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Indeed, the topology is crucial. If $u$ is an element of $L^{2}( \mathbb{R},dx)$ we can show that $-\partial _{x}^{2}$ extends to a non-negative self-adjoint operator $p^{2}$ and we can write \begin{equation*} u(t)=\exp [kp^{2}t]u(0). \end{equation*} In this case $u(t)$ is well defined for all $t$.

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Writing down the formula $$u(x,t)= \sum_{i=1}^{n}b_n e^{k\frac{n \pi }{L}t}\sin (\frac{n \pi x }{L}) \tag{1}$$ [NB: I dropped extraneous "x" in the exponent] is a good idea, but you should connect it to the goal: proving that the $u(x,t)$ does not depend continuously on the initial data $f$. The goal is a bit vague because to make sense of "continuous" we ...

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Since the laplacian is a linear differential operator it can be differentiated like any other linear operator. The way the differential works is the same way as for the product of functions : Let $D(t)$ be a linear differential operator then ...

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$\newcommand{\R}{\mathbb{R}}$Yes. The approximating argument that you use in the usual case of $C^\infty(0, T;\mathbb{R})\subset W^1(0, T;\mathbb{R})$ (that is, convolution against a mollifying family) works verbatim here. For the details you may look in Evans's book on PDEs (chapter on Sobolev spaces, look in the last paragraph "Spaces involving time"), or ...

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First, note that you also have a solution for an eigenvalue of 0 in this case, which is the constant solution X(x)=A. Now, for positive eigenvalues, take the derivative, and you get $X'(x)=-A\sqrt \lambda cos(\sqrt \lambda x)+B\sqrt \lambda sin(\sqrt \lambda x)$ From here, to solve for A and B, you put in your initial values. Typically one of the ...

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If $|B|=0$, then $|A||B|\geq|A\cdot B|$ holds trivially. Else, consider: $$0\leq |A-\gamma B|^2=|A|^2-2\gamma A\cdot B+\gamma^2|B|^2.$$ Now, if you use $\gamma=\frac{A\cdot B}{B\cdot B}$, then the above is translated to $$0\leq |A|^2-2\frac{(A\cdot B)^2}{|B|^2}+\frac{(A\cdot B)^2}{|B|^2}=|A|^2-\frac{(A\cdot B)^2}{|B|^2}\implies |A|^2\geq \frac{(A\cdot ... 1 The obstacle that you encountered is real. Consider one-dimensional situation: for every \lambda<0, the function u(x) = - \cosh (\sqrt{-\lambda} x) satisfies -u''=\lambda u and$$\sup_{[-1,1]} u = -1 > -\cosh (\sqrt{-\lambda}) = \max(u(1),u(-1))$$For eigenfunctions with \lambda<0 we have maximum principle for u^+, u^-, and ... 0 The usual separation of variables in polar coordinates produces$$\phi \left( r,\theta \right) =\sum _{n=0}^{\infty } \left( A_{{n}}\sin \left( n\theta \right) +B_{{n}}\cos \left( n\theta \right) \right) \left( {r}^{n}C_{{n}}+{r}^{-n}E_{{n}} \right) +F\ln \left( r \right)$$Applying the boundary condition at r=R_1 we obtain$$\sum _{n=0}^{\infty ...

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Using separation of variables, you will have two ODE's in terms of time and space, apply initial conditions T(x,0) to the time variant ODE and apply boundary conditions for space variant ODE. In this $c^2 = \frac{1}{k}$ Put T(x,0) the intitial condition and equate a) $$T(x,0) = b_nsin(\frac{n\pi x}{L}) = T_0sin(\frac{n\pi x}{L})$$ Leading to $$b_n = ... 0 The correct way to proceed is to use some regularization method. So using convolution on f, for example, we got a series of smooth functions given by: f_{\epsilon}=f*\phi_{\epsilon} where \phi_{\epsilon} are suitable mollifiers. Then we can define the distributions: <f_{\epsilon},\phi\phi_{j}>=\int f_{\epsilon}\phi\phi_{j}. which in the ... 1 In my opinion, the term radiating is better suited to diffusion equation; I prefer to think of that part of the boundary as being elastic: it resists the motion, but does not make it impossible. This means that the boundary does some nonzero work, and we should include it in the computation of energy. Without the elastic part, the energy functional would ... 0 Assuming t>0 and x \in [a,b] \subset \mathbb{R}, we have:$$ \int^b_a \left( \frac{\partial f}{\partial t} + \frac{\partial \, f g}{\partial x}\right)\omega_1 \mathrm{d}x = 0,$$where \omega_1 is a test function. On the other hand, for the second equation we have to apply 1st Green's identity, which yields:$$\begin{align} \int^a_b \left( ...

