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Helmholtz in cartesian coordinates being $\dfrac{\partial^2 \phi}{\partial x^2} + \dfrac{\partial^2 \phi}{\partial y^2} + k^2 \phi = 0, \tag{1}$ we recall that $\nabla^2 \phi = \dfrac{\partial^2 \phi}{\partial x^2} + \dfrac{\partial^2 \phi}{\partial y^2}; \tag{2}$ since $\nabla^2 u(r,\theta)=\dfrac{\partial^2 u}{\partial ... 0 You can let$U'$be an open set containing$U$so that$\partial U'$is$C^1$. Any function$u \in W^{1,p}_0(U)$extends naturally to a function in$W^{1,p}_0(U')$without changing its norm: just define it to be zero outside$U$. Use the fact that$W_0^{1,p}(U')\subset\subset L^q(U')$to conclude that a bounded sequence in$W^{1,p}_0(U)$, once extended, has ... 0 Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:$\dfrac{dt}{ds}=1$, letting$t(0)=0$, we have$t=s\dfrac{df}{ds}=0$, letting$f(0)=f_0$, we have$f=f_0\dfrac{dx}{ds}=v(f)=v(f_0)$, letting$x(0)=F(f_0)$, we have$x=F(f_0)+v(f_0)s=F(f)+v(f)t$0 It's not clear why you decided to perform these manipulations, but it seems you are considering the energy method. The energy for this equation is $$E(t)=\int_0^l (v_t^2+a^2v_x^2)\,dx$$ which is basically the sum of kinetic and potential energy. Differentiate with respect to time, apply the PDE, and integrate one of the terms by parts: you will get zero. ... 0 HINT: The function$ k(x,y)=\frac{y}{x^2+y^2} $is in fact a special case of Poisson kernel, which is used to solve Laplace equation. More specifically,$k(x,y)$is the Poisson kernel defined on the upper half plane. Appended: Technically the term "heat equation" includes steady-state heat equation, which is the Laplace's equation. So if you choose to ... 1 If an end is isolated, the$x$derivative there must be set to zero: $$u_x(0,t)=u_x(L,t)=0\tag{1}$$ This will imply the conservation of mass: $$\frac{d}{dt}\int_0^L u(x,t) \, dx =\int_0^L cu_{xx}(x,t) \, dx =c (u_x(L,t)-u_x(0,t))=0$$ The physical meaning of$(1)$becomes transparent if we recall that the flow of heat is proportional to the gradient: ... 2 Change variables: $$\xi=x,\quad\eta=x+y.$$ Then $$u_x=u_\xi+u_\eta,\quad u_y=u_\eta.$$ The equation becomes $$u_\eta=f(\xi,\eta-\xi).$$ To solve it just integrate in the variable$\eta$. 1 Your equation is a transport equation. Usually we write it with one variable identified as "time" and the rest identified as space, like this: $$u_t + v u_x = S(t,x,u)$$ where$v$is a fixed velocity and$S$is a source. In the physical situation modeled by this equation, the particles which are currently present are moving at a velocity$v. We can use ... 1 You have $$3x - x^{2} = \sum_{n = 1}^{\infty} c_{n} \sin \bigg( \frac{n \pi x}{3} \bigg)$$ Multiplying both sides by $$\sin \bigg( \frac{m \pi x}{3} \bigg)$$ for orthogonality, we get $$(3x - x^{2})\sin \bigg( \frac{m \pi x}{3} \bigg) = \sum_{n = 1}^{\infty} c_{n} \sin \bigg( \frac{n \pi x}{3} \bigg)\sin \bigg( \frac{m \pi x}{3} \bigg)$$ Integrating ... 1 Using the integral \begin{align} \int_{0}^{L} \sin\left( \frac{m \pi x}{L} \right) \, \sin\left(\frac{n \pi x}{L} \right) \, dx = \frac{L}{2} \, \delta_{n,m} \end{align} then \begin{align} \int_{0}^{L} (3x - x^{2}) \, \sin\left( \frac{m \pi x}{L} \right) \, dx &= \sum_{n=1}^{\infty} B_{n} \, \int_{0}^{L} \sin\left(\frac{m \pi x}{L}\right) \, ... 