1.17 Functional Dependence
Let \(u(x,y)\) and \(v(x,y)\). If \(u = g(v)\) or \(v = h(u)\), then \(u\) and \(v\) are functionally dependent. If functional dependence between \(u\) and \(v\) exists, then we can consider \(f(u,v) = 0\).
\begin {align*} \text {So}\hspace {1cm} \frac {\partial f}{\partial u}\cdot \frac {\partial u}{\partial x} + \frac {\partial f}{\partial v}\cdot \frac {\partial v}{\partial x} & = 0\\\\ \frac {\partial f}{\partial u}\cdot \frac {\partial u}{\partial y} + \frac {\partial f}{\partial v}\cdot \frac {\partial v}{\partial y} & = 0\\ \end {align*}
\[\begin {pmatrix} \dfrac {\partial u}{\partial x} & \dfrac {\partial v}{\partial x}\\\\ \dfrac {\partial u}{\partial y} & \dfrac {\partial v}{\partial y}\\ \end {pmatrix} \begin {pmatrix} \dfrac {\partial f}{\partial u}\\\\ \dfrac {\partial f}{\partial v}\\ \end {pmatrix} = 0 \]
Since we desire a non-trivial solution, the determinant of the coefficient matrix must be zero for functional dependency. \[\text {i.e}\hspace {1cm} \begin {vmatrix} \dfrac {\partial u}{\partial x} & \dfrac {\partial v}{\partial x}\\\\ \dfrac {\partial u}{\partial y} & \dfrac {\partial v}{\partial y}\\ \end {vmatrix} = 0\]
but \(\det A = \det A^T\)
\(\therefore \hspace {0.4cm}\) the equation is equivalent \[ \begin {vmatrix} \dfrac {\partial u}{\partial x} & \dfrac {\partial v}{\partial x}\\\\ \dfrac {\partial u}{\partial y} & \dfrac {\partial v}{\partial y}\\ \end {vmatrix} = 0 \iff J = \frac {\partial (u,v)}{\partial (x,y)} = 0 \]
Determine if the following are functionally dependent. \begin {align*} u & = x + z\\ v & = x + 2z^2\\ w & = x - 4yz - 2y^2\\ \end {align*}
Solution. \begin {align*} \frac {\partial (u,v,w)}{\partial (x,y,z)} & = \begin {vmatrix} \dfrac {\partial u}{\partial x} & \dfrac {\partial u}{\partial y} & \dfrac {\partial u}{\partial z}\\\\ \dfrac {\partial v}{\partial x} & \dfrac {\partial v}{\partial y} & \dfrac {\partial v}{\partial z}\\\\ \dfrac {\partial w}{\partial x} & \dfrac {\partial w}{\partial y} & \dfrac {\partial w}{\partial z}\\ \end {vmatrix} = \begin {vmatrix} 0 & 1 & 1\\ 1 & 0 & 4z\\ 1 & -4z - 4y & -4y\\ \end {vmatrix}\\\\ & = 0 \end {align*}
\(\therefore \hspace {0.4cm} u,v,w\) are functionally dependent, in fact \(w = v - 2u^2\) (check).
Determine if the following are functionally dependent. \begin {align*} f(x,y) & = e^x \sin y\\ g(x,y) & = x + \log \sin y\\ \end {align*}
Solution. \begin {align*} \text {So}\hspace {1cm} \frac {\partial (f,g)}{\partial (x,y)} & = \begin {vmatrix} \dfrac {\partial f}{\partial x} & \dfrac {\partial f}{\partial y}\\\\ \dfrac {\partial g}{\partial x} & \dfrac {\partial g}{\partial y}\\ \end {vmatrix} = \begin {vmatrix} e^x\sin y & e^x\cos y\\ 1 & \cot y\\ \end {vmatrix}\\\\ & = e^x\cos y - e^x\cos y\\ & = 0 \end {align*}
\(f\) and \(g\) are functionally dependent, in fact \((\log f - g = 0)\).
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