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1

Hint. If you translate the boundary condition in cartesian coordinates, then it looks like $$ u(x,y)=x, \quad \text{whenever}\,\,x^2+y^2=1. $$ Also note, that $u(x,y)=x$ is a harmonic function in $\mathbb R^2$.


0

As user_of_math suggested, you should be consistent with your $k$ index.Let $k$ denote the index for time, $i$ for $x$-direction in space and $j$ for $y$-direction in space. Then you use the wrong discretization. It should always be $k+1$ as the time index except for the time derivative since your scheme should be implicit. \begin{align} ...


0

Would it be possible to see how you did it in dimension 1 ? I am consistently stuck because of remaing terms. I only can get: $ \theta ||\Delta u||^2_{L^2} \leq <Lu;-\Delta u> + \text{something annoying} $ Plus, you can refine a little bit what you said. By elliptic regularity (I believe the boundary is supposed to be smooth in this exercise), you ...


0

2 and 3 are easy enough: If $y$ is a point with $f(y)\leq f(x_0)$ then by (strict!) convexity, for all $t\in [0,1]$, $$ f((1-t)x_0+ty)< (1-t)f(x_0)+tf(y)\leq f(x_0). $$ Taking $t\to0$ yields a contradiction to the fact that $f$ has a minimum at $x_0$. Therefore the minimum is global and unique. If $f$ had a minimum at $x$ with $|x|=R$, then for $h$ ...


1

This is a quite classical topic, subject of several graduate courses in a lot of universities. A bunch of applications are in Monte Carlo simulation is solutions of PDE in high dimension (where the finite differences methods become inefficient), in particular in Finance. For advanced extensions to finance, I can recommend you a book from Pierre Henry ...


1

For the variance one has $$Var(X\pm Y) = Var(X) + Var(Y) \pm Cov(X,Y)$$ so if $X$ and $Y$ are uncorrelated, i.e. $Cov(X,Y)=0$ (which does not mean independent, but if they are independent then they are uncorrelated) then $$Var(X-Y) = Var(X)+Var(Y)$$ and hence $$E[(X-Y)^2] = Var(X)+Var(Y) + E[X-Y]^2$$ if you look at wikipedia, having $X\sim ...


0

Use Green's representation on the constant function $1$ to see that $$ \int_{\partial\Omega} \frac{\partial G}{\partial \nu} (x,y) d\sigma(y) =1, \quad \forall x\in B(0,1). $$ On the other hand, by the maximum principle we have $$ G(x,y)\leq C(n)|x-y|^{2-n}, \quad \forall x,y\in \Omega. $$ Now, for every $x\in B(0,1)$ we get $$ \int_{B(0,1)} |x-y|^{2-n} dy ...


1

For the equation $u_{xx} + u_{yy} = 0$ with $u(0,y)=\sin\pi y$, $u(1,y)=0$, and $u(x,0)=u(x,1)=0$ it can be seen that for $u(x,y) = F(x) G(y)$ then \begin{align} G'' + \mu^{2} G &= 0 \\ F'' - \mu^{2} F &= 0. \end{align} The solutions are $F(x) = A_{1} \cosh(\mu x) + A_{2} \sinh(\mu x)$ and $G(y) = B_{1} \cos(\mu y) + B_{2} \sin(\mu y)$. For the case ...


0

I have an answer to this but it is sort of particular to one way of thinking about PDEs. I think other methods yield similar results, but I am not as familiar with how they justify them. So! To be specific, semigroup theory says something like if you want to solve $U_t = A u$ where $A$ is an unbounded operator on a Hilbert space $H$ with domain $D(A)$ ...


3

Your problem can be reformulated as follows (upon the change of notation $a=x, b=y, c=z, d=t$). You are assigning the differential form $$ \omega=f_1(x)dx +f_2(y) dy + f_3(z)dz+f_4(t)dt, $$ which is closed, hence exact on $\mathbb{R}^4$. You want to find a potential function, that is, a function $F=F(x, y, z, t)$ such that $dF=\omega$. One of them is given ...


0

One of the problems here is that each element of your spaces is really an equivalence class of a.e. equal functions and not a genuine function. This is especially a problem in the case $u:(0,T)\to L^2$, because one has trouble selecting for each $t$ some representative function $x \mapsto u(t)(x)$. One way to circumvent this is the following. Observe that ...


1

It converges uniformly in $M\times M$ for each $t>0$. If I remember correctly, a proof is given in Rosenberg's Laplacian on Riemannian manifolds. Chavel's book must also have something on this. In any case the proof is straightforward if you use Weyl's law (Please try).


