# Chapter 10 Applications of Multivariate Calculus

In Chapter 9, we introduced multivariable functions and notions of differentiation. In this chapter, we present several applications of multivariate calculus.

## 10.1 Change of Coordinates

In lots of situations, one may need a change of coordinates, which in general could be a nonlinear transformation. In this section, we focus on the familiar example of the change of coordinates from polar coordinates ($$\vec{x}= (r, \theta)$$) to cartesian ($$\vec{x} = (x, y)$$) and vice versa to illustrate the concepts. We generalise these results at the end of the section.

Let $$x = x(r, \theta)$$ and $$y = y(r, \theta)$$, we have: $x = r\cos{\theta}, \quad y = r \sin{\theta}.$ Conversely, we have for $$r = r(x, y)$$ and $$\theta = \theta(x, y)$$: $r = \sqrt{x^2 + y^2}, \quad \theta = \arctan{(\frac{y}{x})}.$ Using total differentiation, we can obtain a relationship between the vectors of infinitesimal change $$\vec{dx}$$ and $$\vec{dr}$$. $\begin{eqnarray*} dx = \left( \dfrac{\partial x}{\partial r}\right)_\theta dr + \left( \dfrac{\partial x}{\partial \theta}\right)_r d\theta, \\ dy = \left( \dfrac{\partial y}{\partial r}\right)_\theta dr + \left( \dfrac{\partial y}{\partial \theta}\right)_r d\theta. \end{eqnarray*}$ We can write this in vector-matrix form as $\begin{bmatrix} dx\\ dy \end{bmatrix} = \begin{bmatrix} \left( \dfrac{\partial x}{\partial r}\right)_\theta & \left( \dfrac{\partial x}{\partial \theta}\right)_r\\ \left( \dfrac{\partial y}{\partial r}\right)_\theta & \left( \dfrac{\partial y}{\partial \theta}\right)_r \end{bmatrix} \begin{bmatrix} dr\\ d\theta \end{bmatrix}.$ By defining matrix $$J$$ known as Jacobian of the transformation as below we can write this relationship as $$\vec{dx} = J \vec{dr}$$ in short. $J = \begin{bmatrix} \left( \dfrac{\partial x}{\partial r}\right)_\theta & \left( \dfrac{\partial x}{\partial \theta}\right)_r\\ \left( \dfrac{\partial y}{\partial r}\right)_\theta & \left( \dfrac{\partial y}{\partial \theta}\right)_r \end{bmatrix} = \begin{bmatrix} \cos{\theta} & -r\sin{\theta}\\ \sin{\theta} & r\cos{\theta} \end{bmatrix}.$

Similarly for the polar to cartesian transformation we have: $\begin{eqnarray*} dr = \left( \dfrac{\partial r}{\partial x}\right)_y dx + \left( \dfrac{\partial r}{\partial y}\right)_x dy \\ d\theta = \left( \dfrac{\partial \theta}{\partial x}\right)_y dx + \left( \dfrac{\partial \theta}{\partial y}\right)_x dy \end{eqnarray*}$ In vector-matrix form we can writh this using matrix $$K$$ defined below as $$\vec{dr} = K\vec{dx}$$: $K = \begin{bmatrix} \left( \dfrac{\partial r}{\partial x}\right)_y & \left( \dfrac{\partial r}{\partial y}\right)_x\\ \left( \dfrac{\partial \theta}{\partial x}\right)_y & \left( \dfrac{\partial \theta}{\partial y}\right)_x \end{bmatrix} = \begin{bmatrix} \cos{\theta} & \sin{\theta}\\ -\frac{\sin{\theta}}{r} & \frac{\cos{\theta}}{{r}} \end{bmatrix}.$ It is evident that $$K$$, the Jacobian of the transformation from polar to cartersian is equal to the inverse of $$J$$, the Jacobian of the transformation from cartesian to polar. $K = J^{-1}.$

Application 1: Infinitesimal element of length

Consider the curve $$y(x)$$, the infinitesimal element of length $$ds$$, along this curve is $ds^2 = dx^2 + dy^2$ What is the infinitesimal element of length in polar coordinates? Given we have $$\vec{dx} = J\vec{dr}$$, we have: $ds^2 = (dx)^2 + (dy)^2 = \begin{bmatrix} dx & dy \end{bmatrix}\begin{bmatrix} dx \\ dy \end{bmatrix} = \begin{bmatrix} dr & d\theta \end{bmatrix}J^TJ\begin{bmatrix} dr \\ d\theta \end{bmatrix}=(dr)^2 + r^2(d\theta)^2.$

