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I'm looking for general existence of a PDE of the form

$$ f: U \times [0, \delta) \to \mathbb{R}$$

$$\frac{\partial f}{\partial t}(p,t) = F(f(p,t))$$

where $f(p,0)$ is prescribed and $F$ is non-linear but it is known that

$$|F(f) - F(g)| \leq K \|f(\cdot, t) - g(\cdot, t)\|_{C^1(U)}$$

I'm interested in how one would prove short time existence, and also if I can allow for $$ |F(f) - F(g)| \leq K \|f(\cdot, t) - g(\cdot, t)\|_{C^k(U)} $$ with $k \geq 2$ and still get the same existence. I'm struggling to find analogies to previous existence theorems, e.g. those for parabolic systems or using Picard-iterates (which would require a different lipschitz bound)

My PDE is similar in spirit to the following: solve $(\Delta - 1) u = 0$ on $\Omega \subseteq \mathbb{R}^n$ (a smooth convex set with boundary) and $u \big|_{\partial \Omega} = F(\nabla^2 f, \nabla f)$ with $f: \partial \Omega \to \mathbb{R}$. Then

$$\frac{\partial f}{\partial t}(p) = \frac{\partial u}{\partial \nu}(p) $$

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    $\begingroup$ The generalization to $k = 2$ certainly won't work, since the reverse-time heat equation falls into that category. // Also, can you be a bit more precise on what $F$ depends on? Is $F: \mathbb{R}\to\mathbb{R}$ and you literally plugging in $F\circ f$, or is the notation $F(f(p,t))$ more of a functional dependence? The example you gave in the bottom using the Dirichlet to Neumann operator cannot be really written as $F(f)$. $\endgroup$ Commented Feb 2 at 2:45
  • $\begingroup$ Hi @WillieWong, unfortunately it really is a Dirichlet-to-Neumann operator so I'm not sure if there's more dependence. If useful, I'm looking at $\partial_{\nu} u$ where $u = (-\Delta + 1)^{-1} (P(D^2f, Df))$. I think $P= P(v,q) = a^{i} v_i + b^j p_j + c^{ij} v_i p_j$ though. Does this help? $\endgroup$
    – JMK
    Commented Feb 2 at 17:45
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    $\begingroup$ So technically speaking your equation is not even a PDE (it is nonlocal); I am not convinced by the "hyperbolic-pde" tag in this case. (Did you mess up $p$s and $q$s in your definition of $P(v,q)$?) $\endgroup$ Commented Feb 3 at 1:58

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I don't think in the level of generality you are looking at you can say anything useful. Let me give two examples with very contrasting behaviors. Here I am, per your comment, allowing myself to think about non-local operators.

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When your equation is actually hyperbolic, there is hope. The simplest example of this is the following: Let $U = \mathbb{R}$ parametrized by $x$ and consider the transport equation $$ \partial_t f = \partial_x f $$ Indeed the mapping $F$ satisfies $$ |F(f) - F(g)| \leq \|f-g\|_{C^1} $$ For any $C^1$ data $f_0$ we can get a $C^1$ solution $f(x,t) = f_0(x-t)$.

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But your equation may not be hyperbolic. Now let $U = \mathbb{T}^1$. Let $T$ be the Fourier multipier

$$ \widehat{Tg}(k) = |k|^\epsilon \hat{g}(k) $$

for some small positive $\epsilon$. We have (very unsharp, I am being lazy)

$$ \|Tg\|_{L^\infty} \lesssim \|Tg\|_{H^{1/2 + \epsilon}} = \|g\|_{H^{1/2 + 2\epsilon}} \leq \|g\|_{H^1} \leq \|g\|_{C^1} $$

So the desired bound on the RHS holds (noting that $T$ is linear). The equation

$$ \partial_t f = Tf $$

with initial data $f_0$ can be solved via Fourier transform. You have

$$ \hat{f}(k,t) = \exp(|k|^{\epsilon}t) \hat{f}(k) $$

and you see you run into the same instantaneous blow-up problem as the reverse-time heat equation. (For example, there exists a sequence of solutions to the equation $\phi_n$ such that $\phi_n(\cdot,0)$ is uniformly bounded in $C^1$ but for any $t > 0$, the family of functions $\phi_n(\cdot,t)$ is not uniformly bounded even in $C^0$.)

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  • $\begingroup$ Thanks for this instructive answer! $\endgroup$
    – JMK
    Commented Feb 4 at 20:34

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