# Tag Info

1

The diffusion is proportional to the gradient of the concentration: $$-k\,\nabla u.$$ The minus sign comes from the fact that diffusion goes from higher to lower concentration. $k$ depends on the chemical being diffused and on the medium in which the dispersion takes place. If $k$ is constant, we get the equation $u_t-k\,\Delta u=0$. But $k$ may deppend ...

1

The function $1-\cos^2\phi$ vanishes for $\phi=0$ and $\phi=\pi$. Is it odd? Otherwise $C_{\nu}J_{\nu}(k\rho)+D_{\nu}Y_{\nu}(k\rho)$ is just $\nu$th Fourier coefficient in the expansion of $f(\rho,\phi)$ with respect to $\phi$. In other words, to decompose you only need orthogonality of exponentials $e^{i\nu\phi}$, not orthogonality of Bessel functions. ...

0

Yes. This in true in general for any kind of linear equation. If $T$ is a linear operator, the solution of the linear equation $$Tx=f$$ is $$x=x_h+x_p$$ where $x_h$ is the general solution of $Tx=0$ and $x_p$ a particular solution of $Tx=f$. Proof: let $y$ be any solution. Since $T$ is linear, $y-x_p$ is a solution of $Tx=0$.

0

The first eigenvalue of the clamped square plate is sign-changing, for a reference see my answer here.

0

There is no such thing as "the definition of ellipticity". There are definitions of ellipticity, not mutually compatible, each adapted to a particular context (class of PDE problems). Someone working with constant/smooth coefficients will define it differently from someone working with measurable coefficients, who will define it differently from someone ...

3

Hint: $$\nabla (fg)=f\nabla g+g\nabla f$$

1

The last equal sign is just the product rule, we have $$\def\div{\mathop{\rm div}}\def\grad{\mathop{\rm grad}}\div (eD\grad e) = e \div(D\grad e) + \grad e \cdot D\grad e$$ hence $$e\div(D \grad e) = \div(eD\grad e) - \grad e \cdot D \grad e$$ Addendum: For the simplified version with $p$, it's the same, we have $$\div(p \grad p) = \grad p ... 1 You will find what you are looking for in Chapter 3.1 of Gazzola, F., Grunau, H.-Ch., Sweers, G.: Polyharmonic Boundary Value Problems. Lecture Notes 1991. Springer, Berlin (2010). EDIT: To be more specific, you are looking for Theorem 3.8 at page 67. The original proof goes back to Friedrichs, K. Die Randwert- und Eigenwertprobleme aus der Theorie der ... 0 The idea is to work from one side, the uniform continuity of norm 1 yields an arbitrary \delta_1, then invoke the norm equivalence result, we found the choice of \delta works for norm 2. Let f be uniformly continuous w.r.t. \|\cdot\|_1. Then,$$\forall\,\epsilon>0\,\exists\delta_1>0\,\ni\,\forall \ ...

0

Suppose $f$ is uniformly continuous w.r.t $||\cdot||_1$. Using $K||x||_1 \le ||x||_2 \le M||x||_1$. Let $\epsilon >0$. For this you can find $\delta_1 >0$ such that, $\forall \ x,y \in \mathbb{R}^n, ||x-y||_1< \delta_1 \implies |f(x)-f(y)|< \epsilon$ Now use the norm equivalence relation, $||x-y||_2 \le M \ ||x-y||_1 < M \delta_1$. So, ...

-1

The reason is that there is not a transformation such as rM that allows you to reduce the equation of Euler Poisson Darboux to a wave equation in one dimensión. You can make the computations in d=2, the transformation rM does not work. And in fact the proof does depend on the dimension. If you keep Reading Evan's book you will find the correct ...

0

Some pointers $$\partial_t = \frac{dz}{dt}\partial_z = \frac{1}{B}\partial_z\\ \partial_x = \frac{dy}{dx}\partial_y = \frac{1}{A}\partial_y\\ \partial_{xx} = \frac{1}{A}\partial_y\frac{1}{A}\partial_y = \frac{1}{A^2}\partial_{yy}$$ therefore your equation becomes $$C\frac{1}{B}\partial_z v(x,t) + g(v(x,t)-E) = \frac{a}{2R}\frac{1}{A^2}\partial_{yy}$$ ...

2

The symbol $\otimes$ is the tensor product which you can read more about on Wikipedia. If you think of the vectors $u$ and $v$ as column vectors (3x1 matrices if we are talking about 3D space) then the tensor product $u\otimes v$ can be thought of as a square matrix (3x3 in the 3D case) whose entries are all the products of the individual entries of the ...

