Consider a heat equation in one space dimension $$\frac{\partial u(t,x)}{\partial t} = \frac12\Theta(x)\frac{\partial^2u(t,x)}{\partial x^2} \tag{1}$$ where Heavyside function $$ \Theta(x) = \begin{cases} 0,\, x<0 \\ 1,\, x\ge 0 \end{cases} $$ with initial condition $$u(t=0,x) = x_+.$$

Obviously, $u(t,x)=0,\,\forall x<0$. In effect we may consider Eq. (1) as a heat equation with unit conductivity on the positive real axis. If we take this route, how should we specify the boundary condition for this heat equation on the positive axis? Would $u$ be continuous over $x$? Would the right hand side first spatial derivative be equal to the left hand side for positive time $t$, i.e. $\frac{\partial u}{\partial x}(t>0,x=0_+)=0$?

On the other hand, we observe, for the particular initial condition at hand, if we ignore the differentiability of $u(t>0,x)$ with respect to $x$ at $x=0$, $u(t,x)=x_+$ is a solution on the whole $x$-axis.

Further, for $t>0$ fixed, is $u(t>0,x)\in C^\infty(-\infty,\infty)$?

If it is true $\frac{\partial u}{\partial x}(t>0,x=0_+)=0$, I am then puzzled that it seems physically, the heat should diffuse to the left and accumulates at $x=0$ that the temperature $u$ at $x=0_+$ should be positive and higher than $x_+$ immediately to the right of $x=0$. If $\frac{\partial u}{\partial x}(t>0,x=0_+)=0$, it implies that the heat diffuses away to the right of $x=0$ and the temperature immediate to the right of $x=0$ is lower than $x_+$. How should one explain that?


Your question is slightly confusing for me, namely the question: How should we specify the boundary condition $\frac{\partial u}{\partial x}(t,x=0+)$? You do not have any boundary data, so why does it need to be Neumann?

A Neumann boundary condition is imposed in a weak sense by considering the following weak formulation: find $u\in L^2(0,T;H^1(-\infty,\infty))$ that satisfies \begin{align} \int_{-\infty}^\infty (u_t-\Theta(x) u_{xx})v&=\int_{0}^\infty (u_t-u_{xx})v\\ &=\int_0^\infty u_tv+u_xv_x\\ &=gv|_{x=0} \end{align} for all $v$ in $H^1(-\infty,\infty)$ (where $g$ is the Neumann data).

The problem does not come with boundary data because the problem is given on the real line (the only notion of boundary data is that $u\to 0$ as $|x|\to\infty$).

Why not impose Dirichlet data? If you know $u=0$ for $x<0$, why not impose $u=0$ at $x=0$? Then the solution is known by reflecting the fundamental solution, and the solution is continuous, since it must satisfy $u=0$ at $x=0$.

The solution would then be $$u(t,x) = \int_0^\infty(\Gamma(x-y,t)-\Gamma(x+y,t))u(0,y)\,dy,$$ where $\Gamma$ is the fundamental solution given by $$\Gamma(x,t)=\left\{\begin{array}{l l}&\frac{1}{4\pi t}e^{-\frac{x^2}{4t}}&t>0,\\ &0& t\le0.\end{array}\right.$$

Any smoothness properties could then be calculated.

  • $\begingroup$ Yes, the question is not very well posed. I apologize for that. I meant to ask what boundary condition should be imposed on the effective problem on the positive axis. My question is whether $u$ and its first spatial derivative should be continuous, and therefore $0$ across $x=0$, or not. Could you please look into that? I have edited my question. Please review. $\endgroup$ – Hans Dec 28 '15 at 21:51
  • $\begingroup$ you can calculate $u_x$ explicitly to find out if it is continuous across $x=0$, I think it makes sense to consider a Dirichlet BC, since this imposes the continuity of $u$ across $x=0$. $\endgroup$ – Ellya Dec 28 '15 at 22:09
  • $\begingroup$ You missed my question: IS $u(t>0,x)$ continuous with respect to $x$ across $x=0$ --- not that I want to impose the continuity condition but IS it or not? $\endgroup$ – Hans Dec 28 '15 at 22:26
  • $\begingroup$ Sorry, I answered that in my question, and indeed it is, this is because $\Gamma(-y,t)-\Gamma(y,t)=0$. $\endgroup$ – Ellya Dec 28 '15 at 22:27
  • $\begingroup$ But this solution comes about AFTER one has imposed the Dirichlet condition that $u(t>0,x=0_+)=0$. My question is whether one should impose such a boundary condition. We do not know in advance whether this is true. This is an EFFECTIVE boundary condition which should be derived from the original equation with variable conductivity $\Theta(x)$, NOT imposed arbitrarily. $\endgroup$ – Hans Dec 28 '15 at 22:32

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