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Suppose we have the PDE

$\frac{\partial u}{\partial t} - \Delta u = f(x,t)$

with some boundary conditions.

I am confused about what it means to say that the weak time derivative $u_t \in L^2([O,T];H^{-1})$. It makes no sense to me. My teacher has written:

We have $u_t = \Delta u + f(x,t)$ so $u_t \in L^2([O,T];H^{-1})$.

As far as I'm concerned, $u_t$ for a fixed $t$ is a function, not a functional that takes in a $H^1_0$ function and gives out a real number. What am I doing wrong?

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$u_t$ represents the derivative of $u$ wrt to $t$, but as a function of $(t,x)$ -- i.e. $t$ is not fixed. If your teacher did not define the space on the right you should ask him to do that. – user20266 Dec 7 '11 at 18:18
up vote 1 down vote accepted

The reason for this is that $\Delta u \in L^2([0,T];H^{-1})$, as $u \in L^2([0,T];H^1_0)$. Since the right hand side the PDE is in $L^2([0,T];H^{-1})$, the left hand side must be too.

The functional associated to $u_t$ is given by integration by parts:

$$\langle u_t, v\rangle = \int u_t v dx = \int (\Delta u + f)v dx = \int -\nabla u \cdot \nabla v + fv dx,$$

for $v \in H^1_0$. For each $v \in H^1_0$, $\langle u_t,v\rangle$ returns a real number.

EDIT: In general, if $u \in L^2$, then $\partial_{x_i} u \in H^{-1}$ as follows: for any $v \in H^1_0$, $\partial_{x_i} u$ acts on $v$ by the pairing

$$\langle \partial_{x_i} u,v\rangle = -\int u (\partial_{x_i} v) dx.$$

Note that when $u \in H^1$, this pairing agrees, by integration by parts, with the usual $L^2$ inner product

$$ \langle \partial_{x_i} u, v \rangle = \int (\partial_{x_i} u)v dx.$$

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But how is $\Delta u \in L^2([O,T];H^{-1})$? What does it mean to even write $\lVert \Delta u \rVert_{H^{-1}}$ when the Laplacian is just the sum of the 2nd derivatives? There doesn't seem any natural explanation to me. Or does one just take implicitly that $\Delta u$ is a functional such that $\Delta u (v) = \int \Delta u v dx$? – Court Dec 7 '11 at 19:03
Yes that's almost correct. You can think of $\Delta u$ as a mapping $\Delta u : H^1_0 \to \mathbb{R}$, given by the pairing $\langle \Delta u, v\rangle = -\int \nabla u \cdot \nabla v dx$. Since $u$ only has one weak derivative, as it is in $H^1_0$, you have to define the functional $\Delta u$ as I have above and not $\int \Delta u v dx$, although when $u$ is smoother, the two agree by integration by parts. – Jeff Dec 7 '11 at 21:23
Note that $v \in H^1_0$ in what I wrote above. Then the $H^{-1}$ norm of $\Delta u$ is just $\|\Delta u\|_{H^{-1}} = \sup_{\|v\|_{H^1_0}=1} \int \nabla u \cdot \nabla v dx$. Hope this helps. – Jeff Dec 7 '11 at 21:26
Thanks, that helps. – Court Dec 9 '11 at 9:17

A function $u:[0,T]\times \Omega\to\mathbb{R}$ can be viewed from different angles. The most natural one is the one which I have just written: Put in $t$ and $x$ and obtain $x$. However, this is not always appropriate (as you should learn is PDE-classes) and often it makes much more sense, to view them as elements in function spaces, e.g. $u\in L^p([0,T]\times\Omega)$ (this makes a difference, e.g. this view does not allow for point evaluation of $u$).

Another view is the one you need here: $u$ can be viewed as a mapping from $[0,T]$ into a function space. For each $t$ you get a function depending on $x$: $u(t,\cdot):\Omega\to\mathbb{R}$. If you have this view, than you write, e.g. $u:[0,T]\to L^p(\Omega)$ or $u:[0,T]\to H^1(\Omega)$. If you now impose regularity for this mappings you end up with spaces like $L^p([0,T],H^1(\Omega))$. These spaces are called Bochner spaces.

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Thanks for the post. – Court Dec 9 '11 at 9:17

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