Application 2: Infinitesimal element of area in polar coordinate

Infinitesimal element of area is useful in taking integrals of functions of two variables ($$f(x, y)$$) over a domain in the $$(x, y)$$ plane. In cartesian coordinates we have: $dA = dxdy.$ We note that letting $$\vec{dx} = dx \hat{i}$$ and $$\vec{dy} = dy \hat{j}$$ and using the fact that the area of a parallelogram is equal to the magnitude of the cross-product of vectors spanning its edges, we can also write $$dA$$ as: $dA = \lVert\vec{dx}\times\vec{dy}\rVert = \lVert \text{det}\begin{bmatrix}\hat{i} & \hat{j} & \hat{k}\\ dx & 0 & 0\\ 0 & dy & 0 \end{bmatrix}\rVert.$ What is area element for a general transformation? Consider a general transformation in two-dimensions: $x = x(u, v) \quad \text{and} \quad y = y(u, v).$ We have: $\begin{bmatrix} dx \\ dy \end{bmatrix} = J \begin{bmatrix} du \\ dv \end{bmatrix},$ where $$J$$ is the Jacobian of the transformation. We therefore have for $$\vec{du}$$ and $$\vec{dv}$$ in cartesian coordinates: $\vec{du} = J\begin{bmatrix} du \\ 0 \end{bmatrix}, \quad \vec{dv} = J\begin{bmatrix} 0 \\ dv \end{bmatrix}.$ Using the cross-product rule for the area of the parallelogram, we have: $dA' = \lVert \vec{du} \times \vec{dv} \rVert = \lvert \text{det}J \rvert du dv.$ For the polar coordinate we obtain:

$dA' = \lvert \text{det} J \rvert drd\theta = rdrd\theta.$ Note that some texts do not use prime on the transformed area element, although $$dA$$ and $$dA'$$ are not mathematically equal (one can easily check that $$dxdy \ne rdrd\theta$$). However, there is a conceptual equivalence between $$dA$$ and $$dA'$$.

This result generalizes to higher dimensions for a volume element. Given a general transformation $$u$$ in $$n$$ dimensions we have: $\vec{dx}_{n\times 1} = J_{n \times n} \,\vec{du}_{n \times 1},$ where $$J$$ is the Jacobian of the transformation. The infinitesimal volume element in cartesian coordinates is $$dV = \prod_{i=1}^{n} dx_i$$. For the volume element in the transformed coordinates we have $dV' = \lvert\text{det}J\rvert\prod_{i=1}^n du_i.$

## 10.2 Partial Differential Equations (PDEs)

Analogous to an ordinary differential equation (ODE), one can define a partial differential equation or PDE with $$f(\vec{x})$$, $$\vec{x} \in \mathbb{R}^n$$ satisfying: $f(x_1, \cdots, x_n, f, \dfrac{\partial f}{\partial x_1}, \cdots, \dfrac{\partial f}{\partial x_n}, \dfrac{\partial^2 f}{\partial x_i x_j}, \cdots) = 0.$

Consider the following 2 dimensional examples. 1. Laplace Equation for $$u(x, y)$$ (relevant to multiple areas of physics including fluid dynamics): $\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0.$ 2. Wave Equation for $$u(x, t)$$ (describing the dynamics of a wave with speed $$c$$ in one spatial dimension and time): $\frac{\partial^2u}{\partial x^2} -\frac{1}{c^2} \frac{\partial^2u}{\partial y^2} = 0.$ Our discussion of PDEs here will be very brief, but this topic is a major part of your multivariable calculus course and one of the applied elective courses in the second year. If you would like to have a sneak preview and for some cool connections to the Fourier series you saw last term, you can check out this video.

