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0

Coming back to this, I wanted to say that this problem can actually be solved by doing an odd reflection through the origin, as you appear to be trying to solve the heat equation in the half open domain $[0,\infty)$. Consider first letting $v(y,t) = u(y,t) - u_0$. Here I'll be assuming $u_0 \in \mathbb{R}$ since you didn't specify. Then our new $v$ can solve ...


0

Here's a hint. Use the Fourier transform on $\mathbb{R}^2 / (2 \pi \mathbb{Z})^2$ to write $$ u(x,t) = \sum_{n \in \mathbb{Z}^2} \varphi_n(t) e^{i x \cdot n}. $$ Plugging into the PDE then shows that $$ \ddot{\varphi}_n(t) + |n|^2 \varphi_n(t) +a \dot{\varphi}_n(t) =0 \\ \phi_n(0) =0 \text{ and } \dot{\phi}_n(0) = \hat{f}_n = \int f(x) e^{-ix\cdot n}dx. ...


0

Indeed, it is not obvious that the extended function is $C^2$. But you don't have to prove harmonicity by checking the definition at the beginning of the chapter: there are other results you could use. For example, there is a characterization of harmonicity that does not involve $C^2$: Theorem 6 (1st edition) states that a continuous function that has the ...


2

HINT: Note that we have $$\frac{\partial g}{\partial \nu}=\frac{2}{\epsilon}$$ on $\Gamma_{\epsilon}$. Therefore, $$\lim_{\epsilon\to 0}\int_{\Gamma_{\epsilon}}\left(g\frac{\partial\phi}{\partial\nu}-\phi\frac{\partial g}{\partial\nu}\right)\,ds=-4\pi\phi(0)$$


1

If $R$ is large enough $\varphi$ and $\nabla \varphi$ are equal to $0$ on $\Gamma_2$, and so is the RHS in the last formula.


1

Let $u=v^2$ , Then $u_x=2vv_x$ $u_y=2vv_y$ $\therefore2vv_x2vv_y=v^2$ with $v(x,0)=0$ $4v^2v_xv_y=v^2$ with $v(x,0)=0$ $v_y=\dfrac{1}{4v_x}$ with $v(x,0)=0$ $v_{xy}=-\dfrac{v_{xx}}{4v_x^2}$ with $v(x,0)=0$ Let $w=v_x$ , Then $w_y=-\dfrac{w_x}{4w^2}$ with $w(x,0)=0$ $\dfrac{w_x}{4w^2}+w_y=0$ with $w(x,0)=0$ Follow the method in ...


1

I think it would be best to send the authors an email and ask them, since they are the only ones who knows by certain. Nevertheless, let me make a guess: I think that $$ u\wedge M=\min(u,M) $$ and that $$ u\vee M=\max (u,M). $$ I'm not certain of what they want to do, but I guess that they are cutting the sequence $\phi_n$ off at the values $\pm M$ (in ...


1

Note that $\int_{\mathbb{R}} |\nabla u(x)|^2 dx=-\int_{\mathbb{R}}u(x)\Delta u(x)dx=0$. So $\nabla u(x)=0$ and $u$ is identically a constant.


1

You are correct. A measure has no reason to preserve this property. As a matter of fact, the only measures that absolutely continuous wrt Lebesgue measure that preserve this property for all $f,g \in C^1_c(\Bbb{R})$ are scaled Lebesgue measures. Let $g\neq 0$. If $(gh)'=g'h$, then $gh'+g'h=g'h$, or $ gh'=0$, implying $h'=0$ and $h=c$. Or if the question ...


2

$$T=\partial/\partial x + I \partial/\partial y$$ where I is the identity operator. Applying it to a sum: $$T(u+v)=u_x+v_x+u u_y+uv_y+vu_y+vv_y=T(x)+T(y)+uv_y+vu_y;$$ as you can see you get two terms too many!


1

I think the problem is more a problem of basic proof-strategy/proof-writing than really a problem about linearity or differential equations. Therefore, I will try to show how mechanical this kind of proof can be. An interesting lecture on that topic is this blogpost by the Field medalist Tim Gowers. The difference between two solutions of an ...


2

You can use linearity like this: If $Lu_1 = g \tag{1}$ and $Lu_2 = g, \tag{2}$ then $Lu_1 - Lu_2$ $= g - g = 0; \tag{3}$ but, by linearity, $Lu_1 - Lu_2$ $= L(u_1 - u_2), \tag{4}$ so $L(u_1 - u_2) = 0, \tag{5}$ as per request.


