New answers tagged

0

If $u_1, u_2$ are solutions then we know $a{[u_1]}_t+b{[u_1]}_x=0$ and $a{[u_2]}_t+b{[u_2]}_x=0$ Let $u=c_1u_1 + c_2u_2$.   So if $u$ actually is a solution then $a{\big[c_1u_1+c_2u_2\big]}_t +b{\big[c_1u_1+c_2u_2\big]}_x =0$.   Can you show that this is so? In the second case, If $u_1, u_2$ are solutions then we know ...


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Since you already took a course in numerical analysis before, you should be able to follow Gilbert Strang's lectures Computational Science and Engineering I Mathematical Methods for Engineers II


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With your hypotheses, and for the function $f(x)=x$, the Fourier series of sines is equal to $f$ everywhere, except at $x=\pi$, where it is $0$ (as you can check explicitly, at this point, simply substituting $x=\pi$ in the series that you already wrote). More generally, the Fourier series of sines of a function $f$ with the hypotheses that you described ...


0

After dividing through by $1+t$ you see that the characteristic curves satisfy $\frac{dx}{dt}=\frac{x}{1+t}$. This equation can be solved; you get $x=x_0(t+1)$. Then $\frac{dw}{dt}=\frac{e^t}{t+1}$ so $w(t)=u(0,x_0)+\int_0^t \frac{e^s}{s+1} ds$. Finally, you need to express the answer in terms of $t$ and $x$ rather than $t$ and $x_0$. So you solve and get ...


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$$(u_x+bu)_y+a(u_x+bu)=0$$ Let $v=u_x+bu$,then solve two ODEs


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First, why this expression for $c$? If I recall correctly then $$ c=\sqrt{\frac{E}{\rho}}\,. $$ Second, the eigenvalues you found are the eigenvalues of the operator $$ -\frac{d^2}{dx^2}, $$ and the eigenvalues are $\omega_n^2$, it is easier to deal with squares, hence to relate everything you need to find the eigenvalues of $-A$ (you use $A$ in two ...


0

As it turns out the two results for $\psi$ are actually equivalent as by the dispersion relation: $$\frac{1}{\omega}-\alpha=\frac{1}{\omega}\left(1-\alpha\omega\right)=-\omega$$ Alternative method: First, redefine $\theta$ and $\psi$ using a scalar $\phi$: $$\theta=\partial_{\eta}\phi\quad\psi=-\partial_{\tau}\phi$$ This ensures that: ...


2

Such ansatz makes sure that your $A$ is still symmetric. If you just multiply by $M^{-1}$, you will not have a symmetric matrix on the right-hand side of the equation. Here's how the given ansatz works: \begin{align} M\frac{d^2}{dt}X & = KX \\ M^{1/2}\frac{d^2}{dt}\left(M^{1/2}X\right) & = KX \\ \frac{d^2}{dt}\left(M^{1/2}X\right) & = ...


1

The way I view the use of the FT using periodic BCs is that the problem is really about expressing the solution in terms of Fourier series with the "transform" being the Fourier coefficients. The periodic BC's imply that $$u(x,t) = \sum_{n=-\infty}^{\infty} c_n(t) e^{i n \pi x} $$ where $$f(x) = \sum_{n=-\infty}^{\infty} c_n(0) e^{i n \pi x} $$ Thus, ...


1

The notes that you are following should be corrected by erasing the initial fraction: $$ \frac{dE}{dt}(t) = \int_\Omega (u_tu_{tt} + c(x)^2 \nabla u \cdot \nabla u_t + q(x)uu_t)\, dx. $$ In particular, the derivative of $(u_t)^2$ is obtained exactly as you wrote. As for the middle term, you have instead $$ \frac{\partial}{\partial t}|\nabla u|^2 = ...