2

I understood my mistake, and it was caused by misunderstanding the notation. Given an approximate answer \begin{align} u_N = \phi_0 + \sum_{i=1}^N c_i\phi_i \end{align} We cannot say that $l_i = l(\phi_i)$, but \begin{align} l_i = l(\phi_i) - B(\phi_i, \phi_0) \end{align} That way, choosing $\phi_0 = x$ and $\phi_i = x^i(1-x)$, we find \begin{align} ...


1

Hint: Write down everything explicitly. A orthogonal matrix $A$ satisfies $A A^T = I$. Thus $$\sum_{j=1}^n a_{ij} a_{kj} = \delta_{ik},$$ where $\delta_{ik} = 1$ if $i=k$ and $0$ if $i\neq k$. By definition, $$(u \circ A)(x_1, \cdots , x_n ) = u(A(x_1, \cdots x_n)) = u(\sum_{i=1}^n a_{1i} x_i, \cdots , \sum_{i=1}^n a_{ni} x_i)$$ Now take ...


1

Following through on the comment I gave above, note that we can rewrite the integral by reversing the order of summation (Fubini's theorem) and thereby write the inverse transform as $$u(x,t)=\int_{-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(r-x)\xi +i t\xi^3}\:d\xi\right)f(r)\:dr.$$ It remains to show that the integral inside is of the ...


0

My suggestion would be: $-\Delta u = f - c(u)$. Now you are done if you can show that the RHS is in $L^2$ by elliptic regularity. To do this it is of course enough to show that $c(u)$ is in $L^2(\mathbb{R}^n)$. $$\int |c(u)|^2 = \int_{\text{supp}(u)}|c(u)|^2 \le \|c\|^2_{L^{\infty}(\text{supp}(u))}|\text{supp}(u)| < \infty$$ Do you think this work? Let ...


0

Use an explicit example. The function $g$ is not helpful; its contribution dissipates as $t^{-2}$ in three dimensions as it is stretched over a sphere of radius $t$ (this should also be familiar from physics: light intensity, etc). I would take $g=0$ and $h=1$ in the unit ball (tapering off in some way to make $h$ smooth and compactly supported). Looking ...


0

I'm a little bit rusty in this, so I hope I'm not doing any foolish mistakes. Assume $\phi = X(x)Y(y)$, then as you already calculated: $$ Y(y)X''(x) + Y''(y)X(x) = KX(x)Y(y) $$ Assuming neither $X$ nor $Y$ vanish in their domain, this means that, $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = K $$ From which we get that $$ \frac{X''(x)}{X(x)} = ...


1

You seem to infer, from continuity of $u$, that every number between $\inf_{\partial\Omega}u$ and $\sup_{\partial\Omega}u$ is in $u(\partial\Omega)$. That won't work without some connectedness assumption.


-1

I correct your differentiation mistake, $$(\sin^2(\frac{t}{2}))' = 2\sin(\frac{t}{2})(\frac{1}{2})\cos(\frac{t}{2}) = \sin(\frac{t}{2})\cos(\frac{t}{2}).$$


2

The Calculus distance scale factors in spherical coordinates are $s_{r}=1$, $s_{\theta}=r$, $s_{\phi}=r\sin\theta$. So the Laplacian is $$ \Delta=\frac{1}{s_{r}s_{\theta}s_{\phi}}\left[ \frac{\partial}{\partial r}\frac{s_{\theta}s_{\phi}}{s_{r}}\frac{\partial}{\partial r} ...


0

$u$ is a supersolution, i.e. $\Delta u\leq0$, then $\Delta v=-\Delta\ln u=-\frac{1}{u}\Delta u+\frac{|\nabla u|^2}{u^2}\geq0$, i.e. $v$ is a subsolution.


3

In polar coordinates, the separated solutions of the Laplace equation are (up to constant coefficients): $1$ $r^n \cos n\theta$, $n\ge 1$ $r^n\sin n\theta$, $n\ge 1$ $\log r$ $r^n \cos n\theta$, $n\le -1$ $r^n\sin n\theta$, $n\le -1$ The last three have a singularity at the origin, so can't be used in a disk; they are listed here for completeness. ...


0

It's not a tridiagonal matrix; most equations involve five unknowns. It is sparse, but having to flatten a matrix of unknowns into a vector of unknowns means we don't get the nice multidiagonal form that 1D Laplace equation has. Also your system is incomplete: you have fewer equations than unknowns. You need to use the difference scheme at the points of ...


0

If you consider $L=-\Delta +V$ on the linear space of complex functions, then the involution operator of complex conjugation on this space commutes with $L$. If you start with some domain $\mathcal{D}(L)$ which is invariant under this involution on which $L$ is symmetric, then $L^{\star}$ commutes with conjugation. I cannot conceive of a reason that you ...


