# Why are weak solutions to PDEs good enough?

Obviously solutions in $C^k$ are nicer but it seems people are happy with obtaining weak solutions in some Sobolev space that only satisfy the weak formulation. Why should this, in the real world, be seen as a "solution"? After all (unless you are lucky enough to be in the right dimension to apply an embedding theorem), you can't say that when $u \in L^2(0,T;H^1)$ that

$u(x_1,t_1)$ is the temperature of a fuel tank at $x_1$ at time $t_1$

So it is useless right?

• In the case of something like the heat equation. you have elliptic regularity: once you've produced a weak solution in, say, $L^2(0,T; H^1)$, you can show it's actually in Sobolev spaces of all orders, and then Sobolev embedding tells you it's $C^\infty$. In this case it doesn't matter what dimension you're in. – Nate Eldredge Apr 26 '13 at 19:25
• I didn't know that, surely the data $f$ must be in $C^\infty$ for that to work? – michael_faber Apr 26 '13 at 20:44
• Yes, that's right. I was thinking of the homogeneous equation. – Nate Eldredge Apr 26 '13 at 20:45

In other cases, the Sobolev space is sufficient. For instance, in the case of the heat equation, often it is desired to know how much heat energy is in the tank at a given time, what the flux of heat energy out of the domain is, how rapidly it will lose the energy it had to begin with, etc. These are all things for which a $L^2$ in time and $H^1$ in space solution are sufficient. In physical applications, rarely is the specific point value of a function important - it is averages over small patches in both time and space, which are integrals.