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The best introduction is Raymond Elementary Introduction to the Theory of Pseudodifferential Operators CRC Press.

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My application of separation of variables was incorrect: $\Delta u - 1/\delta * u = R(x,y)$, Consider the homogeneous case: $X''Y+XY''-1/\delta*XY=0$ $\frac{X''Y}{XY}+\frac{XY''}{XY}-1/\delta*\frac{XY}{XY}=0$ $\frac{X''}{X}+\frac{Y''}{Y}+1/\delta=0$ Because X is only a function of x and Y is only a function of y, the only way for the above to be true ...

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Part a: $$v = u(x',t')$$ where $$x' = ax - x_0 \\t' = a^2t - t_0$$ Therefore $$v_x = \frac{\partial{u}}{\partial{x'}}\frac{\partial{x'}}{\partial{x}} +\frac{\partial{u}}{\partial{t'}}\frac{\partial{t'}}{\partial{x}} =u_{x'}a + 0$$ Similarly: $$v_{xx} = a^2u_{x'x'} \\v_{t} = a^2u_{t'}$$ and the result follows using $u_{t'}=ku_{x'x'}$. The other ...

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You're right to feel something is wrong: Try giving a detailed argument why $p$ not being a local maximum for $u(x)+\varepsilon|x-p|^2$ implies $p$ not a local maximum for $u$. You're on the right track though. Consider $v_\varepsilon(x)=u(x)+\varepsilon|x|^2$ (without loss of generality we assume that $0\notin \Omega$). The arguments you use give that ...

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Details may depend on the definition of Green's function you are using, but here are the key points: Fix $y$. The function $h(x) = G(\cdot,y)$ is harmonic in $\Omega\setminus \{y\}$ Since $G(x,y)\to-\infty$ as $x\to-\infty$, there is $r>0$ such that $h(x)<0$ when $|x-y|\le r$. Apply the strong form of the maximum principle to $h$ on $\Omega ... 0 Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:$\dfrac{dx}{dt}=x$, letting$x(0)=1$, we have$x=e^t\dfrac{dy}{dt}=-y$, letting$y(0)=y_0$, we have$y=y_0e^{-t}=\dfrac{y_0}{x}\dfrac{dU}{dt}=xU+x^2y=e^tU+y_0e^t$, we have$U(x,y)=f(y_0)e^{e^t}-y_0=f(xy)e^x-xy=F(xy)e^x$0 If$p \in \Omega$suppose$p$is a maximum point of$f$. Then$\frac{\partial f}{\partial x_i}(p) =0$and$\frac{\partial^2 f}{\partial x_i^2}(p) \leq 0 $for every$1\leq i\leq n$thus$\Delta f (p) = \sum_{i=1}^n \frac{\partial^2 f}{\partial x_i^2}(p)\leq 0$1 What I get is $$u(x,t,y) = \mathbf{E} \left[ y(T)e^{-t(T-t)} \right. \left| X(t) = x, Y(t) = y \right],$$ where the processes$X$, and$Yfollow $$dX(t) = rX(t) dt + \sigma X(t) dW(t),\text{ and } dY(t) = h( t,X(t))dt.$$ 2 First of all, we note that by a change of variables \begin{align*} u(x) &= \frac{1}{n \omega_n r^{n-1}} \int_{\partial B(x,r)} u(z) \, dS(z) \\ &= \frac{1}{\sigma_{n-1}} \int_{S_{n-1}} u(x+r \xi) \, dS(\xi) \tag{1} \end{align*} where\sigma_{n-1}$denotes the volume of the ($(n-1)$-dimensional) volume of the sphere$S_{n-1} = \partial ...

