\prime }=2y$ for the rest of the domain. At a transition point, we must have $y^{\prime }=y=2y$ forcing $y=0$. But in either solution, setting $y=0$ forces the solution identically to be the zero function. This actually can be realized completely as a solution in either family, hence the only mixed solution actually lies as a pure solution in both families. The general solution is thus a union of two solution families:

• $y=C{e}^x,C\in \mathbb{R}$
• $y=C{e}^{2x},C\in \mathbb{R}$

Example with mixed solutions

Consider the differential equation:

$(y^{\prime }-2x)(y^{\prime }-3x^2)=0$

We want to find all possible functional solutions for this on all of $\mathbb{R}$.

The solution families corresponding to the individual factors are:

$y=x^2+C_1,y=x^3+C_2,C_1,C_2\in \mathbb{R}$

We now need to consider the possibility of mixed solutions. A mixed solution would be a function that is of the form $x^2+C_1$ on part of the domain and $x^3+C_2$ on the rest of the domain, possibly even transitioning multiple times between the two solution forms. Suppose $x=x_0$ is a transition point. Then, the $y$-values and $y^{\prime }$-values for both functions must agree at the point, so:

$x_0^2+C_1=x_0^3+C_2,2x_0=3x_0^2$

Solving the second equation, we get that $x_0=0$ or $x_0=2/3$.

• Case of a single transition point $x_0=0$: In this case, we get $C_1=C_2$, so the function description for $y$ could be either of these, with one solution of each type for each $C_1\in \mathbb{R}$:

$y=\{\begin{array}{cc}{x}^2+C_1,& x\le 0\\ x^3+C_1,& x>0\\ \end{array}$

or a transition in the opposite sense:

$y=\{\begin{array}{cc}{x}^3+C_1,& x\le 0\\ x^2+C_1,& x>0\\ \end{array}$

• Case of a single transition point $x_0=2/3$: In this case, $C_2=C_1+4/27$, so the function description could be either of these, with one solution of each type for each $C_1\in \mathbb{R}$:

$y=\{\begin{array}{cc}{x}^2+C_1,& x\le 2/3\\ x^3+C_1+4/27,& x>2/3\\ \end{array}$

or a transition in the opposite sense:

$y=\{\begin{array}{cc}{x}^3+C_1+4/27,& x\le 2/3\\ x^2+C_1,& x>2/3\\ \end{array}$

• Case of two transition points $x_0=0$ and $x_0=2/3$: This situation is trickier, The first type of transition is as follows:

$y=\{\begin{array}{cc}{x}^2+C_1,& x\le 0\\ x^3+C_2,& 0<x\le 2/3\\ x^2+C_3,& x>2/3\\ \end{array}$

We get the relations $C_1=C_2=C_3+4/27$, so this becomes:

$y=\{\begin{array}{cc}{x}^2+C_3+4/27,& x\le 0\\ x^3+C_3+4/27,& 0<x\le 2/3\\ x^2+C_3,& x>2/3\\ \end{array}$

The other possibility is:

$y=\{\begin{array}{cc}{x}^3+C_2,& x\le 0\\ x^2+C_1,& 0<x\le 2/3\\ x^3+C_3,& x>2/3\\ \end{array}$

In this case, we get $C_2=C_1$ and $C_3=C_1+(4/27)$, so plugging in, we get:

$y=\{\begin{array}{cc}{x}^3+C_1,& x\le 0\\ x^2+C_1,& 0<x\le 2/3\\ x^3+C_1+4/27,& x>2/3\\ \end{array}$

Thus, we overall have eight solution families, each with one freely varying real parameter (which we will now just call $C$ instead of $C_1,C_2{C}_3$):

