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The characteristic curves solve $\dot{X}=X$, hence $X(t,\xi )=e^t\xi$. By the chain rule, $d[u(t,X(t,\xi ))]/dt=0$. Pulling back along the characteristic flow map givesFor $u_0\in C^1$, this is a classical solution: $u_t=-e^{-t}x{u}_0^{\prime }(e^{-t}x)$ and $u_x=e^{-t}{u}_0^{\prime }(e^{-t}x)$ cancel in the transport equation, and at $t=0$ the initial condition holds. Conversely, constancy along every characteristic curve forces this formula.

A characteristic curve has $x-at$ constant. Backtracking reaches the initial line if $x\ge at$ and the inflow boundary otherwise. HenceValues on the dividing characteristic curve are immaterial for bounded weak solutions.

For a classical solution on the closed quadrant, choose $u_0,f\in C^1([0,\infty ))$ and impose the corner compatibility for constant-speed transportThese make the values and both first derivatives agree across $x=at$; each branch solves the transport equation. They are also necessary for a $C^1$ solution up to the initial and inflow boundaries.

Spatial $C^{\infty }$ regularity for every $t\ge 0$, up to $x=0$, holds precisely for smooth data with all matching jets:Indeed the spatial derivatives of order $n$ at $x=at$ are $u_0^{(n)}(0)$ and $(-1/a{)}^n{f}^{(n)}(0)$. Necessity of smooth $u_0$ follows at $t=0$; smoothness of $f$ on any finite interval follows by reading the boundary branch in a spatial slice at a larger time.

In the bounded-domain setting of the weak maximum principle, let $u_1,u_2$ be two classical solutions with the same forcing and boundary values. Their difference $z=u_1-u_2$ satisfies $Lz=0$ and $z=0$ on the boundary. Apply the weak maximum principle first to $z$ and then to $-z$:Thus $u_1=u_2$, proving uniqueness of the Dirichlet problem. The same proof works on an unbounded domain if the difference and its negative satisfy the required condition at infinity. As literally printed without either restriction, uniqueness need not hold: on $\{x_n>0\}$ the two harmonic functions $0$ and $x_n$ have the same zero boundary values.

For the bounded-domain correction, let $\mathcal{S}$ be the family of Perron subfunctions for the Poisson equation $v\in C^0(\overline{\Omega })$ with $v\le \phi$ on $\partial \Omega$. The subfunction condition is ball comparison: whenever $\overline{B}\subset \Omega$ and a classical solution $H$ of $\Delta H=f$ on $B$ satisfies $v\le H$ on $\partial B$, it also satisfies $v\le H$ throughout $B$. For a $C^2$ function this is the sign condition $\Delta v\ge f$. A superfunction reverses both inequalities.

The Perron solution is the pointwise upper envelope of the subfunctions:This is the Perron method for the Dirichlet problem with the Poisson equation replacing the homogeneous Laplace equation. The power barrier for an exterior cone supplies a global member of $\mathcal{S}$ and a global superfunction, as constructed below. The permitted subfunction-superfunction comparison therefore makes this envelope finite. The standard interior Perron theorem, available in the question, gives $u\in C^2(\Omega )$ and $\Delta u=f$; neither result needs to be proved in this part.

For the corrected bounded-domain problem, the interior theorem for the Perron method for the Dirichlet problem gives a classical solution of the Poisson equation, and the power barrier for an exterior cone argument gives its continuous extension with the required Dirichlet boundary condition. Hence existence holds in $C^2(\Omega )\cap C^0(\overline{\Omega })$.

If $u_1,u_2$ are two such classical solutions, then $w=u_1-u_2$ is a harmonic function with zero boundary trace. The weak maximum principle for elliptic operators, applied to $w$ and $-w$ on the bounded domain, gives $w\le 0$ and $-w\le 0$. ConsequentlyThere is exactly one solution after adding boundedness of $\Omega$. For the printed unbounded-domain version, the half-plane counterexample in part (i) disproves uniqueness; an additional condition at infinity or a suitable restricted solution class would instead have to be specified.

Write $X(t,a)=X(t;0,a)$ and $A(t,x)=X(0;t,x)$. If $u_0\in C^1({\mathbb{R}}^d)$, the unique global classical solution isIt is $C^1$, and the chain rule makes its value constant along every characteristic curve. The same calculation proves classical uniqueness. No boundedness of $u_0$ is needed for this classical statement.

For $u_0\in L^{\infty }$, a bounded weak solution on the positive half-line is defined byfor every compactly supported $C^1$ test function on $[0,\infty )\times {\mathbb{R}}^d$. The analogous backward-time identity defines the solution on all real time. The $u\mathrm{div}F$ term is essential: this is transport, not the conservative equation $u_t+\mathrm{div}(Fu)=0$.