Transforming a PDE under a change of coordinates

We again consider the example of transformation from cartesian to polar coordinates: $u(x, y) \quad \longleftrightarrow \quad u(r, \theta),$ with $$J$$ being Jacobian of the transformation. Using total differentiation we have:

\begin{align*} du & = \left(\frac{\partial u}{\partial x}\right)_y dx + \left(\frac{\partial u}{\partial y}\right)_y dy \\ & = \left(\frac{\partial u}{\partial x}\right)_y\left[\left(\frac{\partial x}{\partial r}\right)_\theta dr + \left(\frac{\partial x}{\partial \theta}\right)_r d\theta\right] + \left(\frac{\partial u}{\partial y}\right)_y \left[\left(\frac{\partial y}{\partial r}\right)_\theta dr + \left(\frac{\partial y}{\partial \theta}\right)_r d\theta\right] \\ & = \left[\left(\frac{\partial u}{\partial x}\right)_y\left(\frac{\partial x}{\partial r}\right)_\theta + \left(\frac{\partial u}{\partial y}\right)_y \left(\frac{\partial y}{\partial r}\right)_\theta \right]dr + \left[\left(\frac{\partial u}{\partial x}\right)_y\left(\frac{\partial x}{\partial \theta}\right)_r + \left(\frac{\partial u}{\partial y}\right)_y \left(\frac{\partial y}{\partial \theta}\right)_r\right]d\theta. \end{align*} Now, by equating the terms in the above and the total derivative of $$u(r, \theta)$$, and also using the definition of $$J$$, we obtain the following: $\begin{bmatrix} \left(\frac{\partial }{\partial r}\right)_\theta \\ \left(\frac{\partial }{\partial \theta}\right)_r \end{bmatrix} u(r, \theta)= J^T \begin{bmatrix} \left(\frac{\partial }{\partial x}\right)_y \\ \left(\frac{\partial }{\partial y}\right)_x \end{bmatrix} u(x, y).$

Example 10.1 (Laplace Equation in polar coordinates)

Laplace equation in Cartesian coordinates is: $\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0.$ In polar coordinates we have: \begin{align*} \dfrac{\partial}{\partial x}\left[ u\right] &= \dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial x} + \dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial x}\\ &=\left[\cos{\theta} \frac{\partial}{\partial r} - \frac{\sin{\theta}}{r}\frac{\partial}{\partial \theta} \right]u. \end{align*}

Now taking second derivative we have: \begin{align*} \dfrac{\partial^2}{\partial x^2}\left[ u\right] &= \left[\cos{\theta} \frac{\partial}{\partial r} - \frac{\sin{\theta}}{r}\frac{\partial}{\partial \theta} \right]\left[\cos{\theta} \frac{\partial}{\partial r} - \frac{\sin{\theta}}{r}\frac{\partial}{\partial \theta} \right]u \\ & = \cos^2\theta \dfrac{\partial^2 u}{\partial r^2} + \dfrac{2\cos \theta \sin \theta}{r^2} \dfrac{\partial u}{\partial \theta} - \dfrac{2\cos \theta \sin \theta}{r} \dfrac{\partial^2 u}{\partial r \partial \theta} + \dfrac{\sin^2 \theta}{r} \dfrac{\partial u}{\partial r} + \dfrac{\sin^2 \theta}{r^2}\dfrac{\partial^2 u }{\partial \theta^2}. \end{align*}

Similarly, we have: $\dfrac{\partial^2}{\partial y^2}\left[ u\right] = \sin^2\theta \dfrac{\partial^2 u}{\partial r^2} - \dfrac{2\cos \theta \sin \theta}{r^2} \dfrac{\partial u}{\partial \theta} + \dfrac{2\cos \theta \sin \theta}{r} \dfrac{\partial^2 u}{\partial r \partial \theta} + \dfrac{\cos^2 \theta}{r} \dfrac{\partial u}{\partial r} + \dfrac{\cos^2 \theta}{r^2}\dfrac{\partial^2 u }{\partial \theta^2}.$

Finally, we have: $\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = \dfrac{\partial^2 u}{\partial r^2} + \frac{1}{r}\dfrac{\partial u}{\partial r} + \frac{1}{r^2}\dfrac{\partial^2 u}{\partial \theta^2} = 0.$ For example, if one looking for function $$u(r)$$ (with no dependence on $$\theta$$) fulfilling the Laplace equation, we could solve the following ODE: $\frac{d^u}{dr^2} + \frac{1}{r} \frac{du}{dr} = 0.$

## 10.3 Exact ODEs

The concept of total differentiation provides an alternative method for solving first order nonlinear ODEs. Consider the given first order ODE:

