Consider the Cauchy problem of finding $u=u(x,t)$ such that $$\frac{\partial{u}}{\partial{t}}+u\frac{\partial{u}}{\partial{x}}=0\text{ for }x\in\mathbb{R},t&gt;0\\u(x,0)=u_0(x),\;x\;\epsilon\;\mathbb{R}$$  which choice(s) of the following functions for $u_0$ yield a $C^1$ solution $u(x,\ t)$ for all $x\in\mathbb{R}$ and $t&gt;0.$   $u_0(x)=\frac{1}{1+x^2}$ $u_0(x)=x$ $u_0(x)=1+x^2$ $u_0(x)=1+2x$   Because characteristic of the given PDE is $U=U_0(x-ut)$ from option 2, 4 only the satisfies the given PDE is i am right
123562
Consider the Cauchy Problem of finding $u(x,t)$ such that    $$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0,x\in\mathbb{R},t&gt;0$$ $$u(x,0) = u_0(x), x\in\mathbb{R}$$ Which choices of the following functions for   $u_{0}$ yield a $C^{1}$ solution $u(x,t)$ for all   $x\in\mathbb{R},t&gt;0$         $u_{0}(x)=\frac{1}{1+x^{2}}$   $u_{0}(x)=x$   $u_{0}(x)=1+x^{2}$   $u_{0}(x)=1+2x$.      If I use the existence and uniqueness theorem for Cauchy problem i get the corresponding determinant is non zero so all are true according to me. But in answer key only option 2nd and 4th is given. Please help me to solve the problem. Thanks a lot.
9814
Solve $$u_{xx}-3u_{xt}-4u_{tt}=0$$ where $u(x,0)=x^2$ and $u_t(x,0)=e^x$.  My workings so far: I have factored the differential equation in the following way: $$(\delta_x-4\delta_t)(\delta_x+\delta_t)=0$$ where $\delta_x=\frac{\delta}{\delta x}$ etc. Now if we let $v$ be the solution to $(\delta_x+\delta_t)u$ then we have the following two equations: \begin{eqnarray*} (\delta_x+\delta_t)u=u_x+u_t=v\\ (\delta_x-4\delta_t)v=v_x-4v_t=0 \end{eqnarray*} Now for $v$ we simply find $$v=h(t+4x)$$ where $h$ is an arbitrary function of one variable. Now what remains is find $u$ such that $$u_x+u_t=h(t+4x)$$ I am stuck here, I thought about making a change of variables $\zeta=x+t$ and $\eta=x-t$ and thus using the product rule to show that $u_x=u_{\zeta}+u_{\eta}$ and $u_t=u_{\zeta}-u_{\eta}$ and thus $u_x+u_t=2u_{\zeta}$ and we need to solve $$u_{\zeta}=h(t+4x)$$ (I left out the factor 2 because $h$ is an arbitrary function). Do I simply integrate now and conclude $$u=f(\zeta)h(t+4x)+g(\eta)=f(x+t)h(t+4x)+g(x-t)$$ This seems wrong to me...  Some help would be greatly appreciated!
27106
The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback!
73964
I couldn't solve this problem, can you help me please?  The Burgers' equation   $$ u_y + uu_x = 0 $$ $ - \infty &lt; x &lt; \infty $ , $ y &gt; 0 $ , $ u(x,0)=f(x) $  My question;  is there any solution $ f(x)= -x $ for all $ y&gt;0 $ , if not, why?  Thank you for your help.
38192
I am trying to prove that, if $f:\mathbb{R}\rightarrow \mathbb{R}$ and $F:\mathbb{R}^2\rightarrow\mathbb{R}$ are differentiable, $F$ satisfying $F(x,f(x))=0$ and $\partial{F}/\partial{y}\ne 0$ then $$f'(x)=-\frac{\partial{F}/\partial{x}}{\partial{F}/\partial{y}}$$  This appears in a section on the chain rule for many-variables functions, but without knowing the details of $F$ I find it very hard to prove much about it.  To the best of my ability to reason about it, I suppose I could say  $$ \frac{\partial{F}}{\partial{x}}=\frac{\partial{F}}{\partial{x}}\frac{\partial{x}}{\partial{x}} + \frac{\partial{F}}{\partial{y}}\frac{\partial{y}}{\partial{x}}$$  but for more than one reason I doubt that what I have written is correct.  Yet I'm not exactly sure what's wrong about it, nor can I guess what else I could do.  I tried finding the partial with respect to $y$ but got nothing more useful looking than this.  I also tried writing out limit definitions and saw no useful path forward through that.
