Weather Map

Using polar coordinates we see that $p(x,y) \; = \; x^3 - x^2y + xy^2 - y^3 - x + y \; = \; (x-y)(x^2 + y^2 - 1) \; = \; r(r^2-1)(\cos\theta - \sin\theta) \; = \; \sqrt{2}r(r^2-1)\cos(\theta+\tfrac14\pi)$ Maximising the function of $r$ , we see that the extreme values of $p$ thus occur when $\theta = -\tfrac14\pi,\tfrac34\pi$ and $r = \tfrac{1}{\sqrt{3}}$ , and the extreme values of $p$ are $\pm\sqrt{\tfrac{8}{27}}$ . Thus the answer is $55.1 \times 2 \times \sqrt{\tfrac{8}{27}} \; = \; 59.98528215 \; \approx \; \boxed{60}$ hectopascals.

We consider the function $p(x,y) = x^3 - x^2 y + x y^2 - y^3 - x + y$ We search for zeros of the gradient $\vec \nabla p = \left( \begin{array}{c} \frac{\partial p}{\partial x} \\ \frac{\partial p}{\partial y} \end{array} \right) = \left( \begin{array}{c} 3 x^2 - 2 x y + y^2 - 1 \\ - x^2 + 2 x y - 3 y^2 + 1 \end{array} \right) \stackrel{!}{=} 0$ Adding both equation for the x- and y-component results $\frac{\partial p}{\partial x} + \frac{\partial p}{\partial y} = 2 (x^2 - y^2) = 0$ Therefore, $y = \pm x$ . We can exclude the case $y = x$ , because this is an äquipotential line $p(x,x) = 0$ , so that there are no extrema. We consider only the case $y = -x$ . The x-component of the gradient results $\begin{aligned} \left.\frac{\partial p}{\partial x}\right|_{y = -x} &= 3 x^2 + 2 x^2 + x^2 - 1 = 6x^2 - 1 = 0 \\ \Rightarrow \quad (x_\pm, y_\pm) &= \pm \left( \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}\right) \end{aligned}$ Evaluation of the function results $p\left(\frac{1}{\sqrt{5}},-\frac{1}{\sqrt{5}} \right) = - \left(\frac{2}{3}\right)^{3/2}, \qquad p\left(-\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}} \right) = \left(\frac{2}{3}\right)^{3/2}$ Therefore, the points correpond to the minimum and maximum, so that $\Delta p = p_\text{high} - p_\text{left} = 2 \left(\frac{2}{3}\right)^{3/2} \approx 1.089$ (Multiplying this with the prefactor 55.1 hPa results a pressure difference of 60 hPa.) The fact that a minimum and a maximum actually exist, can be proved by the eigenvalues of the Hessian matrix: $\begin{aligned} H(x_\pm, y_\pm) &= \left( \begin{array}{cc} \frac{\partial^2 p}{\partial x^2} & \frac{\partial p}{\partial x \partial y} \\ \frac{\partial p}{\partial x \partial y} & \frac{\partial^2 p}{\partial y^2} \end{array} \right) \\ & = \left( \begin{array}{cc} 6x - 2y & -2x + 2y \\ -2x + 2y & 2x - 6y \end{array} \right) \\ &= \pm 4 \left( \frac{2}{3} \right)^{3/2} \left( \begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right) \\ \left| \begin{array}{cc} 2- \lambda & -1 \\ -1 & 2 - \lambda\end{array} \right| &= \lambda^2 - 4\lambda + 3 \stackrel{!}{=} 0 \quad \Rightarrow \quad \lambda = 1, 3 \end{aligned}$ Therefore, the Hessian is for $(x_+, y_+)$ positive definite and for $(x_-, y_-)$ negative definite.