New answers tagged

3

You an observe that your map is injective because, by hypothesis of Lax-Milgram theorem, it satisfies the following property: For each $v\in V$ $\| \mathcal{A}(v)\|\geq \|v\|$ Then if $v\in \ker(\mathcal{A})$ then $\|v\|\leq \| \mathcal{A}(v)\|=\|0\|=0$ So $v=0$, that means $\ker(\mathcal{A})=\{0\}$. Moreover $V$ and $V^*$ have the same dimension, but ...


0

Hint Let $y=z\, e^x$ to get $z'=1$


2

The equation $\frac{dy}{dx}=e^x+y$ can be rewritten as $\frac{dy}{dx}+P(x)y=Q(x)$ where $P(x)=-1$ and $Q(x)=e^x$. The general solution for $y$ can then be found through the well known method of integrating factors.


1

Hint $$\frac{d}{dx}\left(e^{-x}y\right)=1$$


1

Let us follow the method of characteristics. $\frac{\text d}{\text d t}x = x$, letting $x(0) = x_0$ we know $x = x_0e^t$; $\frac{\text d}{\text d t}y = x+y$, letting $t(0) = t_0$ we know $y = (x_0 t + y_0)e^t$; $\frac{\text d}{\text d t}u = 1$, letting $u(0) = u_0$ we know $u = t + u_0$. Since the prescribed data is $u(1,y) = y$, we set $x_0 = 1$ and $u_0 =...


0

In the case where $\mu_1 = 0, \, \mu_2 = 1, f'(0) = 1$ and $f$ is concave down, the travelling front has the asymptotic behavior $V(\xi) \sim e^{-r\xi}$ with $r = \frac{c - \sqrt{c^2-4}}{2}$. This follows from a linearization argument. Since $f(V)$ has the same asymptotic behavior as $V$ at $+\infty$, it follows that $V'$ and $V''$ have the same asymptotic ...


0

Partial answer. WolframAlpha (or with simpler input) gives $$ u(t, x) = 2 c_2 \tanh(t c_1 + x c_2 + c_3) - \frac{c_1}{c_2} $$ to be the solution. I am not quite sure what you mean by your boundary conditions, is it something like $$ \lim_{t \to \pm\infty} u(t,x) = 0 \ \forall x, \quad \lim_{x \to \pm\infty} u(t,x) = 0 \ \forall t $$ ? Sadly, I haven't ...


2

a) Let us follow the method of characteristics. $\frac{\text d}{\text d s}x = x$, letting $x(0) = x_0$ we know $x = x_0e^s$; $\frac{\text d}{\text d s}t = t$, letting $t(0) = t_0$ we know $t = t_0e^s$; $\frac{\text d}{\text d s}u = pu$, letting $u(0) = u_0$ we know $u = u_0e^{ps}$. b) Let us assume that $p=4$. On the unit circle, we have ${x_0}^2 + {t_0}^2 ...


0

The Method. Set $\mathbf x = (x,t)$. Your PDE is $$ \mathbf a \cdot \nabla u= b$$ where $\mathbf a(x,t) = (x,x+t), b\equiv 1. $ The initial data is prescribed along the curve $\mathbf x_0(z) =(1,z)$, which is $$ u(\mathbf x_0(z)) = u_0(z)$$where $u_0(z) = z$. Note that $\mathbf x_0'(z)^\perp \cdot \mathbf a(\mathbf x_0(z)) =(-1,0)\cdot (1,1+z) = -1 $ so ...


2

Let $$D=\partial_t-\partial_x^2$$ i.e. $$D: u \mapsto \partial_tu-\partial_x^2u$$ And $D$ is linear iff $\forall$ scalar $\alpha$ and vectors $u,v$ we have that $$D(u+\alpha v)=Du+\alpha Dv$$ So you just need show that $$\partial_t (u+\alpha v)-\partial_x^2(u+\alpha v)=\partial_t u -\partial_x^2 u+\alpha (\partial_t v - \partial_x^2 v)$$ Using the ...