1

The given initial conditions are probably not enough: specifying some derivatives at just $2$ points is unlikely to be a boundary condition that guarantees uniqueness. You would probably need boundary conditions on at least an open subset of the boundary, like the conditions you are suggesting, and you may need more; consider the example of boundary-value ...


0

The $L^2$ assumptions are not imposed because the assumption $u\in C_1^2(\overline{U}_T)$ supersedes them. This is because $U$ is assumed to be bounded and $T>0$ is fixed, so $\overline{U}_T$ is a compact domain. (You can assume $T$ is finite, because in the case $T=\infty$ you can just keep proving uniqueness for successively larger values of $T$.) Since ...


1

As Ian wrote, $u$ being a weak solution of $u_x+u_y=0$ means that $$\int_{\mathbb{R}^2} u(x,y) (v_x(x,y) + v_y(x,y))\, dx dy = 0\tag{1}$$ for every test function $v$. This follows by formal integration by parts of $$\int_{\mathbb{R}^2} (u_x(x,y)+u_y(x,y)) v(x,y) dx dy = 0\tag{2}$$ If $u(x,y)=f(x-y)$, then integral in $(1)$ becomes as ...


5

This is not possible for all $A\subseteq\mathbb{R}$ of measure $0$. Define $\mathbb{D}[f]$ to be the set of non-differentiable points of $f$. The following result is due to Z. Zahorski, exhibited in this $1946$ paper: Theorem. Given $X\subseteq \mathbb{R}$, there exists $f$ such that $\mathbb{D}[f]=X$ if and only if $X=\mathcal{A}\cup\mathcal{B}$ ...


1

The best way to view this in my opinion is by looking at the actual ODE you get when you solve the characteristic equation. This ODE is what you get along the path $x=x_0+\frac{3}{2}t$. For each $x_0$ you have $v(t)=u(t,x_0+\frac{3}{2}t)$ with $v(0)=f(x_0)$. It satisfies $\frac{dv}{dt}=\frac{1}{2} \cos(x(t))=\frac{1}{2} \cos(x_0+\frac{3}{2}t)$. The solution ...


2

$$3u_x+2u_t=\cos(x)$$ HINT : The equations characteristics are : $$\frac{dx}{3}=\frac{dt}{2}=\frac{du}{\cos(x)}$$ First characteristic : $\frac{dx}{3}=\frac{du}{\cos(x)} \quad\to\quad 3du-\cos(x)dx=0 \quad\to\quad u-\frac{1}{3}\sin(x)=c_1$ Second characteristic : $\frac{dx}{3}-\frac{dt}{2}=0 \quad\to\quad 2x-3t=c_2$ General solution on implicite form, ...


2

You have characteristics given as functions $x(s)$ and $y(s)$ which depend on parameter $s$: \begin{align} x = x(s) &= x_0 + 2s && \implies& x_0 &= x - 2s \label{1}\tag{1} \\ y = y(s) &= y_0 - s && \implies& y_0 &= y + s \label{2}\tag{2} \end{align} The initial condition $u\left(x_0, 0\right) = ...


0

The characteristic equations are : $$\frac{dx}{2}=\frac{dy}{-1}=\frac{du}{0}$$ which leads to the equations of chararacteristics : $$\begin{cases}u=c_1\\x+2y=c_2 \end{cases}$$ The general solution on implicit form with is : $$\Phi\left(u\:,\:x+2y\right)=0$$ where $\Phi$ is any derivable function of two variables. This is equivalent to : $$u=F(x+2y)$$ where ...


1

The reasoning seems to be that, letting $z = x-y$, we can use the chain rule to say that $$\frac{\partial Q}{\partial y} = \frac{\partial Q}{\partial z} \frac{\partial z}{\partial y} = \frac{\partial Q}{\partial z} \left(\frac{\partial}{\partial y}\left(x - y \right)\right) = - \frac{\partial Q}{\partial z} \Longrightarrow \frac{\partial Q}{\partial y} = - ...


