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I'm given the following problem:

Let $\Omega=(-1,1)$ and define $\phi(x)=-x^2+\frac{3}{4}$ and \begin{align} K=\lbrace v\in H^1_0(\Omega) | v(x) \geq \phi(x) a.e. x \in \Omega \rbrace \end{align} Consider the problem: Find $u \in K$ such that for all $v \in K$ it holds that \begin{align} \int_{\Omega} u'(v'-u')dx \geq 0 . \end{align}

First, I would like to show that there exists a unique solution. Then, I would like to compute the solution.

In the lecture we learned that there exists a unique solution if $\phi \in H^1(\Omega)$ such that $\phi \leq 0$ on $\partial \Omega$.

Easy computations show that $\phi(-1)=\phi(1)=-1/4<0$ and that $||\phi||_{H^1}< \infty$, hence the conditions for the existence of a solution are satisfied. (Is that really it, is there more to show?)

I am totally lost for the second part of the question. How could I compute the solution?

sunflower234
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1 Answers1

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Since the obstacle is an even function and you know that the solution $u$ is unique you know that $u$ must also even since $u(-x)$ is also a solution. Hence, it is enough to solve the problem on the interval $(0,1)$.

Let's make the ansatz that $$ u(x) = \begin{cases} \phi(x), &\text{if } x\in (0,\alpha) \\ mx +c , &\text{if } x\in [ \alpha , 1) \end{cases} $$ for some $\alpha \in (0,1)$, $m,c\in \mathbb R$. If we can find a solution $u \in H^1_0(-1,1)$ of this form then we are done since the solution is unique. If you draw a picture I think it is clear why this is a logical guess. I'm also making use of the fact that the solution to the obstacle problem is harmonic (i.e. affine in 1D) in $(-1,1) \cap \{u>\phi\}$. In fact, from regularity theory for the obstacle problem, we know that $u \in C^{1,1} ((-1,1))$ which will make our life easier since we will just find $\alpha,m,$ and $c$ such that this is the case.

Now, since we want $u=0$ at $x=-1, 1 $ (in the trace sense), we need $$c=-m.$$ Next, $$ \lim_{x\to \alpha^-} u(x) = \frac34-\alpha^2 \text{ and }\lim_{x\to \alpha^+} u(x) = m(\alpha-1),$$ so $\frac34-\alpha^2=m(\alpha-1)$ which gives $$ m = \frac{3-4\alpha^2}{4(\alpha-1)}$$ Finally, $$\lim_{x\to \alpha^-} u'(x) =-2\alpha \text{ and } \lim_{x\to \alpha^+} u'(x) =\frac{3-4\alpha^2}{4(\alpha-1)} ,$$ so $$-2\alpha=\frac{3-4\alpha^2}{4(\alpha-1)} $$ which gives that $\alpha = 1/2.$ Substituting these value back into our ansatz gives that $$u(x)= \begin{cases} 3/4-x^2, &\text{if } x\in (-1/2,1/2) \\ 1-\vert x \vert , &\text{if } x\in [-1,-1/2] \cup[ 1/2 , 1] \end{cases}. \tag{$\ast$} $$

Now let us check that $u$ given by $(\ast)$ solves the variational inequality. Since $u(-1)=u(1)=0$ and $u\in C^{1,1}([-1,1])$, we have that $u\in H^1_0((-1,1))$. Moreover, for all $v\in K$, \begin{align*} \int_{-1}^1 u'(x) (v'(x)-u'(x) \, dx &= \int_{-1}^{-1/2}v'(x)\,dx-2\int_{-1/2}^{1/2} x v'(x) \, dx-\int_{1/2}^1 v'(x)\,dx - 4 \int_0^{1/2}x^2 \, dx - \int_{1/2}^1\,dx \\ &=v(-1/2)+2\int_{-1/2}^{1/2} v(x) \, dx -2(\frac12 v(1/2)+\frac12 v(-1/2)) + v(1/2) -\frac23\\ &=2\int_{-1/2}^{1/2} v(x) \, dx-\frac23\\ &\geqslant 2\int_{-1/2}^{1/2} (3/2-x^2) \, dx-\frac23\\ &=1\\ &\geqslant0. \end{align*}

JackT
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