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I have the following boundary value problem $$ \begin{cases} -u'' = \delta_{0} \quad \text{in }(-1,1) \subset \mathbb{R} \\[2ex] u(-1) = 0, \quad u(1) = 0 \end{cases}\tag{A} $$ where $ \delta_{0} \in \mathscr{D}^{''}(-1,1) $ is the Dirac delta in $0$. The goal is to obtain a variational formulation of (A), to demonstrate that there is a unique solution $ u \in H_{0}^{1}(-1,1) $ and to derive the analitic expression of the solution $u$.

I note that $ H_{0}^{1}(-1,1) \subset C([-1,1]) $, so that $ v \in H_{0}^{1} $. The following should hold: $ \langle \delta_{0}, v \rangle = v(0) $, and $ \lvert\langle \delta_{0}, v \rangle\rvert = \lvert v(0) \rvert \leq \displaystyle \max_{x \in [-1,1]} \lvert v(x) \rvert = \lVert v \rVert_{C([-1,1])} \leq M\lVert v \rVert_{H_{0}^{1}(-1,1)} $, where M is a constant. This should help, but here it is where I'm really stuck.

Although I guess this is not a difficult problem, I'm getting stuck with some basic issues. This is a homework problem, and I have very few skills with Hilbert spaces and the application of functional analysis to PDEs. Any guide will be much appreciatted.

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  • $\begingroup$ Welcome to math stack exchange! Please add some information, for example which arguments "u" has and how the Dirac delta function is defined. $\endgroup$
    – Peter
    Dec 21, 2016 at 23:51
  • $\begingroup$ Thanks for the prompt respone, and for correcting my original question; the argument of $u$ is $x$ (that is, $-u''$ is minus the Laplacian of $x$). The Dirac's delta is defined as $ \delta_{0} \in H_{0}^{1} = H^{-1} $. $\endgroup$
    – Susan.
    Dec 22, 2016 at 10:50

2 Answers 2

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If you are after an analytic expression what the variational forumation does allow you to do is consider a subspace of the actual Hilbert space you are interested in, so consider a space $V_N = \mbox{span}\left\{ v_n \right\} \subset V$, and then write your approximate solution $u_n$. So I hope you don't mind if I make a change of variables and work in the interval $[0,1]$ and consider $\delta_{\frac{1}{2}} = \delta(x-\frac{1}{2})$. As others have already mentioned the weak formulation of this problem is $$ \begin{align} \mbox{seek } u \in V, \; \; \int_0^{1}u^{\prime}v^{\prime}dx = \int_0^{1} fvdx, \; \;&\mbox{for all }v\in V, \end{align} $$ Leaving $f$ arbitrary for the moment then in terms of the approximate solution we have a system of integral equations where $u_N$ is represented by a linear combination of these basis vectors $$ \begin{align} u_N = \sum_{j=1}^{N} \beta_j v_j \end{align} $$ $$ \begin{align} \int \left(\sum_{j=1}^{N}\beta_jv_j \right)^\prime v^{\prime}_i&= \int_0^{1} f v_i^{\prime} dx &i=1,\ldots,N. \\ \implies \sum_j\beta_j\int_{0}^{1} v_j^{\prime} v_i^{\prime}dx &= \int_0^{1}fv_i dx, &i=1,\ldots,N \end{align} $$ Now the next careful bit is chosing your functions $v_i$ but a good choice in this instance is $v_i = \sin (i \pi x)$. Then you can use the fact that the first derivatives of these functions are again orthogonal and therefore this system of equations becomes $$ \begin{align} \beta_i &= \frac{2}{\pi^2 i^2}\int_0^{1}f(x)\sin i\pi x dx \\ &= \frac{2}{\pi^2 i^2}\int_0^1 \delta(x-\frac{1}{2})\sin i \pi x dx\\ &= \frac{2}{\pi^2 i^2}\sin \left( \frac{i \pi }{2}\right), \end{align} $$ So an approximate solution is given by $$ u_N = \frac{2}{\pi^2}\sum_{i=1}^{N} \frac{1}{i^2}\sin\left(\frac{i \pi}{2} \right)\sin i\pi x, $$ Now you can get the analytic solution you are after by taking the limit as $N \rightarrow \infty$, I can leave you to try that but here are some suggestive plots of the solutionenter image description here and the negative of the first and second derivativesenter image description hereenter image description here for increasing $N$.

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  • $\begingroup$ Forgot to add that it was important to make sure the choice of approximating subspace satisfied your boundary conditions, so that the the boundary term in the integration by parts disappears $\endgroup$
    – Nadiels
    Dec 22, 2016 at 17:58
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The key idea is this: if $f$ and $g$ are two equal "vectors" (i.e. functions in this case), then $\langle f,v \rangle$ should equal $\langle g,v \rangle$ for every (appropriate) vector $v$. In the case of our functions, we've defined $$ \langle f,v \rangle = \int_{-1}^1 f(x) v(x)\,dx $$ (assuming these functions are real valued). Along these lines, the variational formulation of the problem is that for all (integrable) functions $v(x)$, we should have $$ \int_{-1}^1 u''(x)v(x)\,dx = \int_{-1}^1 \delta_0(x)v(x)\,dx $$ With the same initial condition. Simplifying the right side, we have $$ \int_{-1}^1 u''(x)v(x)\,dx = v(0) $$ and from there, applying integration by parts on the left brings us to the "weak formulation".

So, why do we care? Think about it this way: $\delta_0$ doesn't really make sense as a function, so how could a function $u$ satisfy $-u'' = \delta_0$? On the other hand, integrating with $\delta_0(x)v(x)$ makes perfect sense (or at least, there's a rule that lets us handle it), so the variational form gives us an equation where all the pieces have a literal interpretation in terms of functions.

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  • $\begingroup$ Shouldn't the lower limits be $-1$? Also, there is a minus sign on the left side. The integration by parts also has a bit of a catch to it, related to a constraint on the choices of $v$ which are admissible. $\endgroup$
    – Ian
    Dec 22, 2016 at 1:01
  • $\begingroup$ Thanks for this, I have added further remarks in the question above that might help. $\endgroup$
    – Susan.
    Dec 22, 2016 at 11:26

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