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2

see https://www.youtube.com/watch?v=WnJKI6lKwu0 ..................................... Note: on the question of attempting difficult sounds in languages you do not know: first, a Dutch guy with whom I played soccer for years agreed with me that he would prefer announcers to just say the closest thing they could manage and not overdo it. So, I encourage you ...


1

In addition to the answer by user90090, it can be told that in Sobolev spaces by Adams, as well as in its version by Adams–Fournier, no appropriate details are to be found. The details of dependence of the best Sobolev constant on parameters of the cone condition, usually referred to by experts in the field as well-known, belong to what is now called ...


0

my answer is : firstly,we assume that: $$X=\dfrac{yf_{y}-z}{f_{x}}+x=P+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y=Q+y,Z=z-xf_{x}-yf_{y}=z-R$$ then, $C^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$ ...


0

The fact that the first condition does not hold is not the main reason why the theorem cannot be applied. It can never be applied, because it is basically wrong. Indeed, take $X=\mathbb{R}$ and consider a functional $\varphi\,\colon \mathbb{R}\to\mathbb{R}\,$ of the form $\,\varphi(x)=e^x$.  It is clear that $\,\inf\limits_{x\in\mathbb{R}}{e^x}=0$, so ...


0

Your problem has no solution. It is easy to see by the Riesz–Fredholm theory (see Theorem 1.14, Part (ii), page 10, http://books.google.de/books/about/Nonlinear_Analysis_and_Semilinear_Ellipt.html?id=4O6tfmZtxQAC&redir_esc=y ), since $\lambda_1 = 1$ is the first eigenvalue and $v_1 = \sin t$ is the first eigenfunction, but $$ \int_0^\pi \sin t \sin t \ ...


1

There is a useful recent survey Geometrical structure of Laplacian eigenfunctions by Denis S. Grebenkov and Binh-Thanh Nguyen. Quote: a level set of the first Dirichlet eigenfunction on a bounded convex domain in $\mathbb R^d$ is itself convex [274]. Predictably, the reference is to Bernhard Kawohl's book Rearrangements and Convexity of Level Sets in ...


0

Assume that $\frac{1}{c^2h(t)} \frac{d^2 h}{d t^2}(t) = \frac{1}{\varphi(x)}\frac{d^2 \varphi}{d x^2}(x)=-\lambda(x,t)$, this is the most general assumption you can make at first. But you can differentiate $\lambda(x,t)$ with respect to $x$, using the fact that $\frac{1}{c^2h(t)} \frac{d^2 h}{d t^2}(t) = -\lambda(x,t)$ you conclude that $\lambda_x(x,t)=0$, ...


1

The cylinder (of unit radius, for simplicity) is obtained from the infinite strip $S = \{z\in \mathbb C: |\operatorname{Im}z|<\pi \}$ by identifying its sides. What you need is Neumann Green's function for $S$ (i.e., with zero normal derivative on $\partial S$), because it will remain harmonic after gluing the sides. And to find this function, you can ...


0

a) As specified by your question, the initial condition is $T(x,0) = 0$ and the boundary conditions are $T(0,t) = 0$ ($x=0$ maintained at 0), and $T_x(a,t) = 0$. b) The steady-state solution is reached when $T_t(x,t) = 0$, so: $$ \frac{\partial^2 T_s}{\partial x^2} = -\frac{1}{k} \implies \frac{\partial T_s}{\partial x} = - \frac{x}{k} + ...


0

Following Hans Lundmark's suggestion, we compose 3 maps: $\frac{z+1}{1-z}$, $z^2$, $\frac{-i+z}{i+z}$, to get the map $$h(z)=\frac{(1+z)^2-i(1-z)^2}{(1+z)^2+i(1-z)^2}$$ which maps the upper boundary of the semidisk to the upper boundary of the unit disk, and the lower line segment of the semidisk to the lower boundary of the unit disk. Now we solve the ...


1

You just need to go a bit further. The equation $f_{st}=0$ tells you that $f_t$ is constant with respect to $s$, so it must be a function of $t$: $$f_{t}=G(t).$$ Let the integral of $G$ be $g$. Then this equation tells us that $$f=g(t)+h(s),$$ since the constant of integration must be constant with respect to $t$. Can you finish?


1

This problem is quite tricky especially considering that it doesn't have a strict space boundary. I assume that you're not very familiar with numerical methods, thus I can give you an advice on where one would start the numerical treatment if it had a more strict space bound, for example $[0, T]\times[a,b], T>0$. 1) We discretize time domain using first ...


