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You are being asked to show that if $u_{tt} - c^2u_{xx} = 0$, then function $$v(t,x) = u\left(\frac{t}{c^2t^2 - x^2}, \frac{x}{c^2 t^2 - x^2}\right)$$ has $v_{tt} - c^2 v_{xx} = 0$. You will need to use the chain rule a few times in order show that $v$ solves the wave equation.

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Apply the (symmetry) transformation to a solution and show that the transformed solution solves the PDE too. $$L u = 0 \Rightarrow L T u = 0$$

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$G$ itself is a function, but the derivative $\Delta_y$ is taken in the sense of distributions, so the resulting object $\Delta_y G(x,y)$ need not be a function. (A simpler example: the Heaviside function $H(x)$ is a function, and its derivative in the ordinary sense is zero for all $x\neq 0$ and undefined at the origin. But one can instead interpret the ...

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The whole exercise is connected to $L^1$-$L^\infty$ duality and the missing reflexivity of $L^1$. For the first part, choose e.g. $n=1$, $U=\Bbb{R}$ and $V= (-1,1)$ as well as $u=\chi_{(0,\infty)}$, i.e. the indicator function of the positive reals. We then have (for $h>0$, the argument for $h<0$ is analogous) $$D_1^h u(x)= \begin{cases} 0, & ... 1 Let's assume that x_0 is the largest root of 1+x-x^{1+b}=-0.5x^{1+b}. In particular, 1+x-x^{1+b}\ge-0.5x^{1+b} does not hold if x>x_0>0. If Q(0)\leq x_0, then Q might increases, however, once it reaches x_0, then it must decrease since the right-hand side of the above differential inequality becomes negative. So we may assume that ... 0 Why do you think the expression implies such a lack of independence? Intuitively, the Markov property allows expectations conditional on randomness at a single point in time, t, to be replaced by expectations conditional on all of the randomness at points in time up to and including time t. Thats why \mathfrak{F}_t, which includes all the ... 1 That's not quite right. The general solution to the ODE y'' + y^3 = 0 is y = c \; \text{sn}(c(t - t_0)/\sqrt{2}, i). Since this is a nonlinear equation, you can't take these as a "fundamental set": other solutions will not be linear combinations of these. EDIT: For example, there will be radially symmetric solutions y = f(r) which (in 3D) are ... 0 The integration over such an open interval is defined by taking the limits for the approaches to the boundary. (Since the boundary itself has zero measure what happens there is irrelevant as long as the limits do exist.)$$\int_{(0,1)} f(x)\operatorname d x = \lim_{0<h\to 0}\;\lim_{1>k\to 1}\;\int_h^k f(x)\operatorname d x$$You don't need to ... 0 Here's the real and proper definition of a Lipschitz domain. See the local coordinate as a chage of variable in \mathbf{R}^d. 1 You did a mistake when computing your expression. The variable x multiplies f_{u}(v,u), so that the correct expression is $$f_u(v,u) x (2 a -1)+f_v (v,u)= y$$ Now expressing x and y as function of v, u: $$f_u(v,u) v (2 a -1)+f_v (v,u)= u - a v^2$$ if you set a=\frac{1}{2} you get: ... 1$$\begin{align} u_x + u_y + u &= e^{x + 2y} \\ \implies u_x + u_y &= e^{x + 2y} - u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)\\ \end{align} $$Setting u = u(x(s),y(s)) we find$$\begin{align} \frac{d}{ds} u &= \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial y} \cdot \frac{dy}{ds} \\ &= ...

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You are right, there are not many books on MCF in a Riemannian manifolds. There are basically two reasons: (1) Not much is known in the general case. One early result was given by G. Huisken, which says that if the curvature of the ambient space is pinched and the initial embedding is convex, then the MCF shrinks to a round point (This is a generalization ...

