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Consider equation $$\partial_t u = \partial_x u + \partial_{xx} u - c u + f, \hbox{ on } (t, x) \in (0, \infty) \times \mathbb R$$ with initial condition $u(0, x) = g(x).$

Suppose that $c(t, x)$ and $f(t,x)$ are continuous in $(t, x)$ and $\phi (\cdot) = c(t, \cdot), f(t, \cdot), g(\cdot)$ satisfy $$|\phi|_0 + |\partial_x \phi|_0 + |\partial_{xx} \phi|_0 <K$$ for some $K>0$. [Question.] Is there unique classical solution for the equation with the above conditions?

Remark: I have seen that some conditions for the existence requires at least Holder continuity in $t$ for $c$ and $f$. I want to know if it is still true by dropping Holder $t$-continuity?

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  • $\begingroup$ Solution is not unique if not to require some growth bounds for solutions at infinity. mathoverflow.net/questions/89120/… $\endgroup$
    – Andrew
    Jun 26, 2020 at 17:02
  • $\begingroup$ Yes it's true if to drop Holder $t$-continuity. $\endgroup$
    – Andrew
    Jun 26, 2020 at 17:04
  • $\begingroup$ @Andrew Yes, true. I concentrated on existence which is where regularity enters. $\endgroup$ Jun 26, 2020 at 17:26

2 Answers 2

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In your case the domain of the operator is $UCB^2(R)$ (uniformly continuous and bounded functions up to the second derivative) and $f$ is continuous with values in the domain of the operator. Semigroup theory yields that the mild solution is a classical one, that one you are looking for. In the case of $D^2$, you can also write down the explicit formula for the solution $u$ and check that $u_{xx}$ exists, by differentiating $f$ under the integral. Then one needs an argument for $u_t$...this is the point where I prefer semigroup theory. Another possibility is to approximate $f$ with $f_n$, better in $t$, consider $u_n$ the corresponding solutions and let $n \to \infty$. Then $u_n \to u$, $(u_{n)_{xx}} \to u_{xx}$, using the fundamental solution, and then, by difference, $(u_n)_t$ also converges.

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  • $\begingroup$ It is helpful, I will look for details as you suggested. $\endgroup$
    – kenneth
    Jun 26, 2020 at 15:10
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The first attempt with method of continuity does not go through. This is second attempt based on the hints of other replies. The answer seems to me Yes if the state domain is changed to 1-torus $\mathbb T$ from $\mathbb R$.

  • [Conclsion] There exists $u\in C^{1,2}([0, T]\times \mathbb T)$.

Let $$F(c, u) = F[c] u = \partial_t u - \partial_{xx} u - \partial_x u + cu.$$ We use the following claims. (C1) and (C2) are standard, and (C3) will be proved later.

  • (C1) The operator $F: C^{0,0} \times C^{1,2} \mapsto C^{0,0}.$ is a continuous mapping

  • (C2) If $c, f$ are Holder continuous, then exists unique $u\in C^{1,2}$, s.t. $F(c, u) = f$, i.e. $u = F^{-1}[c] f$ is well defined.

  • (C3)For $c_i, f_i \in C^{1,3}$ satisfying $|c_i|_{0,2} + |f_i|_{0,2} < K$ with $i = 1, 2$, we have
    $$|F^{-1}[c_1] f_1 - F^{-1} [c_2] f_2 |_{1,2} \le\Psi(K, T)( |f_1 - f_2|_{0,2} + |c_1 - c_2|_{0,2})$$ for some strictly increasing continuous function $\Psi(K, T)$ with $\Psi(0, T) = \Psi(K, 0) = 0.$

  • (C4) if $\phi \in C^{0,2}$, then there exists $\phi_n \in C^{1,3}$ s.t. $|\phi_n - \phi|_{0,2} \to 0$. This is the place why the domain is changed to $\mathbb T$. Indeed, one can use polynomial $p_n$ approximate $\partial_{xx} \phi$ by stone-weirstauss and take integral twice to get $\phi_n$.

