For question about weak derivatives, a notion which extends the classical notion of derivative and allows us to consider derivatives of distributions rather than functions.

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Finding Weak Solutions to ODEs

I'm wondering if anyone has a reference to a good set of notes on finding weak (distributional) solutions to ODEs, or has any tips or tricks. For example, $$ xy^\prime=0 $$ has a classical solution ...
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2answers
87 views

How to prove if $u\in W^{1,p}$, then $|u|\in W^{1,p}$?

How to prove if $u\in W^{1,p}$, then $|u|\in W^{1,p}$? Since $|u|\in L_p$, I only need to show weak derivative of $u$ exists and $Du \in L_p$. Can anyone give me some hint? Thanks!
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1answer
101 views

$C_c^{\infty}$ is dense in $W^{k,p}(\mathbb R^n)$

As the title say, I want to prove that $C_c^{\infty}(\mathbb R^n)$ is dense in $W^{k,p}(\mathbb R^n)$ i.e. $\displaystyle{ W^{k,p} (\mathbb R^n) = W_0^{k,p}(\mathbb R^n) \quad (\star)}$. In a book ...
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2answers
584 views

Questions about weak derivatives

There are two definitions of generalized differentiation that seem relevant to the context of PDEs. (That is we generalize what objects can be differentiated but we stay in Euclidean space. There are ...
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2answers
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Does convergence in H1 imply pointwise convergence?

I'm trying to figure out if convergence in $H^1(a,b)$ implies pointwise convergence (by the way: what is the usual name of this space?). It is defined to be Hilbert space of absolutely continuous ...
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1answer
482 views

Is a continuous function with continuous weak derivatives of class $C^1$?

Let $f$ be a real valued continuous function of many variables whose weak derivatives of first order are continuous. Is this function equals a.e. function of class $C^1$ ?
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445 views

about weak derivative ( Sobolev Spaces )

the following afirmation is true ? Consider $\Omega $ a bounded and smooth domain . Let $u \in W^{1,p} ( \Omega)$ ( p>1). Supose that $u \geq 0$. let $\alpha >1$ . Then $\nabla u ^{\alpha} = ...
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1answer
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Sobolev spaces in one-dimensional vs multidimensional

Here in Wikipedia, it is said that in the one-dimensional case, it is enough to assume that the $(k-1)$-th derivative of the function $f$, is differentiable almost everywhere and is equal almost ...
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1answer
87 views

How does integration over $\delta^{(n)}(x)$ work?

For a math paper I need to be able to evaluate $\int_{-a}^{a}\delta^{(n)}(x)\ f(x)\ dx$ for differentiable $f$. I know that it is 'supposed' to equal $(-1)^nf^{(n)}(0)$: $$\int_{-a}^a\delta^{(n)}(x)\ ...
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1answer
609 views

function a.e. differentiable and it's weak derivative

Note - I am just starting to learn about theory of distributions, so this may be a trivial question, if so I'd be grateful for a reference, nevertheless the question is the following: suppose I have a ...
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1answer
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Computing weak derivatives on an open square

I am looking at the computation of weak derivative in the blog https://sunlimingbit.wordpress.com/2012/11/25/one-example-related-to-weak-derivative-2/ For equation (3), I have some confusions. Here ...
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0answers
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Does existence and uniqueness of a classical solution impose uniqueness of weak solutions to a pde?

I wonder if one knows that there exists a unique classical solution of a pde (for instance: Fokker-Planck equation), is one able to conclude that there isn't any weak solution of the pde, which ...
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1answer
352 views

Weak Derivative Heaviside function

I have to prove that the Heaviside function $$ H(x):=\begin{cases} 1 &\mbox{if } x \in [0,+\infty) \\ 0 &\mbox{otherwise}\end{cases} $$ doesn't admit weak derivative in ...
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1answer
163 views

First order weak derivatives of $f(x)=|x|^r$

Let $f(x)=|x|^r$ for a given real number $r$. Show that $f$ has first order weak derivatives on the unit ball $B_1(0)\subset \mathbb{R}^n$ provided that $r > 1-n$. Does anyone have an idea on how ...
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1answer
273 views

Prove Corollary of the Fundamental lemma of calculus of variations

From the Fundamental lemma of calculus of variations we know the following: Let $u \in L_{loc}(a,b)$ and for every $\phi \in C_0^\infty$, $$\int_a^b u(x) \phi(x) dx = 0$$ holds, then $u(x) = 0$ ...