0

It is true that twice differentiable wave solutions defined for all $x$ are of the form $f(x+t)+g(x-t)$, but I think something else applies here. The function $u(x,t) = e^{t-x}\sin(t-x)$ for $0\le x\le t$ and $u(x,t) = 0$ for $t\le x$, $\ t>0$ is probably what was intended. It is a weak solution not differentiable along the ray $x=t$.


1

You are pretty much there. Since $f$ is continuous on $[0,M]$, it follows that $\lim_{\varepsilon\to0^+}\int_\varepsilon^Mf(x)\, dx=\int_0^Mf(x)\, dx$. EDIT: $f$ here is defined as you originally did before you edited the question: $$f(x):=\begin{cases} \frac{\varphi(x)-\varphi(0)}x&\text{if }x>0,\\ \varphi'(0)&\text{if }x=0. \end{cases}$$


1

Try $\frac{u(x+he_i)-u(x)}{h}=\int_{\mathbb{R}^n}\Phi(y)\{\frac{f(x+he_i-y)-f(x-y)}{h}\}dy$. Then $\frac{f(x+he_i-y)-f(x-y)}{h}\rightarrow\frac{\partial f}{\partial x_i}(x-y)$ uniformly(!) in $h$ and of course $u\in C^1$ implies $u\in C^0$. Similarly You can show $u\in C^2$.


2

Check this link to see how to deal with differentiating an integral over an evolving domain: https://en.wikipedia.org/wiki/Time_evolution_of_integrals. We use this to see that $$\partial_t\left(\int_0^tf(x+(s-t)b,s)\,ds\right)=\int_0^t\frac{\partial f(x+(s-t)b,s)}{\partial t}\,ds+f(x,t),$$ now notice that $$\int_0^t\frac{\partial f(x+(s-t)b,s)}{\partial ...


1

The first equation looks like the Eikonal equation in $\Bbb R^{1+1}$, it is solved by $u=\sqrt{x-t}$, you would then plug this into the wave equation to determine $f$. We see that $u_x=x/u$, $u_t=-t/u$, $u_{xx}=(u-x^2/u)/u^2$ and $u_{tt}=-(u+t^2/u)/u^2$ thus$$u_{xx}-u_{tt}=(u-x^2/u)/u^2+(u+t^2/u)/u^2$$ $$=2u/u^2+(t^2-x^2)/u^3$$ ...


1

You can think of a linear functional on a vector space as giving you a coordinate in that space. A continuous linear functional is a coordinate that varies continuously as the vector varies continuously; one is not so much interested in the coordinates that are disconnected from the topology of the space. Representation of a linear functional gives the ...


1

Taking the first order partial derivatives gives $$z_x=2xf'(x^2+y^2)+1\\ z_y=2yf'(x^2+y^2)+1 $$ can you relate these two expressions somehow?


2

The proof looks correct to me, well done! A couple of nitpicks: in the definition of $V \subset \subset U$ you forgot to include that $\overline{V}$ needs to be compact; the very last $\delta$ should be $\delta^p$. The only thing that I would like to add is an explicit definition for $V_k$: $$V_k := \Big\{x \in U : \text{dist}(x, \partial U) > ...


4

Following √Člie Cartan, you want to think of your differential system $dR - R\omega = 0$ on $M\times SO(n)$. This $\mathfrak{so}(n)$-valued $1$-form is integrable, as you said, because of vanishing curvature. The integral manifolds of this differential system will locally be the graphs of functions $R\colon U\to SO(n)$.


2

You have $Lu = u_x + u u_y$, hence: $$L(a u + b v ) = a u_x + b v_x + (a u + b v) \, (au_y + bv_y)$$ Can you take it from here?


0

disclaimer: I use the notation $K'$ to denote the derivative of the function $K$ if it only depends on a single variable. Well, there's a plus sign, so why not assume that $$u(x,y) = X(x)+Y(y)$$ and give it a try? From $u(x,0)=0$ one could tell now that $$X(x) +Y(0) = 0$$ which leads to $$X(x) = -Y(0)$$ The problem is that the derivative thereof is ...


0

This is called the method of characteristics. If it is the first time you ever see this I would suggest that after reading my answer you take a look at the wikipedia page (which I have already linked in the comments). You can of course read more about this in any introductory book to PDEs. A very clear exposition can be found in chapter 3 of Evan's book, ...


0

An alternative to the method of characteristics. I like to call it common sense. Note that the PDE is linear and with constant coefficients. Hence, the structure of the solution is given by $u(x,y) = u_h + u_p$, where $u_h$ is a solution of $u_x + u_y = 0$ and $u_p$ is a particular solution. Trying solutions of the form $u_h = A \exp{(a x+ by)}$ leads to ...


