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2

Let us prove that $2uu_{xx} = 2\frac{\partial }{\partial x}\big ( uu_{x} \big ) -2u_{x}^{2}$. First, we elaborate the first term of the right hand side : $$ 2\,\frac{\partial }{\partial x}\big ( uu_{x} \big ) = 2 \big( u_x \cdot u_x + u \cdot u_{xx}\big) = \boxed{\ 2\left( u_x^2 + u u_{xx}\right) \ } $$ Second, we subtract $2u_{x}^{2}$ from the ...


1

If you start from the left hand side, you can integrate it and use by-part: $$\int 2uu_{xx}dx=2uu_x-2\int u^2_xdx$$ Then differentiate both sides with respect to x and get: $$2uu_{xx}=2\frac{\partial}{\partial x}(uu_x)-2u^2_x$$ as desired. If you start from the right hand side, you can use the product rule: $$2\frac{\partial}{\partial x}(uu_x)-2u^2_x$$ ...


3

Correct for the first two. For the third you need the product rule, which says that $\frac{\partial}{\partial x}(uv)=u_xv+uv_x$ if $u,v$ are functions of $x$. Se we have: $$2\frac{\partial}{\partial x}(uu_x)-2u_x^2$$ $$=2(u_x\cdot u_x+u\cdot u_{xx})-2u_x^2$$ $$=2uu_{xx}$$


1

Just use the product rule on the first term on the right hand side: $2\frac{\partial}{\partial x}(u u_x) = 2u_x u_x+2uu_{xx} = 2u_x^2+2uu_{xx}$.


0

You want a solution $f(u,v)=g(v)$? if that's the case you use that ansatz and find solutions of that form. You will get a ode given by $$ \frac{1}{v}\dfrac{d}{dv}v\dfrac{dg}{dv} = 0 $$ but are you sure you want this? if you want a full solution then use this $$ f(u,v) = g(v)h(u) $$ and you will get a similar equation for $v$ and a another equation for $u$.


0

Actually I've managed to figure out that we just need linear combinations of 1, -2, 1 for (*) evaluated at $S_{n-1}$, $S_n$, $S_{n+1}$ respectively and we will get the required expression.


0

The mean value property tells us (in particular) that for any ball $B_r(p)$, $0=u(p)=\frac{1}{|B_r|}\int_{B_r(p)}u(y)$, WLOG assume that $u(x)>0$ in $B_r(p)\setminus\{p\}$, then $\frac{1}{|B_r|}\int_{B_{r(p)}}u(y)>0$, noting that $u$ is continuous, this contradicts the mean value property. Note that this breaks down for $n=1$, take the example ...


1

Ok, I think I found an example of this. Take a solution of $$(1) \qquad \partial_t u + L u=0$$ Assuming we can justify the following steps, we integrate in $t$ to get $$\int_0^\infty (\partial_t u + Lu) d\tau = 0$$ $$\lim_{t\rightarrow \infty} u(x,t) - u(x,0) + L \int_0^\infty u d \tau = 0$$ $$-u(x,0)+L\int_0^\infty u d \tau = 0,$$ where in the last step I ...


0

There is a contradiction between the conditions : $$u(x,0)=2x \:\:\: \text{and} \:\:\: u_x(0,t)=-2\left( u(0,t)-t \right)$$ The derivation of the first condition gives : $u_x(x,0)=2$ , hense : $$u_x(0,0)=2$$ $u(x,0)=2x$ hense $u(0,0)=0$ Second condition : $u_x(0,t)=-2\left( u(0,t)-t \right)$ hense $u_x(0,0)=-2\left( u(0,0)-0 \right)=-2\left( 0-0 ...


0

Basically we need to show that $$ D (S_n^2 \Gamma_n) = S_n^2 D \Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + \Gamma_{n+1} + \Gamma_{n-1} $$ where $D$ is second order finite difference derivative operator $$ D u_n \equiv \frac{u_{n+1} -2 u_n+ u_{n-1}}{\Delta S^2} $$ Note, that $D$ can be written as $$ D = D_+ D_- = D_- D_+ $$ where $$ D_+ ...


0

I am assuming you are familiar with the setting of your equation i.e. Hilbert spaces. If this is the case, then you can easily show $B[u,u] \geq C\|u\|^2$ by choosing $v = u$ in your bilinear term, using Friedrich's inequality and the assumption that $\alpha > 0$.


2

You can reduce this problem to a more familiar one with the following trick. Consider the problem $$ q_t(x,t) = -au_x(x,t) + \frac{D}{2}q_{xx}(x,t) \quad (0 < t, 0 < x < 2L)\\ q(x,0) = 0, \quad aq(0,t) - \frac{D}{2}q_x(0,t) = f(t) = -aq(2L,t) - \frac{D}{2}q_x(2L,t) $$ and solve it with a standard method. The solution will have a specific symmetry ...


