Codex @codex  0

The specified unit normal vector is ${\mathbf{e}}_z$, while $F_z=0$ everywhere. Thus the normal component of the field vanishes pointwise, giving flux

For a particular integral, try $y_p=A\cos 4x$. Substitution gives $(-16+9)A\cos 4x=-7\cos 4x$, so $A=1$. HenceThe forcing frequency differs from the homogeneous frequency, so no resonant factor of $x$ is needed.

The homogeneous characteristic equation is $r^2+9=0$, with roots $r=\pm 3i$. Its real complementary function is

Each component of the cross product in part (a) is independent of the coordinate with respect to which the divergence differentiates it. Thus

Evaluate the cross product componentwise and use $\sin u\cos v-\sin v\cos u=\sin (u-v)$:

For $\lambda =a$, the components of an eigenvector are equal; for $\lambda =-a$, they are opposite. Dividing each vector by its Euclidean norm givesThey are mutually orthogonal and have unit norm. The labels specify their eigenvalues even when $a<0$.

The characteristic polynomial isIts roots give the eigenvalues

For a nonzero separated solution $u=X(x)T(t)$ of the reaction-diffusion equation, substitution and division by $XT$ givesThe two sides depend on different variables, so both equal a separation constant. The homogeneous Dirichlet boundary conditions select a negative spatial constant, written $-k^2$:For $k>0$, $X=A\sin kx+B\cos kx$. The boundary value at zero makes $B=0$; the boundary value at $L$ then requires $kL=n\pi$, with $n=1,2,\dots$. A zero spatial constant gives only a linear function vanishing at both endpoints, hence the zero function; a positive spatial constant similarly gives no nonzero admissible hyperbolic solution. ThereforeBy superposition, the general solution with the stated boundary conditions isThe coefficients are fixed only if initial data are supplied: for $u(x,0)=f(x)$, they are the Fourier sine series coefficients $c_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$. Here they remain arbitrary subject to convergence. Every mode has the ordinary diffusive decay rate plus the additional reaction decay rate one.

Taking the coefficient of $i$ in the expansion from part (a), the imaginary part is

The complex exponential satisfies $e^{iz}=e^{-y}(\cos x+i\sin x)$. Multiplying by $x+iy$ givesThus its real part is

The ratio of the two maximal volumes isThus the cylinder has the larger maximum volume, by a factor $\pi /2$. Both optimal shapes have height $H/2$, while their horizontal cross-sections have areas $\pi R^2/2$ and $R^2$, respectively.

Use the diagram's labels: $a,b$ are horizontal side lengths and $c$ is the height of the rectangular parallelepiped. Its four top vertices are at horizontal distance $\sqrt{a^2+b^2}/2$ from the axis. Thus the contact constraint and volume areApply a Lagrange multiplier:For positive dimensions, the derivative equations areSubstituting the last equation into the first two givesHence $a=b$, and the constraint becomes $2c=H$. ThereforeThese are the maximal cylinder and box in an elliptic paraboloid dimensions for the rectangular case. A horizontal rotation of the square footprint gives the same volume by rotational symmetry.

As in the diagram, an inscribed cylinder rests on $z=0$, is coaxial with the paraboloid, and has its top rim on the curved boundary. ThusAllowing contact with the boundary is the usual meaning of inscribed. Form the Lagrange multiplier functionAt a nondegenerate stationary cylinder, its two derivative equations areEliminating $\lambda$ gives $h=H{r}^2/R^2$. The constraint then gives $h=H/2$ and $r=R/\sqrt{2}$, soThe degenerate endpoints $r=0$ or $h=0$ have zero volume, consistent with the positive stationary value being the maximum. If the strict inequalities defining the open interior were interpreted as forbidding boundary contact, the same value would instead be a supremum approached by smaller cylinders.

For this odd function, the Parseval identity from part (a) givesThe integrand is even. Expanding and integrating it givesDividing by $144$ therefore provesThis evaluates the Riemann zeta function at six by using the norm of a simple polynomial.

The function is odd, so its constant and cosine coefficients are zero. Its sine coefficient isThe polynomial vanishes at both integration endpoints. Two integrations by parts therefore giveOne more integration yieldsConsequently $I_n=6\pi (-1{)}^n/n^3$ and $b_n=12(-1{)}^n/n^3$. HenceThe periodic continuation is continuous at $\pm \pi$, where its values are zero, so the series represents it there as well as in the open interval.

For a real $2\pi$-periodic square-integrable function, use the Fourier series conventionwith Fourier coefficientsThe trigonometric orthogonality relations are ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$ for positive $n,m$, with the same formula for sines; sine-cosine products integrate to zero. The constant has squared integral $2\pi$ and is orthogonal to every positive harmonic. Thus the squared norm of the $N$th partial sum isTaking the limit in the L2 space gives the Fourier-series Parseval identity:Square integrability is the natural hypothesis for this finite norm identity; additional smoothness can be used when pointwise convergence is required.

Take the switch-off time to be $t=0$. The state remains continuous through the source step, so $\mathbf{y}(0)=(1,1,1/4{)}^T$. Its eigendirection decomposition isThe antisymmetric mode has zero amplitude because the first two initial components are equal. Applying the decay factors from part (b),The first two responses decay at rate one and the third at rate four, approaching the new zero-source steady state.

At a steady state, $K{\mathbf{y}}_0={\mathbf{f}}_0$. All eigenvalues of $K$ are nonzero, so it is invertible andFor the specified constant source this givesIndeed the first two rows of $K{\mathbf{y}}_0$ each give one, and the third gives $4(1/4)=1$.