1

Substitute $\,u(x,t)=v(x,t)+t+xt\,$ and consider the problem $$ \begin{cases} v_{tt}=c^2v_{xx}, \quad 0<x<\pi,\;t>0;\\ v(0,t)=v(\pi,t),\quad t\geqslant 0;\\ v(x,0)=0,\;\;v_t(x,0)=\sin{x},\quad 0\leqslant x\leqslant \pi. \end{cases} $$ Consider the Sturm–Liouville eigenvalue problem $$ X''=\lambda X, \quad 0<x<\pi,\\ X(0)=X(\pi)=0, $$ and ...


2

This is a regular Sturm-Liouville problem. Whether you use Dirichlet or Neumann conditions, there is an infinite set of eigenvalues $$ \lambda_{t,1} < \lambda_{t,2} <\cdots < \lambda_{t,n} < \cdots $$ with corresponding real eigenfunctions $\phi_{j}(x)$ such that $$ \phi_{t,j}''+ ...


1

Let's write the integrand in components: $$ \Lambda\psi(-\partial_2\psi\partial_1\theta+\partial_1\psi\partial_2\theta). $$ Now we integrate by parts (there are no boundary terms because of the periodicity) $$ \int \Lambda\psi(-\partial_2\psi\partial_1\theta+\partial_1\psi\partial_2\theta)=\int ...


1

A little more details as requested (in private email) Sorry I made a mistake above, of course you want $b \ll 1$. If that is the case you can use whats called "regular perturbation theory" (as opposed to singular perturbation theory which is what you would need if for example the $f_{xx}$ term was multiplied by a small parameter). Make the following ...


-1

Hint: substitute the trial function $u(x,t)=A(t)e^{ikx}$ into the PDE to obtain an ODE for $A(t)$. $$\frac{\partial}{\partial t}\left(A(t)e^{ikx}\right)+c\,x\,A(t)e^{ikx}=0\\ \iff e^{ikx}A'(t)+c\,x\,A(t)e^{ikx}=0$$


1

Try: Numerical Solution of Partial Differential Equations: Finite Difference Methods , by G. D. Smith Also, use the open courseware at: MIT Open Courseware


0

Despite being concise, formulation of the "cited" theorem does contain two mistakes. One mistake is just grave: a missing constant factor, depending on $p$, in the right-hand side of the estimate forcing solutions to stay bounded while $p\to 1$ or $p\to \infty$ which is well known to be wrong. Another mistake is not just very grave, it is fantastically ...


0

Firstly you need to use polar coordinates, so you seek the solution: $u(x,y)=u(r(x,y),\theta(x,y))=R(r)\Theta (\theta)$ (here we use separation of variables) If you follow the calculations through, you will be lead to a cauchy euler equation. Can you take it from here? Also $r^2=x^2+y^2$ and $\tan\theta=\frac { y } { x}$.


1

Alberto Calderón and Antoni Zygmund have nothing to do with this theorem that cannot be true without specifying some boundary condition at $\partial B_2$. The missing boundary condition is of course a terrible blunder making it easy, for example, to establish that the subspace of all harmonic functions in $W^{2,p}(B_2)$ is finite-dimensional, which is ...


0

You should set up the problem in the space $W^{1,2}(]0,\pi[)$. Moreover the functional should be: $$\varphi(u):=\int_0^\pi \left(\frac12|u'|^2-\lambda u^2-\frac{u^4}{4}\right)dt$$ which verifies the hypotheses of the Mountain Pass Theorem. Notice that you need $\lambda>0$ to get rid of the space of constant functions. In this way you can prove the ...


1

If $0 < c < 1$ then $1/c > 1$, so if $1 \leq k \leq J$ we have $$ \begin{align} \sum_{j=k}^J jc^{k - j - 1} &= \sum_{j=k}^J j \left(\frac{1}{c}\right)^{1+j-k} \\ &\leq \sum_{j=k}^J j \left(\frac{1}{c}\right)^{1+j-1} \\ &\leq \sum_{j=1}^J j \left(\frac{1}{c}\right)^{j}. \end{align} $$ The first inequality holds holds because the map $x ...


0

At your link to MathWorld, "linearization" is to be understood as a Fréchet derivative of the appropriate nonlinear mapping, which has become quite habitual nowadays. Your second-order PDE is quasilinear, i.e., linear w.r.t. the highest-order derivatives. A strict formal definition of a quasilinear nonlinear PDE is generally being omitted mostly due ...


1

$$\dfrac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}|y|e^{\frac{-(\xi-y)^2}{4t}}dy$$ $$=\dfrac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}ye^{\frac{-(\xi-y)^2}{4t}}dy + \dfrac{1}{\sqrt{4\pi t}}\int_{-\infty}^{0}(-y)e^{\frac{-(\xi-y)^2}{4t}}dy$$ $$=\dfrac{1}{\sqrt{4\pi t}}\int_{0}^{\infty}ye^{\frac{-(\xi-y)^2}{4t}}dy - \dfrac{1}{\sqrt{4\pi ...


