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I want to understand what weak solution actually means. This is a solution which is not as "smooth" and satisfies bi-linear formulation. So, what I am trying to understand is whether it satisfies the original equation at all or how it related to it. Say we constructed in some way $f(x)$ which is a weak solution of $u_t=u_{xx}+u_y$. I can't just plug $f$ and check because it is not twice differentiable in $x$, so what is that then? Can I treat $f_{xx}$ then as a second weak derivative and claim that in that case it satisfies the PDE?

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Consider a function $f \in L^1_{loc}(\mathbb{R})$. Instead of checking the values of $f$ at certain points, which doesn't make sense for members of $L^p$ in general, we can try to determine $f$ by integrating against test functions. This means that if $\phi$ is a smooth, compactly supported function with integral equal to $1$ and whose support is a very small neighborhood of $x_0$ then $$\int_{\mathbb{R}} \, f(x) \phi(x) \, dx$$ should be very close to "$f(x_0)$". Here again, the quotation marks emphasize that $f(x_0)$ is in general nonsense.

This way of evaluating functions leads us to the theory of distributions. Lots of functions $f$, in particular those in $L^2$, induce a linear functional $T_f$ on the space of smooth compactly supported functions (or "test functions") by $$T_f\phi = \int_{\mathbb{R}} \, f(x) \phi(x) \, dx$$ We identify $f$ with $T_f$ because of the first point made above.

The next step is to consider what it means to take the derivative of such a function. We say that $f \in L^2$ has a weak derivative $Df \in L^2$ if $$\int_\mathbb{R} \! Df \phi \, dx = - \int_\mathbb{R} \! f \phi' \, dx$$ for all test functions $\varphi$. In other words, we want the weak derivative to be a function $Df$ which induces a distribution $T_{Df}$ such that $$T_{Df}\phi = -T_f\phi'$$

This is a natural definition, for suppose $f \in L^2$ is differentiable in the classical sense. Then by integration by parts we see that $$\int_\mathbb{R} \! f'(x)\phi(x) \, dx = -\int_\mathbb{R} \! f(x)\phi'(x) \, dx$$ for all $\varphi$ so that the weak and classical derivatives agree. To see that we are truly generalizing, I leave it to you to find the weak derivative of the absolute value function on $[-1,1]$.

All of this can be generalized along similar lines to give weak derivatives of higher orders and to treat functions defined in basically arbitrary domains in $\mathbb{R}^n$.

So basically what we've done here is we're no longer evaluating functions a single points, but at "small open sets". By doing this, functions that are not really differentiable on a small subset can become weakly differentiable, and this weak derivative is defined in a natural way. A weak solution to some PDE just means that the equation interpreted in terms of weak derivatives holds. For example, to say that $f$ is a weak solutions of $$Df = g$$ means that $f$ is weakly differentiable and that its weak derivative as defined above is equal to $g$. Of course, mathematicians didn't start considering weak problems on purpose, it was out of necessity. By considering the functions with in $L^2$ with weak derivative in $L^2$, we can form a Hilbert space of "differentiable" functions which lends itself much more nicely to the analysis of partial differential operators than the typical spaces $C^k$. By the way, an exellent reference for all of this is Partial Differential Equations by Rauch, in particular chapter 2.

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