The profile is an even function, so its Fourier transform is real andFor $\omega \ne 0$, integration by parts givesThereforeAt zero, direct integration gives the area of the triangle, namely two; the displayed quotient has the same limit. In particular $\hat{T}(\omega )\ge 0$ for every real frequency. This is the Fourier transform of a triangular function.

One can also see the nonnegative square structurally: $T=\frac{1}{2}{1}_{[-1,1]}\ast 1_{[-1,1]}$, so the convolution theorem gives $\hat{T}=\frac{1}{2}(2\sin \omega /\omega {)}^2$.

Use the Fourier transform conventionThe absolute integrability of $v$ makes this integral exist for every real $\omega$, with $|\hat{v}(\omega )|\le \Vert v{\Vert }_1$. With this convention the inverse Fourier transform has factor $1/(2\pi )$; maintaining that normalization is essential in part (e).

For a real solution, $u^4\ge 0$, so $\Delta u\ge 0$. The maximum principle for subharmonic functions gives $u\le {\max }_{\partial D}u=0$. The additional nonnegativity condition gives $u\ge 0$. ThereforeThis function satisfies both the equation and boundary condition, so it is exactly the full set of nonnegative solutions. The sign restriction is used alongside the maximum principle, rather than treating the nonlinear equation as a linear uniqueness problem.

If $u_1,u_2$ satisfy the same boundary value problem, their difference $w=u_1-u_2$ satisfies $Lw=0$ in $D$ and $w=0$ on the boundary, by linearity. Part (b)(ii) givesApply it again to $-w$, which also satisfies $L(-w)=0$ and zero Dirichlet boundary conditions. This gives $-w\le 0$. Hence $w=0$ throughout $\overline{D}$, provingThis proves uniqueness whenever a solution exists; existence is the separately supplied assumption. When the drift coefficients vanish, the same argument uses part (a).

The maximum principle for subharmonic functions here states that if $u\in C^2(D)\cap C(\overline{D})$ and $\Delta u=f\ge 0$ in $D$, thenIf $f\le 0$, apply this to $-u$ to obtain the corresponding minimum principle. In particular a harmonic function has both its maximum and minimum on the boundary. Without a sign condition on $f$, this conclusion need not hold.

First suppose $\Delta u>0$. By compactness and continuity, $u$ attains a maximum on $\overline{D}$. At an interior maximum, the first derivatives vanish and the Hessian matrix is negative semidefinite. Its trace is then nonpositive, contradicting $\Delta u>0$. Thus a maximum lies on the boundary.

For the non-strict case put $u_{ϵ}=u+ϵ(x^2+y^2)$, with $ϵ>0$. Then $\Delta u_{ϵ}=f+4ϵ>0$, so the strict case yields, for $z\in \overline{D}$,Let $ϵ\downarrow 0$. This gives $u(z)\le {\max }_{\partial D}u$, proving the theorem. The argument proves the weak maximum principle; it does not claim that every maximizer must be on the boundary when $u$ is constant.

The Picard-Lindelöf theorem gives local existence and uniqueness for an initial value problem when $f$ is continuous on a rectangle around $(a,b)$ and is Lipschitz continuous in its second variable there, with a constant uniform in the first variable. A sufficient hypothesis is that $f$ and $\partial f/\partial y$ are continuous on that rectangle.

More quantitatively, if the rectangle is $|x-a|\le h_0$, $|y-b|\le k$, and $|f|\le M$, a sufficiently small $h>0$ with $h\le h_0$ and $hM<k$ yields a unique differentiable solution on $[a-h,a+h]$ remaining inside the vertical strip. The proof uses the integral equationand Picard iteration, or the Banach fixed-point theorem after shrinking $h$ to make the integral operator a contraction. The continuity and Lipschitz hypotheses must hold near the actual initial point.