Parabolic PDE with no nonnegative solution

Question

I am looking for a simple example of a second-order parabolic linear PDE with locally bounded coefficients on an unbounded domain which admits no nonnegative solutions (except the trivial one).

As a candidate, I am considering: $$(\partial_t - \partial_{xx} - f)u = 0 \tag{*}$$ on the domain $(0,T) \times \mathbb{R}$, where $f \ge 0$ is some rapidly growing function of $x$, such as $f(x) = e^{x^4}$. Let us concentrate on classical solutions and require $u \in C^{1,2}((0,T) \times \mathbb{R}$.

Intuitively, the zero-order term $f$ should be inflating the function $u$ much faster than the second-order term $\partial_{xx}$ can diffuse it, forcing the solution to become infinite immediately. But I do not quite see how to prove this.

As a start, we can see that any nonnegative solution of (*) is a supersolution of the heat equation $$(\partial_t - \partial_{xx}) u \ge 0$$ and so obeys the strong minimum principle. Thus $u$ must be strictly positive everywhere, hence bounded away from zero on compact sets. Using the minimum principle to compare $u$ with solutions of the heat equation, we can get that for all $t \ge \epsilon$, $u(x,t) \ge C e^{-c x^2}$ for some $C,c$.

If I could get some control on the negative part of $\partial_{xx} u$, I could get something like $\partial_t u \ge fu \ge C' e^{x^4}$ for $t \approx \epsilon$. This would imply, by the mean value theorem, that $u(2\epsilon, x) \approx C'' \epsilon e^{x^4}$. Now the heat equation with initial conditions $e^{x^4}$ blows up, which is to say I can find arbitrarily large solutions $v_n \uparrow \infty$ of the heat equation whose values at time $2\epsilon$ are bounded by $e^{x^4}$. By the minimum principle, on $(2\epsilon, T)$, $u \ge v_n$ for all $n$, so $u \equiv \infty$.

I'd be interested in any suggestions to fix this gap, or other approaches. Thanks!

Answer 1

Disclaimer: this is a formal manipulation. It may be illegal, but I don't see it as morally reprehensible. Perhaps replacing $e^{-x^2/\alpha}$ with a compactly supported version of this function can make this rigorous.

Assuming that the solution remains smooth for all times $t<T$, we must have $$E_\alpha(t):=\int_{\mathbb R} u(x,t)e^{-x^2/\alpha}<\infty $$ for all $t<T/2$ and all $\alpha<T/2$ (otherwise the solution blows up before time $T$ even without further input from $f$). The function $E_\alpha$ ought to be smooth for any such value of parameter $\alpha$. Differentiate it with respect to $t$:
$$E_\alpha'(t)= \int_{\mathbb R} u_t (x,t)e^{-x^2/\alpha} = \int_{\mathbb R} u_{xx}(x,t)e^{-x^2/\alpha} + \int_{\mathbb R} u(x,t)e^{x^4-x^2/\alpha}$$ Integrate the first term by parts: $$\int_{\mathbb R} u_{xx}(x,t)e^{-x^2/\alpha} =\int_{\mathbb R} u(x,t)(4x^2/\alpha^2-2/\alpha)e^{-x^2/\alpha} $$ This integral converges because $x^2e^{-x^2/\alpha} \lesssim e^{-x^2/\beta}$ for any $\beta>\alpha$, and we can take $\beta\in (\alpha,T/2)$.

On the other hand, $\int_{\mathbb R} u(x,t)e^{x^4-x^2/\alpha}$ diverges due to your lower bound. Contradiction.

Answer 2

Your example $f(x)=e^{x^4}$ should work. According to what you know so far, whenever $t\ge\epsilon$, $$f(x)u(x,t)\ge e^{x^4}Ce^{-cx^2}\ge c_1 e^{x^4/2}=:g(x). $$ Fix a smooth enough increasing function $h$ approximating the Heaviside function, with $h(x)=0$ for $x<0$, $h(x)=1$ for $x>1$, and find $v(x,t)$ solving $$v_t-v_{xx}=h, \quad v(x,\epsilon)=0,$$ by the standard Duhamel formula. Then one should have $v(x,t)\ge \hat C e^{-\hat cx^2}$ for $t\ge2\epsilon$. Now for each integer $n>0$, because $u_t-u_{xx}\ge g$ and $g(x)\ge g(n)h(x-n)$, comparison yields for $t\ge2\epsilon$ that $$ u(x,t)\ge g(n)v(x-n,t) \ge c_2 e^{n^4/2}e^{-\hat c(x-n)^2} \to\infty $$ as $n\to\infty$. So $u$ cannot be finite.