0

What happens is that the two equations involving divergences are redundant provided that they are satisfied at any given specific time (that is, with this last assumption the two equations follow from the other six). This was shown by Stratton in the 1940's.


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A conservation law has the following structure: $$u_t=-\mathrm{div} (\mathbf{F}(x,y,u))$$ where $\mathbf{F}=(F^1,F^2)^T$ is a vector field that, in principle, can be nonlinear. The last PDE can be rewritten in the following form: $$u_t=-{F^1}_x-{F^2}_y-\mathbf{F}_u\cdot \nabla u$$ In your case we have ${F^1}_x={F^2}_y=0$ and this implies $F^1=f(y,u)$ and ...


1

There is nothing wrong with your computations. The only problem is that $$ \frac{200}{m}\sum_{n=1}^\infty\frac{1}{n}\Bigl(\cos\Bigl(\frac{n\,\pi}{3}\Bigr) - \cos\Bigl(\frac{2\,n\,\pi}{3}\Bigr)\Bigr)\,\sin(n\,x) $$ converges slowly to $f(x)$. Below you have the graphs of the sum for the the first $10$ and $100$ terms.


0

It is a diffusion equation. So I get another solution $ u (t,x) =K \frac{{{e}^{-\frac{{{x}^{2}}}{2\, \left( 2\, t+1\right) }}}}{\sqrt{2\,t+1}} - K $ but I also have a problem with the boundary conditions.


0

You also have to account for the possibility that $\lambda=0$. Then you have $$ T'=X''=0, $$ so you can have $Bx+C$ as another solution. Setting $x=0$ implies that $C=0$, and $x=\pi$ gives $C=1$, but you have to deal with the initial condition separately. Define $U=u-x$, then $U$ satisfies $$ U_t=(1+2t)U_{xx}, $$ but with different BCs: $$ U(0,t)=0, \quad ...


1

Some years ago I had the same question; I found J. Leray. Sur le mouvement d’un liquide visqueux emplissant l’éspace. Acta Mathematica, 63, 7 1934, surprisingly easy to read and absorb. If you don't know French and you've been trained in functional analysis, you could start from P. Constantin and C. Foias. Navier–Stokes equations. Chicago Lectures in ...


1

Firstly, one of your equations is incorrect; it should be $$\frac{du}{ds} = -cu$$ From your first ODE, we find $dt= ds$. Hence the other two ODEs become $$\frac{dx}{dt} = -u^{2}, \quad \frac{du}{dt} = -cu$$ Solving the first gives $$x(t) = -u^{2} t + x_{0} \implies x_{0} = x + u^{2} t$$ The second ODE is separable so we find \begin{align} \ln u &= ...


1

Sketch/Hint: Let me show you one "follow your nose" way to tinker until you reach a conclusion. We want to show that $|F(x_k,t_k) - F(x,t)| \to 0$ as $(x_k,t_k) \to (x,t)$ in $\mathbb R^n \times \mathbb R$. Since we know that $|f(x_k,s)-f(x,s)| \to 0$ as $k \to \infty$ (why?) and $f(x,s)$ is the integrand inside the integral definition $F(x,t)$, we can try ...


0

a) No, the form of the boundary conditions does not have any influence on the form of the PDE. Whatever the boundary conditions, the PDE itself is still autonomous. b) Yes, in this case you obtain an equation for $F(t) = u(s(t),t)$ which depends explicitly on $s(t)$ and its derivatives.


0

A minimal surface problem is an Elliptic PDE. MATLAB's documentation has an example for how to solve this here.


1

This follows immediately from Poincare's inequality: Let $1 \le p < \infty$ and let $\Omega \subset \mathbb{R}^N$ be a connected extension domain for $W^{1,p}(\Omega)$ with finite measure. Let $E \subset \Omega$ be a Lebesgue measurable set with positive measure. Then there exists a constant $C = C(p,\Omega,E) > 0$ such that for all $u \in ...