3

If $\vec u$ is divergence free, then $\nabla \cdot \vec u=0$. Now, we have \begin{align} \vec u\cdot \nabla \vec u&=\nabla \cdot(\vec u\otimes \vec u)\\\\ &=\left(\sum_{i=1}^3\hat x_i\frac{\partial }{\partial x_i}\right)\cdot \left(\sum_{j=1}^3\hat x_ju_j\sum_{j=1}^3\hat x_ku_k \right)\\\\ &=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\hat x_k(\hat ... 2 It means a sort of "dyadic product" (it's really a tensor product, but that's not relevant): in index notation, (u \times v)_{ij} = u_i v_j. $$Therefore, what it means is that (in summation convention)$$ (\operatorname{div}(u \otimes u))_i = \partial_j (u_j u_i ) = (\partial_j u_j) u_i + u_j \partial_j u_i = ((\operatorname{div} u) u)_i + ((u \cdot ...

0

You note that $||f (x)-f (y)||_1 <M\epsilon \Longrightarrow ||f (x)-f (y)||_2 <\epsilon$ and $||f (x)-f (y)||_2 <K\epsilon \Longrightarrow ||f (x)-f (y)||_2 <\epsilon$.

0

Look for a particular solution of the form $e^{A(t-t_0)}C(t)$. Substituting in the equation leads to $$e^{A(t-t_0)}C'(t)=f(t)\implies C'(t)=e^{-A(t-t_0)}f(t).$$

0

Let $\begin{cases}p=t-1\\q=x\end{cases}$ , Then $u_t=u_pp_t+u_qq_t=u_p$ $u_x=u_pp_x+u_qq_x=u_q$ $\therefore u_p-uu_q=0$ Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dp}{ds}=1$ , letting $p(0)=0$ , we have $p=s$ $\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$ $\dfrac{dq}{ds}=-u=-u_0$ , letting ...

0

I would argue as follows. Assume that $n\ge 2$ and let $$T_i f (x)=\int_{\mathbb{R}^n} f(y)\frac{x_i-y_i}{\lvert x-y\rvert^{n}}\chi_\Omega(y)\, dy.$$ Here $\chi_\Omega(y)$ is $1$ if $y\in \Omega$ and $0$ otherwise. The kernel of this operator can be estimated in absolute value by the kernel of the Riesz potential: $$\left\lvert \frac{x_i-y_i}{\lvert ... 0 Partial answer: The Laplacian measures how a function changes “on average” as you move away from a given point. It’s rotationally invariant, so, for example, f_{xx}+f_{yy}=0 describes a property of a function on the Euclidean plane that doesn’t depend on the choice of Euclidean coordinates. 0 You have found the solution such that u(0,x)=x/(1+x^2), but the one you want satisfies u(-1,x)=x/(1+x^2). All you need is a translation in time. 1 Use the associativity of convolution:$$ u(s+t,g)=p_{s+t}\star g=(p_t\star p_s)\star g=p_t\star (p_s\star g)=p_t\star u(s,g)=u(t,u(s,g)). $$2 The singularity at x=0 is removable, provided y\ne 0. Indeed, as x\to 0, we have |x|y\to 0. The vector |x|\tilde x does not have a limit as x\to 0, but its magnitude stays at 1, and \Phi is radially symmetric. So, \Phi(|x|(y-\tilde x))  has a limit as x\to 0 (it's whatever value \Phi has on the unit sphere), and this is used to extend ... 1 Not a big deal. Just use the following chain rules. Consider f(x,t) is a function of independent variables x and t, define \alpha = x + ct and g(\alpha ) = f(x,t) hence by chain rule one can obtain$$\left\{ \matrix{ {{\partial f} \over {\partial x}} = {{dg} \over {d\alpha }}{{\partial \alpha } \over {\partial x}} = {{dg} \over {d\alpha }} ...