$\dfrac{dy}{dx} = \dfrac{-F(x, y)}{G(x, y)}.$

If we have a solution of the ODE in implicit form $$u(x, y) = 0$$, assuming $$u$$ is continuous with continuous derivatives. For total derivative of $$u$$ we have: $du = \left(\frac{\partial u}{\partial x}\right)_y dx + \left(\frac{\partial u}{\partial y}\right)_y dy = 0 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{-\left(\frac{\partial u}{\partial x}\right)_y}{\left(\frac{\partial u}{\partial y}\right)_y}.$ For the solution $$u(x, y) = 0$$ to exist then we need the RHS of the above equation to be equal to the RHS of the ODE, we are trying to solve. This will be the case if $F = \left(\frac{\partial u}{\partial x}\right)_y \quad \text{and} \quad G = \left(\frac{\partial u}{\partial y}\right)_y.$ But, if that is the case, due to the symmetry of the partial second mixed derivatives, we should have: $\frac{\partial u^2}{\partial y\partial x} = \frac{\partial u^2}{\partial x\partial y} \quad \Rightarrow \quad \frac{\partial F}{\partial y} = \frac{\partial G}{\partial x}.$ This is known as the condition of integrability of the ODE and if it is satisfied then there exists $$u(x, y)$$, where $F = \left(\frac{\partial u}{\partial x}\right)_y \quad \text{and} \quad G = \left(\frac{\partial u}{\partial y}\right)_y,$ and then $$u(x, y) = 0$$ is a solution of the first order ODE. We call this kind of ODE exact.

Example 10.2 ( Is the following ODE exact?) $\dfrac{dy}{dx} = \dfrac{-2xy -\cos x \cos y}{x^2 - \sin x \sin y}$

Letting $F(x, y) = 2xy +\cos x \cos y \quad \text{and} \quad G(x, y) = x^2 - \sin x \sin y.$ We can check the condition of integrability and see the ODE is exact: $\frac{\partial F}{\partial y} = \frac{\partial G}{\partial x} = 2x - \cos{x} \sin{y},$ since the ODE is exact, we can look for a solution in implicit form $$u(x, y) = 0$$ such that:

$F(x, y) = \dfrac{\partial u}{\partial x} \quad \text{and} \quad G(x, y) = \dfrac{\partial u}{\partial y}.$ We can do that in two steps. Firstly, we have: $\left(\dfrac{\partial u}{\partial x}\right)_y = 2xy + \cos{x}\cos{y} \quad$ Integrating with respect to $$x$$ and assuming the constant of intgration could depend on $$y$$, we obtain: $u = yx^2 + \cos{y}\sin{x} + f(y).$ Now in the second step, we use $G(x, y) = \dfrac{\partial u}{\partial y} = x^2 - \sin{y}\sin{x} \quad \Rightarrow \quad \frac{df}{dy} = 0.$ This implies $$f$$ is a constant, so we obtain the general solution $$y(x)$$ of the ODE in implicit form: $u(x, y) = yx^2 + \cos{y}\sin{x} + c = 0.$

When the ODE is not exact, sometimes we can find a function (an integrating factor) that will make the equation exact. Given:

$F(x, y)dx + G(x, y)dy = 0,$

is not exact, we look for a function $$\lambda(x)$$ or $$\lambda(y)$$ such that:

$\lambda F(x,y)dx + \lambda G(x, y) dy = 0,$ is exact. Note that an integrating factor can in general be a function of both $$x$$ and $$y$$, but in this case we cannot find an explicit solution for $$\lambda$$, and it is for this reason we can not solve very many ODEs.

Example 10.3 (Is this ODE exact? If not find an integrating factor to make it exact) $\dfrac{dy}{dx} = \dfrac{xy - 1}{x(y-x)}.$

Letting $$F = xy -1$$ and $$G = x^2 - xy$$, we see that the ODE is not exact as: $\frac{\partial F}{\partial y} \ne \frac{\partial G}{\partial x}.$

So, we will try to find a $$\lambda(x)$$ (or $$\lambda(y)$$ that will make the ODE exact). We need to find $$\lambda(x)$$ such that:

$\dfrac{\partial [\lambda(x)(xy-1)]}{\partial y} = \dfrac{\partial [\lambda(x)(x^2 - xy)]}{\partial x}.$ After simplification we obtain: $(x - y)\left[\frac{d\lambda}{dx}x + \lambda\right] = 0.$ As we obtain an ODE for $$\lambda(x)$$ (that does not depend on $$y$$), an integrating factor of the form $$\lambda(x)$$ exists and can be obtained by solving this ODE to be: $\lambda = \frac{c}{x}.$

Now we can solve the exact ODE that is obtained (left as a quiz):

$(y - \dfrac{1}{x})dx + (x - y)dy = 0.$

## 10.4 Sketching functions of two variables

Similar to the sketching of functions of one variable, we will use the following steps in sketching a function of two variables $$f(x, y)$$:

