# Implicit function theorem

Implicit function theorem In mathematics, more specifically in multivariable calculus, the implicit function theorem[a] is a tool that allows relations to be converted to functions of several real variables. It does so by representing the relation as the graph of a function. There may not be a single function whose graph can represent the entire relation, but there may be such a function on a restriction of the domain of the relation. The implicit function theorem gives a sufficient condition to ensure that there is such a function.

More precisely, given a system of m equations fi (x1, ..., xn, y1, ..., ym) = 0, i = 1, ..., m (often abbreviated into F(x, y) = 0), the theorem states that, under a mild condition on the partial derivatives (with respect to the yis) at a point, the m variables yi are differentiable functions of the xj in some neighborhood of the point. As these functions can generally not be expressed in closed form, they are implicitly defined by the equations, and this motivated the name of the theorem.[1] In other words, under a mild condition on the partial derivatives, the set of zeros of a system of equations is locally the graph of a function.

Contents 1 History 2 First example 3 Definitions 4 Statement of the theorem 4.1 Higher derivatives 5 Proof for 2D case 6 The circle example 7 Application: change of coordinates 7.1 Example: polar coordinates 8 Generalizations 8.1 Banach space version 8.2 Implicit functions from non-differentiable functions 9 See also 10 Notes 11 References 12 Further reading History Augustin-Louis Cauchy (1789–1857) is credited with the first rigorous form of the implicit function theorem. Ulisse Dini (1845–1918) generalized the real-variable version of the implicit function theorem to the context of functions of any number of real variables.[2] First example The unit circle can be specified as the level curve f(x, y) = 1 of the function f(x, y) = x2 + y2. Around point A, y can be expressed as a function y(x). In this example this function can be written explicitly as {displaystyle g_{1}(x)={sqrt {1-x^{2}}};} in many cases no such explicit expression exists, but one can still refer to the implicit function y(x). No such function exists around point B.

If we define the function f(x, y) = x2 + y2, then the equation f(x, y) = 1 cuts out the unit circle as the level set {(x, y) | f(x, y) = 1}. There is no way to represent the unit circle as the graph of a function of one variable y = g(x) because for each choice of x ∈ (−1, 1), there are two choices of y, namely {displaystyle pm {sqrt {1-x^{2}}}} .

However, it is possible to represent part of the circle as the graph of a function of one variable. If we let {displaystyle g_{1}(x)={sqrt {1-x^{2}}}} for −1 ≤ x ≤ 1, then the graph of y = g1(x) provides the upper half of the circle. Similarly, if {displaystyle g_{2}(x)=-{sqrt {1-x^{2}}}} , then the graph of y = g2(x) gives the lower half of the circle.

The purpose of the implicit function theorem is to tell us the existence of functions like g1(x) and g2(x), even in situations where we cannot write down explicit formulas. It guarantees that g1(x) and g2(x) are differentiable, and it even works in situations where we do not have a formula for f(x, y).

Definitions Let {displaystyle f:mathbb {R} ^{n+m}to mathbb {R} ^{m}} be a continuously differentiable function. We think of {displaystyle mathbb {R} ^{n+m}} as the Cartesian product {displaystyle mathbb {R} ^{n}times mathbb {R} ^{m},} and we write a point of this product as {displaystyle (mathbf {x} ,mathbf {y} )=(x_{1},ldots ,x_{n},y_{1},ldots y_{m}).} Starting from the given function {displaystyle f} , our goal is to construct a function {displaystyle g:mathbb {R} ^{n}to mathbb {R} ^{m}} whose graph {displaystyle ({textbf {x}},g({textbf {x}}))} is precisely the set of all {displaystyle ({textbf {x}},{textbf {y}})} such that {displaystyle f({textbf {x}},{textbf {y}})={textbf {0}}} .

As noted above, this may not always be possible. We will therefore fix a point {displaystyle ({textbf {a}},{textbf {b}})=(a_{1},dots ,a_{n},b_{1},dots ,b_{m})} which satisfies {displaystyle f({textbf {a}},{textbf {b}})={textbf {0}}} , and we will ask for a {displaystyle g} that works near the point {displaystyle ({textbf {a}},{textbf {b}})} . In other words, we want an open set {displaystyle Usubset mathbb {R} ^{n}} containing {displaystyle {textbf {a}}} , an open set {displaystyle Vsubset mathbb {R} ^{m}} containing {displaystyle {textbf {b}}} , and a function {displaystyle g:Uto V} such that the graph of {displaystyle g} satisfies the relation {displaystyle f={textbf {0}}} on {displaystyle Utimes V} , and that no other points within {displaystyle Utimes V} do so. In symbols, {displaystyle {(mathbf {x} ,g(mathbf {x} ))mid mathbf {x} in U}={(mathbf {x} ,mathbf {y} )in Utimes Vmid f(mathbf {x} ,mathbf {y} )=mathbf {0} }.} To state the implicit function theorem, we need the Jacobian matrix of {displaystyle f} , which is the matrix of the partial derivatives of {displaystyle f} . Abbreviating {displaystyle (a_{1},dots ,a_{n},b_{1},dots ,b_{m})} to {displaystyle ({textbf {a}},{textbf {b}})} , the Jacobian matrix is {displaystyle (Df)(mathbf {a} ,mathbf {b} )=left[{begin{matrix}{frac {partial f_{1}}{partial x_{1}}}(mathbf {a} ,mathbf {b} )&cdots &{frac {partial f_{1}}{partial x_{n}}}(mathbf {a} ,mathbf {b} )\vdots &ddots &vdots \{frac {partial f_{m}}{partial x_{1}}}(mathbf {a} ,mathbf {b} )&cdots &{frac {partial f_{m}}{partial x_{n}}}(mathbf {a} ,mathbf {b} )end{matrix}}right|left.{begin{matrix}{frac {partial f_{1}}{partial y_{1}}}(mathbf {a} ,mathbf {b} )&cdots &{frac {partial f_{1}}{partial y_{m}}}(mathbf {a} ,mathbf {b} )\vdots &ddots &vdots \{frac {partial f_{m}}{partial y_{1}}}(mathbf {a} ,mathbf {b} )&cdots &{frac {partial f_{m}}{partial y_{m}}}(mathbf {a} ,mathbf {b} )\end{matrix}}right]=[X|Y]} where {displaystyle X} is the matrix of partial derivatives in the variables {displaystyle x_{i}} and {displaystyle Y} is the matrix of partial derivatives in the variables {displaystyle y_{j}} . The implicit function theorem says that if {displaystyle Y} is an invertible matrix, then there are {displaystyle U} , {displaystyle V} , and {displaystyle g} as desired. Writing all the hypotheses together gives the following statement.

