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Elliptic and parabolic PDE share many properties. They each, for example, have an associated maximum principle and their value at any point depends on the entirety of the boundary data.

I have been told that estimates for solutions to parabolic PDE typically mirror those for elliptic PDE, but are also more difficult to prove. Since an elliptic PDE can be thought of as the steady state solution to a parabolic PDE, I am tempted to think of parabolic results as more general.

My question: Is there some common method used to get results for elliptic PDE from results for parabolic PDE?

My first guess would be to take $t\rightarrow \infty$ of a solution and, provided you have convergence in some appropriate sense, you get a solution to the associated steady-state problem. However, I haven't found any source that does this.

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Ok, I think I found an example of this. Take a solution of $$(1) \qquad \partial_t u + L u=0$$ Assuming we can justify the following steps, we integrate in $t$ to get $$\int_0^\infty (\partial_t u + Lu) d\tau = 0$$ $$\lim_{t\rightarrow \infty} u(x,t) - u(x,0) + L \int_0^\infty u d \tau = 0$$ $$-u(x,0)+L\int_0^\infty u d \tau = 0,$$ where in the last step I have assumed $u(x,t)$ decays in $t$.

Thus, if we set $u(x,0)\equiv 0$, and let $u$ be the solution to (1), then $$(2) \qquad v(x)=\int_0^\infty u(x,\tau) d\tau$$ solves the steady-state problem $Lv=0$.

Now, if we have a heat kernel representation for $u$, then the integral is (2) can be computed, and we should be able to derive estimates for $v$ from $u$.

Edit: Follow-up questions... What boundary conditions does $v$ satisfy? Can we do this for inhomogeneous equations (i.e., for solutions to $Lv=f$, $f\not\equiv 0$)?

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    $\begingroup$ $v$ would satyisfy BC's that are the integral of the BC's for $u$, this should be fine, for the $Lv=f$ you could have $u_t-Lu=f(x,t)$, since then $Lv=\int_0^\infty Lu=-\int_0^\infty f(x,\tau)=:g(x)$ $\endgroup$
    – Ellya
    Jul 31, 2015 at 18:27
  • $\begingroup$ Notice that you needn't even set $u(x,0)=0$. $\endgroup$
    – Ellya
    Jul 31, 2015 at 18:29
  • $\begingroup$ Thanks, ellya. I found a related question here, for specific case of $L=-\Delta$. $\endgroup$
    – Mit
    Jul 31, 2015 at 18:37

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