1

You can change variables to arrive to the standard heat equation. Consider $$ v(x,t)=e^{-rt}u(x,t). $$ If you multiply the original equation by $e^{-rt}$, you get $$ \partial_t v=\partial_t u e^{-rt}-re^{-rt}u=e^{-rt}\Delta u=\Delta v $$ which is the classical heat equation. Now you can solve by separation of variables easily.


1

I do not understand how you obtained the characteristic $v$. Here is my solution $$\frac{dx-dy}{(x-y)(z-1)}=\frac{dz}{z^2-1}\Rightarrow v=\frac{x-y}{z+1}=c_{2}$$ To find the integral surface which the curve it passes we should parametrize the curve, let $x=t$ then $z=t^2$ and $y=1$. Substituting these values into $c_1$ and $c_2$ we get ...


0

The following counterexample is not fully satisfying ($u$ is unbounded, and $\mu$ is not strictly negative) but perhaps serves as a start: Take $\Omega$ to be the unit disk and set $u(r,\theta) = \left(\frac{3r^2}{1-r^3}, 0\right)$. This vector field diverges as $r\to 1$ but is differentiable on the interior of $\Omega$. But for $\eta = 1-r^3$, we have that ...


1

For the Neumann problem $\,(\ast)\,$ in a bounded domain $U\subset\mathbb{R}^n$, $n\geqslant 2$, satisfying the cone condition, to prove that assumption $$ f\in \{ L^2(U)\,\colon\;\int\limits_{U}f\,dx=0\}\tag{1} $$ implies the existence of a weak solution $u\in H^1(U)$, it is convenient to introduce the space $$ \widetilde{H}^1(U)=\{w\in ...


0

Starting from $$ y_{n+1}=y_{n}+hy_{n+1}(1-y_{n+1}) $$ multiply with $4h$ and rearrange $$ (2hy_{n+1})^2+2(2hy_{n+1})(1-h)=4hy_n $$ to get \begin{align} y_{n+1}&=\frac1{2h}\left(-(1-h)\pm\sqrt{(1-h)^2+4hy_n}\right) \\[0.8em] &=\begin{cases} \dfrac{2y_n}{(1-h)+\sqrt{(1-h)^2+4hy_n}} &\approx \dfrac{y_n}{1-h} &&\text{ for } y_n\approx 0 ...


0

Hint: try to write $$\frac{\partial p(a,t)}{\partial a}+\frac{\partial p(a,t)}{\partial t} $$ as $dp(a(t), t) / dt$ for a certain function $a(t)$.


1

This is definitely the product rule. Note that $$\frac{1}{r} \frac{\partial}{\partial r} \left(r \; k \frac{\partial T}{\partial r} \right)={1\over r}\left(\frac{\partial}{\partial r}(r)\cdot\left(k \frac{\partial T}{\partial r}\right)+\frac{\partial}{\partial r}\left(k \frac{\partial T}{\partial r}\right)(r)\right)=\frac{\partial}{\partial r} \left(k ...


0

Since $\,a,b,c\,$ are continuous, the non-empty subset $V_2$ is closed. Hence the cases $1,3,4$ are to be excluded. The only case left satisfies all the requirements. So you are quite right, and no better way could be found.


2

Since we conjecture $u(x,y)=f(x^2+y^2)$, let us (carefully) apply calculus and verify $-yu_x+xu_y=0$. Note (by chain rule), $$ u_x = f'(x^2+y^2)*2x, \qquad u_y = f'(x^2+y^2) * 2y. $$ Thus, $-yu_x+xu_y=-2xyf'+2xyf'=0$. So, you had it, except a small detail with the chain rule. Another way to have caught the mistake -- $f$ is a function of a single ...


0

Suppose $x=x(t)$ is a characteristic line. Then $\frac{du}{dt} = u_x x' + u_t$. We conclude $x'(t) = 1$ -- the characteristic curves are $x=t+x_0$. However, $u$ is not constant along the characteristic -- it varies in a predictable way. In fact, $u'+f(t+x_0)u=0$ is an easily-solved ODE for $u(t)\equiv u(x(t),t)$, given that we know $f(x)$.


0

Let us try to obtain a counterexample. We assume that $u$ is sufficiently smooth. Let us write the variational definition of the first eigenvalue: $$ \mu_1[u] = \inf_{\eta \in W_0^{1,2}(\Omega)}\frac{\int_\Omega |\nabla \eta|^2 \,dx - \int_\Omega \nabla \cdot (u \eta) \, \eta \,dx}{\int_\Omega |\eta|^2 \,dx}, \quad \left( \eta \not\equiv 0 \right). \tag 1 ...