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I'll denote the Laplace transforms of variables by the corresponding capital letter: for instance, the Laplace transform of $g\left(x,t\right)$ is $G\left(x,s\right)$. The Laplace transform of the system of PDEs is: $$(1')\space\space sG+v\frac{\partial G}{\partial x}=-k_1\left(G-H\right)$$ $$(2')\space\space sH = k_2\left(G-H\right)$$ Solve (2') for $H$: ...

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For part b) we take $u=\sqrt{a}v$ and calculate $$\Delta u = \nabla\cdot(\nabla(\sqrt{a}v)) = \nabla \cdot \left(\frac{v\nabla a + 2 a\nabla v}{2\sqrt{a}}\right) = \\ \frac{1}{2\sqrt{a}}\nabla \cdot \left(v\nabla a \right) + \frac{1}{2\sqrt{a}}\nabla \cdot \left(2a\nabla v \right) + (v\nabla a + 2 a\nabla v)\cdot \nabla\left(\frac{1}{2\sqrt{a}}\right)$$ ...

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It should probably be $\frac{\partial^2f}{\partial x\partial y}(0,0)$ and $\frac{\partial^2f}{\partial y\partial x}(0,0)$. You should check your book (or other material or instructor) for notation conventions as they vary from source to source. In some contexts derivatives with $d$, $\partial$ and $D$ mean the same thing, sometimes not. For example, you ...

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Yes, it would be correct to say that an elliptic boundary value problem always has an underlying elliptic PDE. However, I should warn you that the Wikipedia article Elliptic partial differential equation considers only second-order linear equations in nondivergence form. Ellipticity is defined differently for divergence and non-divergence type linear ...

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It seems you might be working with Evans' book on PDE, and if you've never encountered multivariable calculus before, it might be a good idea to start with that first. Take a simple example: $f(x,y) : \mathbb{R}^2 \rightarrow \mathbb{R}$. Here, $N = 2$. Then $Df = (f_x(x,y), f_y(x,y))$, where $f_x := \frac{\partial f}{\partial x}$, and ditto for $f_y$. ...

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This question need the change variables formula, the degree is actually the cardinal of $u^{-1}(x_0)$, so we have to discuss whether $Du$ is invertible near each of this point. Then change variable to $y=u(x)$.

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We have \begin{align} w^2 Dv &= w^2 D(u/w) \\ &= w^2 \left(\frac{w Du - u Dw}{w^2} \right) \\ &=wDu-uDw. \end{align} So \begin{align} D \cdot (w^2 Dv) &= D \cdot (w Du-uDw) \\ &= w \Delta u-u \Delta w \\ &= w(cu)-u(cw) \\ &=0. \end{align} Thus, $-\operatorname{div}(w^2 Dv)=0$.

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A Dirichlet condition is the prescription of the values of the solution at the boundary. A zero Dirichlet condition means the solution needs to be $0$ there. So you are asked to find the $\theta$ so that for this choice of $\theta$, you have $u(0,t)= 0$ and $u(l,t)=0$. Practically speaking, you set $u(0,t)= 0$, with the function $u$ you have, and solve ...

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For the first part: $L^2(\Omega)\subset L^1_{loc}(\Omega)$, hence indeed you can apply your previous exercise. Then again, what is $(C^{\infty}_0)^\bot$? It is merely the set $\{f\in L^2(\Omega) : \, \forall \phi\in C^{\infty}_0\int_\Omega f \phi =0\}$, which results in $(C^{\infty}_0)^\bot=\{0\}$. Finally, the result "orthogonal complement of the subspace ...