1

By the chain rule, $$\frac{d}{dt}|u + th|^p = \frac{d}{dt}(|u + th|^2)^{p/2} = \frac{p}{2}(|u + th|^2)^{\frac{p}{2}-1}\frac{d}{dt}(|u + th|^2),$$ and $$\frac{d}{dt}(|u + th|^2) = 2(u + th)h.$$ So $$\frac{d}{dt}\bigg|_{t = 0} |u + th|^p = \frac{p}{2}|u|^{p-2}(2uh) = p|u|^{p-2}uh,$$ that is, $$\lim_{t\to 0} \frac{|u + th|^p - |u|^p}{t} = ... 1 Since u\in C^1(\overline U), |u| is Lipschitz, hence absolutely continuous. The fundamental theorem of calculus applies to such functions: the value at x_n=0 is related to the integral of x_n-derivative in the usual way (the other boundary term is zero since \zeta is compactly supported). (From a comment by John Ma) 0 If  (u,v)  is a  C^{1} -solution pair, then it is, in fact, a  C^{\infty} -solution pair. To prove this, let  F_{f}  denote the distribution corresponding to a locally integrable function  f  on  \Bbb{R}^{2} . As$$ {\partial_{x}}(F_{u}) = F_{u_{x}}, \quad {\partial_{y}}(F_{u}) = F_{u_{y}}, \quad {\partial_{x}}(F_{v}) = F_{v_{x}} \quad ...

1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$ $\begin{cases}\dfrac{dx}{ds}=y\\\dfrac{dy}{ds}=-a^2x\end{cases}$ $\therefore\dfrac{d^2x}{ds^2}=\dfrac{dy}{ds}=-a^2x$ $x=C_1\sin as+C_2\cos as$ $\therefore y=C_1a\cos as-C_2a\sin as$ $x(0)=x_0$ , $y(0)=y_0$ : ...

0

Hint: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dz}{dt}=z$ , letting $z(0)=1$ , we have $z=e^t$ $\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0z$ $\dfrac{dx}{dt}=\dfrac{x(z-2y^2)}{z-y^2-2x^3}=\dfrac{x(e^t-2y_0^2e^{2t})}{e^t-y_0^2e^{2t}-2x^3}$

3

Short answers: no, there isn't a separable solution; yes, there are eigenfunctions of the biharmonic. Unfortunately, the biharmonic isn't separable like the Laplacian. Off the top of my head, I don't know of a nice closed-form solution for the eigenfunctions of the biharmonic. The image below is an approximation of the eigenfunction for the smallest ...

0

let the solution is $$u=X.T$$ where $X=f(x)$ and $T=g(t)$ so $$u_t=XT'$$ $$u_x=X'T$$ institute in the equation $$XT'+X'T+3XT=0$$ divide by $XT$ $$\frac{T'}{T}+\frac{X'}{X}+3=0$$ now let $$\frac{T'}{T}=\lambda$$ or $$\frac{T'}{T}=-\lambda$$ sign of $\lambda$ depends on the problem anyhow, I will take the positive to write the solution $$T=C_1e^{\lambda ... 0 Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: \dfrac{dx}{dt}=x , letting x(0)=1 , we have x=e^t \dfrac{dy}{dt}=y+1 , we have y+1=y_0e^t=y_0x \dfrac{du}{dt}=u-1 , we have u(x,y)=F(y_0)e^t+1=xF\left(\dfrac{y+1}{x}\right)+1 u(x,2x-1)=e^x : xF(2)+1=e^x , which is impossible. \therefore There is no ... 0 Yes. This is the functional-analytic formulation of the study of linear PDEs, in which a linear differential operator L is viewed as a linear operator between two appropriate vector spaces. For example, L is a differential operator of order k and u is assumed to live on some domain U, then one might naturally think of considering L as an operator ... 4 We assume that p\ge 0 in \mathscr{D}. Suppose that u and v are distinct solutions to \nabla^2u=p(\vec x)u, for \vec x\in \mathscr{D}, and \frac{\partial u}{\partial n}=g for \vec x\in \partial \mathscr{D}. Then, w=u-v satisfies$$\nabla ^2 w=pw\tag 1$$for x\in \mathscr{D} and$$\frac{\partial w}{\partial n}=0 \tag 2$$for \vec ... 1 Hint... You can simplify the exponential term and write the equation as$$ \frac{dx}{dt} = -x^2 + B + A x_0 \left(\frac{x}{x_0}\right)^C, $$which is a separable variable DE. However the method of solution will depend on the value of C. 0 What do you mean by a maximum principle here? There are lots of versions of this rule. I assume you mean that the absolute maximum and minimum of all functions defined on the same domain and satisfying the same differential equation will happen on the boundary of the domain. I don't think that such a thing exist for this eigenvalue problem. To give some ... 1 First let's find out where the logarithm changes sign. We can rewrite the logarithm to make it$$ -\frac{1}{4\pi} \log{\left( \frac{\lvert x \rvert^2 \big\lvert y-x/\lvert x \rvert^2 \big\rvert^2 }{\lvert x - y \rvert^2}\right)}. $$Next, expand out the terms in the fraction:$$ \frac{\lvert x \rvert^2 \big\lvert y-x/\lvert x \rvert^2 \big\rvert^2 }{\lvert x ...