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Well the first part: $$\int_{\partial B_p}\frac{\partial u}{\partial \nu}\,ds=\int_{B_p}\Delta u\,dx=0$$ (I will leave out the brackets for ease of typing) follows directly from the divergence theorem (also called Gauss's Theorem), which in a nutshell is as follows for $U\subset\Bbb R^n$, and some vector field $F$: $\int_{U}\nabla\cdot F\,dx=\int_{\partial ... 2 Separation of variables works only for homogeneous boundary conditions. Begin by reducing to homogeneous by subtracting off a solution of the PDE which satisfies the boundary conditions. Namely, let$v(x,y) = u(x,y)-1-x. The new function satisfies $$\begin{split} v_{tt} &= c^2 v_{xx}, \quad 0<x<1, \mathrm{and \;}t>0\\ v(x,0) &= 0,\\ ... 1 It is because the left hand side depends only on x while the right hand side depends only on t. The two quantities are equal no matter what choice of x,t you choose. We deduce that the quantities are constant, because that is precisely how you define a constant function; a function that returns the same value everywhere, regardless of choice of input. 0 The formulas with a and b not only rotate the coordinate system, they also change the scale of the axes by a factor of \sqrt{a^2+b^2}, and reverse the relative orientation of the axes (since the determinant of the coordinate change is negative). If you take$$ x'' = \frac{x'}{\sqrt{a^2+b^2}} = \frac{a}{\sqrt{a^2+b^2}} \, x + \frac{b}{\sqrt{a^2+b^2}} \, ... 0 For the equation \begin{align} 2 u_{x} - 3 u_{y} + u - x = 0 \end{align} it is seen that \begin{align} \frac{dx}{2} = \frac{dy}{-3} = \frac{du}{x-u}. \end{align} Solving for thex-y$equation,$2 dy + 3 dx = 0$, leads to$c_{1} = 3 x + 2 y. Consider the second characteristic from \begin{align} dy = \frac{3 \, du}{u-x}. \end{align} This leads toc_{2} = ...

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Yes, there is a unique analytic solution in a (small) neighbourhood of the $x$-axis.

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The derivative of the (finite) sum is the sum of the derivatives. This is the usual sum rule from Calc I. If the series were infinite you would need to check a few things before interchanging derivative with sum.

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Your equation, if expanded, reads: $$v_t + 2A_x v_x + 2A_y v_y + 2A_z v_z = i \, \mathbf{B}(x) \cdot \mathbf{A} \, v.$$ This is a linear 1st order PDE for which the method of characteristics reads: $$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}x}{2A_x} = \frac{\mathrm{d}y}{2A_y} = \frac{\mathrm{d}z}{2A_z} = \frac{\mathrm{d}u}{i \, \mathbf{B}(x) \cdot ... 1 Sure, you should take try to take the inverse Laplace transform of  f(t) . Here is a sketch. for simplicity, I'll flip the conditions so that  \partial_x u|_{x = 0} = 0  and  \partial_x u_{x = L} = f. Let  \psi_k(x,t) = \frac{-1}{k \sin(k)} \cos(k x) e^{-\lambda_k t}  where  \lambda = A k^2 + B  (mod some negative signs). Then you see that  ... 1 To avoid complications I will assume that u,\partial_tu\ge0. Then we have$$ \frac{\partial u_t}{u+u^\alpha}\le C. $$Let F(u)=\int_0^uds/(s+s^\alpha). Then$$ F(u(t))\le C(t-t_0)+F(u(t_0)),\quad t\ge t_0. $$and$$ u(t)\le F^{-1}\bigl(C(t-t_0)+F(u(t_0))\bigr),\quad t\ge t_0. $$Observe that since \alpha>1 F^{-1} will be defined on a finite ... 1 You have, by the method of characteristics:$$ \frac{dx}{y} = \frac{dy}{-x} = \frac{du}{0}. $$If you solve the first equality, you will find:$$x^2+y^2 = C,$$where C is a constant. On the other hand, the last fraction tells you that du = 0 and therefore u = K, being K another constant. Put K as a function of C to have the general solution ... 1 Maple finds solutions of the ODE (with +) in terms of Kummer U and M functions:$$f \left( x \right) = \left( ax+b \right) ^{-{\frac {\sqrt {{a}^{2 }+4\,k}-a}{2a}}} \left( C_{{1}}\;{\text{KummerU}\left({\frac {\sqrt {{a}^{2 }+4\,k}-a}{2a}},\,{\frac {a+\sqrt {{a}^{2}+4\,k}}{a}},\,{\frac {1}{ \left( ax+b \right) ...

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The conductor have to be a continuous function. By continuity the conductor will assume the same value in the whole B (the conductor can't jump when you go to the boundary). Then it have the same value over B(sorry for the english)

