Change of Variables

In mathematics, a change of variables is a basic technique used to simplify problems in which the original variables are replaced with functions of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem.

Change of variables is an operation that is related to substitution. However these are different operations, as can be seen when considering differentiation (chain rule) or integration (integration by substitution).

A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth-degree polynomial:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^6 - 9 x^3 + 8 = 0.}

Sixth-degree polynomial equations are generally impossible to solve in terms of radicals (Abel–Ruffini theorem). This particular equation, however, may be written

$(x^{3})^{2}-9(x^{3})+8=0$

(this is a simple case of a polynomial decomposition). Thus the equation may be simplified by defining a new variable $u=x^{3}$. Substituting x by ${\sqrt[{3}]{u}}$ into the polynomial gives

$u^{2}-9u+8=0,$

which is just a quadratic equation with the two solutions:

$u=1\quad {\text{and}}\quad u=8.$

The solutions in terms of the original variable are obtained by substituting x³ back in for u, which gives

$x^{3}=1\quad {\text{and}}\quad x^{3}=8.$

Then, assuming that one is interested only in real solutions, the solutions of the original equation are

$x=(1)^{1/3}=1\quad {\text{and}}\quad x=(8)^{1/3}=2.$

Simple example

Consider the system of equations

$xy+x+y=71$

$x^{2}y+xy^{2}=880$

where $x$ and $y$ are positive integers with $x>y$. (Source: 1991 AIME)

Solving this normally is not very difficult, but it may get a little tedious. However, we can rewrite the second equation as $xy(x+y)=880$. Making the substitutions $s=x+y$ and $t=xy$ reduces the system to $s+t=71,st=880$. Solving this gives $(s,t)=(16,55)$ and $(s,t)=(55,16)$. Back-substituting the first ordered pair gives us $x+y=16,xy=55,x>y$, which gives the solution $(x,y)=(11,5).$ Back-substituting the second ordered pair gives us $x+y=55,xy=16,x>y$, which gives no solutions. Hence the solution that solves the system is $(x,y)=(11,5)$.

Formal introduction

Let $A$, $B$ be smooth manifolds and let $\Phi :A\rightarrow B$ be a $C^{r}$-diffeomorphism between them, that is: $\Phi$ is a $r$ times continuously differentiable, bijective map from $A$ to $B$ with $r$ times continuously differentiable inverse from $B$ to $A$. Here $r$ may be any natural number (or zero), $\infty$ (smooth) or $\omega$ (analytic).

The map $\Phi$ is called a regular coordinate transformation or regular variable substitution, where regular refers to the $C^{r}$-ness of $\Phi$. Usually one will write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \Phi(y)} to indicate the replacement of the variable Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} by the variable Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} by substituting the value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} for every occurrence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} .

Other examples

Coordinate transformation

Some systems can be more easily solved when switching to polar coordinates. Consider for example the equation

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U(x, y) := (x^2 + y^2) \sqrt{ 1 - \frac{x^2}{x^2 + y^2} } = 0.}

This may be a potential energy function for some physical problem. If one does not immediately see a solution, one might try the substitution

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle (x, y) = \Phi(r, \theta)} given by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle \Phi(r,\theta) = (r \cos(\theta), r \sin(\theta)).}

Note that if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} runs outside a $2\pi$-length interval, for example, $[0,2\pi ]$, the map Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} is no longer bijective. Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} should be limited to, for example Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0, \infty] \times [0, 2\pi)} . Notice how Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = 0} is excluded, for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} is not bijective in the origin (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} can take any value, the point will be mapped to (0, 0)). Then, replacing all occurrences of the original variables by the new expressions prescribed by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} and using the identity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin^2 x + \cos^2 x = 1} , we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(r, \theta) = r^2 \sqrt{ 1 - \frac{r^2 \cos^2 \theta}{r^2} } = r^2 \sqrt{1 - \cos^2 \theta} = r^2\left|\sin\theta\right|. }

Now the solutions can be readily found: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(\theta) = 0} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = 0} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = \pi} . Applying the inverse of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} shows that this is equivalent to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = 0} while Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \not= 0} . Indeed, we see that for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = 0} the function vanishes, except for the origin.

Note that, had we allowed Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = 0} , the origin would also have been a solution, though it is not a solution to the original problem. Here the bijectivity of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi} is crucial. The function is always positive (for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x,y\in\reals} ), hence the absolute values.