0

Start with $v(x,y)=x(x-a)/2$ which satisfies $\nabla^{2}v=1$ and $$v(0,y)=0,\;\;\; v(a,y)=0,\\ v(x,0)=x(x-a)/2,\;\;\; v(x,b)=x(x-a)/2.$$ Then solve for $\nabla^{2}w_1=0$ such that $$w_1(0,y)=0,\;\;\; w_1(a,y)=0,\\ w_1(x,0)=0,\;\;\; w_1(x,b)=x(x-a)/2.$$ It follows that ...

0

Finally figured out how to apply the method of images for this problem, it's similar as described previously in the same book (see page 83). Consider the domain $D = \{(x_1,x_2): x_1 \in (-\infty,\infty), x_2 >0\}.$ We use the method of images to find the Green's function $g(x,\xi)$. Consider the point $(x_1,x_2)$ and it's image point $(x_1,-x_2)$. From ...

0

Spectral methods look for approximations of solutions of an equation as a linear combination of a determined set of fucnctions, ussualy a a complete orthonormal system with respect to some weight. The firs step is to be able to approximate a function in that way . To fix things, fix an interval $[a,b]$ and a complete orthonormal system ...

0

I think your confusion comes from the fact of having $u$ as a function of two variables, i.e. $u=u(x,t)$. So when you derive wrt $t$, $x$ is supposed to be constant. Now for your first case, this is a small example that may clarify the idea: Given $f(x)$, then derivative of $f^2(x)$ is $2* f(x)* f^{'}(x)$ where prome here refers to$\frac{d}{dx}$. So ...

0

You can use the method you mention in the link: if $\gamma(t)=(x(t),y(t))$ and $\gamma'(t)=X\circ\gamma(t)\Rightarrow$ \begin{align*} x'(t)&=x(t)^2+y(t)^2\\ y'(t)&=x(t)y(t)\end{align*}Then you need to use the chain rule/inverse function theorem (you need to assume, as noted in the comments, that the domain of this is $\mathbb{R}^2\setminus\{0\}$, ...

0

A standing assumption in exercises in that chapter is that the coefficients $a^{ij}$ satisfy an ellipticity condition. In particular, the matrix $A=(a^{ij})$ is positive definite, which implies $$\sum_{i,j=1}^n a^{ij} u_{x_i} u_{x_j} = \nabla u^T A\nabla u\ge 0$$ Since also $v\ge 0$ and $\phi''\ge 0$, the conclusion -\int_U \phi''(u) v \sum_{i,j=1}^n ... 0 Your third claim does not require the large amount of regularity your initial data has. You can just follows Evans' proof exactly. Whether or not your domain is bounded in time or not does not matter when you take the limit as t\to0, and your expression for u is only defined for t>0. As an alternative, you could use the Fourier transform for parts ... 2 The characteristics (x(t), y(t), z(t)) of the system, where z(t) := u(x(t), y(t)) , are defined by the ODEs \begin{align*} \dot x &= x & \dot y &= 1 & \dot z &= -zy\\ x(0) &=x_0 & y(0)&=y_0 & z(0) &= u(x_0,y_0) = z_0, \end{align*} where p_0 = (x_0, y_0) is a point in the domain with a known value for u, which ... 2 When you multiply the metric by a factor, the Ricci doesn't scale. One way to see this is to look at its expression in the normal coordinates given here. 0 The book by Evans is a graduate textbook, which will not teach you to solve actual problems if you did not take an undergraduate course in PDE. All of the topics you mentioned (exept for Laplace transform), together with great number of examples, are present, e.g., in Peter Olver's PDE undergraduate textbook. 0 The reference book I normally use for the basics on PDEs is Partial Differential Equations, by Lawrence C. Evans. I think it covers most (maybe all) the topics you mention. 1 Let f be the eigenfunction for Laplacian with eigenvalue \mu. Then\Delta f + \mu f= 0 \Rightarrow \Delta ^2 f + \mu \Delta f =0 \Rightarrow \Delta^2 f - \mu^2 f = 0.$$Thus if the biharmonic equation satisfy maximum principle with \lambda >0, then so is the Laplace equation with arbitrary \mu \neq 0. 2 I would be surprised if this equation had a maximum principle. The prototypical example$$\tag{1}f^{(4)}(x) =f(x), \qquad x \in (0, 2\pi) fails to have one, as it supports the oscillating solution $f(x)=\sin x$. This solution satisfies Dirichlet's boundary conditions. (Neumann's boundary conditions would lead to cosines). Also, by translation and ...

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