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Assume $u\in H^1(U)$ is a bounded weak solution of $$-\sum_{i,j=1}^n(a^{ij}u_{x_i})_{x_j}=0 ~~~in~ U$$

Let $\phi:R\rightarrow R$ be convex and smooth,and set $w=\phi(u)$ Show $w$ is a weak subsolution; that is $$B[w,v]\leq 0$$

for all $v\in H^1_0(U),~v\geq0$ $$B=\int_U \sum_{i,j=1}^na^{ij}v_{x_i}w_{x_j}$$

I used integration by part and eliptic property, actually my problem is that I don't know when $\phi$ is convex $\phi'(u)$ is positive or not?or

$$\int_U \sum_{i,j=1}^na^{ij}u_{x_jx_j}v\phi'(u)dx$$ is positive?

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Please define $B$. – Tomás Dec 30 '12 at 14:43
@Tomás I'm going to guess that $B[w,v]=\int_U a^{ij} w_{x_i} v_{x_j}$. And yes, user54688 should add something to the post, like the assumptions on $a^{ij}$ in addition to what tomasz said. – user53153 Dec 31 '12 at 3:03
$\phi$ convex implies that $\phi''\geq 0$. Try to write $B[w,v]$ in terms of $\phi''$ and see what happens. – Tomás Jan 1 '13 at 15:07
If this is taken from an exercise of the Evans textbook, then $B$ is defined, according to the author, as $$B[w,v]=\int_U \sum_{i,j=1}^n a^{ij}w_{x_i} v_{x_j} \, dx.$$ (Just saying.) – Cookie Feb 15 '15 at 7:32
up vote 1 down vote accepted

Im gonna use here the Einstein summation convention and $\frac{\partial v}{\partial x_i}=v_i$. Also, Im gonna assume ellipticity on $(a_{ij})$ and that $w\in H^1$, because that only with your hypothesis this is not always true. Note that $(v\in C_0^1(U),\ v\geq 0)$\begin{eqnarray} B[w,v] &=& \int_U a_{ij}v_iw_j \nonumber \\ &=& \int_U a_{ij}\phi'(u)u_iv_j \nonumber \\ &=& \int_U a_{ij}u_i(\phi'(u)v)_j-\int_U(a_{ij}u_iu_j)v\phi''(u) \end{eqnarray}

To conclude, you have to show that $\phi'(u)v\in H_0^1$ and $-(a_{ij}u_iu_j)v\phi''(u)\leq 0$, then you use the fact that $C_0^1$ is dense in $H_0^1$. Can you do this?

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The real pros of Einstein summation use sub-superscript pairing... – user53153 Jan 2 '13 at 20:13
Please @PavelM, explain me the difference. – Tomás Jan 2 '13 at 21:11
I was mostly kidding, but anyway: it is useful and often necessary to distinguish between a vector space $V$ and its dual $V^*$, even though in finite dimensions they are isomorphic. So one has a basis $(e^i)$ for $V$ and a basis $(f_i)$ for $V^*$. The coordinates are set in opposite way: $x=x_i e^i$ and $f=f^i x_i$. This helps one visually identify the expressions that are invariant under change of basis: $f^i x_i$ is invariant while $x_i x_i$ is not. // As Wikipedia says, when working with fixed basis on $\mathbb R^n$ one may identify $V$ and $V^*$, but on manifolds this would make a mess. – user53153 Jan 2 '13 at 21:29
Also, – user53153 Jan 2 '13 at 21:30
Ok thank you @PavelM . Im really learning a lot from you and I really apreciate your efforts to help people improve their skills. – Tomás Jan 2 '13 at 21:40

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