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I believe you can find a brief literature review and general proof in a delightful paper by McKean and Singer: "Curvature and the Eigenvalues of the Laplacian" by H.P. McKean and I.M. Singer. Journal of Differential Geometry 1 (1967) 43-69. They note (as you observed) that Kac conjectured but did not prove the estimate $$Z(t) = ... 1 The first hint states that the temperature does not change in steady state. In symbols, u_t=0. Thus, the equation u_t=ku_{xx} becomes u_{xx}=0. This is an ODE that we can solve by simply integrating twice to get u(x) = ax+b for some constants a and b. Now, as the boundaries are insulated, we must have u_x(0)=u_x(\ell)=0. Since u_x=a, we ... 1 Hint: Employ the identity$$\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B),$$within c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x). 0 While U^0_m = U_0(mh) is a natural choice for the discrete initial condition it isn't the only one. There are situations in which other discrete initial conditions might be better and the definition you gave takes that into account. Consider e.g. the case that U_0 is highly oscillating, compared to your grid size: U^0_m = U_0(mh) would reflect the ... 0 Euler equations is a term for a class of PDEs from fluid dynamics. Their prominent common feature is absence of viscosity. The PDEs within this class can be classified further as: Incompressible fluid (constant density, zero divergence of velocity) or compressible fluid (all others) Fluid with free surface (think of water waves splashing) or without ... 4 For K \subset \mathbb{R}^n compact, consider the space$$\mathcal{D}_K := \{ \varphi \in \mathcal{D} : \operatorname{supp} \varphi \subset K\}$$endowed with the seminorms$$\lVert\varphi\rVert_k = \sup \{ \lvert D^k\varphi(x)\rvert : x \in \mathbb{R}^n\}$$for k \in \mathbb{N}^n. It is straightforward to show that \mathcal{D}_K is then a Fréchet ... 2 The function u^-(x,y) = f(bx-ay) solves au_x+bu_y=0, while u^+(x,y) = f(bx+ay) doesn't, for the same reason that x^+=3 solves 4x-12=0, while x^-=-3 doesn't - it's just a matter of plugging in. In the equation 4x-12=0, the symbol x represents a number. Any solution to the equation is some number with the property that, if you plug that ... 0 I think it will be much easier for you to assume:$$ p(x,y,t) = \frac{1}{t}\phi(\xi), \\ \xi=\frac{1}{t}(x^2+y^2) $$so that  \xi  is not squared. Then it is working out much nicer for me. EDIT: Here is how I did it. Assuming the above, one finds:$$ \frac{\partial p}{\partial t} = - \frac{\phi }{t^2} + \frac{\xi^2}{t} \phi' \\ \frac{\partial ...

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Both uniqueness and finite propagation speed follow by an energy method just like in the one-dimensional case. For the sake of simplicity the following is for $c=1$, the same argument works for arbitrary $c>0$. Fix some point $x_0 \in \mathbb{R}^n$ and time $t_0 > 0$, and define $$E(t) = \int_{B(x_0, t_0-t)} \left( u_t(x,t)^2 + |\nabla u (x,t)|^2 ... 1 I think you need to at least combine the result of existence & uniqueness with the result of regularity. For existence & uniqueness w.r.t a solution u\in H_0^1(\Omega), I would suggest the first existence theorem in Evans book, chapter 6.2, look for Lax-Milgram. But I think you may need to assume in addition that 0<\alpha\leq \gamma, that is, ... 2 Plug u(x,t)=T(t)\,X(x) into the heat equation. Obtain an equation where on the left hand side you have T and T' and on the right hand side X and X'. If a function that depends only on t is equal to a function that depends only on x, what type of functions can they be? Obtain a second order ordinary differential linear equation for X and ... 0 For the homogeneous equations, 3 and 4, convert any one of them to laplace space, or simply write 3 or 4 in laplace space. then use separation of variable, to obtain two separate equations in x and y. multiply the new equations by x and y respectively to put them in a form solvable using the confluent hypergeometric function (Kummer function of the first ... 1 I would be very surprised if this would not be in Gilbarg/Trudinger "Elliptic Partial Differential Equations of Second Order" - I would start looking in Chapter 6 "Classical Solutions" (in fact they distinguish between "classical solutions" that are C^2 and "strong solutions" that are W^{2,p}). An example added by another user: The following result is ... 3 Symmetry dictates that the solution should satisfy u(a/2,y)=V_0/2 for 0<y<b. Thus, we first, consider the problem$$\Delta u=0, \: u(0,y)=u(x,0)=u(x,b)=0, \text{ and } u(a/2,y)=V_0/2.$$This problem can be solved by separation of variables and Fourier series; we obtain$$ u(x,y) = \frac{1}{2} V_0 \sum _{n=0}^{\infty } \frac{4 \, ...