1

In the heat equation $u(t,x)$ is a function of the two variable $t$ and $x$ that represents the temperature at time $t$ in the point $x$. So $u(0,x)=f(x)$ is the function that represents the initial temperature at point $x$, and it is a function of $x$ only. Your notation simply indicates this $f(x)$ with the symbol $u_0(x)$.


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Because of the form of the heat equation you can always add an arbitrary first-degree polynomial in $x.$ That freedom is used to satisfy the inhomogeneous boundary conditions (temperature at the end points known but not necessarily zero) when a hypothetical solution with homogeneous boundary conditions $t(0)=t(1)=0$ is supposed known.


1

$$ \partial_S = \frac{\partial x}{\partial S}\partial_x = \frac{1}{S}\partial_x $$ To get second derivitve you apply the above twice so you get $$ \partial_{SS} = \frac{1}{S}\partial_x\left(\frac{1}{S}\partial_x\right) = -\frac{1}{S^3}\frac{\partial S}{\partial x}\partial_x + \frac{1}{S^2}\partial_{xx} $$ Using the first derivarive with $S$ you find that ...


1

The integral of $\tau$ is $\tau^2/2$, not $\tau$. It should be $$ x=\frac{\tau^2}{2}+s-\frac12. $$


1

Another approach would be to consider how both sides scale under the transformation $x = \lambda y$, $\lambda>0$. By the change of variables and the chain rule, it becomes $$ \Lambda \int_{\lambda^{-1}\Omega} \frac{u^2}{\lambda^p |y|^p} \, \lambda^n\mathrm{d}y \leq \int_{\lambda^{-1}\Omega} \lambda^2 |\nabla u|^2 \lambda^n\, \mathrm{d}x $$ Thus, if ...


1

My intuition on why it is not complete: it is already complete under the usual norm $\|f\|_{\sup}+\|\nabla f\|_{\sup}$. The Sobolev norm too measures both $u$ and $\nabla u$ in a similar fashion (the version you wrote down is equivalent to $\|f\|_{L^2}+\|\nabla f\|_{L^2}$) but with supremum norms replaced by $L^2$ norms, which makes the Sobolev norm strictly ...


0

Take $C^1_0= C^1_0((-1,1))$ and look at $u(x)= |x-1|$. It's not difficult to see that $||u||_{1,2}$ is finite and that there is a sequence $u_n \in C^1_0((-1,1))$ which converges to $u$ in that norm. (Define $u_n(x)=u(x)$ if $|x|>\frac{1}{n}$ and $=-\frac{1}{2} n x^2 +b_n $ for $|x|\le \frac{1}{n}$ with $b_n$ chosen so that this is $C^1$ near $\pm ...


1

Your solution is practically there. You're right that the evenly spaced nested balls is what's missing. Fix $y \in \Omega$ and fix some multi-index $\alpha$ with $| \alpha | = k$. Let $d = dist(y, \partial \Omega)$. Let $d_{0} = \frac{d}{|\alpha|}$. Then consider the ball $B = B(y, d_{0})$. Let $\alpha_{1}$ be a multi-index such that $|\alpha_{1}| = k-1$ ...


0

$f(x)=|\sin (\pi n x/L)|$ for a fixed value of n. You need to calculate $$\int_0^L f(x) \cos( m x) dx $$ and $$\int_0^Lf(x) \sin (m x) dx$$ for each $m\in \{0\}\cup N$ in order to calculate the co-efficients.


1

Yes, the Laplace equation in two dimensions is often written as $u_{z\bar z}=0$, which simplifies some computation. And yes, a harmonic function can be locally represented as the sum of a holomorphic and antiholomorphic functions. In a simply-connected domain this representation is also global; but in general multi-valued functions may enter the picture. For ...


