1.17 Functional Dependence
Definition 1.17.1. Two functions \(u(x,y)\) and \(v(x,y)\) are functionally dependent on a region if there is a function \(\Phi \), not identically zero on any open set, with \(\Phi (u(x,y),v(x,y)) = 0\) throughout the region; for example, \(u = g(v)\) or \(v = h(u)\).
Dependent functions carry the same information: knowing one determines the other. The Jacobian detects this.
Theorem 1.17.2. If \(u\) and \(v\) have continuous partial derivatives and are functionally dependent through a differentiable \(\Phi \) whose gradient is never zero, then \[\frac {\partial (u,v)}{\partial (x,y)} = 0\] throughout the region. Conversely, if this Jacobian vanishes identically on an open region in which \(\nabla u\neq \mathbf 0\), then \(u\) and \(v\) are functionally dependent there.
Proof of the first part. Differentiating \(\Phi (u,v) = 0\) with respect to \(x\) and \(y\), \[\Phi _u\,u_x + \Phi _v\,v_x = 0,\qquad \Phi _u\,u_y + \Phi _v\,v_y = 0,\] that is, \[\begin {pmatrix} u_x & v_x\\ u_y & v_y \end {pmatrix} \begin {pmatrix} \Phi _u\\ \Phi _v \end {pmatrix} = \begin {pmatrix} 0\\ 0\end {pmatrix}.\] The vector \((\Phi _u,\Phi _v)\) is never zero, so this homogeneous system has a non-trivial solution, and its determinant must vanish. That determinant is the transpose of the Jacobian matrix, and \(\det A^{\mathsf T} = \det A\), so \(\frac {\partial (u,v)}{\partial (x,y)} = 0\). □
For the converse, the condition \(\nabla u\neq \mathbf 0\) allows \(u\) to be used as a new coordinate near each point. The vanishing Jacobian then says that \(v\) does not change along the level curves of \(u\), so \(v\) is a function of \(u\) alone. The same results hold for \(n\) functions of \(n\) variables, with the \(n\times n\) Jacobian.
Example 1.17.3. Determine whether the following are functionally dependent: \[u = y + z,\qquad v = x + 2z^2,\qquad w = x - 4yz - 2y^2 .\]
Solution. \[\frac {\partial (u,v,w)}{\partial (x,y,z)} = \begin {vmatrix} 0 & 1 & 1\\ 1 & 0 & 4z\\ 1 & -4z - 4y & -4y \end {vmatrix} = 0 - 1\,(-4y - 4z) + 1\,(-4z - 4y) = 0 .\] So \(u\), \(v\), \(w\) are functionally dependent. To find the relation, note that \(v - w = 2z^2 + 4yz + 2y^2 = 2(y + z)^2 = 2u^2\), so \(w = v - 2u^2\).
Example 1.17.4. Determine whether \(f(x,y) = e^x \sin y\) and \(g(x,y) = x + \ln \sin y\) are functionally dependent on the region \(0<y<\pi \).
Solution. \[\frac {\partial (f,g)}{\partial (x,y)} = \begin {vmatrix} e^x\sin y & e^x\cos y\\ 1 & \cot y \end {vmatrix} = e^x\sin y\cot y - e^x\cos y = 0 ,\] so \(f\) and \(g\) are functionally dependent; indeed \(\ln f = x + \ln \sin y = g\).
Example 1.17.5. Show that \(u = x + y\) and \(v = x - y\) are independent, and find all \((x,y)\) at which \(u = x^2 + y^2\) and \(v = xy\) fail the Jacobian test.
Solution. For the first pair, \(\frac {\partial (u,v)}{\partial (x,y)} = \begin {vmatrix} 1 & 1\\ 1 & -1\end {vmatrix} = -2\neq 0\), so they are independent everywhere. For the second, \[\frac {\partial (u,v)}{\partial (x,y)} = \begin {vmatrix} 2x & 2y\\ y & x\end {vmatrix} = 2(x^2 - y^2),\] which vanishes only on the lines \(y = \pm x\). Since it is not identically zero, \(u\) and \(v\) are independent, but on those lines they are locally related: for \(y = x\), \(u = 2x^2 = 2v\). These are exactly the lines where the mapping \((x,y)\mapsto (u,v)\) fails to be locally invertible, folding the plane over.
Example 1.17.6. Show that \(u = \frac {x - y}{1 + xy}\) and \(v = \tan ^{-1}x - \tan ^{-1}y\) are functionally dependent, and find the relation.
Solution. \[u_x = \frac {(1 + xy) - (x - y)y}{(1 + xy)^2} = \frac {1 + y^2}{(1 + xy)^2}, \qquad u_y = \frac {-(1 + xy) - (x - y)x}{(1 + xy)^2} = -\frac {1 + x^2}{(1 + xy)^2},\] and \(v_x = \frac {1}{1 + x^2}\), \(v_y = -\frac {1}{1 + y^2}\), so \[\frac {\partial (u,v)}{\partial (x,y)} = -\frac {1 + y^2}{(1 + xy)^2(1 + y^2)} + \frac {1 + x^2}{(1 + xy)^2(1 + x^2)} = 0 .\] By the subtraction formula for the tangent, \(\tan v = \frac {x - y}{1 + xy} = u\).
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