The same inverse-flow formula defines the unique weak solution, with $\Vert u(t){\Vert }_{\infty }=\Vert u_0{\Vert }_{\infty }$, equality almost everywhere, and a representative continuous into $L^1$ on each compact spatial set. For a direct uniqueness argument, change variables $x=X(t,a)$ in the weak identity and put $w(t,a)=u(t,X(t,a))$. Using $J_t=(\mathrm{div}F)(t,X)J$ givesinitially for $C^1$ compactly supported $\zeta$. Since $J>0$ and $J,J_t$ are continuous on compact sets, approximation in the spatial parameter permits $\zeta =\eta /J$: both $\zeta$ and ${\zeta }_t$ are uniformly approximable there, which are the only derivatives left in this identity. Hence $\int w{\eta }_t+\int u_0\eta (0)=0$ and $w=u_0$ almost everywhere. This also verifies existence by reversing the argument, and avoids assuming second derivatives of the $C^1$ field. Local $L^1$ continuity follows by approximating $u_0$ by continuous functions on compact flow tubes.

This is the weak transport solution under characteristic coordinates. The weak-strong uniqueness principle says that a weak solution in the specified class agrees with a classical solution having the same data for as long as that classical solution exists. Here it holds globally in the bounded class whenever the classical initial data are also bounded.

For a bounded $C^1$ classical solution and a smooth convex entropy $\eta$, let its entropy flux for a scalar conservation law satisfy ${\psi }^{\prime }=f^{\prime }{\eta }^{\prime }$. The chain rule givesIntegration against a nonnegative test function, including its initial boundary contribution, gives equality in the entropy inequality. Approximate a convex piecewise $C^1$ entropy by smooth convex functions on the bounded state range. Their values and flux antiderivatives converge uniformly, so the integrated equality passes to the limit. Therefore every bounded classical solution is an entropy solution. The bounded qualifier is required by the definition's $L^{\infty }$ membership; an arbitrary unbounded $C^1$ solution would not satisfy that membership condition.

Conversely, the affine entropies ${\eta }_{+}(z)=z$ and ${\eta }_{-}(z)=-z$ are both convex, with fluxes $f$ and $-f$. Their inequalities force the same linear weak integral to be both nonnegative and nonpositive. It is therefore zero for every nonnegative test, hence for every test by subtraction of nonnegative compactly supported tests. Thus $\boxed{\text{entropy solution}⟹\text{weak solution}.}$ No assumption that the flux itself is convex is used.

Set $V(x)=(1+x^2{)}^{-1}$ and $A_t={\int }_0^{t}V(B_s)ds$. On a fixed horizon $T$, the Itô formula and Itô product rule giveThe drift vanishes by the partial differential equation. This makes $M$ a continuous local martingale. Since $0\le A_t\le T$,The right side is a finite deterministic bound under the stated boundedness assumption, so the dominated convergence theorem removes a localizing sequence and $M$ is a true martingale. If the derivatives are also bounded, the Itô isometry alternatively proves the stochastic integral is square-integrable because $\mathbb{E}{\int }_0^T{e}^{2A_s}|u_x(T-s,B_s){|}^{2}ds\le T{e}^{2T}\Vert u_x{\Vert }_{\infty }^2$.

For the bounded parabolic differentiability class, distinguish global bounds from bounds on each finite time slab. Only the latter are needed here; a single global bound on the classical solution $u$ would conflict with its requested long-time growth in part (c).

For the true martingale of part (a), $M_0=u(T,x)$ and $M_T=f(B_T)e^{A_T}$. Taking expectations gives the Feynman-Kac formula with a bounded potential:The right side is finite since its absolute value is at most $e^t\Vert f{\Vert }_{\infty }$. This also proves uniqueness among classical solutions bounded on every finite time slab: applying the formula to their difference with zero initial data gives zero. The positive sign of the potential in the partial differential equation explains the positive exponent; changing it to a negative exponent would solve a different equation.

The infinite occupation time of one-dimensional Brownian motion impliesFor a start $x\ne 0$, first wait until $B$ hits zero, which happens in finite time by recurrence of one-dimensional Brownian motion, and apply the Strong Markov property. The repeated exit-and-return argument in Question 4(c) then proves the infinite occupation integral. Since $f\ge C>0$, the Feynman-Kac formula gives $u(t,x)\ge C{\mathbb{E}}_x{e}^{A_t}$. The random variables $e^{A_t}$ increase to infinity almost surely; the monotone convergence theorem therefore yields the intended positive-potential growth from Brownian recurrence:Under global boundedness, no such solution exists when the initial function is bounded below by a positive constant. The PDF prints $C_b^{1,2}({\mathbb{R}}_{+}\times \mathbb{R})$; if the subscript means bounded over the whole product, its hypotheses with $f\ge C>0$ are incompatible, rather than an example of growth within that class. The precise correction is to require $u$ and its indicated derivatives to be bounded on $[0,T]\times \mathbb{R}$ for every finite $T$. The derivation above then gives the requested limit. Already $f\equiv 1$ satisfies the printed initial-data assumption but cannot have a globally bounded classical solution, by this formula. Also the PDF's $u(x,t)$ in this clause reverses the argument order from $u(t,x)$ used in the equation; the limit above keeps the equation's time-first convention.

