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I'm working with Evans PDE book and I can't understand this:

Let $U\subset \mathbb{R}^n$ open and $u\in L^{2}([0,T], H_0^1(U))$ with $u' \in L^2([0,T] , H^{-1}(U))$ and now we consider the mollifications of $u$ and $u'$.

Why this is correct?

For $\epsilon, \delta >0$, $$\frac{d}{dt} \|u^{\epsilon}(t)- u^{\delta}(t) \|^2_{L^2(U)} = 2\langle u^{\epsilon'} - u^{\delta'},u^{\epsilon} - u^{\delta}\rangle $$

ÝTAN
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1 Answers1

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Once you convolve $u, u'$ with say a gaussian, they become smooth $\forall t > 0$ (with compact support). Then this equality comes down to "moving the derivative inside the integral": \begin{eqnarray*} \frac{d}{dt} \|u^{\epsilon}(t)- u^{\delta}(t) \|^2_{L^2(U)} &=& \frac{d}{dt} \int_U |u^{\epsilon}(t)(x) - u^{\delta}(t)(x) |^2 dx \\ &=& \int_U \frac{d}{dt} |u^{\epsilon}(t)(x) - u^{\delta}(t)(x) |^2 dx \\ &=& \int_U 2(u^{\epsilon'}(t)(x)- u^{\delta'}(t)(x))(u^{\epsilon}(t)(x)- u^{\delta}(t)(x)) dx \\ &=& 2\langle u^{\epsilon'}(t) - u^{\delta'}(t),u^{\epsilon}(t) - u^{\delta}(t)\rangle \\ \end{eqnarray*} You can move the derivative inside due to Leibniz' Integral Rule.

muaddib
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  • In some books I found this proof, but I'm confuse because what's mean the product $(u^{\epsilon'} - u^{\delta'})(u^{\epsilon} - u^{\delta})$? This is a "product" of elements of a vector space that not have a defined product. – ÝTAN Aug 07 '15 at 17:25
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    @Inzit $H_0(U)$ usually is a set of real (or complex) valued functions: $f : \mathbb{R}^n \to \mathbb{R}$ (or $\mathbb{C}$). I left out the $x$'s in the above but that is the product of two real (or complex) numbers. I'll assume its real and add in the x's. Updated. If you want the range to be a vector space $V$ then replace that product with the inner product on the space $V$. – muaddib Aug 07 '15 at 17:37
  • Thanks for your help, now all is clear... Just I was confused with the concepts – ÝTAN Aug 07 '15 at 18:01