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0

The equation $$ R''(r)+2rR'(r) - \mu rR(r) = 0 $$ can be written as $$ (rR(r))''-\mu(rR(r))=0. $$ Therefore, if you substitute $S(r) = rR(r)$, you get $S''-\mu S=0$, which is a familiar equation. You have the wrong condition for insulated. Insulated means that the normal derivative of the heat distribution at the ...


1

Uniqueness for Laplace's equation on a domain $D$ (with, say, piecewise-differentiable boundary $\partial D$) with, say, Dirichlet boundary conditions $u=f$ on $\partial D$ is easy: suppose there are two such solutions, $u$ and $v$. Take $$ -\nabla^2 (u-v) = 0, $$ multiply by $u-v$ and integrate over $D$, and use the divergence theorem: $$ 0 = -\int_D ...


0

You could try these books: - An Introduction to Ordinary Differential Equations by James C Robinson - Differential Equations by Richard Bronson and Gabriel Costa These are the books I used to use and they are pretty good, and your course, from your description, looks a lot like my first year differential equations course (I tried looking at your dropbox ...


2

Hints: To show that $u \to 0$ as $t \to \infty$ or $|x| \to \infty$, use dominated convergence to move the limit under the integral sign. To show that $u$ is integrable, use Fubini's theorem To show that $u$ is smooth, use differentiation under the integral sign.


0

From a basic ODE class you should know that the solution to $y'' = \lambda y$ is either ( $\lambda <0$) $$y = A \sin (\sqrt {-\lambda} t) + B \cos ( \sqrt {-\lambda} t ) $$ ( $\lambda =0$ ) $$y = A t + B $$ ( $\lambda >0$) $$ y = A \exp (\sqrt \lambda t) + B \exp ( -\sqrt \lambda t )$$ Use the boundary conditions to see what case you fall into, this ...


0

This is a vague question and I can sadly only give a vague answer in a few sentences....$\Gamma$-convergence is a very powerful tool when it comes to a specific type of problems that are obtained as limits of certain types of other problems. One formulates solutions of PDEs as critical points of specific functionals. $\Gamma$-convergence of these ...


0

This follows from the very simple fact: If a function $h(x)$ is odd, then $$ h(0) = 0 . $$ To prove this, note that $h(-x) = - h(x)$ by definition of odd, so if we put $x = 0$, we get $h(0) = - h(0)$. This implies $h(0) = 0$. Conclusion: An odd function automatically satisfies the boundary condition $h(0) = 0$. For this reason, if you want to find a ...


1

Note that the solution of $-\Delta u = f$, with homogeneous dirichlet BC's is given by $$u(x)=\Phi*f=\int_\Omega\Phi(x-y)f(y)\,dy$$ where $\Phi$ is the fundamental solution of the Laplace equation, and (in 2D) $$\Phi(x)=\frac{\ln(|x|)}{2\pi}.$$ Note that in the sense of distributions, $-\Delta\Phi(x)=\delta(x)$, and so $$-\Delta u = ...


1

You should be looking for $f(x)=g(\|x\|)$, there shouldn't be a squared term, this is because the Laplace operator is rotation invariant, so you can seek a radial solution. Now applying $\Delta$ to $g$, you will end up with an ODE in the variable $r=\|x\|$, which you can solve to find the fundamental solution of the Laplace equation. Note that ...


0

Here is the answer provided by the author of the original article, where I saw this notation first: $\dot{H^{-1}}$ is the homogeneous $H^{-1}$ norm. It is the dual space of $\dot{H^{1}}$, the homogeneous $H^1$ norm. The homog $H^1$ norm of $g$ is $\int |\nabla g|^2 dx$. i.e., no zeroth order term. (So it is not really a norm, but a seminorm. The ...


0

What you have is actually not a Laplace Equation, but rescaled Poisson's Equation because it has non-zero right-hand side, i.e. it is non-homogeneous. The coefficient 9 before the $U_{xx}$ term can be easily replaced with 1 by appropriate change of variables. The general solution $U$ of any linear non-homogeneous differential equation can be found as a sum ...


1

you are using the variable $t$ in two different way. you are also forgetting that a solution of $\frac{dx}{d\tau} = -y$ is not $x = -\tau y$ because $y$ is also a function of $\tau.$ i will use a new variable $s$ instead. so we have $$\frac{dt}{ds} = t^2, \frac{dx}{ds} = -y, \frac{dy}{ds} = x \to \frac{d^2x}{ds^2} + x = 0 $$ so that $$x = a\cos(s-b), y = ...


