Question 12.11.2: Consider the system of two nonlinear equations: u² + v = xy;......
Consider the system of two nonlinear equations:
u² + \nu = xy; u\nu = −x² + y²
(a) What has the counting rule to say about this system?
(b) Find the differentials of u and \nu expressed in terms of dx and dy. What are the partial derivatives of u and \nu w.r.t. x and y?
(c) The point P = (x, y, u, \nu) = (1, 0, 1,−1) satisfies the system. If x = 1 is increased by 0.01 and y = 0 is increased by 0.02, what is the new value of u, approximately?
(d) Calculate u^{\prime\prime}_{12} at the point P.
(a) There are four variables and two equations, so there should be two degrees of freedom. Suppose we choose fixed values for x and y. Then there are two equations for determining the two remaining variables, u and \nu. For example, if x = 1 and y = 0, then the system reduces to u² = −\nu and u\nu = −1, from which we find that u³ = 1, so u = 1 and \nu = −1. For other values of x and y, it is more difficult to find solutions for u and \nu. However, it seems reasonable to assume that the system defines u = u(x, y) and \nu = \nu(x, y) as differentiable functions of x and y, at least if the domain of the pair (x, y) is suitably restricted.
(b) The left- and right-hand sides of each equation in the system must be equal functions of x and y. So we can equate the differentials of each side to obtain d(u² + \nu) = d(xy) and d(u\nu) = d(−x² + y²). Using the rules for differentials, we obtain
2u du + d\nu = y dx + x dy
\nu du + u d\nu = −2x dx + 2y dy
Note that by the invariance property of the differential stated in Section 12.9, this system is valid no matter which pair of variables are independent. We want to solve the system for du and d\nu. There are two equations in the two unknowns du and d\nu of the form
A du + B d\nu = C
D du + E d\nu = F
where, for instance, A = 2u, C = y dx + x dy, and so on. Using Eq. (3.6.3), or standard elimination, provided that \nu \neq2u^{2}, we find that
x={\frac{c e-b f}{a e-b d}}\ \ {\mathrm{and}}\ \ y={\frac{a f-c d}{a e-b d}},\ \ {\mathrm{provided~that}}\ a e\not=b d (3.6.3)
\mathrm{d}u={\frac{2x+y u}{2u^{2}-\nu}}\,\mathrm{d}x+{\frac{x u-2y}{2u^{2}-\nu}}\,\mathrm{d}y, \mathrm{d}\nu={\frac{-4x u-y \nu}{2u^{2}-\nu}}\operatorname{d}\!x+{\frac{4u y-x \nu}{2u^{2}-v}}\operatorname{d}\!y
From the first of these two equations, we obtain immediately that
u_{1}^{\prime}={\frac{2x+y u}{2u^{2}-\nu}}\quad{\mathrm{and}}\quad u_{2}^{\prime}={\frac{x u-2y}{2u^{2}\nu}}Similarly, the partial derivatives of \nu w.r.t. x and y are the coefficients of dx and dy in the expression for d\nu. So we have found all the first-order partial derivatives.
(c) We use the approximation u(x + dx, y + dy) ≈ u(x, y) + du. At point P, where (x, y, u, \nu) = (1, 0, 1,−1), and so u_{1}^{\prime}={\frac{2}{3}},u_{2}^{\prime}={\frac{1}{3}}, we obtain
\quad\quad\quad\quad\quad\quad\quad\quad=1+{\frac{2}{3}}\cdot0.01+{\frac{1}{3}}\cdot0.02
\quad\quad\quad\quad\quad\quad\quad\quad=1+{\frac{4}{3}}\cdot0.01
\quad\quad\quad\quad\quad\quad\quad\quad\approx1.0133
Note that in this case, it is not easy to find the exact value of u(1.01, 0.02).
(d) We find u^{\prime\prime}_{12} by using the chain rule as follows:
u_{12}^{\prime\prime}=\frac{\partial}{\partial y}(u_{1}^{\prime})=\frac{\partial}{\partial y}\left(\frac{2x+y u}{2u^{2}-\nu}\right)=\frac{(y u_{2}^{\prime}+u)(2u^{2}-\nu)-(2x+y u)(4u u_{2}^{\prime}-\nu_{2}^{\prime})}{(2u^{2}-\nu)^{2}}At the point P where (x, y, u, \nu) = (1, 0, 1,−1), we obtain u^{\prime\prime}_{12}= 1/9.