# Tag Info

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If $(0,0)$ is considered a boundary point, with $u(0,0)=1$ as a boundary condition, then there is no solution of this boundary value problem among the conditions. Indeed, a slightly generalized maximum principle says that if $u$ is a bounded harmonic function in a bounded domain $\Omega$ and $u\le 0$ holds on $\partial \Omega$ except for finitely many ...

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Maybe you were looking for this: http://web.uvic.ca/~monahana/ode_2_flowchart.pdf

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Yes,if we write $$y=f(x)$$ then tangent line $$Y-f(x)=f'(x)[X-x]$$then $$X=-\dfrac{f(x)}{f'(x)}+x,Y=f(x)-xf'(x)$$ so $$f(x)-xf'(x)+x-\dfrac{f(x)}{f'(x)}=C$$ Both sides of the x derivation $$f'(x)-f'(x)-x[f'(x)]^2+1-\dfrac{[f'(x)]^2-f(x)f''(x)}{(f'(x))^2}=0$$ $$\Longrightarrow xf''(x)=\dfrac{f(x)f''(x)}{[f'(x)]^2}$$ case one $f''(x)=0$ then we have ...

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Yes sure one can find the solution in the continuous or Holder spaces. Way out is that we have the fundamental solution (Newton Potential) E(x,ξ) for Laplace equations, one solution to such nonlinear PDEs when f=f(x,u) or f=f(x,u,∂_i u) is given by u^*=∭_Ω E(x,ξ) f(ξ,u(ξ),∂_i u(ξ))dξ in a bounded set Ω ⊆ R^3 Which can be easily verified to be in the ...

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In the 5th displayed equation, it should read $f'(t) \in H^s$. That's what $C^1$ means - one derivative with respect to the variable $t$. Suppose $\theta:[0,\infty) \to [0,\infty)$ satisfies $\theta(s)/(1+s) \in [\frac12,2]$, but $\theta'(s)$ varies a lot. Consider $\hat f(t,\xi) = \theta((1+t) |\xi|)^{-s-3/2-\epsilon}$ for some small $\epsilon>0$. ...

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The easiest way to show the coincidence of the two solutions is to verify that both are weak solutions of the class $L^2(0, T; H^1_0(U))$. For problem (HP), a function $u\colon\, (0,T)\to H^1_0(U)$ is called a weak solution of the class $u\in L^2(0, T; H^1_0(U))$ if $u$ satisfies the integral identity \begin{align*} -\!\!\int\limits_{Q_T}\! ...

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The Poisson kernel provides the solution of the harmonic Dirichlet problem: $$\left\{\begin{array}{ccc} \Delta u=0 \quad\text{in}\,\,\Omega, \tag{1}\\ u=f \quad\text{on}\,\,\,\partial\Omega. \end{array} \right.$$ That is $$u(x)=\int_{\partial\Omega} P(x,y)\,f(y)\,dy,$$ satisfies $(1)$. In particular $u(x)=\int_{\partial\Omega} P(x,y)\,dy,$ corresponds ...

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So, let $D(H)=C_0^{\infty}(\Omega;\mathbb{C})$, and assume additionally that $C_{\alpha}\in C(\overline{\Omega};\mathbb{C})$. Denote by $\overline{H}$ the minimal closed extension of $H$. The graph $\mathcal{G}(\overline{H})$ of $\overline{H}$ is a closure in $L^2(\Omega;\mathbb{C})\times L^2(\Omega;\mathbb{C})$ of the graph $\mathcal{G}(H)$ that consists ...

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An equation of a line in the plane can be written in parametric scalar form, $x=x_0+at$, $y=y_0+ bt$ parametric vector form, $\vec r=\vec r_0+t\vec v$. Here $\vec v$ is a direction vector, its components are $a$ and $b$ from the preceding equation. implicit scalar form, $\alpha x+\beta y=c$ implicit vector form, $\vec r \cdot \vec n = c$. Here $\vec n$ ...

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This $$|D_iu(y)| \leq \frac nR \sup_{\partial\Omega}|u| \leq \frac n {d(y,\partial\Omega)} \sup_\Omega |u|\tag{\ast}$$ is wrong for $R < d(y,\partial\Omega)$ and harmonic $u \not\equiv 0$. The first of the two inequalities is okay if $u$ has decent boundary values on $\partial\Omega$ or we replace $\partial\Omega$ with $\partial B_R(y)$ or $\Omega$ ...

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I don't know which book you are reading, but I think if you know the reason for this result, you don't need counterexamples to help you understand. Actually, if $u\in C^1$, the given PDE could be transformed to the divergence form. Then you can multiply a $v \in C^{\infty}_0$ to change this equation to its weak form. After that, you may use your bilinear ...

