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Use the gradient flow of $-f$, namely $\dot{\mathbf{r}}=\nabla f$; any positive change of speed leaves the same trajectory. At the starting point,so the path initially points up and left, with slope $dy/dx=-1/2$. At every regular point it crosses the level sets orthogonally, because the gradient is normal to the contours. Along it,until a stationary point is approached.

The red curve in the plot first rises above $y=1$, then bends down towards the maximum $(0,1)$. It stays on the side $x>0$ and approaches the maximum without crossing it in finite gradient-flow time. Indeed $\dot{x}=-4x(1+y+y^2)/D^2$ preserves the sign of $x$, and the critical level increases from its starting value $0$ towards $1$. The requested curve is an ascending trajectory, not a contour or a straight segment joining the initial and final points.

The zero level set is especially simple:The two axes and the unit circle divide the plane into regions where the signs alternate. Inside the circle, $f$ has the opposite sign to $xy$; outside it has the same sign. Closed level curves surround the two interior local minima and two interior local maxima; the zero curves cross at the five saddle points of a scalar function. Farther out, the nonzero level curves are open within the displayed square.

At a regular constrained extremum, every tangent vector to the level set $g=A$ gives zero directional derivative of $f$. Thus $\nabla f$ is perpendicular to the tangent plane, and must be a scalar multiple of its nonzero normal vector $\nabla g$. The Lagrange multiplier equations areOne solves these four equations for $x,y,z,\lambda$, evaluates $f$ at the candidates, and classifies or compares the values as required. Points where $\nabla g=0$, and any boundary or limiting cases of the feasible domain, require separate checks because the regularity argument does not cover them.

For a differentiable objective $f$ and regular constraint $g=r$, a constrained stationary point satisfieswhere $\lambda$ is a Lagrange multiplier. Solve these equations for both the coordinates and $\lambda$.

Geometrically, the gradient $\nabla g$ is normal to the constraint level set when $\nabla g\ne 0$. Every feasible infinitesimal displacement $\mathbf{v}$ is tangent to that level set, so $\nabla g\cdot \mathbf{v}=0$. At a constrained stationary point, the directional change $\nabla f\cdot \mathbf{v}$ must also vanish for every such tangent displacement. Thus $\nabla f$ is normal to the same tangent plane and is a scalar multiple of $\nabla g$. This is the geometric basis of the Lagrange multiplier method.

For a second constraint $h=s$, with linearly independent vectors $\nabla g$ and $\nabla h$, the feasible tangent directions are perpendicular to both. Their normal space is the linear span of these two gradients, so introduce a second Lagrange multiplier:These equations find candidates; deciding whether they are maxima, minima or saddle points requires further analysis. Constraint points where the required gradient independence fails must be examined separately.

Now $h=e^x-e^y-x+y$ and the origin has stationary value zero. Near it,The level curves therefore resemble hyperbolas near the saddle point of a scalar function. Along the $x$-axis $h=e^x-x-1>0$ for $x\ne 0$, while along the $y$-axis $h=1-e^y+y<0$ for $y\ne 0$. The zero level set includes the exact diagonal $x=y$ and another branch crossing it at the origin; this second branch is tangent to $y=-x$ but is curved away from the origin. The right panel of the contour figure displays these features.

In the translated coordinates $u=x-1$, $v=y+1$, the zero level set is the pair of lines $u=0$ and $v=0$. The function is positive in the quadrants $uv>0$ and negative in those with $uv<0$, and tends to zero as $u^2+v^2\to \infty$. Thus its nonzero contours form closed curves around the four off-centre stationary points.

For an explicit description of the contour plot, use polar coordinates $u=r\cos \theta$, $v=r\sin \theta$:At fixed direction within a quadrant, the magnitude increases until $r=\sqrt{2}$ and then decreases towards zero. Also $|u|e^{-u^2/2}\le e^{-1/2}$, with equality exactly when $|u|=1$, and similarly for $v$. Consequently $|f|\le e^{-1}$; each positive level below $e^{-1}$ encloses a positive peak in each positive quadrant, with the analogous negative contours in the negative quadrants.