0

The solution is $$ y(x,t) = \left( e^{-\frac{b}{a}x}y_l + e^{-\frac{b}{a}x}\frac{c}{b}\left( 1 - e^{b(t+\frac{x}{a})} \right)\right)\sigma\left(t + \frac{x}{a}\right) - \frac{c}{b}\left(1-e^{bt}\right) - \mathcal{L}^{-1}\{ F(x,s) \}, $$ where $$F(x,s) = \frac{1}{as}e^{\frac{s-b}{a}x} \int\limits_0^x e^{\frac{b-s}{a}\xi}f(\xi) \text{d}\xi, $$ with $y_l := ...


1

Let $y(x,t) = w(x,t) + \phi(x)$ where $\phi(x) = a x^2 + b x + c$. From this it is seen that $\phi'(x) = 2 a x + b$, $\phi''(x) = 2a$ and \begin{align} \partial_{t}^{2} w = c^2 \, \partial_{x}^{2} w + 2a c^2 + L. \end{align} In order to cancel the $L$ term let $2 a c^2 + L = 0$ which leads to \begin{align} \phi(x) = - \frac{L \, x^{2}}{2 \, c^{2}} + b \, x + ...


2

I found the solution: Since $\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$, $(x, y) ∈ (0, 1)×(0, 1)$ The function u(x,t) must have $e^{x−y}$ From the boundary conditions: $u(x, 0) = 0$ then u(x,t) must be 0 for x=0, so we assume u(x,t) must have $e^{x−y}*x$ $u(x, 1) = 0$ then u(x,t) must be 0 for x=1, so we assume u(x,t) must have ...


1

You have to use the chain rule in the last step. Putting in some extra parentheses for emphasis, you have: $$D_t (\phi(x-ct)) = (\nabla \phi)(x-ct) \cdot (-c)$$ which is a scalar.


0

Maybe you want to take a look to this paper: http://arxiv.org/abs/1104.4345


1

The general form of a harmonic function in an annulus is $$ u(r,\theta) = (A_0 +B_0\log r)+ \sum_{n\ne 0} r^n (A_n \cos |n|\theta+B_n\sin |n|\theta)\tag{1} $$ The stated boundary conditions are: $$ 0 = (A_0 +B_0\log 2)+ \sum_{n\ne 0} 2^n (A_n \cos |n|\theta+B_n\sin |n|\theta) \tag{2}$$ and $$ \sin\theta = (A_0 +B_0\log 4)+ \sum_{n\ne 0} 4^n (A_n \cos ...


2

Let's use the area form of the mean value property: For any $r>0,$ $$|u(0)| = |\frac{1}{\pi r^2}\int_{D(0,r)}u\,dx\,dy\,| \le \frac{1}{\pi r^2}\int_{D(0,r)}|u|\,dx\,dy$$ $$ \le \frac{1}{\pi r^2}\int_r^r\int_r^r|u(x,y)|\,dx\,dy\le \frac{1}{\pi r^2}\cdot C\cdot 2r = \frac{2C}{\pi r}.$$ Let $r\to \infty$ to see $u(0)=0.$ Clearly we could repeat the ...


2

Let $v(x) = x_1^2 + \cdots +x_n^2$. Then $v \ge 0$ and $\Delta v = 2n$. If $u \ge 0$ and $\Delta u = -1$ then $u + v/2n \ge 0$ and $\Delta(u + v/2n) = 0$. A nonnegative harmonic function is constant, so that $u + v/2n = C$ for some constant $C$. This leads rather quickly to a contradiction since $v(x) \to \infty$ as $|x| \to \infty$.


0

By method of characteristics: $u_x +u_y=2u$, and we know $du=u_xdx+u_ydy$, so $\frac {du}{dx}=u_x+\frac{dy}{dx} u_y$ and $\frac {dy}{dx}=1$ as shown by your analysis. So $u_x+u_y=\frac {du}{dx}$, substitute this to original equation we get:$\frac {du}{dx}=2u$, the. You can do the integration to get $ln(u)=2x+C$, which $C$ is a constant based on your ...


0

We are looking for the function $w=w(x,t)$, as I understand. Equations for the characteristics are $$\frac{dt}{ds}=1,\quad\frac{dx}{ds}=4,$$ then $$\frac{dx}{dt}=4,$$ which gives you the system of characteristics $$x-4t=c,$$ where $c\in R$ is a constant. Because $w$ has to be constant along the characteristics, every function $$w(x,t)=F(x-4t),$$ where $F\in ...


