# Variational formulation of a PDE with a Dirac's Delta?

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.

• Welcome to math stack exchange! Please add some information, for example which arguments "u" has and how the Dirac delta function is defined. Dec 21, 2016 at 23:51
• 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 $\delta_{0} \in H_{0}^{1} = H^{-1}$. Dec 22, 2016 at 10:50

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 solution and the negative of the first and second derivatives for increasing $N$.
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.
• 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.