1.14 Implicit Function Theorem
The general theorem says when a system of \(m\) equations can be solved for \(m\) of the variables in terms of the others, and the condition is that a Jacobian does not vanish. It contains Theorem 1.13.1 as the case \(m = n = 1\).
Theorem 1.14.1 (Implicit function theorem). For \(i = 1,2,\dots , m\), let the functions \(F_i(x_1,\dots ,x_n,\ y_1,\dots ,y_m)\) be defined in a neighbourhood of the point \[P_0 = (x_1^0,\dots ,x_n^0,\ y_1^0,\dots , y_m^0)\] and have continuous first partial derivatives there. Suppose that the \(m\) equations \(F_i = 0\) are all satisfied at \(P_0\), and that \[\frac {\partial (F_1,\dots ,F_m)}{\partial (y_1,\dots ,y_m)} \neq 0 \quad \text {at}\quad P_0 .\] Then in some neighbourhood of \((x^0_1,\dots , x^0_n)\) there is a unique set of continuous functions \[y_i = f_i (x_1, \dots ,x_n), \qquad i = 1,\dots , m,\] with \(f_i (x^0_1,\dots ,x^0_n) = y_i^0\) for each \(i\), such that \[F_i\big (x_1,\dots , x_n,\ f_1(x_1,\dots ,x_n),\dots , f_m(x_1,\dots ,x_n)\big ) = 0\] throughout that neighbourhood. The functions \(f_i\) have continuous first partial derivatives, given by the formulas of the previous subsection.
The theorem is a statement about linear approximation. Near \(P_0\) the equations are approximately linear in the \(y_i\), with coefficient matrix \(\left (\frac {\partial F_i}{\partial y_j}\right )\), and a linear system can be solved uniquely exactly when its determinant is non-zero. The proof, which belongs to a course in analysis, shows that the non-linear terms do not spoil this near \(P_0\).
Example 1.14.2. Let \(F(x,y,z) = x^2y^3 + 4xyz^8 - 5xyz^4 + z^2 + 1\). Show that the equation \(F = 0\) defines \(z\) as a function of \(x\) and \(y\) in a neighbourhood of the point \((2,-1,1)\), and compute \(\frac {\partial z}{\partial x}\) and \(\frac {\partial z}{\partial y}\) there.
Solution. First, the point lies on the surface: \(F(2,-1,1) = -4 - 8 + 10 + 1 + 1 = 0\). Next, \[F_z = 32xyz^7 - 20xyz^3 + 2z,\qquad F_z(2,-1,1) = -64 + 40 + 2 = -22 \neq 0 .\] So by Theorem 1.14.1, for \((x,y)\) in a neighbourhood of \((2,-1)\) there is a unique function \(z = f(x,y)\) with \(f(2,-1) = 1\) and \(F(x,y,f(x,y)) = 0\). With \[F_x = 2xy^3 + 4yz^8 - 5yz^4 \to -4 - 4 + 5 = -3,\qquad F_y = 3x^2y^2 + 4xz^8 - 5xz^4 \to 12 + 8 - 10 = 10\] at the point, \[\frac {\partial z}{\partial x} = -\frac {F_x}{F_z} = -\frac {-3}{-22} = -\frac {3}{22},\qquad \frac {\partial z}{\partial y} = -\frac {F_y}{F_z} = -\frac {10}{-22} = \frac {5}{11} .\]
Example 1.14.3. Let \[F = 2x^2 + y^2 + z^2 -zw = 0,\qquad G = x^2 + y^2 +2z^2 + zw- 8 = 0 .\]
- 1.
- Show that \(z\) and \(w\) can be written as functions of \(x\) and \(y\) in a neighbourhood of \((1,1,1,4)\).
- 2.
- Find \(\frac {\partial z}{\partial x}\) and \(\frac {\partial w}{\partial x}\) there.
Solution.
- 1.
- At \((1,1,1,4)\), \(F = 2 + 1 + 1 - 4 = 0\) and \(G = 1 + 1 + 2 + 4 - 8 = 0\), and \[\frac {\partial (F,G)}{\partial (z,w)} = \begin {vmatrix} F_z & F_w\\ G_z & G_w \end {vmatrix}=\begin {vmatrix} 2z - w & -z\\ 4z + w & z \end {vmatrix} = (2z - w)z + z(4z + w) = 6z^2 ,\] which is \(6\neq 0\) at the point. By Theorem 1.14.1, \(z\) and \(w\) are functions of \(x\) and \(y\) near \((1,1)\).
- 2.
- By the formulas of the previous subsection, \[\frac {\partial z}{\partial x} = -\frac {1}{6z^2}\begin {vmatrix} F_x & F_w\\ G_x & G_w \end {vmatrix} = -\frac {1}{6z^2}\begin {vmatrix} 4x & -z\\ 2x & z \end {vmatrix} = -\frac {6xz}{6z^2} = -\frac {x}{z},\] which is \(-1\) at \((1,1,1,4)\), and \[\frac {\partial w}{\partial x} = -\frac {1}{6z^2}\begin {vmatrix} F_z & F_x\\ G_z & G_x \end {vmatrix} = -\frac {1}{6z^2}\begin {vmatrix} 2z-w & 4x\\ 4z + w & 2x \end {vmatrix} = \frac {12xz + 6xw}{6z^2} = \frac {x(2z + w)}{z^2},\] which is \(6\) at the point.
Example 1.14.4. For the system of Example 1.14.3, find \(\frac {\partial z}{\partial y}\) and \(\frac {\partial w}{\partial y}\) at \((1,1,1,4)\), and use them to estimate \(z\) and \(w\) at \(x = 1\), \(y = 1.03\).
Solution. With \(F_y = 2y\) and \(G_y = 2y\), \[\frac {\partial z}{\partial y} = -\frac {1}{6z^2}\begin {vmatrix} 2y & -z\\ 2y & z\end {vmatrix} = -\frac {4yz}{6z^2} = -\frac {2y}{3z},\qquad \frac {\partial w}{\partial y} = -\frac {1}{6z^2}\begin {vmatrix} 2z - w & 2y\\ 4z + w & 2y\end {vmatrix} = \frac {2y(z + w)}{3z^2},\] which are \(-\frac 23\) and \(\frac {10}{3}\) at the point. So for \(y = 1.03\), \[z\approx 1 - \frac 23(0.03) = 0.98,\qquad w\approx 4 + \frac {10}{3}(0.03) = 4.10 .\]
Example 1.14.5. Show that the equation \(x^2 - y^3 = 0\) does not satisfy the hypothesis of Theorem 1.13.1 for \(y\) at the origin, but that for \(x\) it fails even more seriously.
Solution. With \(F = x^2 - y^3\), \(F_y = -3y^2 = 0\) at the origin, so the theorem does not apply to \(y\). Nevertheless \(y = x^{2/3}\) is a continuous solution through the origin; it is just not differentiable there. Solving for \(x\), \(F_x = 2x = 0\) at the origin too, and now there is no continuous solution at all near \(y = 0\): for \(y<0\) there are no points of the curve, and for \(y>0\) there are two, \(x = \pm y^{3/2}\). A vanishing Jacobian does not always mean that no solution exists, but it removes the guarantee.
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