1

By taking a linear sum of the $2$ linear equations for $u_x$ and $u_y$, and using the given fact that not both of $a$ and $b$ are $0$ so $a^2 + b^2 \neq 0$, this means the factor of $u_{x'} = 0$. Thus, this shows that $u$ has no component of $x'$. As such, $u = f(y')$ for some function $f$. Since $y' = bx - ay$, this gives that $u = f(bx - ay)$. As for what ...


1

You have to solve $u_{x'} = 0$. The PDE $u_{x'}(x', y') = 0$ has the general solution $u(x', y') = f(y')$. Now you just have to make the transformation undone, i. e., replace $x'$ and $y'$. You obtain $u(x, y) = f(bx - ay)$ since $y' = bx - ay$.


1

1) For any real number $c$ the following is true: $u(x,y)=u(x',y')$ whenever $(x,y)$ and $x',y')$ satisfy the equation s$bx-ay=c$ and $bx'-ay'=c$. This is what is meant by sayning that the solutio is consatn t on the lines $bx-ay=$ constant. 2). Define $f$ as follows: $f(t)=u(x,y)$ where $(x,y)$ is any pair such that $bx-ay=t$. [One such choice is $y=0,x=\...


3

An opening remark: The original PDE can actually be reduced to the heat equation by the substitution $u(x,t)=e^{-at} f(x,t)$. The observation that $a=\pi^2$ tends to produce $u(x,t)\to b\sin \pi x$ as $t\to \infty$ therefore amounts to the claim that $f(x,t)$ tends to behave like $e^{-\pi^2 t}\sin \pi x$ for large times. At one level, this is what we expect ...


1

Using separation of variables, assume a solution of the form $$u(x,t)=X(x)T(t)$$ such that $$u_x(x,t) = X'(x)T(t)$$ $$u_{xx}(x,t) = X''(x)T(t)$$ $$u_t(x,t) = X(x)T'(t)$$ $$u_{tt}(x,t) = X(x)T''(t)$$ Substitution into the PDE $u_{xx} = u_{tt}$ yields: $$X''(x)T(t) = X(x)T''(t)$$ $$\dfrac{X''(x)}{X(x)} = \lambda = \dfrac{T''(t)}{T(t)}$$ for some $\...


1

Why do you think that the problem in unsolvable? Are you solving the problem through separation of variables? Let $u(x,t)=X(x)T(t)$. Substitution into the PDE produces $$X(x)T''(t)-X''(x)T(t)=0$$ where by seperating the variables, we obtain $$\frac{X''(x)}{X(x)}=\frac{T''(t)}{T(t)}=-\lambda$$ for some constant $\lambda$. We have $$X''(x)+\lambda X(x)=0\...


0

A possible proof goes like follows. First, note that it is sufficient to prove that if $u=0$ a.e. on a Borel subset $E$ of $\Omega$ then $\nabla u=0$ a.e. in $E$. Now, it is possible to show [EG, Theorem 4.4] that if $u\in W^{1,1}(\Omega)$ then $\nabla u=0$ a.e. on the zero-level set $\{u=0\}$. But this concludes the proof because if $u=0$ a.e. in $E$ then $...


0

The classic lecture "Lectures on Gas Theory": https://www.amazon.com/dp/B00A73IPJ8 Stanford Encyclopedia of Philosophy: Boltzmann’s Work in Statistical Physics and this seems worth a read... Numerical Solution of Boltzmann's Equation


2

The substitution is just exploiting a convenience. I find it actually quite surprising how many seemingly difficult PDEs can be reduced to something much more manageable (heat, wave, potential equations) with a simple transformation of coordinates, translation of solution, etc. The only point of this substitution was to make data on the boundary zero. So ...


1

The idea is that, since we prefer having trivial initial data on $\Gamma_D$ (due to the Fourier transform method), we want to have the following decomposition of the solution $u$: \begin{align*} u=u^*+w, \end{align*} where \begin{align*} -\nabla^2u^*=x(1-x) \quad &in \quad \Omega\\ u^*=0 \quad &on \quad \Gamma_D\\ \frac{\partial u^*}{\partial x}=0 \...