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I may answer my question by myself because I find a nice answer in Ringstrom's book "The Cauchy problem in general relativity". In Chapter 4, he shows me a energy: $$E_k=\frac{1}{2}\sum_{|\alpha\le k|}\int_{\mathbb{R}^n}[(\partial^\alpha\partial_t u)^2+|\nabla \partial^\alpha u|^2]dx$$ For the linear case, of course, this energy is conserved, so according ...

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The computations become a little cleaner if you introduce $w=u-v$, which also solves the heat equation and has zero boundary conditions. You have shown that $$\frac{d}{dt}\int_S w^2 \le -\int_S |\nabla w|^2 \tag{1}$$ What is needed is to relate the right hand side to $\int_S w^2$. This is what the Poincaré inequality does: $$\int_S w^2 \le C \int_S ... 1 A typical Interspecific competition between two species a and b can be represented by the following equations:$$\frac{da}{dt} = \lambda_{1}a [1-\frac{a}{k}]$$This is the logistic population growth model in the absence of competing species b. Now ,$$\frac{da}{dt} = \lambda_{1}a [1-\frac{a}{k_{1}} - \frac{\beta_{12}b}{k_{1}}]$$is the logistic population ... 2 So far I have assumed that f\in L^2(\mathbb R) is (essentially) bounded Bad idea. For one thing, this is not true in general, for another, L^\infty on a space of infinite measure is quite far from L^2. Let's review the situation. The scaling in$$\phi_t(x) = \frac{1}{\sqrt{4\pi t}} \exp\left\{ - \frac{x^2}{4t}\right\} $$is arranged so that ... 2 This "result" is false. The rule of thumb is that spaces characterized by a supremum condition, such as$$\sup_{0<h<T} \frac{1}{h} \int_0^{T-h}\lVert u(t+h)-u(t) \rVert_{L^2}^2 <\infty \tag{1}$$are non-separable spaces of Lipschitz/Hölder type, in which smooth functions are not norm-dense. This is quite unlike H^1. As a concrete ... 1 This is not a complete answer to your question. So one may not be satisfied to it. Hint: Write it done in its full expansion and see. It is a combination of parabolic and hyperbolic partial differential equation. Number of initial and boundary conditions will depend on the region where you have modeled your equations, co-ordinate system you are using, ... 1 The family g_t is a well-known convolution kernel. Most often, one considers the limit t\to 0, when g_t converges (in the sense of distributions) to the Dirac measure, but the limit t\to +\infty is also interesting (up to insignificant normalisation, g_t is the heat kernel). When looking at the limiting behaviour for convolutions with a ... 1 Denote z=h\lambda, and solve the second order equation in z:$$ 1+z+\theta z^2<1,  1+z+\theta z^2>-1, $$Then, sub z=h\lambda and find \theta. You will find an interval within wich you method will be stable. 2 Let \begin{cases}\alpha=5x+y\\\beta=5x-y\end{cases} , Then u_x=u_\alpha\alpha_x+u_\beta\beta_x=5u_\alpha+5u_\beta u_{xx}=(5u_\alpha+5u_\beta)_x=(5u_\alpha+5u_\beta)_\alpha\alpha_x+(5u_\alpha+5u_\beta)_\beta\beta_x=5(5u_{\alpha\alpha}+5u_{\alpha\beta})+5(5u_{\alpha\beta}+5u_{\beta\beta})=25u_{\alpha\alpha}+50u_{\alpha\beta}+25u_{\beta\beta} ... 0 H^{-1}=(H^1_0)^*, so −\Delta:H^1_0(Ω) \rightarrow H^{−1}(Ω) means -\Delta u \in H^{−1}(Ω) for any u \in H^1_0(Ω), and that is obviously right. Because -\Delta u \in H^{−1}(Ω) means (-\Delta u,v) is a scalar, for any v \in H^1_0(Ω). this also satisfies the definition of weak solution. Therefore, \Delta u \in L^2(\Omega), you mean this ... 0 Yes, it could be true. Actually, The maximum principle does not depend on the boundary conditions. It's really up to the sign of c (Lu=-\sum a^{ij}u_{x_i x_j}+\sum b^iu_{x_i}+cu). About the more specific proof, I think you can see Evan's book Partial differential equations or Gilbarg and Trudinger's book or others. 1 Since there is no information about how \nabla u changes with respect to t, we can't hope to obtain any cancellation in |\nabla u(t+h,x) - \nabla u(t,x)|. Just estimate it by |\nabla u(t+h,x)| + |\nabla u(t,x)|. Using the notation U(t) = \|\nabla u(t,\cdot)\|_{L^2} and the Cauchy-Schwarz inequality, we estimate the given integral by ... 0 Thanks to Christopher's answer, I found that I did the last step wrong. I would like to write down the correct formula here. I got correctly$$ u(x,-t)=v(x,t)=\frac{\partial}{\partial t}\left(\frac{1}{4\pi t}\int_{B(x;|t|)}g(y)dS(y) \right) $$for t>0. What I want is the formula for u(x,t) for t<0. Let q=h(t)=-t. Then by the chain rule, I ... 