For real $\lambda >0$, squaring the two nonnegative moduli gives $|z-i{|}^2={\lambda }^2|z+i{|}^2$. In Cartesian coordinates this isAssume $\lambda \ne 1$ and put $h=(1+{\lambda }^2)/(1-{\lambda }^2)$. Dividing by $1-{\lambda }^2$ yields $x^2+y^2-2hy+1=0$. Since $z{z}^{\ast }=x^2+y^2$ and $b{z}^{\ast }+b^{\ast }z=2\mathrm{Re}(b{z}^{\ast })$, the requested complex equation hasIndeed these choices make the mixed term $-2hy$. Completing the square givesThe Circles of Apollonius therefore haveThe radius is positive for the stated range. No spurious point at $z=-i$ has been introduced: at that point the undivided equation would require $4=0$. Positive $\lambda <1$ gives a circle above the real axis, while $\lambda >1$ gives one below it.

A quotient with nonzero denominator has zero complex modulus exactly when its numerator is zero. ThusAt that point $z^{\ast }-i=-2i\ne 0$, so the solution is admissible.

The original PDF places a complex modulus around the quotient; those bars are essential. Its denominator excludes $z=-i$. Using $|z^{\ast }-i|=|z+i|$, the case $\lambda =1$ becomes $|z-i|=|z+i|$. In Cartesian coordinates,This locus is the real axis, the perpendicular bisector of the points $i$ and $-i$ in the complex plane. The excluded denominator point does not lie on it.

The polynomial factors asThus the repeated polynomial root is $x=2$, of multiplicity two. Its repetition can also be checked from $p(2)=p^{\prime }(2)=0$ and $p^{\prime \prime }(2)=14\ne 0$.

The sum-of-cubes polynomial factorization isMultiplication cancels the two mixed cubic terms and leaves precisely the two cubes.

Each Cartesian component of $a\mathbf{r}$ contributes $a$ to the divergence, while the constant vector contributes zero. Hence

Since $\mathbf{r}\cdot \mathbf{r}=x^2+y^2+z^2$, its gradient is

Adding the steady and transient parts givesIntegrating the sine across the channel gives ${\int }_0^1\sin (n\pi x)dx=[1-(-1{)}^n]/(n\pi )$. ThusFor even $n$ the signed flow rate is zero, because positive and negative velocity lobes cancel. For odd $n$ the bracket is two. Both expressions vanish initially and approach their steady values exponentially; $G=0$ gives the trivial zero solution throughout.

Separate $\tilde{u}=f(t)g(x)$. On a nonzero product region,A decaying temporal factor requires $\lambda >0$, with $f=C{e}^{-\nu \lambda t}$. The zero spatial boundary values give $g=A\sin (\sqrt{\lambda }x)$ and $\sin \sqrt{\lambda }=0$, hence $\lambda =j^2{\pi }^2$ with $j=1,2,\dots$. Zero or negative separation parameters give no nonzero spatial mode with both boundary values zero.

The initial profile is already exactly the $j=n$ sine eigenfunction. Its coefficient therefore determines a single separated solution:The exponential tends to zero because $\nu >0$. More generally sine orthogonality would determine a sum of such modes, but all other coefficients vanish here. This is the transient of the heat equation with a single stationary sine forcing.

The steady profile has no time derivative. Substituting $u=\tilde{u}+u_s$ and using its steady equation cancels the forcing, givingThe initial condition isThe resulting heat equation is independent of $G$, as are its boundary conditions. Its initial amplitude still depends on $G$; removing a source from the evolution equation does not remove it from the whole problem.

In a steady state, $\nu u_s^{\prime \prime }=-G\sin (n\pi x)$. Since the second derivative of the sine is $-(n\pi {)}^2$ times itself,The zero boundary values at $x=0,1$ force $A=B=0$. Therefore

Introduce a second multiplier $b$ and useThe stationary equations become $-\log x_i-1-\lambda -b{y}_i=0$. Writing $a=e^{-1-\lambda }>0$ givesThe two constraints determine the constants throughEquivalently, $a=1/Z(b)$ with $Z(b)={\sum }_i{e}^{-b{y}_i}$, and $-d\log Z/db=Y$. This is the maximum entropy distribution with a prescribed finite mean.

