I have theese two problems:

(I) Find the unique distribution $u \in H^1(-1,1)$ such that $-u'' = \delta_0$ and $u(-1) = u(1) = 0$.

(II) Let $c > 0$, find the unique distribution $u \in H^1(\mathbb{R})$ such that $-u'' + cu= \delta_0$

Because we want to use Lax-Milgram's theorem, solving theese problem we began like this:

$(I)$ $u \in H^1(-1,1)$ satisfies $-u'' = \delta_0$ in the distributional way if and only if for all $\varphi \in C_c^\infty(-1,1)$

$$\begin{cases} u \in H^1_0(-1,1)\\[8pt] \int_{-1}^1 u' \varphi' \ dx = \varphi(0) \end{cases}$$

$(II)$ $u \in H^1(\mathbb{R})$ satisfies $-u'' +cu= \delta_0$ in the distributional way if and only if for all $\varphi \in H^1(\mathbb{R})$ $$ \begin{cases} u \in H^1(\mathbb{R}) \\[8pt] \int_{-1}^1 u'\varphi' +cu\varphi \ dx = \varphi(0) \end{cases}$$

My questions are:

Why in $(I)$ we took $\color{red} {u \in H^1_0(-1,1)}$ and not $u \in H^1(-1,1)$ ?

Why in $(II)$ we took $\color{red}{\varphi \in H^1(\mathbb{R})}$ and not $\varphi \in C_c^\infty(\mathbb{R})$ ?

What is the general approach to be sure to choose "the right space for the right function" in this kind of problems?

I'm studying a course of a master's degree (Italy) in mathematics and as you may observe in my last questions on this site, this course was really linked to the solutions of the Dirichlet problem.

Introducing the "weak derivateive" of a function we mentioned that distributions live in spaces where there is not a normal topology to define continuity (taken for example from a norm) but we did not go deeper.

We defined the Sobolev space $W^{k,p}(\Omega)$ as follows: $$W^{k,p}(\Omega) := \{u\in L^p(\Omega) \ : \ D^\alpha u \in L^p(\Omega) \ \text{for all}\ \alpha : |\alpha| \le k\}$$ where $D^\alpha u$ is used as a notion of the distributional derivative of $u$, that in our definition is the distribution that acts on every $\varphi \in C^\infty_c(\Omega)$ as $$\langle D^\alpha u , \varphi \rangle := (-1)^{|\alpha|}\langle u , D^\alpha \varphi \rangle$$

We define $H^k(\Omega) := W^{k,2}(\Omega)$ and $W^{k,p}_0(\Omega)$ is the closure of $C^\infty_c(\Omega)$ inside $W^{k,p}(\Omega)$

Here I state Lax-Milgram theorem:

Let $H$ be an Hilbert real space and let $a:H\times H \to \mathbb{R}$ be a bilinear, continous and coerced form. Then for all $F\in H^*$ there exists one and one only $\bar{u} \in H$ such that $$a(\bar{u},v) = F(v) \qquad \forall v \in H$$ If $a$ is also symmetric then $\bar{u}$ satisfies $$\frac{1}{2}a(\bar{u},\bar{u}) - F(\bar{u}) = \min_{v\in H} \Big\{\frac{1}{2}a(v,v) + F(v)\Big\}$$

  • $\begingroup$ For the first question, you take $u\in H^1_0$ to satisfy the condition $u(-1)=u(1)=0$. For $u\in H^1$, this is not necessarily true and you won't have a unique solution. For the second, the only reason I can think of is that theorems for existence/uniqueness often hold for bi-linear forms over $H\times H$, with $H$ a Hilbert space. $\endgroup$ – Rem Jan 28 at 17:13
  • $\begingroup$ @Rem But for any $u \in H^1(-1,1) = W^{1,2}(-1,1)$ there exists a continous representative $\bar{u}$ so why the border conditions are not necessarily true for $\bar{u}$? $\\$ And about your second sentence we are taking the bilinear form on $u,v \in H^1(-1,1)$ and this is an Hilbert space, doesn't it? I think I am missing something $\endgroup$ – Gabrielek Jan 29 at 10:25
  • $\begingroup$ @Gabrieled Why would $\bar u \in C^0(-1,1)$ satisfy $\bar u(-1)=\bar u(1) = 0$? I am not sure to I understand your question. For the second question, you solve the equation in the distribution's sense, i.e. you want to define a bi-linear form $B(u,v)$ such that $B(u,v)=<\delta_0, v>_{\mathcal D', \mathcal D}$. Typically, you then apply Lax-Milgram theorem to prove existence and uniqueness. But this theorem works for $B:H\times H\mapsto \mathbb R$. That is one of the reason you would take $\varphi\in H^1(\mathbb R)$ and not as a smooth function. $\endgroup$ – Rem Jan 29 at 11:30
  • $\begingroup$ I am really new to this kind of math and I think I have not understood very much about this topic. Thank you for your time. $\endgroup$ – Gabrielek Jan 29 at 13:41
  • 1
    $\begingroup$ I think this is enough, at least it seems to be what I wanted. $\endgroup$ – Teresa Lisbon Feb 3 at 11:37