2

(I would have just commented, but I don't yet have enough reputation) The solution is $u = xy$. Uniqueness tells us that this is the only solution. The "quickest" way (if you have been doing this for a little bit) is by inspection, which is a fancy name for an educated guess. You get better at doing this with simple PDE's as you work more with pdes. ...


2

Hints: Write the partial derivatives as operators, as follows: $$ u_{xx} = \frac{\partial^2}{\partial x^2} u $$ $$ u_{xt} = \frac{\partial^2}{\partial x\,\partial t} u $$ $$ u_{tt} = \frac{\partial^2}{\partial t^2} u $$ so the PDE looks like this: $$ \left(\,3\frac{\partial^2}{\partial x^2} - 10\frac{\partial^2}{\partial x\,\partial t} - ...


1

Perhaps it would be wiser to use the following expansion $$E(f+h)=E(f)+dE(h)+o(h),$$ from what you have calculated, we see that $$\int \nabla f\cdot\nabla h+\frac{1}{2}\nabla h\cdot\nabla h=E(f+h)-E(f)=dE(h)+o(h).$$ Since the the last term on the LHS is $o(h)$, we obtain $dE(h)=\int\nabla f\cdot\nabla h$. Note that this is the Frechet derivative at the ...


0

The line $x=0$ is a characteristic of this PDE. Indeed, the equation says that when $x=0$, the derivative $u_t$ is zero, which means $u$ is constant on this line. The reason you haven't found this characteristic when solving the ODE $$ \frac{dt}{dx} = \frac{1}{2tx^2} $$ is that by putting $x$ in the denominator, we already lost the solution $x=0$. This ...


1

The methodology in the posted question was correct and gives a way forward. Its implementation had some mistakes which we resolve here. We begin with $$\frac{\partial^2 f(x,y)}{\partial x\partial y}=e^{x+2y} \tag 1$$ and integrate $(1)$ with respect to $x$ to obtain $$\frac{\partial f(x,y)}{\partial y}=e^{x+2y}+C_1(y) \tag 2$$ where $C_1(y)$ is an ...


1

Linear PDEs may not have solutions. Hans Lewy constructed such an example about sixty years ago, and it probably surprised just about everyone in the field at the time. ODEs are much nicer in that regard. http://www.jstor.org/stable/1970121?&seq=1#page_scan_tab_contents


0

QDE has one Independent variable, say x. Solution is y(x) PDE has more than one independent variables say x1,x2,...Xn): solution is y(x1,x2,..xn). Partial derivatives are in the equation.A partial derivative differentiates with respect to one independent variable (say x3)while holding the other independent variables constant.


2

Both are differential equations (equations that involve derivatives). ODEs involve derivatives in only one variable, whereas PDEs involve derivatives in multiple variables. Therefore all ODEs can be viewed as PDEs. PDEs are generally more difficult to understand the solutions to than ODEs. Basically every big theorem about ODEs does not apply to PDEs. It's ...


2

Well, given a linear ODE, the set of solutions form a vector space with finite dimension. However, a linear PDE (like the heat equations) has a set of solution that form a vector space with infinitely many dimensions. To see that, one may consider the ODE $$ y'=-ay(t), $$ with solution, $$ y(t)=e^{-at}y_0, $$ (so, the vector space is one dimensional) ...


3

$$J=\int_{\mathbb{R}^n}(x_1^2+\ldots+x_n^2) e^{-\|x\|^2/4}\,d\mu = n\int_{\mathbb{R}^n}x_1^2 e^{-x_1^2/4}e^{-(x_2^2+\ldots+x_n^2)/4}\,d\mu $$ by symmetry. Since $\int_{-\infty}^{+\infty}x^2 e^{-x^2/4}\,dx = 4\sqrt{\pi}$ and $\int_{-\infty}^{+\infty}e^{-x^2/4}\,dx = 2\sqrt{\pi}$, by Fubini's theorem it follows that: $$ (4\pi)^{-n/2}\int_{\mathbb{R}^n}\|x\|^2 ...


0

As the commenters said, the subscript $0$ indicates vanishing on the boundary in the Sobolev space sense: that is, it can be approximated in the Sobolev norm by smooth functions with compact support in $\Omega$. The reason we use $H^1_0$ instead of $H^1$ in the definition of $H^{-1}$ is explained in The dual of the Sobolev space $W^{k,p}$.


