Take the base-ten logarithm of both positive sides. The exponentiation rule for logarithms gives $x{\log }_{10}2=y$, hence

The same chain rule gives $u_{xx}=u^{\prime \prime }(\xi )$ and $u_y=u^{\prime }(\xi )/\sqrt{y}$. Substituting these and part (a) into the partial differential equation givesThus every twice-differentiable profile $u(\xi )$ gives a solution on $y>0$.

For real coordinates, work on $y>0$. The chain rule gives ${\xi }_y=y^{-1/2}$ and ${\xi }_{yy}=-\frac{1}{2}{y}^{-3/2}$. ThereforeThe derivatives on the right are with respect to $\xi$. Behaviour at $y=0$ requires separate regularity of the composed function.

Here $y=0$, the unit normal vector is ${\mathbf{e}}_y$, and $\mathbf{F}\cdot {\mathbf{e}}_y=x^2$. The square has $-1/2\le x,z\le 1/2$, so its outward flux is

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.