Changing the space of test functions in PDEs

Question

Let $V \subset H \subset V'$ be a Hilbert triple.

We can define a weak derivative of $u \in L^2(0,T;V)$ as the element $u' \in L^2(0,T;V')$ satisfying $$\int_0^T u(t)\varphi'(t)=-\int_0^T u'(t)\varphi(t)$$ for all $\varphi \in C_c^\infty(0,T)$.

Then we define the space $W = \{u \in L^2(0,T;V) : u' \in L^2(0,T;V')\}$. We know for example that for $u, v \in W$, $$\frac{d}{dt}(u(t),v(t))_H = \langle u'(t), v(t) \rangle + \langle v'(t), u(t) \rangle.\tag{1}$$

Now suppose I change my space of test functions and define a weak derivative as an element $u' \in L^2(0,T;V')$ satisfying $$\int_0^T u(t)\varphi'(t)=-\int_0^T u'(t)\varphi(t)$$ for all $\varphi \in C_c^1(0,T)$.

Define $\tilde W = \{u \in L^2(0,T;V) : u' \in L^2(0,T;V')\}$ where the derivative is now with respect to these new test functions. How is this related to $W$? Do properties like (1) still hold?

I think yes, by uniqueness of weak derivatives. But I wanted to check in case I missed something.

Answer 1

The derivative you define is the distributional derivative, which can be defined on $C^\infty_c$ or $C^1_c$. But then, you restrict to weak derivatives $u' \in L^2(I;V')$, which means, you allow test functions even from from $H^1(I;V)$, which is the class for which the left hand side of the definition of your derivative makes sense.

Let for a moment be $V=\mathbb R$. Let $u$ be the Heaviside function with jump at zero. Then, its distributional derivative according to your first and second definition is the Dirac functional $\delta(x)$, which is not a weak derivative in your sense. By both of your definitions, you can even compute its distributional derivative $\delta'(x)$ with both your definitions. The next derivative $\delta''(x)$ is only defined on $C^2_c$ and thus would not be caught by your second definition. But as soon as a derivative is defined as a functional on $C^1_c$, it will be the same on $C^\infty_c$, since the latter is a subspace.

Answer 2

The main property that you would want in your weak derivative is that it defines a closed, unbounded operator on $L^2$. Suppose you have an unbounded operator $A$ (for you it would be the derivative) defined on a dense domain $D(A)$ in $L^2$. As such, $A$ need not be closed (it's graph in $L^2\times L^2$ is not a closed subspace). So the next step is to consider its closed extensions (operators whose graphs are closed subspaces of $L^2\times L^2$ which contain the graph of $A$ itself). Among these, there exists a minimal closed extension $\bar{A}$ with domain $D(\bar{A})\supseteq D(A)$ (its graph is the closure of the graph of $A$).

So, when it comes to the minimal closed extension $\bar{A}$, it doesn't matter which domain you pick for $D(A)$ as long as it produces the same $\bar{A}$. Two unbounded operators $A_0$ and $A_1$ (say the derivative on $C^\infty_c$ or $C^1_c$) define the same minimal closed extension, $\bar{A}_0 = \bar{A}_1$, if for instance the graph of $A_1$ is contained and dense in the graph of $\bar{A}_0$ (as it happens to be the case in your example).