2 The line where you perform u(x,t) = v(x,-t) = \ldots  is not correct. You're not allowed to just replace the variable inside the differentiation. What you really want is to calculate the derivative, and then evaluate at a particular point. To better illustrate the incorrect substitution you did, consider the function f(x) = 1 for all x. Then it is ... 0 Yes. To be explicit, let T:W\to C be a bounded linear operator (it need not be injective). For every linear functional f\in C^* we have f\circ T\in W^*. Since u_n converge weakly, f(T(u_n))\to f(T(u)). But this says precisely that T(u_n) converge to T(u) weakly. 1 Yes. If the embedding j \colon W \hookrightarrow C is continuous when the two spaces are endowed with their respective norm topologies, it is also continuous when both spaces are endowed with their weak topologies, and that means weakly convergent nets are mapped to weakly convergent nets. In particular, if the weakly convergent net is a sequence, its ... 1 You can derive such an inequality from the Gagliardo-Niremberg inequality. 0 You can not have an inequality of the form$$ |u|_{H^1}^2\le C(|u|_{H^2}+\|u\|_{L^2}) $$Assume that this is true and replace u by Mu, where M>0 constant. Then the left hand side is {\mathcal O}(M^2) while the right-hand side is {\mathcal O}(M). In your argument, try instead M\sin kx, for k,M large. 0 In fact the real result should be u(x,t)=\dfrac{f(x+t)+f(x-t)+g(x+t)-g(x-t)}{2} , where f(x)=\begin{cases}x^3-x&\text{for}~|x|\leq1\\0&\text{for}~|x|\geq1\end{cases} and g(x)=\begin{cases}-\dfrac{2}{3}&\text{for}~x\leq-1\\x-\dfrac{x^3}{3}&\text{for}~|x|\leq1\\\dfrac{2}{3}&\text{for}~x\geq1\end{cases} 0 In fact the real result should be u(x,t)=\dfrac{f(x+t)+f(x-t)+g(x+t)-g(x-t)}{2} , where f(x)=\begin{cases}x^3-x&\text{for}~|x|\leq1\\0&\text{for}~|x|\geq1\end{cases} and g(x)=\begin{cases}-\dfrac{2}{3}&\text{for}~x\leq-1\\x-\dfrac{x^3}{3}&\text{for}~|x|\leq1\\\dfrac{2}{3}&\text{for}~x\geq1\end{cases} 0 just looking for a nice way to attack this. Would you like a pony with that? It's a nonlinear PDE. How would one go about solving, or describing the solutions By reading the existing literature on Fisher's equation (which is what it is, as pointed out in comments). It is known to have traveling wave solutions for various speeds, and — ... 0 Once I had similar question. If you are given function g on manifold M which integral onver M is nonzero. Than there is no solution of \delta d f = g. You can do this by contradiction. Assume that you have such a solution than:$$ 0\neq \int_M \star g = \int_M d{\star}df = \int_{\partial M} \star df = 0 $$because \partial M = 0. This answer is ... 0 It holds also on sufficiently smooth Riemannian manifolds, and in fact for more general partial differential operators (even pseudo-differential operators.) See for example Hebey, Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities. 1 Try going over the derivation of D'Lambert's solution again (i.e use the change of variables r=x+t and s=x-t to show that the PDE really says \frac{\partial ^2u}{\partial r \partial s} = 0, then integrate with respect to r and s, plug in the boundary conditions). You should see that not much changes when you have piecewise initial conditions, you ... 2 p_k is only real valued for functions u\in\mathscr S (Schwartz' space of rapidly decreasing smooth functions). However, restricted to that space p_k is indeed a norm since for all x\in \mathbb R^n you have$$|u(x)|\le (1+|x|^2)^{k/2} |D^0u(x)| \le p_k(u).$$The reason for calling it nevertheless a seminorm is probably that the general theory of ... 2 No. Take \Omega=(0,1), one-dimensional. Let v(t,x)=x^{-1/3}, i.e., independent of t. And u(t,x)=x^{1/3}, also independent of t. All assumptions are satisfied. But v(u')^2 is a multiple of  x^{-1}, not integrable over \Omega. Subsequent integration of \infty over t still gives \infty. The problem is that you are multiplying |\nabla ... 1 Let me start with$$ K(t)=\frac 12\int_0^1 u_{t}^2dx. $$We have$$ \frac{dK}{dt}(t)=\int_0^1u_tu_{tt}dx=\text{(due to the equation)}=\int_0^1u_tu_{xx} dx. $$Now use the integration by parts$$ \int_0^1u_tu_{xx}dx=u_tu_{x}|_0^1-\int_0^1 u_{x}u_{xt}dx=-\frac 12\frac{d}{dt}\int_0^1u_x^2\,dx=:-\frac{dT}{dt}(t). $$Hence we have that$$ ...