1

This is true, and here is an approach: $$ \varphi\circ(u+h)-\varphi\circ u = h \int_0^1 \varphi'\circ(u+th)\,dt $$ On the right, the Hölder norm of the integral is controlled by the derivatives of $\varphi$ and the Hölder norm of $u+th$. It remains to use multiplicative property: $[uv]_\alpha\le C[u]_\alpha[v]_\alpha$. The latter holds because ...


0

Just follow the method of characteristics, which in your case reads: $$dt = dx = \frac{du}{-u/x}.$$ From 1st and 2nd we have $t-x = c_1$ is a characteristic of the PDE. From 2nd and 3rd we would have that $u = c_2/x$ is the other characteristic curve. Put $c_2$ as a function of $c_1$ and you are done! Cheers!


1

so I look for $f(x)$ Don't look for it in your solution: look in the statement of the problem. Is it given there (like $f(x)=3\sin^2 x$ or something)? If there is no concrete initial condition, there cannot be an concrete solution: the formula that you have for $u$ at the beginning of the post is the answer. Putting the series with $B_n$ into the ...


1

So you are clear about the displayed equation that says $$ {4(1)^2\cdot 6\over \pi^2 N} <0.01 $$ ? If so, they are simply solving the inequality for the smallest integer $N$ which makes it true: \begin{align*} {4(1)^2\cdot 6\over \pi^2 N} <0.01 &\implies {24\over \pi^2 N}<0.01\\ &\implies {\pi^2 N\over 24}>100\\ &\implies ...


1

This is a nonlinear equation, so its classification as elliptic, parabolic, or hyperbolic depends on what function you linearize around. For example, if you linearize around $u(x,y)=y$, the equation is elliptic exactly where $y>0$. If you linearize around $u(x,y)=x$, then it's elliptic where $y>x^2$. Many different regions are possible. Since you ...


1

The function $g$ represents the rate of heat flow through the boundary; in physics terms, its units are different from the units of $u$. Thus, $M = \max\{u_0, g\}$ is never going to be useful. If the net flow into the domain is positive, the temperature inside will grow without bound as $t\to\infty$ at a linear rate. Indeed, $$ \frac{d}{dt}\int_\Omega ...


0

For example, it can model a process of heating a rod with time-depending heat conductivity coefficient (it can either rise or decay), for example, there is a chemical reaction in a rod, and a material (consequently, conductivity coefficient) changes with time. The boundary condition tells us that on one side of rod the temperature is constant, and on another ...


1

Green's identity says that $$\int_D(f\Delta f+|\nabla f|^2)=\int_{\partial\Omega}f\frac{\partial f}{\partial \nu}.$$ Now take $f=u$ and by the assumption that $\Delta u-ku=0$ in $D$, we have $$\int_D(ku^2+|\nabla u|^2)=\int_{\partial\Omega}u\frac{\partial u}{\partial \nu}.$$ Dirichlet boundary condition says that $u=0$ on $\partial\Omega$, and Nerumann ...


0

Note that we get the following relation: $$ g(y) = \frac{y^2}{2} - f(y) $$ so we can simplify $$ u(x,y) = \frac{y^2}{2} - g(ye^{-x}) = \frac{y^2}{2} - \frac{y^2e^{-2x}}{2} + f(ye^{-x}) $$


1

We first claim that if a $C^0(\bar{\Omega})$ function $u$ satisfies the following local mean value inequality $$u(y)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(y)}u ds \text{ for all } R\leq \delta$$ then maximal principal applies, i.e. $u$ attains its maximum at $\partial \Omega$. Assume our claim now. Let $\tilde{u}$ be the harmonic function ...


0

Plug equation 6.4 into the pde with $f=0$ and determine which values of $K$ satisfy it (hint: they're complex). Add these solutions (damped waves travelling in opposite directions) in amounts which match the periodic boundary conditions at $x=0$ and $x=1$. Then apply the same principle to the discretised version of the pde and compare the two solutions to ...


0

Note that \begin{align*} \frac{1}{2L}\int_{-L}^L \mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}} \mathrm{e}^{\mathrm{i} \frac{-m \pi x}{L}} \mathrm{d} x = \delta_{n,m} \end{align*} I.e. the basis functions $\mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}}$ are orthogonal. So note that for $N$ finite, we have \begin{align*} & \left \| \sum_{n = 0}^N C_n ...


0

There is a difference between classification linear and nonlinear PDEs. The above inviscid Burgers' eqaution is a nonlinear PDE, so you can't prove it as hyperbolic by using the formula ,$B^2 - 4AC > 0$, which is meant for linear PDEs. To define the type of nonlinear PDEs for hyperbolic type, you need to find eigenvalues and corresponding eigenvectors ...