1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$ $\dfrac{dz}{dt}=1$ , letting $z(0)=z_0$ , we have $z=z_0+t=z_0+x$ $\dfrac{dy}{dt}=2z=2z_0+2t$ , letting $y(0)=f(z_0)$ , we have $y=f(z_0)+2z_0t+t^2=f(z-x)+2(z-x)x+x^2=f(z-x)-x^2+2xz$ , i.e. $z=x+F(x^2-2xz+y)$ ...


1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$ $\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$ $\dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$ ...


0

Let $u=e^v$ , Then $u_x=e^vv_x$ $u_y=e^vv_y$ $\therefore(e^vv_x)^2+(e^vv_y)^2=(e^v)^2$ with $v(x,0)=0$ $e^{2v}v_x^2+e^{2v}v_y^2=e^{2v}$ with $v(x,0)=0$ $v_x^2+v_y^2=1$ with $v(x,0)=0$ $v_y^2=1-v_x^2$ with $v(x,0)=0$ $v_y=\pm\sqrt{1-v_x^2}$ with $v(x,0)=0$ $v_{xy}=\mp\dfrac{v_xv_{xx}}{\sqrt{1-v_x^2}}$ with $v(x,0)=0$ Let $w=v_x$ , Then ...


0

I think the statement is strange, because clearly a holomorphic function is only defined on even dimensional spaces. I do not think the statement would work for $\mathbb{R}^{3}$, for example. For the reference, the standard one I know is Donaldson's book Riemann Surfaces.


0

EDIT: Sorry I don't know why I was so sloppy with capital versus lowercase letters, I intended to make capitals random variables but I was sloppy and did it inconsistently in some places. I might edit it at some point. This follows the structure of the derivation of the regular B-S equation by Thayer Watkins. I guess the risk-free interest rate is $0$? Let ...


1

The most straightforward, but tedious, way is to generate six equations by computing derivatives of the original equations wrt $x$ and $t$. You can then express them in matrix form $$ \begin{bmatrix} A_{11} & A_{12} & A_{13} & A_{14} & A_{15} & A_{16} \\ A_{21} & A_{22} & A_{23} & A_{24} & A_{25} & ...


1

This is true for $p\geqslant 1$. The case $p=1$, being a limiting one, requires more efforts. It will be sufficient to establish the rule $$ \partial_j(uv)=u\partial_j v+v\partial_j u,\quad j=,\dots,n,\tag{1} $$ for the weak derivatives $\partial_j$ just in case $p=1$, with the equality $(1)$ understood in a weak sense, i.e., as an integral identity $$ ...


1

Maple gives: $$\rho(x,t)=-\dfrac{\sin(x)}{\sin(x)\cdot t-1},$$ so yes $c_1=\dfrac 1{\sin(x)}$.


1

The sentence "We look for solutions $u$ in the form $u(x, t) = T(t)X(x)$" in the notes is a little misleading. What they are looking for is a solution of the form $u(x,t)=\sum_n T_n(t) X_n(x)$ where the functions $X_n(x)$ are chosen as before (the homogeneous case) while the functions $T_n(t)$ have to be chosen so as to take care of the right-hand side ...


1

This is an easy consequence of a more general addition formula for spherical harmonics: \begin{align}P_l\left(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\cos(\varphi_1-\varphi_2)\right)=\\ =\frac{4\pi}{2l+1}\sum_{m=-l}^l Y_l^m(\theta_1,\varphi_1)\overline{Y_l^m(\theta_2,\varphi_2)}.\tag{1} \end{align} Namely, set $\theta_1=\theta_2=\theta$, ...


0

To find the limit as $t \to \infty$, it is equivalent to find the steady state solution. So $$ \frac\partial{\partial x}\left(f(x)\frac{\partial\phi_\infty}{\partial x}\right) = 0,$$ which implies $$ f(x)\frac{\partial\phi_\infty}{\partial x} = C, $$ where $C$ is a constant to be determined, which implies $$ \frac{\partial\phi_\infty}{\partial x} = \frac ...


2

It is advantageous to set $r(x)^2:= x_1^2 + \dots + x_n^2$, then $f(x) = r(x)^{2-n}$, $\partial_{x_1}f = (2-n)r^{1-n} \partial_{x_1}r$. You get shorter expressions for derivatives. Also there is no need for induction. Just compute the partial derivatives, and sum up.


1

First, forget the fact that $\rho$ is a function of two variables and just look at it as a function of $t$. If I tell you that $\sin x$ is a constant $A$ and $\cos x$ just some constant $B$, how would you solve the ODE $$\frac{d\rho}{dt}=A\rho\\\rho(0) = B,$$ where $\rho$ is just a function of time? Once you do that, plug $A$ and $B$ for what they are and ...



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