0

While I don't have an example, like uranix provided, here's a potentially more intuitive way to look at the CFL number. The CFL number limits how large your time step $\Delta t$ can be, for your explicit numerical method to remain stable. The key here is that you are choosing the CFL number (usually, as mentioned, around .5). Therefore, not only is it ...


0

The proof that you are looking for can be found in the page 258 of the Evans book:


1

If you compare your expression with $$ u(x,t) = E^Q\left[ \int_t^T e^{- \int_t^r V(X_\tau,\tau)\, d\tau}f(X_r,r)dr + e^{-\int_t^T V(X_\tau,\tau)\, d\tau}\psi(X_T) \Bigg| X_t=x \right] $$ where $dX = \mu(X,t)\,dt + \sigma(X,t)\,dW^Q$ (this is pasted from the wikipedia article ), we see that in your case the process $X_t$ is simply $B_t$ so $\mu =0$ and ...


0

Observe that $a_n + b_n + c_n = 1,\; a_n, c_n < 0$. Suppose that $V_n^m$ is negative and is a local minimum. $$ V_n^m < 0\\ V_{n+1}^m \geq V_n^m\\ V_{n-1}^m \geq V_n^m. $$ Then $$ V_n^{m+1} = a_n V_{n-1}^m + b_n V_n^m + c_n V_{n+1}^{m} = -a_n (-V_{n-1}^m) + b_n V_n^m - c_n (-V_{n+1}^m) \leq -a_n (-V_n^m) + b_n V_n^m - c_n (-V_n^m) = V_n^m. $$ Since ...


0

lulu's comment sums it up nicely. Shreve mentions this is a backward parabolic PDE, and for it to be well-defined we must specify boundary conditions. From a purely mathematical viewpoint, those can be almost anything we want. The condition $$\lim_{x \rightarrow +\infty} c(t,x)- (x- e^{-r (T-t) K}) =0 \text{ for all $t \in [0, T]$}$$ is purely an economic ...


1

1) also can be justified by that $H^1$ is a reflexive Banach space, so from the bounded sequence you can extract a weakly convergent subsequence with a limit in $H^1(\Omega)$ and because the embedding operator is compact that means that it maps weakly convergent sequence to strongly convergent, i.e $u(.,t_{k_l})\rightarrow w$ strongly in $L_2$ (here, we ...


0

If we consider the simplified problem $$-\Delta u = f\quad\mbox{in}\quad\Omega,$$ $$u|{\Gamma_D}=0,$$ $$\partial_nu|_{\Gamma_N}=h,$$ Then the weak formulation is to find $u\in H^1_E(\Omega)=\{v\in H^1(\Omega):v|_{\Gamma_D}=0\}$ such that $$(\nabla u,\nabla w)=(f,w)+(h,w)_{\Gamma_N}$$ for all $w\in H^1_E(\Omega)$. By the integrability assumptions on $f,w$, ...


0

I think what you've done is a terrible detour that consists mostly of undoing things you did before. The one substantial step involved is substituting $x=\pi$ into the differential equation, which you could have done directly: $$ u_{xt}(\pi,t)+u(\pi,t)u_{xx}(\pi,t)=h(t)=0\;. $$ If I may take the liberty to quote you: That's it. Seems pretty straightforward ...


1

A parabolic operator with constant coefficients is a linear transformation away from the heat operator, so it is a natural guess that the fundamental solutions should be similar. I will use this idea to find the fundamental solution. (If you just want to see the solution, see the last line.) Take two positive definite symmetric $n\times n$ matrices $A$ and ...


1

$$\frac {\partial K(x,y)}{\partial x}y - \frac {\partial K(x,y)}{\partial y}x =0$$ $$\frac {\partial K(x,y)}{x\partial x} - \frac {\partial K(x,y)}{y\partial y} =0$$ Let $X=x^2$ ; $Y=y^2$ ; $K(x,y)=H(X,Y)$ $$\frac {\partial H(X,Y)}{\partial X} - \frac {\partial H(X,Y)}{\partial Y} =0$$ The general solution of this PDE is well-known : $$H(X,Y)=H(X+Y)$$ any ...


0

Combine (2) with (1): $$\frac {\partial K(x,y)}{\partial x}x + \frac {\partial K(x,y)}{\partial y}y+2k(x,y) = 0$$ $$\frac {\partial K(x,y)}{\partial x}(x + \frac {y^2}{x})+2k(x,y) = 0$$ The rest depends on what the difference between k and K is. $$\frac{\frac {\partial K(x,y)}{\partial x}}{k(x,y)}=\frac{-2}{(x + \frac {y^2}{x})}$$ and: $$\ln( ...


