This is a super simple estimate and I have been banging my head against the wall that I can't figure it out. I am trying to estimate the $||\cdot||_{H^k}$ norm of $u_t+uu_x=0$ and can't figure out the nonlinear term.

I'm reading along with a paper, my goal is $\frac{d}{dt}||\partial_x^ku||_{L^2}^2\lesssim||u_x||_{L^\infty}||\partial_x^k u||_{L^2}^2$. Is the following valid for estimating the nonlinear term (assuming $u\in C_0^\infty(\textbf{R})$)? \begin{align*} \left|\int_{\mathbf{R}}\,\partial_x^k{u}\partial_x^k(uu_x)\,dx\right| &= \left|(-1)^k\int_{\mathbf{R}}\,(uu_x)\partial_x^{2k}(u)\,dx\right| \\ &= \left|\int_{\mathbf{R}}\,(uu_x)\partial_x^{2k}(u)\,dx\right| \\ &\leqslant ||u_x||_{L^\infty}\left|\int_{\mathbf{R}}\,u\partial_x^{2k}u\,dx \right| \\ &= ||u_x||_{L^\infty}\left|(-1)^k\int_{\mathbf{R}}\,(\partial_x^{k}u)^2\,dx \right| \\ &=||u_x||_{L^\infty}||\partial_x^ku||_{L^2}^2 \end{align*}

Seems like cheating and I am wondering if there is an inner product identity I am unaware of which makes this smoother. I am doubting myself because I thought I had it earlier and realized that I had made a pretty bad mistake. Thanks!

  • $\begingroup$ Why not $u\in\mathcal C_0^\infty(\mathbb R^2)$ ? :) $\endgroup$ Commented Feb 6, 2020 at 2:11
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    $\begingroup$ @MaximilianJanisch Oops, thats a typo in the "what I am trying to prove" part. Sorry for confusion. And the second to last line should be an equality. Thank you. $\endgroup$
    – krc
    Commented Feb 6, 2020 at 2:23

1 Answer 1


Your inequality step is not valid. What is valid is $|\int f g| \leq\|f\|_{L^{\infty}} \int|g|$ i.e. a special case of Holder's inequality. An easy counterexample is some $g$ with $\int g=0$ like sin on $[0,2 \pi]$ (then extended to be o if you wish for an example on $\mathbf{R}$, and smooth examples also exist), with $f=x$ on $[0,2 \pi]$

Instead, you can proceed as follows (all integrals are over $\mathbf R$): $$ \int\partial^k_xu\partial_x^k(uu_x) \\ = \int \partial_x^ku \sum_{r=0}^k\binom{k}{r}( \partial_x^{k-r}u) (\partial_x^{r+1}u) \\ = \int \partial_x^ku \sum_{r=0}^{k-1}\binom{k}{r}( \partial_x^{k-r}u) (\partial_x^{r+1}u) + \int \partial_x^ku \cdot \partial_x^{k+1}u\cdot u \\ $$ There are two kinds of terms; one term with $k+1$ derivatives on one $u$ and the others with strictly less. For the term singled out, integration by parts to move one of the $k+1$ derivatives gives

$$I:=\int \partial_x^ku \cdot \partial_x^{k+1}u\cdot u = -\int \partial_x^{k+1}u \cdot \partial_x^{k}u\cdot u - \int \partial_x^ku \cdot \partial_x^{k}u\cdot u_x \\ \implies 2I = -\int \partial_x^ku \cdot \partial_x^{k}u\cdot u_x \\ \implies |I| \le \frac12\|u_x\|_{L^\infty} \|\partial_x^ku\|_{L^2}^2.$$

For the other terms, define $$ J_r:= \int |\partial_x^ku || \partial_x^{k-r}u ||\partial_x^{r+1}u| $$ Note that $r+1,k-r$ are now both numbers in $1,2,\dots k$, and if one of them is $1$, the other is $k$. Thus, for $r=0$, $$ J_0 = J_{k-1} = \int |\partial_x^k u|^2 |u_x| \le \|u_x\|_{L^\infty} \|\partial_x^k u\|_{L^2}^2. $$ Else, $r+1$ and $k-r $ are strictly between $0$ and $k$. We have $$ J_r= \int |\partial_x^ku || \partial_x^{k-r}u ||\partial_x^{r+1}u| \le \|\partial_x^k u\|_{L^2} \|\partial_x^{k-r-1} u_x\|_{L^2} \|\partial_x^{r} u_x\|_{L^\infty}$$ Now we apply Gagliardo-Nirenberg: in the case $p\in[2,\infty]$, $j\in(0,K)$ and dimension 1, it says that $$ \|\partial_x^j v\|_{L^p} \lesssim \|\partial_x^K v\|_{L^2}^{\theta_{p,j,K}} \|v\|_{L^\infty}^{1-\theta_{p,j,K}} ,\quad \theta_{p,j,K} = \frac{j-1/p}{K-1/2}$$ we set $K=k-1$ and $v=u_x$. Then \begin{align} \|\partial_x^{k-r-1} u_x\|_{L^2} \|\partial_x^{r} u_x\|_{L^\infty} \lesssim \|\partial_x^k u\|_{L^2}^{\theta_{2,k-r-1,k-1} + \theta_{\infty,r,k-1}} \| u_x\|_{L^\infty} ^{2-\theta_{2,k-r-1,k-1} - \theta_{\infty,r,k-1}} \end{align} since $$ \theta_{2,k-r-1,k-1} + \theta_{\infty,r,k-1}= \frac{k-r-1 - \frac12}{k-1-\frac12} + \frac{r}{k-1-\frac12} = 1 $$ we arrive at $$ \|\partial_x^{k-r-1} u_x\|_{L^2} \|\partial_x^{r}u_x\|_{L^\infty} \lesssim \|\partial_x^k u\|_{L^2}\|u_x\|_{L^\infty}$$ and therefore $$ J_r = \int |\partial_x^ku || \partial_x^{k-r}u ||\partial_x^{r+1}u| \lesssim \|\partial_x^k u\|_{L^2}^2 \|u_x\|_{L^\infty}$$ summing over all the terms gives the required inequality: $$ \left| \int\partial^k_xu\partial_x^k(uu_x)\right| \le |I| + \sum_{r=0}^{k-1} \binom{k}r J_r \lesssim_k \|\partial_x^k u\|_{L^2}^2\|u_x\|_{L^\infty}.$$

May I ask what paper this is from? I know that the book of Heinz-Otto Kreiss on the Navier Stokes equation covers the Burger's equation (its a whole chapter) but I checked and they don't do it this way. Also, if something confuses you, please ask.

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    $\begingroup$ Thank you very much! I was worried that my Holder inequality was dubious. It is a paper by Hunter, Ifrim, Tataru, and Wong: math.ucdavis.edu/~hunter/papers/bh_modified.pdf. Its quite cool, or at least the parts I understand are quite cool! $\endgroup$
    – krc
    Commented Feb 9, 2020 at 18:07

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