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Let's consider the one-dimensional ODE: $$ u_{,xx}(x)+1=0,\quad \forall x\in ]0,\ell[$$ together with the two Dirichlet boundary conditions $u(0)=0$ and $u(\ell)=0$.

The corresponding weak formulation is: Find $u\in H_0^1(0,\ell)$ such that $\forall v\in H_0^1(0,\ell)$: $$\int_0^\ell v_{,x}(x)u_{,x}(x)\,\mathrm{d}x=\int_0^\ell v(x)\,\mathrm{d}x$$

My question is on the fact that the test function $v$ has to belong to $H_0^1(0,\ell)$. I am aware that if $v\in H^1(0,\ell)$, then the weak formulation involves boundary terms in $vu_{,x}$ that are unknown, hence we specify $v\in H_0^1(0,\ell)$ to somehow get rid of them but:

  • How do we know that, by doing so, we are not modifying the equivalence between the strong and weak formulations?
  • If we rewrite the weak formulation as follows: Find $u\in H_0^1(0,\ell)$ such that $\forall v\in H^1(0,\ell)$: $$\int_0^\ell v_{,x}(x)u_{,x}(x)\,\mathrm{d}x=\int_0^\ell v(x)\,\mathrm{d}x+v(x)u_{,x}(x)\Big|_0^\ell$$ it cannot be solved, but why exactly? Is there some kind of operator that cannot be inverted (this is what happens in the discretized version)
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To answer your second question: If $u$ belongs merely to $H_0^1(0,\ell)$, you have no chance to define $u_{,x}(0)$ and $u_{,x}(\ell)$ in a reasonable manner. So your weak definition is not well-defined.

To give an example: It is easy to verify that $u$ defined by $$u(x) = x^{2/3} \, (1-x)$$ belongs to $H_0^1(0,\ell)$ and its weak derivative is $$u_{,x}(x) = \frac23 \, x^{-1/3} \, (1-x) - x^{2/3}.$$ This weak derivative, however, has a singularity at $x = 0$ (and it would be easy to add another one at $x = \ell$).

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  • $\begingroup$ What would be the well-defined form of the weak formulation then? Isn't the proposed weak formulation the usual one for Poisson's equation? $\endgroup$ – pluton Nov 7 '15 at 14:25
  • $\begingroup$ The correct weak formulation is the one with trial and test function in $H_0^1$, as you have mentioned in your question. Is it not possible to use test functions from $H^1$. $\endgroup$ – gerw Nov 7 '15 at 18:34
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The strong formulation implies the weak formulation; this follows after integrating by parts. To show that the weak formulation actually implies the strong formulation for functions $u$ with two derivatives that have sufficient regularity, note that the weak formulation implies that $$\int_0^lu_{xx}v=-\int_0^lv_xu_x=-\int_0^lv,$$ which implies that $$\int_0^l(u_{xx}+1)v=0$$ for all $v\in H_0^1(0,l)$. If again $u_{xx}$ is sufficiently regular, there exists a sequence $(v_n)$ in $C_c^{\infty}(0,l)$ that approximates $u_{xx}+1$ in the $L^2$ norm; this will imply that $$\int_0^l(u_{xx}+1)^2=\lim_{n\to\infty}\int_0^l(u_{xx}+1)v_n=0.$$ Therefore, $u_{xx}+1=0$, which is the strong formulation.

For your second question, usually what happens here is that, to obtain coercivity, we want to control the $H_0^1$ norm of a function $v$ just by the $L^2$ norm of $u_x$; this follows from Poincare's inequality. But, Poincare's inequality does not hold for general $H^1$ functions.

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    $\begingroup$ A small comment for the OP related to this answer: coercivity is the same notion as positive definiteness (or negative definiteness, depending on your sign convention) from linear algebra. In both finite and infinite dimensions it is a sufficient condition for invertibility. So not having coercivity is at least analogous to an inverse not existing in finite dimensions. However, the connection between invertibility and eigenvalues is more delicate in infinite dimensions, since you can have maps which are injective but not surjective, as well as maps which are surjective but not injective. $\endgroup$ – Ian Nov 6 '15 at 12:02
  • $\begingroup$ @Ian: If the bilinear form is not coercive, it still can be invertible. Moreover, by the approach of the OP, the bilinear form (if it would be well-defined) would map $H_0^1(0,\ell) \times H^1(0,\ell) \to \mathbb{R}$, so we even cannot speak of 'coercivity'. $\endgroup$ – gerw Nov 6 '15 at 18:13
  • $\begingroup$ @detnvvp: I disagree with your last paragraph. The issue is that the weak formulation of the OP is not well-defined, see my post. The problem is not the missing coercivity. $\endgroup$ – gerw Nov 6 '15 at 18:14
  • $\begingroup$ @gerw I said "at least analogous". I didn't say equivalent, and suggested that the situation is indeed complicated. $\endgroup$ – Ian Nov 7 '15 at 15:46

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