0

You can't actually determine the coefficient for $xy$ term, since all Taylor expansion of $u_{i,j}$ are essentially expansions by $x$ or $y$, without any mixed derivative term. Moreover, that term is not needed for the partial derivatives $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial x}$, since its derivatives simply vanish on the stencil ...


1

I would convert that into a system of two equations. Let $v = U_t$ and $w = -U_x$. Then you have a following pair of equations $$ v_t + (1 + \epsilon w^2) w_x = 0\\ w_t + v_x = 0 $$ These equations can be rewritten in conservative form as $$ \frac{\partial \mathbf Z}{\partial t} + \frac{\partial \mathbf F(\mathbf Z)}{\partial x} = 0 $$ with $\mathbf Z = (v, ...


0

There is a typo in the question, the correct expression is $$ \frac{d}{dt}\int\limits_{\mathbb{R}^n}\phi |\nabla u|^2=-2\int\limits_{\mathbb{R}^n}\phi(\Delta u)^2 $$


0

I would search method of characteristics. If you use that method the differential relation is $$\frac{dy}{x}=\frac{dx}{y}=\frac{df}{0}$$ $$ydy=xdx$$ So there are two invariants along the characteristics $$y^2-x^2$$ And $f$ itself, so we set one constant equal to a function of the other constant $$f(x,y)=\varphi(y^2-x^2)$$ Now plug this back into the ...


2

If you note $\|\cdot \|_\infty$ the sup ess norm, you have : $$\| u - \tilde{u} \|_{\infty} \leq \| u - u_k\|_{\infty}+ \| u_k - \tilde{u} \|_{\infty}$$ Now let $\epsilon > 0$, As $u_k \to \tilde{u}$ uniformly, there exist $k_1$ such that $\forall k > k_1, \ \| \tilde{u} - u_k\|_{\infty} < \frac{\epsilon}{2}$ As $u_k \to u$ in $W^{1,2}_0$, it ...


0

$$\frac{\partial F(z,t)}{\partial t} = \alpha \frac{\partial F(z,t)}{\partial z} + \beta F(z,t) + \gamma$$ Search for the general solution : Let $F(z,t)=\varphi(z,t)-\frac{\gamma}{\beta}$ $$\frac{\partial \varphi(z,t)}{\partial t} = \alpha \frac{\partial \varphi(z,t)}{\partial z} + \beta \varphi(z,t) $$ Let $\varphi(z,t)=e^{G(z,t)}$ $$\frac{\partial ...


0

The heat equation is $\alpha \Delta u = \partial_t u$. So if $u$ is harmonic, $\nabla^2 u = \Delta u = 0$, we are saying the system is in equilibrium, $\partial_t u = 0$, and thus there is no net heat flow in or out of the region $\Omega$. Proving this result uses an equivalent form of the directional derivative, the divergence theorem, and the fact that ...


0

Every non-negative harmonic function in the upper half plane can be represented as $$ h(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{yd\rho(t)}{(t-x)^{2}+y^{2}}+Ay $$ where $A$, $D$ are positive constants and $\rho(t)$ is a non-decreasing function on $\mathbb{R}$ for which $$ \int_{-\infty}^{\infty}\frac{d\rho(t)}{1+t^{2}} < ...


0

Weak convergence in $H=H^1_0$ implies strong convergence in $L^2$ because $H^1_0$ is compactly embedded in $L^2$, see: https://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem


1

Following the hint by Ian, let $v=u+1$. The function $v$ satisfies the PDE $\Delta v=v$, and is positive on the boundary of $\Omega$. So, if $v$ was negative somewhere, its (negative) global minimum would be attained in the interior... but the Laplacian can't be negative at an interior minimum.


0

This appears to be an example of a heat equation. Let $u(x,t)$ denote the temperature at $x$ at a given time $t$ of a rod of length $l$. We may thus describe the heat distribution as $$\frac{\partial u}{\partial t} = a^2\frac{\partial^2 u}{\partial x^2}$$ where $a^2$ is known as the thermal diffusivity, which depends on the material which the bar is made ...


1

Short answer It should be well-posed for any domain satisfying $0<R_{min} < R_{max}$ and $0 \le \theta_{min} < \theta_{max} \le \pi$. Unfortunately, and probably due to some error, the answer seems to depend on whether I analyze this problem in 3D (assuming axisymmetry) or in 2D (on a plane of constant $\phi$).... Long answer To be well-posed, ...


2

Take a look at Couple stress theory for solids


0

Please do not use the same symbol for your function and test functions. You are confusing yourself. We assume $u\in H$ and $(u_n)\subset C_c^1$ such that $u_n\to u$ in $H$, which in particular means that $u_n\to u$ in $L^2$ and $\partial_i u_n\to v_i$ in $L^2$ for each $i=1,\ldots, N$. Then we observe that, for any $\varphi\in C_c^1$, $$ \int_\Omega ...