1

Since you know Einstein field equations, then you must have heard of metric tensor and curvature tensor and their relationship to the field equations. In local coordinates (normal coordinates) we can make the first derivative of the metric tensor to vanish at a given point but not the second derivative simultaneously. But curvature tensor is a function of ...


3

Ten is simply the number of distinct components of a second order symmetric tensor in a space of dimension four. Writing these equations in tensor form enables us to write EFE as a single equation instead of ten, just as $\vec F=m\ddot{\vec x}$ is a single vector equation written in place of three scalar equations.


1

Hint Use the method of separating variable as follows $$u(x,t)=f(t)g(x)$$to solve the PDE.


1

Take $v_n(t, x) = f_n(t, x) = \sum_{k = 1}^n \theta_k(t) \lambda_k^{-1/2} \varphi_k(x)$ composed of eigenpairs of a Laplacian (with summable $\lambda_k^{-1}$) and the indicator function $\theta_k$ of the temporal interval $[k, k + 1]$. I think the sequence $(v_n, f_n)$ satisfies 1. & 2. uniformly but not "Ineq. (?)".


3

Buy uniqueness of Fourier series expansions the expression for $u(x,0)$ gives $b_2=1,b_4=-\frac 1 3$ and $b_k=0$ for all other $k$.


2

Yes there is. Heuristically, the "mean value over $\mathbb{R}^n$" of our function should give the value of our function at any particular point. Since the integral is finite and $\mathbb{R}^n$ is not, the function vanishes necessarily. This can be made rigorous as follows. Supposing $u(p) = y \neq 0$ for some $p \in \mathbb{R}^n$, then the mean value ...


1

What I would do is to multiply both sides by $\psi(x,y)$, where $\psi(x,y)$ is a compactly supported, smooth, bump function. Use integration by parts twice on the term involving $\nabla^2$. Now choose a suitable sequence of such functions $\psi_n(x,y)$, where the support goes to infinity in the $y$ direction as $n \to \infty$, and where the suppose in the $...


0

As your equation is linear, you can determine the characteristic curves by solving $$ \dot x=a(x,y,z)=z\\ \dot y=b(x,y,z)=-z\\ \dot z=c(x,y,z)=z^2+(x+y)^2 $$ or $$ \frac{dx}{z}=-\frac{dy}{z}=\frac{dz}{z^2+(x+y)^2} $$ As you found, the first two equations combine to $$x+y=c_1.$$ Combining the first and third equations gives $$\frac{zdz}{z^2+c_1^2}=dx\implies ...


3

Not all formulas are useful for practical purposes. For example, Leibnitz's formula states that $$ \frac{\pi}{4} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} \cdots,$$ but the series converges very slowly. A much faster method for computing $\pi$ is the fixed point iteration given by $$ x_{n} = g(x_{n-1}), \quad n = 1,...


3

Usually, you don't need a numerical solution if you have the analytical solution. Occasionally, your analytical solution may contain special functions, and comparison with a numerical solution could be used to validate the implementation of the special functions, but this is a stretch. Typically, numerical solutions depend on a parameter (mesh size, time ...


2

Hint: In the case $N=1$ and $b=0$, we have $d w'' + aw = 0$, which solutions are $$ w(x) = A \cos kx + B \sin kx , \qquad k = \sqrt{\tfrac{a}{d}} \, . $$ The boundary conditions $w(x_\min) = 0 = w(x_\max)$ yield the system $$ \begin{pmatrix} \cos kx_\min & \sin kx_\min \\ \cos kx_\max & \sin kx_\max \end{pmatrix} \begin{pmatrix} A\\ B \end{pmatrix} = ...


1

The idea is to separate $$ \Psi(t, x) = \color{blue}{e^{-\beta t}}\color{red}{u(x)} \tag{1} $$ where $\beta$ is a constant. With this in mind, Schrodinger's equation becomes $$ \color{blue}{e^{-\beta t}}\left[-\frac{\hbar^2}{2m}\frac{{\rm d}^2 \color{red}{u}}{{\rm d}x^2} + \frac{1}{2}m\omega^2 x^2 \color{red}{u}\right] = \color{red}{u}\left[i\hbar \frac{{\...