$\newcommand{\d}[1]{\underline{\mathrm{\large{#1}}}}$ $\color{#db6}{\d{On\ picking\ the\ right\ space\ for\ the\ right\ function}}$

Equality in distribution is interpreted in terms of pointwise equality for functions in $C_c^{\infty}$. The "right space" for a function is picked by seeing its role in the resulting equation phrased for $C_c^\infty$ functions, suitably changed so as to facilitate the use of Lax-Milgram lemma. Suitable changes are performed via transferring derivatives to elements of $C_c^{\infty}$. However, note that the transfer of derivative should be done , while keeping in mind that we need a bilinear form on a space $H$. So the transfer of derivative is done in such a way that a bilinear form is definable and we can use the Lax-Milgram lemma. Usually that space $H$ is a Sobolev space. So the transfer of derivative is (usually) done so that we get the same amount of derivatives on both the test function and the variable function.

Let's take an example to illustrate this. Suppose you had a function $f$, and you wanted to find a $u$ such that $\Delta u = f$. Let's say all this is on a bounded open set $\Omega$, and for boundary conditions we want $u = 0$ on $\partial \Omega$. Classically, you'd demand that $f \in C(\Omega)$ and $u \in C^2(\Omega) \cap C(\overline{\Omega})$.

For the weak formulation : we will have $\int_{\Omega} \phi \Delta u = \int_{\Omega} \phi f$ for all $\phi \in C_c^{\infty}(\Omega)$. Using Green's theorem (an integration by parts for general domains), you can change the RHS to $-\int_{\Omega} \nabla \phi \nabla u = \int_{\Omega} \phi f$ for all $\phi \in C_c^{\infty}(\Omega)$. So now there's one derivative on both $\phi$ and $u$.

We stop now to think about what Lax-Milgram would look like : let's first start with $f$. We usually insist that $f \in L^2(\Omega)$ : this almost never changes for two reasons. One is that there's no derivative on $f$ so it doesn't need to be constrained, and second , the right hand side becomes a continuous functional on $C_c^\infty(\Omega)$ (continuous with respect to the topology you have not discussed, so I won't go into why it's continuous), so we can extend using density to $H_0^1(\Omega)$ if we need.

And we do need to : by continuity, if $\int_{\Omega} \nabla u \nabla \phi = \int_{\Omega} f \phi$ for all $\phi \in C_c^\infty(\Omega)$, then density gives that the same holds for all $\phi \in H_0^1(\Omega)$. But now, $\phi \in H_0^1(\Omega)$, and we need a bilinear form involving $u$ and $\phi$, so it is natural to take $u \in H_0^1(\Omega)$ as well, just so that the bilinear form $B(f,g) = \int_{\Omega} \nabla f \nabla g$ is well-defined on $H_0^1(\Omega) \times H_0^1(\Omega)$ and we can think about Lax-Milgram.

Note that if $u \not \equiv 0$ on $\partial \Omega$, then we can't take $u \in H_0^1(\Omega)$, because a necessary condition for membership in $H_0^1(\Omega)$ is that it should vanish on the boundary, because it is a (a.e. pointwise) limit of functions that have compact support so vanish on the boundary (This particular assertion is actually beyond your reach, since you don't know the topology on $C_c^{\infty}(\Omega)$. Take it at face value). Then we cannot use the Lax-Milgram lemma. We'd have to use something else, but most likely the bilinear form in use will be the same.

So that's how we choose our spaces : make sure that we are transferring derivatives to create the same Sobolev space in which both $u$ and $\phi$ can freely reside, and then we can think about Lax-Milgram.

$\color{blue}{\d{The\ first\ question}}$

In our example, much the same happens.

Indeed, note that we want $-u'' = \delta_0$. Following integration with test functions, we land up with : $$ \forall \phi \in C_c^\infty(-1,1) \quad ; \quad \int_{-1}^1 u'\phi' = \phi(0) $$

(I omit the $dx$ at the end of the integral, to focus on the integrand. This is quite common in PDE) Now , by density of $C_c^\infty$ we get that the same holds for $\phi \in H_0^1(\Omega)$. Now, to ensure that Lax-Milgram comes into play, we insist on $u \in H_0^1(\Omega)$, which doesn't conflict because $u(-1)=u(-1) = 0$ so $u$ is $0$ on the boundary. (Refer to second last paragraph of previous section)

Our bilinear form is then quite obviously $B(f,g) = \int_{-1}^1 f'g'$, and the $F \in (H_0^1(\Omega))^*$ will be given by $F(\phi) = \phi(0)$. Then we can frame our question as : does there exist a unique $u \in H_0^1(-1,1)$ such that $B(u,\phi) = F(\phi)$ for every $\phi \in H_0^1(\Omega)$?