2

Matrix has icomplete rank, so the system is solvable only if $$ \operatorname{rank} M = \operatorname{rank} \begin{pmatrix}M \;\big|\; f\end{pmatrix}. $$ Due to truncation error, the last may not hold (but would hold, if you're using a conservative approximation, I suppose). You can use QR decomposition to deal with that. Suppose $M = QR$ where $Q$ is ...


0

1. Does anybody know a text that works out this problem under Neumann conditions? I do not. However, I've had a similar problem and think the following may be useful to you: G. Barton, Elements of Green’s Functions and Propagation: Potentials, Diffusion and Waves (Oxford Science Publications, Oxford, UK, 1989). Green Function of the Laplacian for the ...


0

The function $w(\cdot,0)$ is identically zero so it's gradient is also identically zero.


1

NOTE: the text below is not perfect rigorous. If you want, you can suggest improvements in comments. Ok, let's start. We have the equation $$\frac{\partial f}{\partial t} + \mu\frac{\partial f}{\partial x} + \lambda[f(t, x + 1) - f(t, 1)]=0\tag1\label1$$ Following Fourier, we use separation of variables (in fact, this will lead us to Fourier transform). Let ...


1

The first-order equation $$\frac{{\partial} y}{{\partial} x}-\frac{1}{v} \frac{{\partial} y}{{\partial} t}=0$$ is also an important PDE, known as the transport equation. All its solutions consist of the initial values moving to the left with velocity $v$ (with the $+$ sign they would move to the right). This is not what we expect a vibrating string to do: if ...


0

The assumption in the book is that $u\ge 0$ everywhere on the boundary and $u>0$ somewhere on the boundary. Also, $u\in C^2(U)\cap C(\overline{U})$. Your argument is correct. Here's a restatement in different words: Suppose that $u=0$ at some interior point. By the strong maximum principle applied to $-u$, it follows that $u\equiv 0$. This contradicts ...


0

Robert Lewis has the right line. $fu$ is a multiplication operator (and is self-adjoint). If $f$ is bounded, then $\Delta+f$ is bounded and self-adjoint so it has a spectral decomposition. The decomposition varies continuously with $f$, no? That is, the only way for it to be the same is if $f$ and $g$ differ on a set of measure zero?


3

You can adapt your proof by using $$V_{\epsilon}(x,t) = u(x,t) + \epsilon \left( |x|^2 -\sup_{x\in\Omega} |x|^2 \right) $$ Then you have $$\Delta \left( |x|^2 -\sup_{x\in\Omega} |x|^2 \right) = -2n < 0 \leq c\left( |x|^2 -\sup_{x\in\Omega} |x|^2 \right) $$


1

Take any $g\in C^\infty[0,\infty)$ such that $g=1$ on $[0,1)$, $g=0$ outside of $[0,2)$ and $0\leq g\leq 1$. Define $\phi(x)=g(|x|)$, then $\partial_{x_2} \phi(x_1, 0)$ is just the tangential derivative on a circle of radius $|x_1|$ and center $0$, which is zero since our function is radial.


0

Just look at the definition: since $\eta(x)=0$ for all $x \in \mathbb{R}^N$ with $|x| \geq 1$, it follows that $\operatorname{supp}\eta \subset \overline{B(0,1)}$. Hence $\eta \in C^\infty_c(\mathbb{R}^N)$.


0

The two forms "are equivalent" (that is, the two functions are the same) whether $$u_1(x,t)-u_2(x,t)=u_\infty(x)=0.$$ It seems this equality is valid because, taking $b>t$, we get \begin{align} u_1(x,t)&=\int_{t}^\infty\partial_s u(x,s)\ ds\\ &=\int_{t}^b \partial_s u(x,s)\ ds+\int_{b}^\infty\partial_s u(x,s)\ ds\\ &=\int_{t}^b \partial_s ...


0

Well, one might try following scheme. Let $$ I^{n}_j = I(t_n, \omega_j), \quad S^n = S(t_n), \quad F^n = F(t_n), \quad t_n = n \tau, \quad \omega_j = j h $$ Note that $I(t,\omega) = 0$ when $\omega > t$. So let's limit the computational domain to $0 \leq t \leq T, 0 \leq \omega \leq L \geq T$. Then $$\begin{aligned} \frac{S^{n+1} - S^n}{\tau} &= ...


0

@Yongyong's answer is very good indeed. Here I just want to point out the idea behind this remark is that $W^{1,1}$ is not a reflexive space. (Note $W^{1,p}$ is reflexive for any $1<p<\infty$). Hence, when we try to use weak compactness to extract a weak subsequence, the limiting function may not lie in $W^1{1,1}$, and hence the theorem fails. If you ...



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