0

I don't think my statement about any homogeneity of $Z$ and $\Xi$ is necessarily correct. In any case this is immediate. Let $Z_\lambda$ and $\Xi_\lambda$ denote $Z(\lambda t,x, \xi/\lambda), \ \Xi(\lambda t,x, \xi/\lambda)$ respectively. By definition \begin{align} \partial_t Z_\lambda &= \lambda (\partial_\xi p)(Z_\lambda, \Xi_\lambda)\\ \partial_t \...


1

Let us consider that $f$ and $g$ are scalar fields, which depend on the position $x = (x_1,x_2,x_3)^\top \in \Bbb R^3$. Moreover, we assume that the vector $v = (v_1,v_2,v_3)^\top \neq 0$ does not depend on $x$. The PDE $v\cdot \nabla f = g$ reads $$ \sum_{i=1}^3 v_i \frac{\partial}{\partial x_i} f(x) = g(x) \, . $$ The method of characteristics for the ...


1

Hint: Start by defining the functionals $$ Y_1(q_1,q_2,u):=\int_\mathbb{R}e^{c\xi}(u(\xi,t)-V_1(\xi,t)-V_2(\xi,t))V'(\xi-q_1)d\xi, $$ $$ Y_2(q_1,q_2,u):=\int_\mathbb{R}e^{c\xi}(u(\xi,t)-V_1(\xi,t)-V_2(\xi,t))V'(\xi-q_2)d\xi. $$ Then you have to prove that $Y_1(0,0,V_1+V_2)=Y_2(0,0,V_1+V_2)=0$. Once you have all of this, you have to compute $$ \frac{\partial ...


1

This claim can be found e.g. in §3.4 p. 27 of (1). One should note that $u(x,0) = \phi(x)$ is the initial condition of this Cauchy problem (missing in OP). The standard definition of weak solutions leads to $$ \int_0^\infty \int_{-\infty}^\infty [u_t + f(u)_x] v\, \text dx \text dt = 0 $$ for all test functions $v$ with compact support. Using Fubini's ...


1

Of course, one can go back to the general definition of weak solutions, as proposed in OP (see related post for complements). I will propose a solution which may be more in the spirit of the book (Example 10 p. 336). The derivatives of the discontinuous function $u(x,t) = H(x-ct)$ viewed as a distribution are \begin{aligned} u_x &= \delta(x-ct) , & ...


2

Using d'Alembert's formula, we have $$ \begin{aligned} u(x,t) &= \tfrac{1}{2}\left[ u(x-ct,0)+u(x+ct,0) \right] \\ &= \tfrac{1}{2}\left[ (x-ct)^2 {\bf 1}_{x\geq ct} + (x+ct)^2 {\bf 1}_{x\geq -ct} \right] \end{aligned} $$ i.e. $$ u(x,t) = \left\lbrace\begin{aligned} &0 && \text{if}\quad x < {-c}t \\ &\tfrac12 (x+ct)^2 && \...


0

There is an anther approach. Let $f^{k}_{1}=(x-2y)^k (k=0,1,\cdots,n)$$, \varphi(f^{k}_1)=0.$For a fixed $ k(\geq 1),$We only need to find $f^{k}_{t}(t=1,2\cdots,k),$such that $\varphi (f^{k}_{t+1})=f^{k}_{t},$ $$\varphi(f^{k}_{1},f^{k}_{2},\cdots,f^{k}_{k+1})$$$$=(f^{k}_{1},f^{k}_{2},\cdots,f^{k}_{k+1})\begin{pmatrix} 0 & 1 & & & \\ &...


0

Given that $$u(x,t) = \frac {1}{2}\big(u(x-ct,0) + u(x + ct,0)\big) = \frac {1}{2}\big(u(x-ct) + u(x + ct)\big)$$ You should break up the analysis into three cases substituting $(x−ct)$ and $(x+ct)$ for $x$ to evaluate $$u(x-ct) = \left\{\begin{matrix} 0 & \text{for}~x<ct, \\ (x-ct)^2 & \text{for}~x\geqslant ct \end{matrix}\right. $$ $$u(x+...