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No they are not the same, though I don't know what a mild solution is. But this paper's title strongly implies that they differ. Gotta say, I'm pretty shocked google didn't answer this one for me. Also, insert joke about mild solutions being when the heat equation doesn't get too hot for me to be comfortable in jeans.

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There is a general way to endow the (set theoretic) product of two Hilbert spaces with a norm that makes it a Hilbert space - $||\cdot ||_{H \times K}=\sqrt{||\cdot ||_H^2+||\cdot ||_K^2}$. You can check that this does in fact make $H \times K$ a Hilbert space. You could also check that the topology given by this norm coincides with that given by the ...

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Of course it is not. Reformulated for Fourier images, your question turns rather trivial. Indeed, denote by $\hat{u}$ the Fourier image of $u$, i.e., $$\hat{u}(\xi,y)=F[u]=\int\limits_{-\infty}^{+\infty}u(x,y)e^{-i\xi x}dx,$$ and notice that $\hat{u}(\xi,y)=\hat{g}(\xi)e^{-|\xi|y}=F[P_y\ast g]$, while $$... 0 Since \varphi is just Lipschitz, second order weak derivatives of \varphi(u) may not exist no matter how smooth be a solution u. That is why u is to be treated only as a weak solution. So, let u\in H^1\bigl(\Omega \times (0,T))\bigr) be a weak solution of the problem, i.e., let u possessing traces u|_{t=0}=u_0 and u|_{\partial\Omega}=0 ... 1 Note: There is a difference between f(x,y) and f(x+y). The first is a function of two variables and the second is a function of one variable z=x+y. For example$$f(x+0)=f(x)=x^2\implies f(x+y)=(x+y)^2\\f(x,0)=x^2\implies \, \text{nothing} $$Also there is a meaning of \frac{\partial f(x,y)}{\partial x} but it doesn't mean any thing for f(x+y) ... 1 For a bounded domain \Omega with \,\partial\Omega\in C^1, it is obviously true if \,u\in C(\overline{\Omega}) and x_0\, is a removable singular point. Of course, it cannot be true whenever a singularity x_0\, be non-removable, e.g., take \,\Omega=\{x\in\mathbb{R}^2\,\colon\,\,|x|<1\}\, with \,x_0=0, and consider an example of \,u(x)= x_1 ... 1 Let's be careful: The second condition leads to:$$\left.\frac{\partial u}{\partial t}\right|_{t=0}=-cf'(x)+cg'(x)=-2cf'(x)=\frac{x}{(1+x^2)^2}$$Thus$$f(x) = \frac{1}{4c} \cdot\frac{1}{1+x^2}+c_1,$$where c_1 is a constant, as$$ f'(x)=-\frac{1}{2c}\cdot\frac{x}{(1+x^2)^2}. $$0 To answer the question as clarified in the comments: Yes, your formulation is fine. (Though numerical stability I know nothing about.) The idea is the following: just take the Laplacian of the first equation, since we know that \Delta u = 0 you get$$ \Delta\Delta w = 0  which is the biharmonic equation. It is well-known that as a fourth order ...

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Consider the following statement, which does not involve any PDE: this is multivariable calculus only. If $f:\mathbb R^n\to \mathbb R$ is $C^1$ smooth and $f(0,0,0)=0$, then for every $R>0$ there exists $C>0$ such that $|f(z)|\le C|z|$ whenever $|z|\le R$. Proof: by the mean-value inequality we have $|f(z)-f(0)|\le |z|\sup_{|\xi|<R} ... 1 The intuitive answer is that you take a solution that goes around$a$times with$0 \lt a \lt 1$. Now run the clock backwards. Unless the solution blows up to$\infty$, you can just continue backwards in time. Each time you pass the initial angle, read off the angle and velocity and you have an initial condition that will circle a given number of times. ... 1 The principle symbol arise naturally when you take the Fourier transform, where the symbol appears the top order multiplier. So the choice of including$i$is immaterial since then$i^{m}$is a constant. The important thing is the property of$\sigma(D)$(like whether it is elliptic, hyperbolic, invertible, etc), and that would not be changed by multiplying ... 2 You are correct. Taking the partial with respect to$t$is akin to treating$x$as a constant. If$x$is fixed at$0$the function$f(t) = u(0,t)$is a constant, hence the derivative$f'(t) = \frac{\partial u}{\partial t}\mid_{x=0}=0\$

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