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Suppose we have the following one-dimensional generalized heat equation:

$$u_t(x,t)=g(x,t)\Delta u(x,t), \quad x\in \mathbb{R},t\in(0,\infty).$$

I need to prove that this equation is ill-posed, for some initial data and some particular $g(x,t)$. Is there any literature on these equations? I have found loads on equations like $$u_t=Δu(x,t)+f(x,t)$$

with different $f(x,t)$, but nothing where the Laplacian is multiplied by another function.

Thanks a lot guys!

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  • $\begingroup$ Why do you "need to" prove this? $\endgroup$ – Michael Renardy Apr 6 '16 at 15:56
  • $\begingroup$ Is just a first step for a a more complicated problem I should solve. I just dont know how to express the solution $u(x,t)$ when I got this $g(x,t)$. I know how to deal with if for example i got some constant $\kappa$ where the sign of $\kappa$ would say if the problem is ill/well posed. However I do not have a constant, but a more generic function $g(x,t)$. $\endgroup$ – user105554 Apr 6 '16 at 16:02
  • $\begingroup$ For example I could suppose $g(x,t)<0$ well defined on an interval $(-a,a)$ and for a short time $t>0$. Then the problem is ill posed as for the backward heat equation? Thanks Michael. $\endgroup$ – user105554 Apr 6 '16 at 16:06
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Let me write an answer which is in fact too long for a comment: the heat equation itself is ill-posed locally in space. Consider the fundamental solution of $\partial _t-∆_x$ ($t$ is the time variable, $x$ is the space variable in $\mathbb R^d $)$$ E(t,x)=H(t)(4π t)^{-d/2}\exp{-\frac{\vert x\vert^2}{4t}}. $$ That function is locally integrable (is also a tempered distribution), with support $\{t\ge 0\}$ and is $C^\infty$ on $(\mathbb R_t\times\mathbb R^d_x)\backslash\{(0,0)\}$. As a result we have for $x_0\not=0$, a smooth function $E$ such that $$ \partial _t E-∆_x E=0,\quad\text{on $\{(t,x)\in \mathbb R\times B(x_0,\vert x_0\vert)\}$},\quad \text{supp } E\subset\{t\ge 0\}, $$ violating unique continuation locally.

If you require the equation to be globally satisfied wrt the space variable, you do have well-posedness results for the heat equation. However the above example shows that you should be more precise about the notion of well-podedness that you want to use, in particular with respect to localization.

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  • $\begingroup$ I want for example that it is ill posed in the sense that the solution to the equation has Fourier coeffiecients which grow up superexponential, and therefore it cant be in any Sobolev space (like for the Backwards heat equation). How ever when I have the function $g(x,t)$ , I do not know how to compute the solution (by Fourier methods for example for the Backwards Heat equation). Thank for your comment! $\endgroup$ – user105554 Apr 7 '16 at 14:08
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If your particular function $g(x,t)$ is identically $-1$, then it becomes $$u_t(x,t)+\Delta u(x,t) =0\ \ \ x∈ℝ,t∈(0,∞)$$ which is equivalent to the backward heat equation. And it is known that this is not in general well-posed.

Sincerely,

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  • $\begingroup$ Yes of course. But I have a more general problem: what kind of conditions you need to impose on g(x,t) so that the problem is ill-posed. Supose for example g(x,t) is negative on some interval (-a,a) and time (0,t'). Then is the problem ill-posed? How do you prove it ( Fourier techniques)? Thanks in advanced $\endgroup$ – user105554 Apr 6 '16 at 10:52
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If $g(t,x)$ is bounded smooth and say $g(t,x)\le-1$ on a strip $0\le t\le T$, then one could solve the problem backwards starting from $t=T$ with rough data. By general properties of parabolic equations the solution becomes smooth for $t<T$. Reversing time in the usual direction, you have constructed smooth data at $t=0$ and a solution which becomes rough at $t=T$.

This answers in a rather weak sense your question; I guess you are interested in stronger forms of ill-posedness. Probably playing with the previous idea (e.g., repeating at a sequence of times $T_n\to0$ and summing the resulting solutions) it should be possible to produce reasonable data for which no local smooth solution exists, but one should check the details.

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  • $\begingroup$ Piero thanks for your answer! Can you give me some references where I can look at this tipe of results? How can you solve the problem backwards? Is there any explicit way to compute the solution u(t,x) like for the case where g(t,x) was a constant? $\endgroup$ – user105554 Apr 11 '16 at 8:34
  • $\begingroup$ Backwards, this is a standard parabolic equation. Any book on parabolic equations gives existence and regularity results of this type. Explicit formulas are not available in general. In the case the coefficient does not depend on $t$ you have very strong estimates, see any book on Heat Kernels e.g. Davies' $\endgroup$ – Piero D'Ancona Apr 11 '16 at 14:44
  • $\begingroup$ In this case, coefficients depends on both variables, time and space. That is one a my main problems, that I didnt found any reference for such equations. However thanks for the reference. I will take a look!! $\endgroup$ – user105554 Apr 11 '16 at 14:50
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The semi-linear case $\partial_tu+Lu=f(u,\nabla u)$, where $L$ is an elliptic linear operator (like the Laplacian $-\Delta$), has a rather simple philosophy. One follows the guidelines of the Cauchy-Lipschitz theory for ODEs. If $a\in X$ is the initial data, one rewrites the Cauchy problem as an integral equation $$u(t)=e^{-tL}a+\int_0^t e^{(s-t)L}f(u(s),\nabla u(s))ds=:Nu(t).$$ When $T>0$ is small enough and the Banach space $X$ is appropriate, one proves that $N$ is a contraction in some ball $B(a;r)$ of $C(0,T;X)$. Then Picard's theorem tells that there is a unique fixed point $u$ ; this is the local solution.

By appropriate, I mean that the operators $S_t:=e^{-tL}$ form a strongly continuous semi-group over $X$. In particular $S_t\in{\mathcal L}(X)$ and $$\|S_t\|_{{\mathcal L}(X)}\le Ce^{\omega t},\qquad\forall t>0$$ for suitable constants $C$ and $\omega$. In practice, we may deal with a scale of Banach spaces $X_s$, like the Sobolev spaces $H^s$, and we have $$\|S_t\|_{{\mathcal L}(X_s,X_r)}\le Ct^{\alpha(s-r)}e^{\omega t},\qquad\forall t>0$$ for some $\alpha>0$, which is related to the order of $L$.

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