To classify the stationary points, the Hessian matrix isAt $u=v=0$ it is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(1,-1)$ is a saddle point. At $u,v\in \{-1,1\}$ it is $-2fI$. Therefore $(2,0)$ and $(0,-2)$ are strict local maxima, and $(2,-2)$ and $(0,0)$ are strict local minima. These are also global maxima and minima by the bound just proved.

The gradient isIt is perpendicular to each nonzero level set and points towards increasing $f$. Near the two positive local maxima, the Hessian matrix is $-2e^{-1}I$, so the gradient arrows point inward. Near the two negative local minima, the Hessian matrix is $2e^{-1}I$, so they point outward. These arrows are shown on the figure in part (c).

At the saddle point, $\nabla f\approx (v,u{)}^T$. Along $v=u$, the arrows point away from the centre on both sides; along $v=-u$, they point towards the centre on both sides. On the positive $u$ axis they point up, on the negative $u$ axis down, on the positive $v$ axis right, and on the negative $v$ axis left. This specifies the four intervening sectors of the gradient pattern.

The Cauchy-Riemann equations give $\nabla v=(-u_y,u_x)$. ThereforeThe gradient is normal to a regular level set, so these curves intersect orthogonally wherever their gradients are nonzero. At a stationary point, where both gradients vanish, a regular curve tangent need not exist; the algebraic identity remains valid.

For a holomorphic function, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. ConsequentlyA gradient is normal to its level set, so these identities give orthogonal level curves wherever the gradients are nonzero. At a critical point where $f^{\prime }=0$, both gradients vanish and a regular tangent to a level set need not exist; the geometric orthogonality statement is restricted to regular intersections.

The Cauchy-Riemann equations are $\boxed{u_x=v_y,u_y=-v_x}$. They give the gradient identityAt a regular intersection, where $f^{\prime }(z)\ne 0$, the two nonzero gradients are normal vectors to the two level sets, so their tangent directions are perpendicular. This is the orthogonality of regular real and imaginary level curves.

The regularity qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }|$, so at a critical point both normals vanish and the curves need not have unique tangents. For example $f=z^2$ gives diagonal lines for $u=0$ and coordinate axes for $v=0$, all meeting at the origin. Thus the conclusion is $\boxed{\text{right-angle intersection wherever }{f}^{\prime }(z)\ne 0}$, rather than an unrestricted claim at critical points.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

For a specified posterior probability $1-\alpha$, an interval $I=({\theta }_1,{\theta }_2)$ is a highest posterior density interval ifand every included point has posterior density at least as large as every excluded point, apart from boundary and null-set choices. Equivalently, $I$ is a level set $\{\theta :\pi (\theta \mid x)>k\}$, with possible choices on its boundary, whose posterior probability is $1-\alpha$.

This selects the most densely supported values of the statistical parameter. When the relevant level set is one interval and has no plateau of positive probability, it is the shortest credible interval of that posterior probability. With several separated peaks, the corresponding highest density region can instead be disconnected, so an interval need not exist. The definition is in the stated parameter coordinate: posterior density ordering can change under a nonlinear change of parameter.

A likelihood-ratio test rejects for small $\lambda$. The derivative of the natural logarithm of its variable factor isThus $t^n(1-t{)}^m$ increases up to $n/(n+m)$ and decreases thereafter, tending to zero at both endpoints. Every nontrivial lower level set of $\lambda$ therefore consists of two tails. For a prescribed test size $0<\alpha <1$, the exact critical region isWriting $F_{n,m}$ for the cumulative distribution function of the beta distribution with parameters $(n,m)$, the required two equations areThe first makes both cutoffs have the same likelihood ratio; the second calibrates the null hypothesis rejection probability. Equivalently, the second left-hand side isAs the common likelihood ratio threshold increases from zero to one, this probability increases continuously and strictly from zero to one; hence the equations determine unique cutoffs. The two tail probabilities need not each equal $\alpha /2$ when $n\ne m$. Exact calibration holds for every common rate because the given null distribution of $T$ is independent of that rate. Boundary equalities have zero probability.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