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It's a 1-D Fokker-Planck equation with a Diffusion proportional to $x^2$ and $0$ drift term. You can equally consider it as an imaginary time 1-D Schroedinger equation for a unit-charged particle coupled to an classical vector potential $A =\text{contstant} \times x$, choosing the Coulomb gauge. By the way, this problem has a well known analytical ...

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I'm 11 months late, so I don't know whether or not you care about this anymore, but your argument for the first and the last case seems legit. Only for the second case where $a=1$, note that the nonlinear equation $x'=x^{a}$ reduces to the simple linear equation $x'=x$, and as I'm sure you know, $e^{t}$ is the unique solution to this equation.(Note that to ...

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This is a very good question, and the solution is actually might be much harder then you expect. Actually you could even relax the condition of $f$ to be bounded and locally Holder continuous (with exponent $\alpha\leq 1$) and you will have $u\in C^2(\Omega)$ and $\Delta u=f$ in $\Omega$. I am not going to write the solution complete here but just refer to ...

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The first Laplacian you mention (sometimes called the Laplace-Beltrami operator) acts on scalar functions, that is, functions $S^2 \to \mathbb{R}$. The de Rham (a.k.a. Hodge) Laplacian acts on differential forms. In particular, the de Rham Laplacian acts on zero-forms, which are precisely scalar functions $S^2 \to \mathbb{R}$, on which it agrees with the ...

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Here is answer to your Question .Complete course of PDE are available there http://nptel.ac.in/courses.php THese lectures are there in youtube channel 'nptel" but contents and syllabus can be seen from link above

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Kahn Academy is very good with teaching Differential Equations. You should check it out: Kahn Academy

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The best way to learn PDE by yourself, I would say buy Evans book, and read chapter 2,5,6. It contains everything you mentioned in your post and presented in a very nice way. It won't be a easy task to learn elliptic PDEs by your own, but if you meet problems or difficulties, you are welcome to post it here and we will do our best to help you. :)

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The integral is finite for all $\alpha\in\mathbb{R}$. Observe that although $\log r$ is unbounded near $r=0$, we always have $\lim_{r\to0^+}|\log r|^{2\alpha}r=0$.

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It is sufficient to require $u\in W^{1,\infty}(\Omega)\cap H^1_0(\Omega)$. Suppose this. Take $v\in H^{-1}(\Omega)$ and $w\in H^1(\Omega)$. The notation $uv$ is kind of sloppy, I guess you meant something like $$(uv)(w) := v(uw) \quad w\in H^1_0(\Omega).$$ Then $$|uv(w)| = |v(uw)| \le \|v\|_{H^{-1}} \|uw\|_{H^1(\Omega)} \le \|v\|_{H^{-1}} ( ... 0 Evan's hint: For all v\in H$$ \int_0^T (v,u_k)\leq C||v|| T. $$Since L^2(0,T;H) is Hilbert, the assumption u_k \rightarrow u weakly in L^2(0,T;H) reads $$\int_0^T (v,u(t)) dt=\lim_{k\to\infty}\int_0^T(v,u_k(t)), \,\,\forall v\in L^2(0,T;H)$$ Consider the particular case, v\in L^2(0,T;H) as v= w with w\in H ... 0 We may consider some simple versions first, for example assume u is smooth with compact support, and$$Lu=-\sum_{ij=1}^n (a^{ij}u_i)_j Then use integration by part, now there is no boundary terms: \begin{align*} (Lu, \Delta u) &=\int_{U}-\sum_{ij=1}^n (a^{ij}u_i)_j \Delta u dx \\ &= \sum_{ij=1}^{n}\int_{U} D(a^{ij}u_i)\cdot (Du_j) dx \\ &= ...

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