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For heat equations in $D,\mathbb{D}\subset \mathbb{R}^{2}$ with zero boundary:

$$u_{t}=\Delta u, u|_{\partial D}=0\text{ and }u(x,0)=f(x),$$ $$v_{t}=\Delta v, v|_{\partial \mathbb{D}}=0\text{ and }v(x,0)=f(g(x)),$$

where $f\in L^{1}(D)$ and conformal mapping $g: \mathbb{D}\to D$, are there any relations between $u$ and $v$?

Q1 To prove or disprove that for fixed t there is any transformation $F_{t}$ s.t. $F_{t}(u,g)(\cdot,t)=v(\cdot,t)$.

In the Dirichlet problem ($\Delta u=0,u|_{\partial D}=0$) we have $v(x)=F(u,g)(x):=u(g(x))$. So maybe some transformation F s.t. $F(u)=v$ eg. $$v(x,t)=\phi_{1}(x,t)\cdot u(\phi_{2}(x,t),\phi_{3}(x,t)).$$

Q2 In my case $f(x):=G_{D}(x,y)$ the Green function for domain D and by conformal invariance $G_{D}(g(x),g(y))=G_{\mathbb{D}}(x,y)$.

Attempts

1)Disproving the $F(u,g)(x,):=u(g(x),t)$. Computing heat eqn for $u(g(x),t)$ gives

$$\Delta (u(g(x),t))=\Delta u(g(x),t) |\nabla g|^{2}=\partial_{t}(u(g(x),t))|\nabla g|^{2}$$ and so only if $|\nabla g|^{2}=1$ we get that it satisfies the heat equation over $\mathbb{D}$.

2)Disproving any linear transformation. One is by taking computing the heat equation for $\phi_{1}(x,t)\cdot u(\phi_{2}(x,t),\phi_{3}(x,t))$. But I think the following identifies the core problem: $\lambda_{D,k}>\lambda_{\mathbb{D},k}$ for domain $vol(D)=vol(\mathbb{D})$ (conjectured for arbitrary k).

. The Dirichlet energy is conformal invariant and so consider eigenpairs $(f_{D,k},\lambda_{D,k})$ normalized in $H^{1}(D)$ i.e. $\int_{D}|\nabla f_{k}|^{2}=1$. Since $$\int_{D} |\nabla (f_{k}(x))|^{2}dx=\int_{g(\mathbb{D})} |\nabla (f_{k}(x))|^{2}dx=\int_{\mathbb{D}} |\nabla f_{k}(g(x))|^{2}| \nabla g(x)|^{2} dx=\int_{\mathbb{D}} |\nabla (f_{k}(g(x)))|^{2} dx$$

this basis is conformally invariant i.e. $f_{D,k}(g(x))$ is a $H^{1}(\mathbb{D})$-basis. By the representation formula wrt to heat kernel $g_{heat,D}(x,y,t):=\sum e^{-\lambda_{D,k}t}\lambda_{D,k}f_{k}(x)f_{k}(y)$: $$u(g(x),t)=\int_{D}g_{heat,D}(g(x),y,t)f(y)dy=\int_{\mathbb{D}}g_{heat,D}(g(x),g(y),t)f(g(y))|\nabla g(y)|^{2}dy=\sum e^{-\lambda_{D,k}t}\lambda_{D,k}f_{k}(g(x))\int_{\mathbb{D}} f_{k}(g(y))f(g(y))|\nabla g(y)|^{2}dy.$$

whereas

$$v(x,t)=\sum e^{-\lambda_{\mathbb{D},k}t}\lambda_{\mathbb{D},k}f_{D,k}(g(x))\int_{\mathbb{D}} f_{D,k}(g(y))f(g(y))dy.$$

So we are left with $\lambda_{D,k},\lambda_{\mathbb{D},k}$ that cannot be related by any linear transformation of arguments t and x.

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  • $\begingroup$ You have to restrict somehow the possible dependence of $F$ on $u$. Otherwise the answer is trivial: define $F$ as follows: take $u$ compute it at $t=0$, plug $g$ and solve the heat equation. This DEFINES SOME operator which does what you ask. $\endgroup$ – Alexandre Eremenko Jun 15 '17 at 6:09
  • $\begingroup$ I propose the requirement that $F(u,g)(x,t)$ depends only on $u(.,t)$, that if $F$ takes only values of $u$ at the moment $t$ as the input, not the whole $u$. $\endgroup$ – Alexandre Eremenko Jun 15 '17 at 6:11
  • $\begingroup$ @Eremenko agreed. But I like the idea of maybe relating them via Duhamels principle and having one as initial condition for the other like by repeated conformal mapping. I will try to formulate this better so it makes more sense. $\endgroup$ – OOESCoupling Jun 16 '17 at 20:41

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