Vary the coupling in the opening example. The eigenvectors $e_{+}=(1,1{)}^T/\sqrt{2}$ and $e_{-}=(1,-1{)}^T/\sqrt{2}$ form a fixed orthonormal eigenbasis, with eigenvalues $3+t$ and $3-t$. Writing a vector as $u{e}_{+}+v{e}_{-}$ givesFor $|t|<3$, both directions cost energy and $A_t$ is a positive-definite matrix. At $t=3$, the quadratic form is $6u^2$: the entire line spanned by $e_{-}$ has zero energy, and $A_t$ is a singular positive semidefinite matrix. At $t=-3$, the flat line is instead spanned by $e_{+}$. For $|t|>3$, the quadratic form is an indefinite quadratic form; one direction lowers the energy without bound as its displacement grows. At $t=4$, for instance, $q_t(e_{+})=7$ while $q_t(e_{-})=-1$.

The geometric transition is precise. For $|t|<3$, $q_t=1$ is an ellipse. At $t=3$, it consists of the two parallel lines $u=\pm 1/\sqrt{6}$, extending freely in the zero-energy direction. For $t>3$, it is a hyperbola. The boundary case retains nonnegative energy but loses both the bounded positive level sets and the unique zero-energy vector.

For $\dot{x}=\alpha x-y+y^3$, $\dot{y}=-x$, the energy $H=x^2-y^2+y^4/2$ satisfies $\dot{H}=2\alpha x^2$. At $\alpha =0$, its compact level sets give two families of closed periodic orbits around $(0,\pm 1)$ and an outer family around both; $H=0$ is a figure-eight homoclinic orbit pair through the saddle point at the origin. The two minima of $H$ are center equilibria for zero damping, stable foci for small negative $\alpha$, and unstable foci for small positive $\alpha$. For $\alpha \ne 0$, strict monotonicity along nonconstant solutions excludes periodic orbits and homoclinic orbits.

The stationary points of this polynomial are the origin and $(a,a)$, $(-a,-a)$. Its Hessian matrix at the origin has eigenvalues $\pm 4a^2$, so the origin is a saddle point. At each of the other points the eigenvalues are $8a^2$ and $16a^2$, giving strict local minima. The identityshows that they are also the only global minima, with value $-2a^4$.

For $-2a^4<c<0$, the level set $g=c$ has two closed components, one surrounding each minimum. The zero level set has two lobes meeting at the origin with the coordinate axes as tangent directions. For $c>0$, there is one closed outer component. These statements also follow in polar coordinates from the quadratic equation in $r^2$, together with the absence of other stationary points.

The zero level set of $F_R$ is the union of the coordinate axes and the circle of radius $R$. The gradient isIts nine stationary points are the origin, the four axis points at distance $R$, and the four points with $x^2=y^2=R^2/4$. The factored gradient proves completeness: a zero coordinate gives the five axis cases; two nonzero coordinates give a linear system in $x^2,y^2$.

The Hessian matrix has diagonal entries $6xy$ and off-diagonal entries $3(x^2+y^2)-R^2$. At the origin its eigenvalues are $\pm R^2$; at the four unit-direction axis points they are $\pm 2R^2$. These are saddle points of a scalar function. At the two interior points with equal coordinate signs, the eigenvalues are $R^2,2R^2$, giving strict local minima of value $-R^4/8$. Opposite coordinate signs give the negative eigenvalues and strict local maxima of value $R^4/8$.

On the zero circle, $\nabla F_R=2xy(x,y{)}^T$. Hence its direction is radial outward for $xy>0$, inward for $xy<0$, and zero at the four axis crossings. This simple geometric sign rule links the zero level set to the gradient sketch without requiring numerical differentiation.