1

the characteristics are given by $$\frac{dx}{dt} = x, \frac{dy}{dt} = y, x(0) = 1, y(0) = b $$ which gives $$x = e^t, y = be^t, b = \frac yx, t=\ln x.$$ along the characteristics we have $$\frac{du}{dt} = u(1+b)e^t, u = 1, t=0 \to \int_1^u\frac{du}{u} = (1+b)\int_0^t e^t\, dt$$ integrating we get $$\ln u=(1+b)(e^t-1). u(x, y) = ...


0

$$\int_{-\infty}^{\infty}f(x)e^{i\omega x}dx=\int_{-a}^{a}1\,e^{i\omega x}dx+\int_{|x|>a} 0\,e^{i\omega x}dx=2a\,\frac{\sin(\omega a)}{\omega a}$$ Note that the integral $\int_{|x|>a} 0\,e^{i\omega x}dx=0$. Also note, that in the integral, the symbol $x$ is a "dummy" variable only. That is to say, we can replace $x$ with $\bar x$ as a dummy ...


-1

The subspace of constants is orthogonal to that spanned by the sine functions. You will not get a sine expansion converging to a constant. All sines are zero-mean, as are $\mathcal{L}^2$ sums of them.


0

The original problem that was posted has no unique solution. For example, letting $a_1=T_0$ and $a_n=0$ for $n\ne 1$ provides one solution. Letting $a_3=-T_0$ and $a_n=0$ for $n\ne 3$ provides another. There are an infinite number of possible solutions. Thus, either there isn't a unique one or the problem is mis-stated. If it has been mis-stated, ...


0

I would guess that it means that if $N$ is the normal vector to the plane, then the dot products $N \cdot \alpha$ and $N \cdot \beta$ have opposite signs. In other words, $\alpha$ and $\beta$ are on opposite sides of the plane.


0

A useful observation: if $\phi$ is harmonic, then $r\frac{\partial\phi}{\partial r}$ is also harmonic. This follows, for example, from $$r\frac{\partial\phi}{\partial r} = \left(\frac{d}{dt}u(tx)\right)\bigg|_{t=1}$$ where $u(tx)$ is harmonic for every $t$. Or you could just write $r\frac{\partial\phi}{\partial r}$ as $xu_x+yu_y+zu_z$ and compute the ...


1

Your question is incomplete, one need to read page 46 of Sneddon's book to figure out what the problem really want. To summarize, what the book want is start from an equation of the form $$z = f(u)\quad\text{ where }\quad u = \frac{xy}{z}$$ derive a PDE for $z$ which doesn't involve the function $f(u)$ explicitly. The tool you need is chain rule for ...


0

I try this today. $p= \dfrac{\partial z}{\partial x}$ and $q= \dfrac{\partial z}{\partial y} $ So $ p= \dfrac{\partial f}{\partial x }\left( \dfrac{yz-pxy}{z^2} \right) $ and $ q= \dfrac{\partial f}{\partial y }\left( \dfrac{xz-pxy}{z^2} \right) $ Is that right? After that, I try many minupulations... All attempts, I needed divided by ...


0

After looking through some textbooks from fluid mechanics I found a similar and problem and it seems that this type of problem is solved by adopting a certain type of PDF and performing a little calculus involving Dirac's delta. PDF definition Firsltly, we assume we assume that the deterministic version of our problem has a solution: \begin{equation} ...


0

Let me show it with a particular example. Consider the semilinear heat equation $$ u_t-\Delta u=u^p,\quad x\in\Omega\subset\mathbb{R}^n,\quad t>0, $$ with initial value $u(x,0)=u_0(x)\ge0$ and Dirichlet boundary conditions $u(x,t)=0$ for $x\in\partial \Omega$, $t\ge0$. Here $\Omega$ is a bounded smooth domain and $p>1$. Then, for appropriate $u_0$, ...


2

This is really an ODE, it suffices to make two arbitrary integration constants in the general solution arbitrary functions of $x$. To solve this ODE, rewrite it as $$\frac{f_{yy}}{f_y}=\frac12\frac{f_y}{f}\qquad \left(\ln f_y\right)_y=\frac12 \left(\ln f\right)_y,$$ which gives $\ln f_y=\frac12\ln f+ \mathrm{const}$ or, in other words, $f_y=2C_1\sqrt f$. ...