• $y=x^2+C,C\in \mathbb{R}$
• $y=x^3+C,C\in \mathbb{R}$
• $y=\{\begin{array}{cc}{x}^2+C,& x\le 0\\ x^3+C,& x>0\\ \end{array},C\in \mathbb{R}$
• $y=\{\begin{array}{cc}{x}^3+C,& x\le 0\\ x^2+C,& x>0\\ \end{array},C\in \mathbb{R}$
• $y=\{\begin{array}{cc}{x}^2+C,& x\le 2/3\\ x^3+C+4/27,& x>2/3\\ \end{array},C\in \mathbb{R}$
• $y=\{\begin{array}{cc}{x}^3+C+4/27,& x\le 2/3\\ x^2+C,& x>2/3\\ \end{array},C\in \mathbb{R}$
• $y=\{\begin{array}{cc}{x}^2+C+4/27,& x\le 0\\ x^3+C+4/27,& 0<x\le 2/3\\ x^2+C,& x>2/3\\ \end{array},C\in \mathbb{R}$
• $y=\{\begin{array}{cc}{x}^3+C,& x\le 0\\ x^2+C,& 0<x\le 2/3\\ x^3+C+4/27,& x>2/3\\ \end{array},C\in \mathbb{R}$

Factoring using quadratic formula: example with mixed solutions plus partly stationary solutions

Consider the differential equation with independent variable $x$ and dependent variable $y$:

$(y^{\prime }{)}^2+y^2=1$

This is a first-order second-degree autonomous differential equation.

Rewrite as:

$(y^{\prime }{)}^2-(1-y^2)=0$

This does not factor in purely polynomial terms. However, if we allow the use of radicals, we can rewrite it as:

$(y^{\prime }-\sqrt{1-y^2})(y^{\prime }+\sqrt{1-y^2})=0$

The two factor differential equations are:

$y^{\prime }=\sqrt{1-y^2},y^{\prime }=-\sqrt{1-y^2}$

Each of them is first-order first-degree autonomous, and in particular separable. The solutions are:

$\arcsin y=x+C_1,\arcsin y=-x+C_2,C_1,C_2\in \mathbb{R}$

Also, both equations have stationary solutions $y=1$ and $y=-1$.

The following is the general solution and includes all the nowhere stationary functional solutions.

$y=\sin (x+C),C\in \mathbb{R}$

If we want to include somewhere stationary solutions, then the space of solutions becomes complicated. Basically, we could have a function that is an increasing sinusoidal function for part of the domain, then stationary at 1 for a while, then a decreasing sinusoidal function till it reaches -1, then stationary at -1 for a while, and so on.

Theory

Differentiability assumptions and mixed solutions

One way of trying to eliminate mixed solutions is to impose higher order differentiability conditions on the solutions. The rationale is that mixed solutions involve transitioning from one solution type to another. This transition is often not going to be completely smooth, so there may be some derivative that is undefined at the point of transition. Imposing differentiability conditions on the solutions eliminates such mixed solutions.

For instance, in the example above:

$(y^{\prime }-2x)(y^{\prime }-3x^2)=0$

all the eight solution families (pure and mixed) are once continuously differentiable on all of $\mathbb{R}$. The pure solution families ($x^2+C,x^3+C$) are infinitely differentiable everywhere. However, we can check that none of the mixed solution families is twice differentiable. The reason is that at both the potential transition points (0 and 2/3) the second derivatives do not match:

• At 0, the second derivative of $x^2+C$ is 2 and the second derivative of $x^3+C$ is 0.
• At 2/3, the second derivative of $x^2+C$ is 2 and the second derivative of $x^3+C$ is 4.

So, in this case, a twice continuous differentiability requirement (i.e., solutions are required to be $C^2$ functions) on the solutions eliminates all the mixed solution families. We may not always be this lucky. For instance, the differential equation:

$(y^{\prime }-4x^3)(y^{\prime }-5x^4)=0$

admits a transition point 0 where the first three derivatives of both the pure solution families agree, which means that even a requirement of thrice continuously differentiable (i.e., $C^3$) on the solution functions would still allow for mixed solutions with transition at 0. The other transition point (4/5), on the other hand, is eliminated by a twice continuous differentiability requirement.

A $C^4$ requirement (i.e., four times continuously differentiable) would eliminate both the possible transition points, and hence all the mixed solutions, leaving only the two pure solution families.