1 No, because already forn=1$we have that the Sobolev space$H_0^2$is a subset of your space. 0 Are the sets$E$edges of the triangulation? if so I don't see why you are applying your Lagrange multipliers on the edges (unless this is just a method for determining a constant that I haven't seen before), you want them on the whole domain so that you can impose the integral of a variable to be zero. Also$q$is fully determined by the system, it is$p$... 0 The equation is Quasilinear if it is nonlinear in any derivative of$u$that is not the highest order derivative possible. It must be linear in the highest order derivative. 0$m^2f$would be zero if$m=0$, all the assumption requires is that$f$becomes negative somewhere, which it does: https://www.wolframalpha.com/input/?i=J_0 I guess it depends on what you mean by$"<<"$, and also I assume you just want there to exist some$z_0$such that this holds (because there is no$z_0$in your equation). 1 Similar to Fourier trigonometric series solution of the heat equation, the general solution for Bessel equation can be written in terms of series of Bessel functions. The final solution will be written as an (infinite) sum over$m$of Bessel Functions of the first kind and Bessel Functions of the second kind. Now, if you are confused about change of ... 1 suppose the eigenvalues of$B$are$\lambda_1, \lambda_2, \cdots, \lambda_n.$then the eigenvalues of$I + tB$are$1 + t\lambda_1, 1 + t\lambda_2, \cdots, 1 + t\lambda_n.$we will use the fact that the determinant of a matrix is the product of its eigenvalues. now we have $$det(I + tB) = (1 + t\lambda_1) (1 + t\lambda_2) \cdots (1 + t\lambda_n) = ... 4 Hoping I'm not completely off-base here! I think that mjqxxx had it right in his comment, i.e. that D \det(I) refers to the linear map from M_n(\Bbb C) \to \Bbb C which is the derivative of \det: M_n(\Bbb C) \to \Bbb C at the point I \in M_n(\Bbb C); then D\det(I) B makes sense as the directional derivative of \det in the B direction at the ... 0 To see why C^1 is not enough, consider$$f(x) = |x|^{3/2}\cos (1/x^{1/4}),\, x\ne 0,\,\, f(0) = 0.$$Then f\in C^1(\mathbb {R}), hence \phi(x,y)= y-f(x) \in C^1(\mathbb {R}^2). Define \Omega = \{\phi < 0\}. Then \Omega is the domain below the graph of f. Good to draw a picture here. Suppose there exists a closed ball B such that B\cap ... 1 The characteristic curves satisfy:$$\frac{dy}{dx}=\frac{b\pm\sqrt{b^2-ac}}{a}$$where a,b,c are coefficients of U_{xx}, U_{xy}, U_{yy}, respectively. With your example, you should get$$\frac{dy}{dx}=\pm\sqrt{-xy}$$Solving these two ODE's you should see the following change of variables:$$\xi=y^{1/2}+\frac{1}{3}(-x)^{3/2}\\ ... 1 This is a quasilinear PDE, and it can be solved using the method of characteristics. 0 Hint: The function $$g(x) = \int_{\partial B_1(0)} P(x, y) dA_y = \int_{\partial B_1(0)} 1 \cdot P(x, y) dA_y$$ satisfies$\Delta g = 0$in$B_1(0)$and$g(y) = 1$on$\partial B_1(0)$. What can$g$be? 1 Any function on a connected topological space$X$that is locally constant on$X$is globally constant on$X.