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Note that your second relation is false. The equation $dy/y = dz/z$ implies that $C_2 = y/z$ and hence: $$u = f(C_1,C_2) = f(\log{y} -x , y/z),$$ since the characteristics method also tells us that the fractions are also equal to $du/0$ and hence $u = \text{const}$. You can check that this solution satisfies the PDE taking into account that: $$... 1$$U_{xy}+U_y=e^{-x}$$Interation relatively to y gives : U_x+U=e^{-x}y+f(x) any derivable function f(x) Let U(x,y)=e^{-x}F(x,y)$$U_x+U=e^{-x}F_x=e^{-x}y+f(x)F_x=y+e^xf(x)=y+g(x)$$Integration relatively to x gives : F=xy+G(x)+h(y) any derivable functions G and h$$U(x,y)=e^{-x}\left(xy+G(x)+h(y)\right)$$or ... 3 V_x+ V = {\rm e}^{-x} implies ({\rm e}^{x}V)_x=1. 1 Consider u(x,t)=X(x)T(t). Then u_{xx}=X''(x)T(t) and u_t=X(x)T'(t). Substitute in the original equation to get X(x)T'(t)=17X''(x)T(t). So \dfrac{T'(t)}{T(t)}=17\dfrac{X''(x)}{X(x)}=-\lambda. So this gives T'(t)+\lambda T=0 and X''(x)+\dfrac{\lambda}{17}X=0. Can you proceed from here. I think its trivial now. 1 Folland defines previously in his book the convolution ... with Borel measures And surface measure is an example of a Borel measure, so the aforementioned definition applies. Without having a book with me, I'll venture a guess: it is$$(f*\mu) (x) = \int_{\mathbb R^n} f(x-y)\,d\mu(y)$$Let's look at the one-dimensional case first. Here the surface ... 1 When the integrand in a double integral over a rectangle is of the form f(x)g(y), you are in luck: the integral splits as$$ \int_c^d\int_a^b f(x)g(y)\,dx\,dy = \int_a^b f(x) \,dx \int_c^d g(y) \,dy $$since when performing integration with respect to y, we can move f(x) outside the integral. So, the double integral you have is the product of ... 0 Put v = u_y, so you get$$v_x + v = e^x$$every solution to this is of the form v(x,y)=h(x,y) + p(x,y), where h satisfies the homogeneous equation$$h_x + h = 0 and $p$ is any solution to the original equation. $p(x,y) = \frac12 e^{x}$ is a solution to the original equation (from guessing; multiples of $e^x$ are a good guess as these differentiate to ...

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Put $u=\exp(-x)w$, then your equation becomes $w_{xy}=\exp(2x)$ with the general solution $w(x,y)=f(x)+g(y)+y\exp(2x)/2$, where $f$ and $g$ are arbitrary smooth functions.

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The general solution of $-3U_x+4U_y=0$ obviously is $U(x,y)=F(\frac{x}{3}+\frac{y}{4}) \space$ where $F$ is any derivable function. I don't understand what you mean "incorporate the curve". If you are talking of a boundary condition on the curve $y=x+1$, please, make clear what is the boundary condition. With the condition $U(x,y=x+1)=3x\space$ one ...

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I am a little suspicious of your solution to the homogeneous problem, since $u(x, y) \equiv 0$ is a solution to this problem too (with the boundary conditions) --- is there a reason in particular that this is not the unique solution to your problem? Have you actually computed the $A_n$? As for your harder question: how to solve the inhomogeneous problem: ...

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I'm writing this answer, due to no replies. This would be my approachment, and we can can discuss it if you like. Let's say we have the following PDE: $1\cdot u_x+ 0\cdot u_y=1$ We write $x,y,u$ as functions of $(t,s)$, such that $x=x(t,s), y=y(t,s), u=u(t,s)$. In this form, $t$ is the variable that parametrizes the curve and $s$ is the variable which ...

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I think the principal reason is Laplacian is the simplest second order elliptic operator available. So once one prove something non-trivial for the Laplacian, it is useful in other settings as well by consider a generalized Laplacian over the manifold. It is natural to consider a differential operator of any order in general(especially if one works with ...

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Yes, integration by parts on $\mathbb R^n$ means integrate by parts on $B(0,R)$ and then let $R\to\infty$. Often this is done without writing anything down, when the fact that the boundary term, the integral over $\partial B(0,R)$, tends to zero is taken as "obvious". (The unfortunate fact is that this is often more obvious to the writer than to the ...

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=2$ , letting $x(0)=0$ , we have $x=2t$ $\dfrac{dy}{dt}=-3$ , letting $y(0)=y_0$ , we have $y=y_0-3t=y_0-\dfrac{3x}{2}$ $\dfrac{dU}{dt}=2(x-U)=4t-2U$ , we have $U=2t-1+F(2y_0)e^{-2t}=x-1+F(3x+2y)e^{-x}$ If $U$ is for $U(x,y)$ : $U(x,x^2)=f(x)$ : ...

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$ $\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0x$ $\dfrac{dz}{dt}=z$ , letting $z(0)=z_0$ , we have $z=z_0e^t=z_0x$ $\dfrac{du}{dt}=0$ , letting $u(0)=f(y_0,z_0)$ , we have ...

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There is no need of the additional relationshp to find the general solution of the PDE. Systematic changes of functions and variables leads to the result below :

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