3

The existing answers provide good reasons towards the question in the title, but from the perspective of a geometer I feel the applications in physics aren't quite as convincing. It's true that singular phenomena that arises in for example conservation laws requires a suitable notion of a generalised solution, but why is it also useful for geometric problems?...


0

At first, in the definition, we are taking derivatives along directions tangent to this submanifold when we are standing on it, and we could taking derivatives along any directions elsewhere. Thus, an object is conormal to a surface (submanifold) means that, along this submanifold, this distribution is smooth, and possible non-smoothness arise in the normal ...


1

So if we have $u_n\in L^2(0,T;L^2(\Omega))\cap L^\infty(0,T;H^{-1}(\Omega))$ and $u_n\to u$ in $L^2(0,T;H^{-1}(\Omega))$, can we deduce the weak convergence of subsequence $u_{n_k}(t)$ to $u(t)$ in $L^2(\Omega)$ for a.e $t$? $\Omega$ is bounded domain in $\mathbb R^3$.


0

Here's a sketch. I have not finished the last step yet, but maybe this will be useful anyhow... I will replace $d$ in your notation by $n$ to avoid confusion with derivatives and differentials. Also, the exponent of the denominator should be $n+s$ for this question to make sense, and the limit must be one sided from below. The key idea is that since the ...


1

Yes, it is the case. I think there are two different ways to approach this problem. I think the second one is more direct if you already know the behavior of the operator $-\Delta$ (without the potential $f\,$). Let's start by applying some spectral analysis (this is why I prefer the second approach). First of all, note that $-\Delta+f$ is essentially self-...


0

The method of characteristics provides the solution in the domain $t\geq a \geq 0$. The characteristic curves along which the information $\frac{\text d}{\text d t} y_k = f_k y_k$ propagates are the straight parallel lines $x = t-t_0$ with $t_0 \geq 0$. Hence, we have $$ y_1(t,a) = c_1 \exp\left(\int_{t-a}^t f_1(s)\, \text d s\right) = c_1 \exp\left(\int_{0}^...


0

First order PDEs can generally be solved by the method of characteristics. To solve your first PDE using this method, put $$ \phi_z(x) = y_1(z + x, x)\ . $$ Then \begin{eqnarray} \frac{d\phi_z}{dx}(x)&=&\frac{\partial y_1}{\partial t}(z+x,x)+\frac{\partial y_1}{\partial a}(z+x,x)\\ &=& f_1(z+x)\,\phi_z(x)\ . \end{eqnarray} This is a first-...


0

Well, I hope this doesn't come off as snarky, but why should we expect that $$x^2 +1 =0$$ should have solutions? And why should we abandon the meaning of "squaring" that we all first learned for real numbers and adopt $$(a,b)^2 = (a^2-b^2, 2ab)$$ It's not a perfect analogy but I think it's rather similar to your questions about PDE solutions.


1

Let us first summarise what you have: You have an equicontinous and uniformly bounded sequence $u_m: [0,T]\to L_\sigma^2(\Bbb R^n)$ and an ONB $\phi_i$ of $L^2_\sigma(\Bbb R^n)$. Then define: $$\theta_{mi}:[0,T]\to \Bbb R, \quad t\mapsto (u_m(t),\phi_i).$$ Now for any fixed $i$ this gives a bounded family of equicontinuous functions. By the diagonal ...


3

No, there are no solutions $f$ where the function $f$ is not an affine function. To see this, note that differentiation of the identity $|\nabla f|^2=1$, once by $x$ and another time by $y$, $$|\nabla f|^2=f_x^2+f_y^2\equiv1$$ implies $$\langle\nabla f,(f_{xy},f_{yy})\rangle = 0$$ and also $$\langle\nabla f, (f_{xx},f_{xy})\rangle = 0$$ where $\langle\cdot\...


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