$* The open convex result mentioned above gives the locally constant property in this problem, and the connectedness of$U$finishes it. *The proof is a typical "the good set is nonempty and clopen, therefore it's everything" argument. 0 Simply use your discretization again to have: $$\frac{M^{n+1}_0 - M^n_0}{\Delta t} = c + f^n_0,$$ from wich you can determine$M^{n+1}_0$as a function of what happened before in the boundary, substituting this value in your finite difference scheme for$i = 0$. For instance, you will have: $$\frac{M^{n+1}_i - M^n_i}{\Delta t} = \frac{1}{\Delta x^2} ... 3 F will still be a constant map. Indeed, U being a connected open subset of \mathbb{R}^n, it is path connected (here is a proof). Let x_0,x_1\in U be arbitrary. There is a path \gamma:[0,1]\to U such that \gamma(0)=x_0 and \gamma(1)=x_1. Now, I is compact and \gamma is continuous, so \gamma(I) can be covered by finitely many open balls ... 3 Careful: This is not true for any \Omega. For example let u_k(x,y) = (-1)^ky in the upper half plane \Omega of \mathbb {R}^2. Then all u_k vanish on \partial \Omega, hence converge uniformly there, but the u_k converge nowhere in \Omega. This will be true for all bounded \Omega however. There it's just the maximum principle: ... 0 Since you assume that u \in C^2, \Delta u is bounded near y, so$$ \left| \int_{B_\rho(y)} v\Delta u \right| \le M \int_{B_\rho(y)} |v| $$but this tends to 0 as \rho\to 0, since v \in L^1. In other words$$ \int_{\Omega \setminus B_\rho(y)} v\Delta u = \int_{\Omega} v\Delta u - \int_{B_\rho(y)} v\Delta u \to \int_{\Omega} v\Delta u.$$(It ... 0 The inequality seems to be false as written. If we take n=1, so that B_1 is the interval (-1,1), and a_{11} = 1 so that the ellipticity condition holds with \Lambda = \lambda = 1, and let u(x) = x and \eta(x) = 1-x^2, then direct calculation shows that the left side of the desired inequality is 16/15 and the right side is 8/15. (I am not ... 1 Your initial thoughts are correct here. Let \{\lambda_n\} be a minimizing sequence in (16) converging to \lambda_0. Then x_{\lambda_n} \to x_{\lambda_0}, while E_{\lambda_0} = \bigcup_{n = 1}^{\infty}E_{\lambda_n}, with E_{\lambda_n} \subset E_{\lambda_{n+1}}. Moreover, u(x) < u(x_{\lambda_n}) for each x \in E_{\lambda_n}. Now let x \in ... 1 Actually, f(u(x))-f(0) is equal to -cu(x), not to -c. Indeed,$$-cu(x) = u(x)\int_0^1 f'(su(x)) \, ds = \int_0^1 \frac{d}{ds} \left(f(su(x))\right) \, ds = f(u(x))-f(0)$$where the second equality is the chain rule. (Keep in mind that in general, f'(2x) means f' evaluated at 2x, not the derivative of f(2x).) So,$$-\Delta u-f(u)+f(0)-f(0) ... 0 A partial answer, but perhaps others will contribute... Quoting from http://www.researchgate.net/post/What_is_the_difference_between_essential_boundary_conditions_and_natural_boundary_conditions "The essential boundary conditions are imposed on the functions in the space where the minimzation of the energy functional is made or the weak formulation is ... 