0

HINT: Consider \begin{align*} m(x,t) = \begin{cases} w\left(\frac{x+1}{\sqrt{D_1}},t\right), \mbox{if} \ x\in(-1, 0)\\ w\left(\frac{x+1}{\sqrt{D_2}},t\right), \mbox{if} \ x\in(0, 1) \end{cases} \end{align*} What PDE does m satisfy?


0

A function $u$ is in this space if $$\int_0^T\|u(t)\|_{L^1(K;\mathbb R^N)}<\infty,$$ for all $K\subset\mathbb R^N$ that are compact.


0

The equation $$-\frac{x}{t^2}v'+f'(v(x/t))\frac{1}{t}v'=0$$ factors. Apparently, we are supposed to ignore the constant solution $v'=0$. Then the equation simplifies to $$f'(v(x/t)) = x/t$$ which, after introducing the notation $s=x/t$, is exactly $v(s)=(f')^{-1}(s)$.


0

$$ \partial_{xx}u(x,y) + \frac{1}{x^2}\partial_{yy}u(x,y)+\frac{1}{x}\partial_x u(x,y) = 0 $$ lets assume a form of $u(x,y) = x^2v(x,y)$ we find $$ \partial_{yy}v(x,y) + x^2\partial_{xx}v(x,y) + 5x\partial_x v(x,y) + 4v(x,y) = k $$ you can transform to $\phi(x,y) = 4v(x,y) -k$ (assuming $k$ is a const) then we have $$ \partial_{yy}\phi(x,y) + ...


3

Remark This is actually more general than it may seem. Consider an equation $$Tf-\Delta f=0\tag1$$ where $T$ is an operator acting on the time variable. (For instance if $T=\partial_t^2+2\partial_t$, the equation is called the telegrapher's equation). Let $h$ be the Green's function of this equation. We can define $$v(x,t)=\int h(s,t)u(x,s)\mathrm ds$$ ...


0

Analytic solving of the PDE : $$u_y + uu_x=1$$ Thanks to the method of characteristics : $$\frac{dy}{1}=\frac{dx}{u}=-\frac{du}{1}$$ On the characteristic curves : \begin{cases} dx+udu=0\:\:\rightarrow\:\: 2x+u^2=c_1\\ dy+du=0\:\:\rightarrow\:\: y+u=c_2\\ \end{cases} The general solution can be expressed on the form : $$\Phi\left(2x+u^2\:,\: ...


0

From what I understand you discretize a PDE into smaller basis functions and solve each one and sum them up to find an approximate solution. Space is partitioned into smaller volumes, the finite elements. The base functions $\Psi^{(e)}_i$ are used to approximately represent the solution $f$ over the finite element $e$. $$ \left. f(x) \right\vert_e ...


0

Quick answer: you can use the solutions you found for the homogeneous equation (where $k=0$) to construct the so-called Green's function. Using that, it's easy to solve the PDE for general $k$; see Wikipedia or MathWorld. Note that your equation is of the form $\mathcal{L} u = f$, with $f = k$ and \begin{equation} \mathcal{L} u = \partial_x^2 u + ...


-1

Analytic solving of the PDE : $$\partial_t u - \alpha u\partial_xu=0$$ Thanks to the method of characteristics : $$\frac{dt}{1}=\frac{dx}{-\alpha u}=\frac{du}{0}$$ On the characteristic curves : \begin{cases} u=c_1\\ \alpha c_1 dt +dx=0\:\:\rightarrow\:\: \alpha u t +x=c_2\\ \end{cases} The general solution can be expressed on the form : ...


1

To answer your edit: once you have found the correct domain of definition for the Laplacian, you have that $(I-t\Delta)^{-1} = \sum \limits _{k=0} ^\infty t^k \Delta^k$. Since $\int (\Delta u) v = \int u (\Delta v)$ (on test functions, at least), you may use induction and prove that $\int (\Delta^k u) v = \int u (\Delta^k v)$, so $\int (\sum \limits _{k=0} ...



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