Let $T:H^1(U)\to L^2(\partial U)$ be the trace map from the Sobolev trace theorem. The weak Robin problem for a uniformly elliptic operator uses all of $H^1(U)$ as the test space, since a Robin boundary condition does not require a zero trace. DefineThe weak formulation isFor a classical solution, integration by parts produces the boundary term $-{\int }_{\partial U}(\sum a^{ij}{u}_{x_i}{\nu }_j)v$. Substituting the Robin boundary condition replaces it by ${\int }_{\partial U}\beta uv$, giving the displayed identity.

Conversely, if $u\in C^2(\overline{U})$ satisfies the identity, tests $v\in C_c^{\infty }(U)$ first show $(L+\lambda )u=f$ in the sense of distributions. Subtracting this interior equality after integration by parts givesEvery smooth boundary function is the trace of a smooth function on $\overline{U}$. The boundary residual is continuous, so this proves the boundary equation pointwise. Thus the formulations agree for $C^2$ solutions. If $f$ is continuous the interior equation holds pointwise; for merely $f\in L^2$, equality with that given representative of $f$ is understood almost everywhere.

Taking the Fourier transform of the equation in the sense of distributions givesThe Fourier multiplieris everywhere positive. Define $\hat{u}=\hat{f}/m$. Sincewe haveThus $u\in H^{s+4}$ exists and obeys the required estimate. If two solutions existed, their difference would have $m\hat{u}=0$; positivity of $m$ proves uniqueness.

A classical solution needs continuous derivatives through order four. Applying the Sobolev embedding theorem with $k=4$ to $u\in H^{s+4}$ shows that this is guaranteed whenor equivalently

Suppose a $C^1$ classical solution existed and set $w=u+iv$. The system is preciselywhich are the Cauchy-Riemann equations for $w$ as a function of the complex number $z=x+it$. Hence $w$ is holomorphic near $z=0$. Every holomorphic function is real analytic, so its restrictionto the real axis is real analytic near zero. Its real part and imaginary part show that both $f$ and $g$ must be real analytic, contradicting the hypothesis. Therefore no such $C^1$ solution exists.

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelöf theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

First choose a test function $v\in C_c^{\infty }(U)$. The weak formulation and integration by parts giveThe fundamental lemma of the calculus of variations implies $-\Delta u=f$ pointwise because $u\in C^2(\overline{U})$ and $f$ is continuous. The Dirichlet boundary condition on ${\Gamma }_1$ already follows from $u\in V$ and continuity of $u$.

For arbitrary $v\in V$, Green's first identity and the interior equation now reduce the weak identity toThe traces of smooth members of $V$ can be chosen freely on compact subsets of ${\Gamma }_2$. Another application of the fundamental lemma of the calculus of variations, now on the boundary, gives ${\partial }_{\nu }u=0$ pointwise on ${\Gamma }_2$. Hence $u$ is a classical solution of the complete mixed boundary value problem.

The characteristic flow map solves the ordinary differential equationand henceAlong this characteristic curve, the chain rule givesThe value is therefore constant, and tracing $(t,x)$ back to time zero gives the classical solutionDirect differentiation verifies both the linear transport equation and its initial value.

For every compactly supported $C^1$ test function $\phi$ on $[0,\infty )\times \mathbb{R}$, define a weak solution by the identityThe extra $\phi$ appears because ${\partial }_x(x\phi )=x{\phi }_x+\phi$. This identity is obtained from the linear transport equation by integration by parts in time and space.

Conversely, if $u$ and $u_0$ have the stated $C^1$ regularity, choosing test functions supported away from $t=0$ shows in the distributional sense that $u_t+x{u}_x=0$. Continuity makes the equation pointwise. Integrating that pointwise equation by parts in the displayed identity leavesfor all boundary test functions. The fundamental lemma of the calculus of variations gives $u(0,x)=u_0(x)$, so $u$ is a classical solution.

A continuous function $v$ is a Perron subfunction for $\Delta u=f$ when it lies below every classical solution of that Poisson equation on a relatively compact open ball whenever it lies below that solution on the ball's boundary. For $C^2$ functions, this is equivalent to $\Delta v\ge f$. A superfunction reverses the comparison and has $\Delta v\le f$ when $C^2$. Adding a boundary upper bound gives the family whose supremum defines the Perron method for the Dirichlet problem.