2

GENERAL DEVELOPMENT: First we establish the equivalence of the two forms for the Navier-Stokes Equations given in the OP. To do this, we use straightforward product rule differentiation to show that $$\begin{align} \frac{\partial \rho \vec v}{\partial t}=\frac{\partial \rho }{\partial t}\vec v+\rho \frac{\partial \vec v }{\partial t} \tag 1 \end{align}$$ ...


2

They are both correct. Remember you also have the divergence free condition: $$ \frac{\partial v}{\partial z} = 0 .$$ Note, this makes the 1D incompressible NS equation quite boring.


0

Yes. Since the PDE has a solution that depends on a Heaviside function, it is discontinous in some parts of its domain, therefore it is not smooth. Being so, the solution does not pass on the stability criteria for a classical (or strong) solution, and so the solution is weak.


1

It seems to me that a definition of "regularity" really depends on where it is being applied, for instance the regularity theory of weak solutions refers to the extra differentiability a function may have, due to it solving a PDE. The regularity a solution can inherit depends on the properties of the problem, i.e., the smoothness of a domain boundary, the ...


2

It will be helpful to view the "factorization" technique as separation of variables. Assume $\,x = \eta+\xi ,\,$ and $\,t = \eta-4\xi .\,$ Then $$ \begin{cases} x = \eta + \xi\\ t = \eta-4\xi \end{cases} \implies \begin{cases} \dfrac{\partial u}{\partial \xi} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial \xi} + \dfrac{\partial ...


0

Here is a possible explicit calculation: Let $A$ be the area of the unit sphere $S(0,1)=\{x\in\mathbb{R}^n;\ |x|=1\}$. Employing polar coordinates (see Theorem 2.49 in the Folland's book) we conclude that $$\int_{B(0,\varepsilon)}1\ dx=\int_0^\varepsilon\int_{\partial B(0,1)} r^{n-1}\ dS\ dr=A\int_0^\varepsilon r^{n-1}\ dr=A\frac{\varepsilon^n}{n}.$$ ...


1

Rearrange the weak formulation and you have $$\int_\Omega\nabla u\cdot\nabla v=\int_\Omega (f-u_t)v,$$ with the necessary assumption that $f\in L^2(0,T;L^2(\Omega))$, $u(t)$ can be considered to be the weak solution of Laplace's equation with right had side in $L^2(\Omega)$, for a.e. $t\in(0,T)$. Elliptic regularity indeed justifies that either $u(t)\in ...


1

I believe the $c_j$'s are as smooth as the coefficients of the PDE and the right hand side. If you look at theorem 1, (I will use your notation of $c_k$'s rather than Evans' $d_k$), we have a linear system of ODE $$c^{k'}_m(t)+\sum_{l=1}^me^{kl}(t)c^l_m(t)=f^k(t),$$ for which there exists a unique absolutely continuous function $c_m=(c^1_m,\ldots,c^m_m)$ ...


1

In distribution theory, any function can be differentiated by applying the derivative to the other side So if $U$ is a distribution and $f$ a test function $$ <D_i u, f > = -< u, D_i f>. $$ This is a definition of $D_i u$ . Any bounded measurable function, $g,$ can be interpreted as a distribution via $$ <g, f> = \int g(x) f(x) dx $$ for ...


1

By $y_k$ do you mean $y(k,t)$? What you have is a PDE with only time derivatives, and so the general solution is given by $$y(k,t)=f_1(k)\cos(k^2+Aka')t+f_2(k)\sin(k^2+Aka')t,$$ for general functions $f_1,f_2$, which would be determined if you had an initial condition for $y$ and $\frac{\partial y}{\partial t}$.


1

Your problem is the classical inverse problem formulated by Helmholtz in 1887. This problem is, as far as I know, completely solved by Mayer and Hirsch in 1897. If you want a complete and modern study of this problem, you can read the book of Olver "Applications of Lie groups to differential equations". In this book, the multidimensional case (PDE) with ...


0

Hint: The directional derivative of $f$, in the direction of vector $\vec u$, is just: $$\nabla f\cdot \vec u,\quad\text{or}\quad \nabla f\cdot \frac {\vec u}{\lVert\vec u\rVert}$$ (there are different conventions, according to context).


1

The amplitude gets smaller due to numerical dissipation that smooths the solution. Take more points to reduce the effect or switch to a method of higher order (but beware of the oscillations). Peaks are just rendering artifacts due to sampling. Also inspect your code for errors, since $\alpha = 1/2$ means that the solution is travelling to the right while ...


7

The equation $u_t + v u_x = 0$ describes the evolution of a density of particles each of which is moving with a velocity of $v$. As a result, for positive $v$, information is transported from left to right, and for negative $v$, information is transported from right to left. So if you want to impose a boundary condition on the right, then information must be ...



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