# Tag Info

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First, you should not believe in anything in mathematics, in particular weak solutions of PDEs. They are sometimes a useful tool, as others have pointed out, but they are often not unique. For example, one needs an additional entropy condition to obtain uniqueness of weak solutions for scalar conservation laws, like Burger's equation. Also note that there ...

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Reason 1. Even if you actually care only about smooth solutions, it some cases it is much easier to first establish that a weak solution exists and separately show that the structure of the PDE actually enforces it to be smooth. Existence and regularity are handled separately and using different tools. Reason 2. There are physical phenomena which are ...

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Let's have a look at the Dirichlet problem on some (say smoothly) bounded domain $\Omega$, i.e. $$-\Delta u=f \text{ in } \Omega\\ u=0~ \text{ on } \partial \Omega$$ for $f \in \text{C}^0(\overline{\Omega})$. Then, Dirichlet's principle states a classical solution is a minimizer of an energy functional, namely $E(u):=\dfrac{1}{2}\int_\Omega \left|\nabla ... 15 People can maybe talk more generally but I have a really simple example (but helpful in my opinion): Not all waves are differentiable. We want all waves to satisfy the wave equation (in some sense). That sense is weak. 14 A mollifier is a function$f$that you convolve with another function$g$to get a function which is "close" to$g$but "nicer". For instance$g$might be a general$L^1$function and$g*f$might be a smooth, compactly supported approximation to$g$. Really a mollifier is not one function but a sequence, or even sometimes a one-parameter continuous family. ... 14 Absolutely nothing in physics is completely described by a PDE, if you look at a sufficiently small resolution, because space and time are not continuous. (Since the OP has said in a comment that he doesn't know much physics, google for "Planck length" for more information.) However almost everything in physics is described at a fundamental level by ... 13 The viscosity solutions was first introduced in the context of the Hamilton-Jacobi equation by the vanishing viscosity method. It would be difficult to apply the notion of distributional weak solution in this context, because the derivatives occur inside a nonlinear function. Further complications would arise because a distributional weak solution is not ... 11 There are a few related concepts here: 1) A bump function is a term for a smooth function with compact support. The set of all bump functions forms a vector space. If these functions are on$\mathbb{R}^n$, then it is often denoted$C_c^\infty(\mathbb{R}^n)$. In distribution theory, this is what is most commonly referred to when one refers to test functions, ... 9 In the elliptic case, these two notions are equivalent. The proof (which is not trivial at all) can be found in the paper H. Ishii, "On the equivalence of two notions of weak solutions, viscosity solutions and distribution solutions", Funkcial Ekvac. Ser. Int. 38 (1) (1995) 101–120.(pdf) Probably the same proof can be extended to the parabolic case (... 9 It seems to me that a definition of "regularity" really depends on where it is being applied, for instance the regularity theory of weak solutions refers to the extra differentiability a function may have, due to it solving a PDE. The regularity a solution can inherit depends on the properties of the problem, i.e., the smoothness of a domain boundary, the ... 8 I like Hairer's notes, but of course these are lecture notes. And some introductory books can be easier to read since there are more explanations. There are basically three approaches to analysing SPDEs: the “martingale (or martingale measure) approach”, the “semigroup (or mild solution) approach” and the “variational approach”. Hairer is using the ... 7 It is a fact that not all physical problems have smooth solutions. Often this situation arises from a set of conservation laws that are expressed mathematically by applying such laws to a finite control volume to obtain an integral equation. Then we let the size of the control volume go to zero and arrive at some PDEs if the flow is smooth. But then we ... 5 The operators$S$and$T$and selfadjoint on their respective domains. These have to do with even and odd extensions. An even extension of$f \in \mathcal{D}(S)$is in$H^{2}(\mathbb{R})$, and an odd extension of$f \in \mathcal{D}(T)$is in$H^{2}(\mathbb{R})$. So the operators$S$and$T$can be expressed in terms of the Fourier cosine and sine transforms: ... 