Differentiation

The chain rule is used to simplify complicated differentiation. For example, consider the problem of calculating the derivative

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac d {d x}\sin(x^2).}

Writing

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \sin u \quad \text{and} \quad u = x^2 }

we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \frac d {dx} \sin(x^2) = \overbrace{ \frac {dy}{dx} = \frac{dy}{du}\, \frac{du}{dx} }^{\text{This part is the chain rule.}} = \left( \frac d {du} \sin u \right) \left( \frac d {dx} x^2 \right) \\[8pt] = {} & \big( \cos u \big) (2x) = \cos(x^2) \cdot 2x. \end{align} }

Integration

Difficult integrals may often be evaluated by changing variables; this is enabled by the substitution rule and is analogous to the use of the chain rule above. Difficult integrals may also be solved by simplifying the integral using a change of variables given by the corresponding Jacobian matrix and determinant. Using the Jacobian determinant and the corresponding change of variable that it gives is the basis of coordinate systems such as polar, cylindrical, and spherical coordinate systems.

Differential equations

Variable changes for differentiation and integration are taught in elementary calculus and the steps are rarely carried out in full.

The very broad use of variable changes is apparent when considering differential equations, where the independent variables may be changed using the chain rule or the dependent variables are changed resulting in some differentiation to be carried out. Exotic changes, such as the mingling of dependent and independent variables in point and contact transformations, can be very complicated but allow much freedom.

Very often, a general form for a change is substituted into a problem and parameters picked along the way to best simplify the problem.

Scaling and shifting

Probably the simplest change is the scaling and shifting of variables, that is replacing them with new variables that are "stretched" and "moved" by constant amounts. This is very common in practical applications to get physical parameters out of problems. For an n^th order derivative, the change simply results in

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^n y}{d x^n} = \frac{y_\text{scale}}{x_\text{scale}^n} \frac{d^n \hat y}{d \hat x^n}}

where

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \hat x x_\text{scale} + x_\text{shift}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \hat y y_\text{scale} + y_\text{shift}.}

This may be shown readily through the chain rule and linearity of differentiation. This change is very common in practical applications to get physical parameters out of problems, for example, the boundary value problem

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu \frac{d^2 u}{d y^2} = \frac{d p}{d x} \quad ; \quad u(0) = u(L) = 0}

describes parallel fluid flow between flat solid walls separated by a distance δ; μ is the viscosity and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d p/d x} the pressure gradient, both constants. By scaling the variables the problem becomes

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2 \hat u}{d \hat y^2} = 1 \quad ; \quad \hat u(0) = \hat u(1) = 0}

where

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \hat y L \qquad \text{and} \qquad u = \hat u \frac{L^2}{\mu} \frac{d p}{d x}.}

Scaling is useful for many reasons. It simplifies analysis both by reducing the number of parameters and by simply making the problem neater. Proper scaling may normalize variables, that is make them have a sensible unitless range such as 0 to 1. Finally, if a problem mandates numeric solution, the fewer the parameters the fewer the number of computations.

Momentum vs. velocity

Consider a system of equations

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} m \dot v & = - \frac{ \partial H }{ \partial x } \\[5pt] m \dot x & = \frac{ \partial H }{ \partial v } \end{align} }

for a given function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(x, v)} . The mass can be eliminated by the (trivial) substitution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(p) = 1/m \cdot p} . Clearly this is a bijective map from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} . Under the substitution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v = \Phi(p)} the system becomes

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \dot p & = - \frac{ \partial H }{ \partial x } \\[5pt] \dot x & = \frac{ \partial H }{ \partial p } \end{align} }

Lagrangian mechanics

Given a force field Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi(t, x, v)} , Newton's equations of motion are

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m \ddot x = \varphi(t, x, v).}

Lagrange examined how these equations of motion change under an arbitrary substitution of variables Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \Psi(t, y)} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v = \frac{\partial \Psi(t, y)}{\partial t} + \frac{\partial\Psi(t, y)}{\partial y} \cdot w.}

He found that the equations

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{ \partial{L} }{ \partial y} = \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial{L}}{\partial{w}} }

are equivalent to Newton's equations for the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L = T - V} , where T is the kinetic, and V the potential energy.

In fact, when the substitution is chosen well (exploiting for example symmetries and constraints of the system) these equations are much easier to solve than Newton's equations in Cartesian coordinates.

Licensing

Content obtained and/or adapted from:

• Wikipedia under a CC BY-SA license