Statement of the theorem Let {displaystyle f:mathbb {R} ^{n+m}to mathbb {R} ^{m}} be a continuously differentiable function, and let {displaystyle mathbb {R} ^{n+m}} have coordinates {displaystyle ({textbf {x}},{textbf {y}})} . Fix a point {displaystyle ({textbf {a}},{textbf {b}})=(a_{1},dots ,a_{n},b_{1},dots ,b_{m})} with {displaystyle f({textbf {a}},{textbf {b}})=mathbf {0} } , where {displaystyle mathbf {0} in mathbb {R} ^{m}} is the zero vector. If the Jacobian matrix (this is the right-hand panel of the Jacobian matrix shown in the previous section): {displaystyle J_{f,mathbf {y} }(mathbf {a} ,mathbf {b} )=left[{frac {partial f_{i}}{partial y_{j}}}(mathbf {a} ,mathbf {b} )right]} is invertible, then there exists an open set {displaystyle Usubset mathbb {R} ^{n}} containing {displaystyle {textbf {a}}} such that there exists a unique continuously differentiable function {displaystyle g:Uto mathbb {R} ^{m}} such that {displaystyle g(mathbf {a} )=mathbf {b} } , and {displaystyle f(mathbf {x} ,g(mathbf {x} ))=mathbf {0} ~{text{for all}}~mathbf {x} in U} . Moreover, denoting the left-hand panel of the Jacobian matrix shown in the previous section as: {displaystyle J_{f,mathbf {x} }(mathbf {a} ,mathbf {b} )=left[{frac {partial f_{i}}{partial x_{j}}}(mathbf {a} ,mathbf {b} )right],} the Jacobian matrix of partial derivatives of {displaystyle g} in {displaystyle U} are given by the matrix product:[3] {displaystyle left[{frac {partial g_{i}}{partial x_{j}}}(mathbf {x} )right]_{mtimes n}=-left[J_{f,mathbf {y} }(mathbf {x} ,g(mathbf {x} ))right]_{mtimes m}^{-1},left[J_{f,mathbf {x} }(mathbf {x} ,g(mathbf {x} ))right]_{mtimes n}} Higher derivatives If, moreover, {displaystyle f} is analytic or continuously differentiable {displaystyle k} times in a neighborhood of {displaystyle ({textbf {a}},{textbf {b}})} , then one may choose {displaystyle U} in order that the same holds true for {displaystyle g} inside {displaystyle U} . [4] In the analytic case, this is called the analytic implicit function theorem.

Proof for 2D case Suppose {displaystyle F:mathbb {R} ^{2}to mathbb {R} } is a continuously differentiable function defining a curve {displaystyle F(mathbf {r} )=F(x,y)=0} . Let {displaystyle (x_{0},y_{0})} be a point on the curve. The statement of the theorem above can be rewritten for this simple case as follows: Theorem — If {displaystyle left.{frac {partial F}{partial y}}right|_{(x_{0},y_{0})}neq 0} then for the curve around {displaystyle (x_{0},y_{0})} we can write {displaystyle y=f(x)} , where {displaystyle f} is a real function.

Proof. Since F is differentiable we write the differential of F through partial derivatives: {displaystyle mathrm {d} F=operatorname {grad} Fcdot mathrm {d} mathbf {r} ={frac {partial F}{partial x}}mathrm {d} x+{frac {partial F}{partial y}}mathrm {d} y.} Since we are restricted to movement on the curve {displaystyle mathrm {d} F=0} and by assumption {displaystyle {tfrac {partial F}{partial y}}neq 0} around the point {displaystyle (x_{0},y_{0})} (since {displaystyle {tfrac {partial F}{partial y}}} is continuous at {displaystyle (x_{0},y_{0})} and {displaystyle left.{tfrac {partial F}{partial y}}right|_{(x_{0},y_{0})}neq 0} ). Therefore we have a first-order ordinary differential equation: {displaystyle partial _{x}Fmathrm {d} x+partial _{y}Fmathrm {d} y=0,quad y(x_{0})=y_{0}} Now we are looking for a solution to this ODE in an open interval around the point {displaystyle (x_{0},y_{0})} for which, at every point in it, {displaystyle partial _{y}Fneq 0} . Since F is continuously differentiable and from the assumption we have {displaystyle |partial _{x}F|

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