Now, let $c, f\in C^{0,2}$ be given. Then, since $C^{1,3}$ is dense in $C^{0,2}$, there exists $c_n, f_n \in C^{1,2}$, s.t. $$|c_n -c|_{0,2} + |f_n -f|_{0,2} \to 0$$ and $$|c_n|_0 + |f_n|_0 \le 2(|c|_0 + |f|_0) := K.$$ We denote $u_n = F^{-1} [c_n] f_n$. Then, $u_n$ is Cauchy in $C^{1,2}$ since by (C3) $$|u_n - u_m|_{1,2} \le \Psi(K, T) (|f_n - f_m|_{0,2} + |c_n - c_m|_{0,2}).$$ So there exists $u\in C^{1,2}$ s.t. $|u_n - u|_{1,2} \to 0$. Now we can verify $u$ is the solution by checking $$F(c, u) = \lim_n F (c_n, u_n) = \lim_n f_n = f.$$ In the above, we used (C1).

The remaining part is the proof of (C3). For $c_i, f_i \in C^{1,3}$ satisfying $|c_i|_{0,2} + |f_i|_{0,2}< K$ with $i = 1, 2$, $u_n = F^{-1} [c_n] f_n$ for $n=1 , 2$ is a classical solution by (C2) and $v_n (t, x) = u_n(T - t, x)$ has probability representation of the form $$v_n(t, x) =\mathbb E \Big[ \int_t^{T} \exp\{- \int_t^{s} c_n(r, X^{t,x}(r)) dr\} f_n(s, X^{t,x}(s) )ds\Big] $$ where $$X^{t, x} (s)= x + (t-s) + W(s) -W(t).$$ By direct approximation, one can have $$|v_1 - v_2|_0 \le KT^2 e^{KT}(|f_1 - f_2|_0 + |c_1 - c_2|_0) .$$ This also holds for \begin{equation} \label{eq:01} |u_1 - u_2|_0 \le KT^2 e^{KT}(|f_1 - f_2|_0 + |c_1 - c_2|_0) := \Psi(K, T)(|f_1 - f_2|_0 + |c_1 - c_2|_0) . \end{equation} Next, we can check that, by (C2) $\bar u_n = \partial_x u_n$ is the classical solution of $$\partial_t \bar u_n = \partial_x \bar u_n + \partial_{xx} \bar u_n - c_n\bar u_n - \partial_x c_n \cdot u_n + \partial_x f_n$$ with $\bar u_n(0, x) = 0.$ Similarly, we have $$|\partial_x (u_1 - u_2)|_0 \le \Psi(K, T)(|(- u_1\partial_x c_1 + \partial_x f_1) - (- u_2\partial_x c_2 + \partial_x f_2)|_0 + |c_1 - c_2|_0).$$ Combined with the earlier estimation on $|u_1 - u_2|_0$, we have $$|u_1 - u_2|_{0, 1} \le \Psi(K, T) (|f_1 - f_2|_{0, 1} + |c_1 - c_2|_{0,1}).$$ Using exactly the same approach, we have $$|u_1 - u_2|_{0, 2} \le \Psi(K, T) (|f_1 - f_2|_{0, 2} + |c_1 - c_2|_{0,2}).$$ Together with original equation $\partial_t u = ...$, we have final estimation $$|u_1 - u_2|_{1, 2} \le \Psi(K, T) (|f_1 - f_2|_{0, 2} + |c_1 - c_2|_{0,2}).$$ This completes the proof of (C1).

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  • $\begingroup$ I am not convinced since $L_0$ maps $C^{1,3}$ to $C^{0,1}$. $\endgroup$ Jun 26, 2020 at 11:39
  • $\begingroup$ agreed, my proof is wrong. $\endgroup$
    – kenneth
    Jun 26, 2020 at 12:38

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