We are in Lax-Milgram territory! All we need is that $B$ be a bilinear, continuous, coerced form. I won't explain why this happens in detail : continuity follows since $|B(u,v)| \leq \|u\|_{H_0^1} \|v\|_{H_0^1}$ (follows from the definition of the norm in $H_0^1$), and coercivity follows from the Poincare inequality, from which one gets that there is a constant $C>0$ such that $B(u,u) \geq C\|u\|^2_{H_0^1}$. Thus, by using the Lax-Milgram lemma, one sees that a $u$ satisfying the given conditions exists and is unique.

EDIT(UNRIGOROUSLY DONE) : To find $u'' = \delta_0$, we need to "integrate" $\delta_0$ twice, keeping track of the initial conditions. Let $v$ be such that $v(x) = v(-1) + \int_{-1}^x \delta_0(x)dx$. Note that $\int_{-1}^x \delta_0(x)dx = 0$ if $x<0$ and $1$ if $x \geq 0$. Therefore, we get that $v(x) = v(-1)$ for $x<0$ and $v(-1)+1$ for $x \geq 0$.

Finally, $u(x) = u(-1) + \int_{-1}^x v(x)dx$ so $u(x) = u(-1) + v(-1)(x + 1)$ for $x <0$ and $u(x) = u(-1) + v(-1) + xv(-1)+x$ for $x \geq 0$. We need to ensure that $u(-1)=u(1)=0$. From that, setting $x=1$ we get $u(1) = 2v(-1) +1 = 0$ so $v(-1) = \frac{-1}{2}$.

Thus, $u(x) = \frac{-(1+x)}{2}$ for $x<0$ and $\frac{x-1}{2}$ for $x \geq 0$.

So that's how solve problems using the Lax-Milgram lemma. Guessing solutions requires "physical intuition" on how $\delta$ functions are integrated.

$\color{red}{\d{The\ second\ question}}$

Here things get a little thicker at the weak formulation stage : we have $$ \int_{-1}^1 [u'\phi' + cu\phi] = \phi(0) \tag{1} $$

for every $\phi \in C_c^\infty(\mathbb R)$. Now we want to apply Lax-Milgram, so we want to ensure that $u$ and $\phi$ are picked from the same space so we can let the LHS of the above equation be a bilinear form.

What is that Hilbert space? At first, it's a toss up between $H_0^1(\mathbb R)$ and $H^1(\mathbb R)$. But the point is, that $\mathbb R$ is unbounded, so it has no boundary. Therefore, we expect that $H_0^1(\mathbb R) = H^1(\mathbb R)$ i.e. that $C_c^\infty(\mathbb R)$ is in fact dense in $H^1(\mathbb R)$. This is a remarkable fact that is true!

From that fact and continuity, if equation $(1)$ was true for all $\phi \in C_c^\infty(\mathbb R)$ it automatically is true for all $\phi \in H^1(\mathbb R)$. Then we realize that $u \in H^1(\mathbb R)$ is also the case (once again, because the only two possible spaces we could have are in fact equal to each other, there's no doubt about this one).

Thus, our Hilbert space here is $H^1(\mathbb R)$, on which the bilinear form $B(f,g) = \int_{-1}^1 [f'g' + cfg]$ is well defined. We then define a functional $F$ on $H^1(\mathbb R)$ by $F(h) = h(0)$. Lax-Milgram is set when we reframe our problem as : we want a $u$ such that $B(u,\phi) = F(\phi)$ for every $\phi \in H^1(\mathbb R)$.

We need to check that $B$ is continuous and coercive. Continuity is pretty clear from elementary bounds. Coercivity is also clear since $B(u,u) \geq c\|u\|^2_{H_0^1}$ and $c>0$. Thus Lax-Milgram tells us there's a unique $u$ satisfying the conditions, and we are done.

We can find an explicit $u$ which satisfies the equation, using cues from classical treatment of the equation $u''+cu = f$. However, I don't wish to go into this because the priority of the post was understanding the Lax-Milgram lemma usage and the various Sobolev spaces in play. I hope I have done some justice.

  • $\begingroup$ Wow. That's majestic! You really answered what I asked. Just one question about the solution you gave to the problem with the unrigorous writing: why do you give that solution? What we got in "class" (at home cause of covid lol) is that $$u(x) = \begin{cases} 1/2(x+1) \quad \text{if} \quad x \in (-1,0) \\ -1/2(x-1) \quad \text{if} \quad x\in (0,1)\end{cases}$$ I understood what you meant but also "your $u$" does not satisfies $u(-1) = u(1) = 0$. Did I misunderstand something? $\endgroup$ – Gabrielek Feb 3 at 19:30
  • $\begingroup$ ooh, I was not quite careful there! I will edit that part. But the rest is quite good. $\endgroup$ – Teresa Lisbon Feb 3 at 19:30
  • $\begingroup$ Absolutely good. Thank you! I surely must practice in this new field of math. $\endgroup$ – Gabrielek Feb 3 at 19:33
  • $\begingroup$ @Gabrielek Edited. I had to check the boundary conditions. I got an answer different from yours, so can you please check my working? I think it is correct. You may also use the bounty if you are satisfied with the answer. Thanks again. $\endgroup$ – Teresa Lisbon Feb 3 at 19:49

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