0

it's me again. As you can see, I commented a few times because I had not found it yet. I just realised my problem: I was thinking of $h(t)$ as entirely arbitrary. But it isn't. $h(t)$ is only defined for $t>0$! That's important, because we can set it to zero elsewhere. If we do that, the BC that removes $C(\tau)$ is not a problem. Consistency requires ...


1

You have $\cos(2x) = \dfrac12(e^{2ix}+e^{-2ix})$ and $\sin x = \dfrac1{2i}(e^{ix}-e^{-ix})$, so $$ \cos(2x)+\sin x = \frac12(e^{2ix}+e^{-2ix}) + \frac1{2i}(e^{ix}-e^{-ix}). $$ Later comment: Since the point is to write the function as a linear combination of $\{e^{inx}\}_{n=-\infty}^\infty$, once you've written the expression above, you're done. You wrote ...


1

As I said in comments, I do not find the conclusion $\Delta u(y)=0$ justified. Here is a different approach. Let $v=\Delta u$. Since both $u$ and its normal derivative vanish on the boundary, Green's second identity implies $$\int_\Omega (v\Delta u - u \Delta v) = 0 $$ which simplifies to $$\int_\Omega (\Delta u)^2 = 0 $$ Thus, $u$ is harmonic. Since it ...


2

The second Green's identity says that $$\int_\Omega f_1\Delta f_2-\int_\Omega f_2\Delta f_1=\int_{\partial\Omega}f_1\frac{\partial f_2}{\partial\nu}-\int_{\partial\Omega}f_2\frac{\partial f_1}{\partial\nu}.$$ Putting $f_1=u$ and $f_2=\Delta u$, we have $$\int_\Omega u\Delta (\Delta u)-\int_\Omega (\Delta u)^2=\int_{\partial\Omega}u\frac{\partial (\Delta ...


0

since $m=\inf_{N-B}w\gt 0$, on $\partial B$, $m\le w$, and so $\overline{w}|_{\partial B}=m$, as $w$ is superharmonic it is continuous. The $\inf$ of two continous functions is continuous, thus $\overline{w}$ is continuous.


1

In regards to question 2, when considering the case when $\lambda<0$, set $\lambda=-\alpha^{2}$ as $\alpha^{2}$ is strictly positive (I'm assuming we're in $\mathbb{R}$) and hence $\lambda$ is strictly negative. Although the poster Semsem didn't do the $\lambda>0$ case, in this particular instance you would set $\lambda=+\alpha^{2}$.


0

$F''-\lambda\,F=0$ gives $m^2-\lambda=0$, not $m^2-\lambda\,m=0$.


1

The method used there is separation of variables. We suppose that we can separate $u(x,t)$ as a product of two functions $G(t)$ and $F(x)$. Then we find the derivatives in the differential equation as $$\frac{\partial^2 u}{\partial t^2}=F\ddot{G} \\ \frac{\partial^2 u}{\partial x^2}=F''G$$ we then substitute in the given PDE $$F\ddot{G}=c^2 F''G\implies ...


0

Following up on comments by Daniel Fischer: The boundary values are rotationally symmetric, hence so will be $u$ You don't actually need a proof of this statement (since your task is to find one solution, not all of them) so I would rather say: The boundary values are rotationally symmetric, hence we should look for a solution that is rotationally ...


0

It is sufficient to verify the mean value property for sufficient small balls for every $x\in \Omega^+ \cup T \cup \Omega^-$. This is clearly true for $x$ such that $x_n>0$, since we start with a function that is harmonic in $\Omega ^+$. By reflection, the function is also harmonic on $\Omega ^-$. Finally, when $x_n=0$, the integral over the upper half ...


1

Begin by differentiating the product, $$ \frac{\partial}{\partial x_i} \left( \frac{v'(r)}{r} \cdot x_i \right) = \frac{\partial}{\partial x_i} \left( \frac{v'(r)}{r} \right) \cdot x_i + \frac{v'(r)}{r} \cdot 1= \dots, $$ then use the chain rule. What you forgot is that $x_i$ depends on $r$.


1

Green's idenitity says that $$\int_{\Omega}u\Delta u+\int_\Omega |\nabla u|^2=\int_{\partial\Omega} u(\nabla u\cdot n).$$ Since $u=0$ on $\partial\Omega$ by assumption, the right hand side is zero. On the other hand, $\Delta u=u^3$ in $\Omega$ by assumption, the left hand side is equal to $$\int_{\Omega}u^4+\int_\Omega |\nabla u|^2.$$ Combining all these, ...



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