1 If$xis analytic, \begin{align} T(t)x & = x+\int_{0}^{t}T(s)Ax ds \\ & = x + (s-t) T(s)Ax|_{s=0}^{t}-\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax -\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax +\left.\frac{t^{2}}{2!}A^{2}x\right|_{s=0}^{t}+\int_{0}^{t}\frac{(s-t)^{2}}{2!}T(s)A^{3}xds ... 3 Actually, one can say that an ODE is a special case of a PDE. The variables (i.e. those parameters which partial derivatives exist with respect to them) can be 1 or more. Once there is only one variable, partial derivatives would not meaningful, as there is only derivatives with respect to one variable. Hence, everything method working with a PDE should be ... 2 No, the use ofL^2$inner product is appropriate here. In this subject it's not unusual to borrow structure from different function spaces at once. E.g., later in the section you will see statements like "$u_j\rightharpoonup u$weakly in$H^1_0$and also$u_j\to u$strongly in$L^2$". This use of$L^2$product is consistent with the way in which we ... 0 I found a property that I think solves the problem without having to use chain rule and is easier:$\frac{df}{dt}=\frac{\partial f}{\partial t} + \{f, H\}$where$\{, \}$is the Poisson Bracket.$\{f,H\} = \nabla f^T J \nabla H$As property,$\{F, H\} = 0 $if and only if F is an integral of$\dot{z}=J\nabla H(t,z)$, and we know that H is an integral of ... 1 Let $$j(x,y)=\left\{\begin{array}{lll}\exp\big((1-x^2-y^2)^{-1}\big) & \text{if} & x^2+y^2<1,\\ 0 & \text{if} & x^2+y^2\ge 1. \end{array}\right.$$ Then$j\in C^\infty(\mathbb R^2)$. A function in$H^1(\mathbb R^2)\setminus L^\infty(\mathbb R^2)$is the following $$f(x,y)=j(x,y)\log\big(\log(1+x^2+y^2)\big).$$ 1 A standard example would be something like $$f(x,y)=\log\log(1+(x^2+y^2)^{-1}) \varphi(x^2+y^2)$$ or $$f(x,y)=\left[-\log(x^2+y^2)\right]^{1/4} \varphi(x^2+y^2)$$ where$\varphi$is smooth, compactly supported, and equal to$1$in a neighbourhood of zero. The rationale here being that$H^1(\mathbb{R}^2)$is embedded into$\operatorname{BMO}(\mathbb{R}^2)$, ... 0$u_{rr}+\dfrac{u_r}{r}+\dfrac{u_{\theta\theta}}{r^2}=0$Let$\begin{cases}r_1=\ln r\\\theta_1=\theta\end{cases}$, Then$u_r=u_{r_1}(r_1)_r+u_{\theta_1}(\theta_1)_r=\dfrac{u_{r_1}}{r}=e^{-r_1}u_{r_1}$... 2 The rule of thumb is that you try to substitute appropriate general functions in your ODEs and thus determine the solution. For more detail see derivation of characteristic equation of a linear differential equation. As for the cases of different values of$K$, the answers can be derived using common sense. First, we take a notion of the boundary and ... 0 Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:$\dfrac{dy}{dt}=1$, letting$y(0)=0$, we have$y=t\dfrac{dz}{dt}=2$, letting$z(0)=z_0$, we have$z=z_0+2t=z_0+2y\dfrac{dx}{dt}=1+\sqrt{z-y-x}=1+\sqrt{z_0+t-x}$Let$u=z_0+t-x$, Then$x=z_0+t-u\dfrac{dx}{dt}=1-\dfrac{du}{dt}$... 0 Hint: Consider$X(x)=a\sin(\lambda(x-1))+b\cos(\lambda(x-1))X(1)=0$:$b=0\therefore X(x)=a\sin(\lambda(x-1))X'(x)=a\lambda\cos(\lambda(x-1))X'(0)-X(0)=0$:$a\lambda\cos(-\lambda)-a\sin(-\lambda)=0\lambda\cos\lambda+\sin\lambda=0\tan\lambda=-\lambda$Hence for$\dfrac{\partial u}{\partial t}=\dfrac{\partial^2u}{\partial x^2}$... 0 Define $$w(x)=\frac{1}{2}(x_1-x_1^2) \max_{S}f^+, x=(x_1,x_2).