1

$$ f_yf \left(\frac{f_y}{f} -2\frac{f_{yy}}{f_y}\right) = 0 $$ Thus we can obtain $$ \dfrac{\partial}{\partial y}\ln f -2\dfrac{\partial }{\partial y}\ln f_y = 0 $$ Or $$ \ln \left(\frac{f}{f_y^2}\right) = g(x) $$ So you get $$ f = g_2(x) f_y^2\implies f_y = \sqrt{\frac{f}{g_2(x)}} $$


0

Hint: set $f(x,y)=g(x)h(y)$ and try and find polynomial solutions for $h(y)$.


0

..and here's how you arrive at the polar coordinate version: $$ \begin{align} & u := f(r(x,y)), \\ & u_x = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}, \\ & u_{xx} = \left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}\right)_x = \frac{\partial^2 f}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 + ...


1

You are almost there: By choosing $v = 0$, we see that $L^*(p) \geq -L(0)$ for all $p \in \mathbb{R}^n$. The superlinearity of $L$ implies that for each $p\in \mathbb{R}^n$ there is $R > 0$ such that for all $v \in \mathbb{R}^n$ with $|v| > R$, \begin{equation} p \cdot v - L(v) < -L(0). \end{equation} We can therefore safely ignore any $v$ with ...


3

Numerical Partial Differential Equations I and II by Thomas have become my go-to references. They cover all the major topics (FD, FE and FV) and have lots of exercises.


2

This book is very good: Numerical Solution of Partial Differential Equations Morton and Mayers


-2

If function $G \in L^1(R^d)$ we can perform Fourier transfer of divergence. This end up in: $$0 = \digamma[\nabla(\nabla G + xG)] = i<k \cdot \hat{F}>$$ where $\hat{F} = ik\hat{G} +i\nabla\hat{G}$ and $\hat{G} = \digamma[G]$ Now, $\hat{F}$ vector is parallel to $k$, i.e. we do not have any perpendicular components that vanish in scalar product $<k ...


0

Consistency conditions can be translated to: The set of possible solutions have to satisfy the ode/pde. Only trajectories that are described via the ode/pde (+ic's/bc's) are candidates for solutions. Example: The following (dynamical) optimization problem requires the measured output $y$ to follow the reference trajectory $y^*$. The consistency condition ...


0

The limit should be $(A+B)/2$. Given $\epsilon > 0$, there is $N$ such that $|\varphi(x) - A| < \epsilon$ for $x \ge N$ and $|\varphi(x) - B| < \epsilon$ for $x \le -N$. The integral over $(-\infty, -N]$ is then within $\epsilon$ of $$\dfrac {B}{2\sqrt{\pi t}} \int_{-\infty}^{-N} \exp\left(\frac{-|x-y|^2}{4t}\right)\; dy $$ which goes to $B/2$ ...


4

Let $v = u^4$ and rewrite the PDE in terms of $v$ to get: $$ v^{-3/4} v_t = \Delta v. $$ Let $\;v = A(t) B(x,y,z)\;$ and separate variables to get: $$ A' = 4 A^{7/4} c \quad \mbox{and} \quad B^{3/4} \Delta B = c \quad \mbox{for a constant } c . $$ The DE in $A$ has the solution $\;A(t)^{3/4} = -\frac{4/3}{4 c t + k}\;$ for a constant $k$. So when $t$ goes to ...


1

$$\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}=0$$ $$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0$$ This implies that $$ \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}\right)$$ $$= \frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$ And $$ ...


0

See section 6 of http://www.ewp.rpi.edu/hartford/~ernesto/F2004/IFEM/Notes/w03.pdf. It's about as general as you can get for Laplace on a cylinder.


2

The sum $\Sigma(\cdots)$ is an asymptotic series as $t \to \infty$, so the first two terms of the expansion of $u(x,t)$ as $t \to \infty$ are simply the two terms which do not decrease exponentially, namely $$ u(x,t) \sim \underbrace{4t}_\text{first} + \underbrace{2x-10}_\text{second} + \cdots $$ as $t \to \infty$. Now, the first two terms of the ...


0

This means that if the point $(x,y,u) \in C$, then $f(x,y,u) = 0$.


3

Don't have time to write a long answer, but the short answer is because of boundary conditions. That is, in your first list of equations, you have the boundary condition $u(t,0)=0$. You don't have this boundary condition for the equations involving $w$. However, because you've chosen the initial conditions of $w$ carefully, the initial conditions you are ...


3

The closure of a bounded subset of $\mathbb R^n$ is compact. A continuous function on a compact set is bounded.


1

Differentiating the first equation with respect to $x$:$$\frac{\partial u}{\partial t} +g\frac{\partial \eta}{\partial x}=0\implies\frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$ and the second with respect to $t$: $$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0\implies \frac{\partial^2\eta}{\partial ...



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