For nonconstant $y_i$, a strictly positive solution exists exactly when ${\min }_i{y}_i<Y<{\max }_i{y}_i$. The mean's derivative with respect to $b$ is $-{\mathrm{Var}}_b(y)<0$, so $b$ is unique in that case. At either extreme mean no strictly positive solution exists. If all $y_i=y_0$, feasibility requires $Y=y_0$ and the extra constraint is redundant: $x_i=1/n$, while $a,b$ are not separately identifiable. The negative entropy Hessian ensures that any feasible interior stationary point is the unique constrained maximum.

Use $\mathcal{L}=-{\sum }_i{x}_i\log x_i-\lambda ({\sum }_i{x}_i-1)$. Differentiating with respect to each positive $x_i$ givesEvery $x_i$ is therefore equal; normalization forcesThe Hessian matrix of $f$ is diagonal with entries $-1/x_i$, so its quadratic form is strictly negative for every nonzero perturbation. In particular it is negative along the tangent space ${\sum }_i\delta x_i=0$, making this the unique global maximum on the positive probability simplex. This is the maximum entropy distribution on a finite set, using natural logarithms.

At a regular constrained stationary point, the change in $f$ along every allowed tangent direction must vanish. Thus its gradient is normal to the constraint curve, giving the Lagrange multiplier equationsprovided $\nabla g\ne 0$. Solve these equations for $x,y,\lambda$, then classify the candidates using the constraint or a constrained second-derivative test. Points where the constraint gradient vanishes require separate analysis.

For $n$ variables and $r$ regular independent constraints $g_j=c_j$, use one multiplier per constraint:There are $n+r$ scalar equations in the $n+r$ unknowns. Independence of the constraint gradients is the regularity condition; boundary points must be treated separately when present.

The given identity immediately yields $V_2=V_1-V_0\sin \omega t$. Thus only the fundamental sine coefficient changes: $V_0/2$ becomes $-V_0/2$. No new integral is needed. The first five nonzero terms areThis also follows from translation of a Fourier series: a half-period delay multiplies the $n$th harmonic by $(-1{)}^n$, leaving the constant and even harmonics unchanged and reversing the only odd harmonic present.

Put $s=\omega t$, so the nonzero half-cycle is $0\le s\le \pi$. The constant coefficient isThe sine coefficients satisfyby sine orthogonality. For the cosine coefficients, the product-to-sum identity gives, when $n\ne 1$,The case $n=1$ is zero directly. HenceThe first five nonzero terms, counting the constant as the zero-frequency term, areThe general coefficient formula specifies every remaining term. This is the Fourier series of a half-wave rectified sinusoid; the waveform is continuous at the joins, so the series gives its actual value there.

The sine and cosine functions are orthogonal over a complete period. For positive integers $m,n$,while every mixed sine-cosine integral and every integral of a nonconstant single harmonic is zero. These identities follow from the product-to-sum formulas and the integer-period endpoint values.

Multiply the Fourier series by $\cos (m\omega t)$ and integrate. All terms except $a_m{\cos }^2(m\omega t)$ vanish, giving $T{a}_m/2$. Repeating with sine isolates $b_m$. ThusIntegrating without a harmonic also gives $a_0=(2/T){\int }_0^T{V}_1(t)dt$. The factor $1/2$ on the constant term in the chosen convention is essential.

One period advances the fundamental phase by $2\pi$, so the angular frequency is

The largest angular-frequency magnitude is ${\omega }_3=\sqrt{(2+\sqrt{2})k/m}$. With overall amplitude $A$, the corresponding displacements areThus the two outer masses move together, while the middle mass moves oppositely with $\sqrt{2}$ times their amplitude. All three have the same period $2\pi /{\omega }_3$.

The pairwise dot products of the normalized vectors areTheir squared norms are respectively $(1+2+1)/4$, $(1+1)/2$, and $(1+2+1)/4$, all one. Hence they form an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices, because $D$ is symmetric and its three eigenvalues are distinct.