5 For$u$to be a classical solution, both$u$and the coefficients in the partial differential operator$P$in the LHS of your first formula need to be regular enough so that$P$does not map u away from$C^0(\Omega)$, for in this case you can integrate by parts the LHS of your second formula. The path from weak solutions to classical solutions relies on two ... 5 We can reverse engineer such a PDE by using ODE and the method of characteristics. Let's start with the ODE part. Consider the function$f: \mathbb{R} \to \mathbb{R}$given by $$f(x) = \begin{cases} -(x+\alpha)^2 & \text{if }x \le -\alpha \\ 0 & \text{if }-\alpha < x < \alpha \\ (x-\alpha)^2 & \text{if } x \ge \alpha \end{cases}$$ for ... 5 This is Hardy's Inequality. You may consider first for a$f \in C_c^1(\mathbb{R}^n)$(where,$n \ge 3) and apply integration by parts as follows, \begin{align} (n-2)\int_{\mathbb{R}^n} \frac{f^2}{|x|^2} \,dx = \int_{\mathbb{R}^n} f^2\operatorname{div}\left(\frac{x}{|x|^{2}}\right)\,dx &= -2\int_{\mathbb{R}^n} \frac{(x\cdot \nabla f)f}{|x|^{2}}\,dx \tag{... 4 I will assume we have good boundary behaviour (at the boundary\partial \Omega$and at$\infty) so I can freely integrate by parts. Consider \begin{align} |h|^2 &= \left| - \Delta w - i c \partial_1 w + \left( \frac{c^2}{2} + 2\right) w \right| ^2 \\ & = \left| \Delta w \right|^2 + \left| c \partial_1 w \right|^2 + \left| \left( \frac{c^2}{2} +... 4 H^2-regularity of p is valid if \Lambda is a bounded, polyhedral set, see chapter 4 of Grisvard's book "Elliptic problems in nonsmooth domains". 4 A function u is harmonic if and only if for every x in the domain u(x) = \frac1{|B(x,\epsilon)|}\int_{B(x,\epsilon)}u(y)dy+o(\epsilon^2). $$This is also true without the error term o(\epsilon^2), but it is important for comparison. Similarly, u is p-harmonic if and only if it satisfies the asymptotic mean value property$$ u(x) = \frac{p-2}{p+... 4 JustDroppedIn's answer works fine foru\in L^1$. Here is why: Since$\Delta u=3u\in L^1$, it is a tempered distribution. Hence we can test against functions$\phi\in\mathscr{S}(\mathbb{R}^n)$. If we do so, we get $$\langle \phi,\widehat{\Delta u}\rangle=\langle \Delta \hat \phi,u\rangle=-4\pi^2\langle \widehat{|\cdot|^2\phi},u\rangle=-4\pi^2\langle \phi,|\... 4 To the excellent longer answers above I will add a short one: weak solutions in a conveniently-chosen (and in particular, finite-dimensional) function space can often be explicitly computed, whereas strong solutions often cannot (even if one can prove a solution must theoretically exist). Computability has obvious and immense practical importance. Of course,... 4 The existing answers provide good reasons towards the question in the title, but from the perspective of a geometer I feel the applications in physics aren't quite as convincing. It's true that singular phenomena that arises in for example conservation laws requires a suitable notion of a generalised solution, but why is it also useful for geometric problems?... 3 |A_2| is not a product, but it can be estimated by one. Note that$$ \zeta |D_k^h Du| |D_k^h u| + \zeta |D_k^h Du| |Du| + \zeta |D_k^h u| |Du| \leq (\zeta|D_k^h Du|+\zeta|D_k^h u|)(|D_k^h u|+|Du|). $$Then one can apply Cauchy's theorem (or its \epsilon-version ab=(a\sqrt\epsilon)(b/\sqrt\epsilon)\leq\frac\epsilon2a^2+\frac1{2\epsilon}b^2) and (a+b)^2\... 3 A reference I have found very useful is the book "Second Order Elliptic Equations and Elliptic Systems'' by Ya-Zhe Chen and Lan-Cheng Wu (English translation). The proofs are complete and well organized. In chapter 9, theorem 2.6 the theorem is proven when k=\nabla \cdot F and F is C^{0,\alpha}. The result also holds when k\in L^p and p>n and ... 3 Yes, just notice that u=(-\Delta +1)^{-1} f=G_2 * f, where \hat{G}_2(\xi)=(1+|\xi|^2)^{-1} is the Bessel potential and it follows from properties of G_2 (in particular that it is an L^1 function) that it maps L^\infty to L^\infty (see "Modern Fourier Analysis" by Grafakos for a proof of this fact). 3 Let N_t:\mathbb{R}^N\to \mathbb{R}, t>0, be the function defined by$$N_t(x)=(4\pi t)^{-N/2}e^{-|x|^2/4t}.$$Since$$\int_{\mathbb{R}^N} e^{-a|x|^2}dx=\left(\frac{\pi}{a}\right)^{N/2},\tag{1}\label{1}$$we can see that N_t\in L^1(\mathbb{R}^N) and \|N_t\|_{ L^1(\mathbb{R}^N)}=1. We know that S(t)v=N_t\ast v. From Young's Inequality, we have$$\... 3 We can get ridd of the source term by defining $$f(x,y,z,t) = xy - ct + u(x,y,z,t)\tag{1}$$ so that the PDE for$u$becomes $$ax^2u_{xx} + bx u_x + u_t = 0\tag{2}$$ We can then perform a change of variables$z = \log(x)\$ to get $$au_{zz} + (b-a) u_z + u_t = 0\tag{3}$$ which is linear second order homogenous PDE with constant coefficients and you can ...

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