$$ Here$s^+=s$if$s\ge0$and$s^+=0$if$s<0$. Then, in$S$$$-\Delta (w-u)=\max_{s}f^++\Delta u=\max_{s}f^+-f\ge0,$$ and on$\partial S$,$w-u\ge0$. By the Maximum Principle,$w-u\ge 0$in$S$or $$u\le w=\frac12(x_1-x_1^2) \max_{S}f^+.$$ Since$x_1\in(-1,1)$, we have ... 2 Just plugin the definition of$T$and$Q_\sigma$. We have $$u = \sigma Tu \iff Tu = \frac{u}{\sigma}$$ that is, by definition of$T$iff$v := \frac{u}\sigma$is the (unique) solution of $$a^{ij}(x,u,Du)D_{ij}v + b(x,u,Du) = 0, \quad v|_{\partial \Omega} = \phi$$ Let here$v = \frac u\sigma$, this gives $$\frac 1\sigma a^{ij}(x,u,Du) D_{ij}u + ... 1 For H^1- functions, this does not hold in general. It is basically the same argument that L^2- functions generally do not vanish at \infty, you just have to take such a function and integrate it, for example a bump function where the bumps get thinner when you go outside. For H^2-functions however, it does hold: Take$$\int_0^a ... 1 Note that, with the fact that$a \neq b$, $$\frac{\log(b)}{\log(\frac{b}{a})} - \frac{\log(a)}{\log(\frac{b}{a})} = \frac{\log(b) - \log(a)}{\log(b) - \log(a)} = 1$$ This tells that $$\frac{n\pi\log(b)}{\log(\frac{b}{a})} = n\pi +\frac{n\pi\log(a)}{\log(\frac{b}{a})}$$ and that $$\cos(\frac{n\pi\log(b)}{\log(\frac{b}{a})}) = ... 0 First note that$$ x\phi''(x)+\phi'(x)+\frac{\lambda}{x}\phi(x) = 0 $$can be written as$$ -(x\phi'(x))' =\frac{\lambda}{x}\phi(x) $$Now suppose that \phi is a solution of the above for which \phi(1)=\phi(2)=0. Then$$ \lambda\int_{1}^{2}\phi(x)^{2}\frac{1}{x}dx = -\int_{1}^{2}(x\phi'(x))'\phi(x)dx \\ = \left\{ ... 1 Because$(\Delta f,f) \le -\lambda_1\|f\|^{2}$for$f\in\mathcal{D}(-\Delta), then \begin{align} \frac{d}{dt}\|T(t)f\|^{2}& =(\Delta T(t)f,T(t)f)+(T(t)f,\Delta T(t)f) \\ & \le -2\lambda_1(T(t)f,T(t)f)= -2\lambda_1\|T(t)f\|^{2},\;\; f \in\mathcal{D}(-\Delta) \end{align} Hence, $$\frac{d}{dt}\left( e^{2\lambda_1 ... 0 Let u(x,t)=\sum\limits_{n=1}^\infty C(n,t)\sin\dfrac{n\pi x}{l} so that it automatically satisfies u(0,t)=u(l,t)=0 , Then \sum\limits_{n=1}^\infty C_t(n,t)\sin\dfrac{n\pi x}{l}=-\sum\limits_{n=1}^\infty\dfrac{n^2\pi^2}{l^2}C(n,t)\sin\dfrac{n\pi x}{l}+g(x,t) \sum\limits_{n=1}^\infty\biggl(C_t(n,t)+\dfrac{n^2\pi^2}{l^2}C(n,t)\biggr)\sin\dfrac{n\pi ... 2$$ u(x,t) = V(x)\mathrm{e}^{st +ikx} $$and$$ u_t =u_{xx} + \mu u $$we find$$ su = \left(V''\mathrm{e}^{st +ikx} +2ikV'\mathrm{e}^{st +ikx}-k^2u\right) + \mu u $$thus$$ \left(V'' +2ikV'\right)\mathrm{e}^{st +ikx}+(\mu -k^2-s)u = 0 $$This is what I would of done without the answer you have shown, make the terms in the bracket zero.$$ \mu -k^2-s = 0 ... 1 Solve $$v_t-v_{xx}=0,\quad v(0,t)=v(\ell,t)=0,\quad v(x,0)=f(x)$$ and $$w_t-w_{xx}=g(x,t),\quad w(0,t)=w(\ell,t)=0,\quad w(x,0)=0.$$ Theu=v+w$. Standard separation of variables gives$v$in the form $$v(x,t)=\sum_{n=1}^\infty a_n\,e^{-\bigl(\tfrac{k\,\pi}{\ell}\bigr)^2\,t}\,\sin\frac{k\,\pi\,x}{\ell}.$$ To find$w$develop$g$in a Fourier series ... 2 The change $$u(t,x)=e^{ax+bt}\,v(t,x)$$ for appropriate choice of$a,b\in\mathbb{R}\$ will transform the equation into $$v_t=v_{xx},\quad v(0,x)=e^{-ax}\,g(x).$$