• Check continuity and find singularities.
• Find asymptotic behaviour $\lim\limits_{x, y \rightarrow \pm \infty} f(x, y) \quad \text{and} \quad \lim\limits_{\vec{x} \rightarrow \vec{x}_{sing}} f(\vec{x}).$
• Obtain some level curves, for example $$f(\vec{x}) = 0$$.
• Find stationary points: minimum, maximum, saddle points

Stationary points for functions of two variables

Reminder that for a function of one variable $$f(x)$$, we find stationary points by setting the first derivative to zero: $f(x^*) = 0.$ Then, using Taylor expansion of $$f(x)$$ near $$x^*$$, we see that one can use the sign of the second derivative of $$f(x)$$ at $$x^*$$ to decide if the stationary point is minimum (if $$\frac{d^2f}{dx^2}(x^*)>0$$) or maximum (if $$\frac{d^2f}{dx^2}(x^*)<0$$).

Using a similar approach for the functions of two variables $$f(x, y)$$, we have stationary points are the points where tangent plane at $$x^*$$ is parallel to $$(x, y)$$ plane: $\frac{\partial f}{\partial x} (\vec{x}^*) = \frac{\partial f}{\partial y} (\vec{x}^*) = 0.$ Then, the type of stationary point can be determined using the Taylor expansion around the stationary point $$\vec{x^*}$$ and by the Hessian matrix. $f(\vec{x}^* + \Delta \vec{x}) = f(\vec{x}^*) + \left[ \dfrac{\partial f}{\partial x}(\vec{x}^*) \quad \dfrac{\partial f}{\partial y}(\vec{x}^*) \right] \Delta \vec{x} + \dfrac{1}{2} \Delta \vec{x}^T\begin{bmatrix} \dfrac{\partial^2 f}{\partial^2 x}(\vec{x}^*) & \dfrac{\partial^2 f}{\partial x \partial y}(\vec{x}^*) \\ \dfrac{\partial^2 f}{\partial y \partial x}(\vec{x}^*) & \dfrac{\partial^2 f}{\partial^2 y}(\vec{x}^*) \end{bmatrix}\Delta \vec{x}.$ Given the fact that the first partial derivatives of $$f(\vec{x})$$ are zero at the stationary points, we have: $f(\vec{x}^* + \Delta \vec{x}) - f(\vec{x}^*) = \dfrac{1}{2}\Delta\vec{x}^TH(\vec{x}^*)\Delta\vec{x}.$ Under the assumption of continuity, we have the symmetry of the second mixed partial derivatives: $\frac{\partial^2 f}{\partial y\partial x} = \frac{\partial^2 f}{\partial x \partial y},$ therfore the Hessain is symmetric ($$H = H^T$$), which implies that $$H$$ is diagonalizable and it has real eigenvalues $$\lambda_1$$ and $$\lambda_2$$. So, there exists a similarity transformation $$V$$ that diagonalise the Hessian: $V^{-1}HV = \Lambda = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}.$ Also as $$H$$ is symmetric, we have $$V^{-1} = V^T$$. So, we have: $f(\vec{x}^* + \Delta \vec{x}) - f(\vec{x}^*) = \dfrac{1}{2}\Delta\vec{x}^T\left[ V((\vec{x}^*)\Lambda(\vec{x}^*)V^T(\vec{x}^*)\right]\Delta\vec{x}.$ Now, if we let $$\Delta\vec{z} = V^T(\vec{x}^*) \Delta\vec{x}$$, we have: $\Delta f = f(\vec{x}^* + \Delta \vec{x}) - f(\vec{x}^*) = \dfrac{1}{2} \Delta \vec{z}^T\Lambda(\vec{x}^*)\Delta \vec{z} = \dfrac{1}{2}\left[ (\Delta z_1)^2\lambda_1 + (\Delta z_2)^2\lambda_2\right].$ Given this expression, we can use the sign of the eigenvalues to classify the stationary points.

• If $$\lambda_1, \lambda_2 \in \mathbb{R}^+$$, we have $$\Delta f > 0$$ as we move away from the stationary point, suggesting $$\vec{x}^*$$ is a minimum.

• If $$\lambda_1, \lambda_2 \in \mathbb{R}^-$$, we have $$\Delta f < 0$$ as we move away from the stationary point, suggesting $$\vec{x}^*$$ is a maximum.

• If $$\lambda_1 \in \mathbb{R}^+$$ and $$\lambda_2 \in \mathbb{R}^-$$, we classify $$\vec{x}^*$$ is a saddle point, since $$\Delta f$$ can be positive or negative depending the direction of $$\Delta \vec{x}$$.

We note that, we could use the trace ($$\tau$$) and determinant $$\Delta$$ of the matrix $$H$$ to know the sign of the eigenvalues with explicityly calculating the eigenvalues as done in section 7.3 in the analysis of the 2D linear ODEs. In particular, $$\Delta > 0, \tau>0$$ ($$\Delta > 0, \tau<0$$) suggests eigenvalues are positive (negative) and we have a minima (maxima). $$\Delta < 0$$ indicates a saddle point stationary point.

Example 10.4 (Sketch the following function of the two variables) $u(x,y) = (x-y)(x^2 + y^2 -1)$

We note the function is continuous and there are no singularities. The asymptotic behavior is that $$u(x, y) \rightarrow \pm \infty$$ as $$x, y \rightarrow \pm \infty$$.

Next, we find the level curves at zero $u(x, y) = 0 \quad \Rightarrow \quad x = y \quad \text{and} \quad x^2 + y^2 - 1 = 0$ In the next step, we obtain the stationary points.
$\dfrac{\partial u}{\partial x}(\vec{x}^*) = \dfrac{\partial u}{\partial y}(\vec{x}^*) = 0$ $\dfrac{\partial u}{\partial x} = (x^2 + y^2 - 1) + 2x(x - y) = 0,$ $\dfrac{\partial u}{\partial y} = -(x^2 + y^2 - 1) + 2y(x - y) = 0.$ Adding these two equations we obtain $$x^* = y^*$$ or $$x^* = -y^*$$.

1. $$x^* = y^* \quad \Rightarrow \quad 2{x^*}^2 - 1 = 0 \quad \Rightarrow \quad P_1 = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), \quad P_2 = (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}).$$

2. $$x^* = -y^* \quad \Rightarrow \quad 6{x^*}^2 - 1 = 0 \quad \Rightarrow \quad P_3 = (\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}), \quad P_4 = (-\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}).$$

We classify the stationary points using the Hessain:

$H(\vec{x}) = \begin{bmatrix} 6x-2y & 2y-2x \\ 2y - 2x & 2x - 6y \end{bmatrix}$ Now we use the determinant ($$\Delta$$) and trace ($$\tau$$) of matrix $$H$$ at each stationary point to classify each stationary point:

$P_1 = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \, \Rightarrow \, H(P_1) = \begin{bmatrix} 4\frac{1}{\sqrt{2}} & 0 \\ 0 & - 4 \frac{1}{\sqrt{2}} \end{bmatrix} \, \Rightarrow \, \Delta <0 \, \Rightarrow \, P_1 \text{ is a saddle point.}$ $P_2 = (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}) \, \Rightarrow \, H(P_2) = \begin{bmatrix} -4\frac{1}{\sqrt{2}} & 0 \\ 0 & 4 \frac{1}{\sqrt{2}} \end{bmatrix} \, \Rightarrow \, \Delta <0 \, \Rightarrow \, P_2 \text{ is a saddle point.}$ $P_3 = (\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}) \, \Rightarrow \, H(P_3) = \begin{bmatrix} \frac{8}{\sqrt{6}} & -\frac{4}{\sqrt{6}} \\ -\frac{4}{\sqrt{6}} & \frac{8}{\sqrt{6}} \end{bmatrix} \, \Rightarrow \, \Delta >0, \tau>0 \, \Rightarrow \, P_3 \text{ is a minimum.}$ $P_4 = (-\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}) \, \Rightarrow \, H(P_4) = \begin{bmatrix} -\frac{8}{\sqrt{6}} & \frac{4}{\sqrt{6}} \\ \frac{4}{\sqrt{6}} & -\frac{8}{\sqrt{6}} \end{bmatrix} \, \Rightarrow \, \Delta >0, \tau<0 \, \Rightarrow \, P_4 \text{ is a maximum.}$

Given the location and stability of the stationary points, we complete our sketch of

$u(x,y) = (x-y)(x^2 + y^2 -1),$ as seen in Figure 10.1, by sketching some level curves, specifying the sign of the funcion and the location of the stationary points. Figure 10.1: The contour plot and sketch of function $$u(x,y) = (x-y)(x^2 + y^2 -1)$$