63709
This is a reworking of a previous question  which was marked as a duplicate. Some nice folks have referred me to solutions to similar problems. I still have a couple of questions, since one of the solutions used the Poisson kernel, which is not the desired solution in the text, because the Poisson kernel has not been introduced yet. The other solutions were for problems that were similar, but whose solutions were only partially applicable, as I understand them.  I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:     If $u$ is harmonic and bounded in $0 &lt; |z| &lt; \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|&lt;\rho$ when $u(0)$ is properly defined.   My Solution So Far:   Here's what I've got for a non-Poisson kernel solution: We want to find a harmonic conjugate for $u$ in the punctured disk, and then we can apply big Picard to show that $0$ is a removable singularity. The problem is that the punctured disc is not simply connected.   If we take $\sigma = (u \circ \exp)(z): L \to \mathbb{C}$, where $L$ is the left half-plane, now we have $\sigma$ harmonic, and defined in a simply connected set $L$, so it has a harmonic conjugate $\tau$. Now we wish to take $v = (\tau \circ \log):\{0 &lt; |z| &lt; \rho\}\to \mathbb{C}$, but in order to do so we need $\tau$ to be $2\pi i$-periodic. This is where my proof breaks down.  One of the similar problems which readers referred me to used the following strategy: note that since $\sigma$ is $2\pi i$-periodic, and $\sigma, \tau$ satisfy the Cauchy-Riemann equations, we have that $\nabla\tau(z) = \nabla\tau(z + 2 \pi i)$ for all $z\in L$, or in other words, $\tau(z) - \tau(z+2\pi i)$ is a constant function, say, equal to $\alpha \in \mathbb{R}$. Then, the solution suggests, we can set $$g(z) = u + iv - \frac{c}{2\pi}z,$$  in order to make $g$ $2\pi i$-periodic, and then we can use $(g\circ \log)$ on $\{0 &lt; |z| &lt; \rho\}$. But the reason why this won't work for my purposes is: now we have changed $u$! On the other hand, we can't simply subtract $\frac{c}{2\pi}\Im{(z)}$, and try to leave $u$ untouched, because now $g$ is not analytic. How can we get around this?  Another strategy entirely might be to try to use the mean-value property. Ahlfors discusses that a lot in the preceding section, and his main example is the punctured disc. Since $u$ is harmonic, we know that the average over circles centered at zero is a constant...
46018
I was toying with equations of the type $f(x+\alpha)=f&#39;(x)$ where $f$ is a real function. For example if $\alpha=\frac{\pi}{2}$ then the solutions include the function $f_{\lambda,\mu}(x)=\lambda cos(x+\mu)$. Are there more solutions?  On the other hand, if I want to solve the equation for any $\alpha$, I can assume a solution of the form $f(x)=e^{\lambda x}$, and find $\lambda$ as a complex number that enables me to solve the equation...   I was wondering: is the set of the solutions of dimension 2 because the derivative operator creates one dimension and the operator $\phi: f(x)\mapsto f(x+1)$ adds another one? Is there some litterature about this kind of equations?  Please satisfy my curiosity if you can... Thanks!
62425
I was thinking about the following problem : Find out which of the following option(s) is/are correct? The differential equation $xu_{x}+yu_{y}=2u$ satisfying the initial condition $y=xg(x),u=f(x)$, with  (a)$f(x)=2x,g(x)=1$ has no solution, (b)$f(x)=2x^2,g(x)=1$ has infinite number of solutions, (c)$f(x)=x^3,g(x)=x$ has a unique solution, (d)$f(x)=x^4,g(x)=x$ has a unique solution.  My Attempt: Using lagrange's method, we see that $\frac{dx}{x}=\frac{dy}{y}=\frac{du}{2u} $ gives $y/x=c_1,$ and $y=\sqrt uc_2$ ,where $c_1,c_2$ being constants. Now we have to apply to the given initial conditions  and  check the given options.Here ,I observe that $y=xg(x)=x.1=x$,[as $g(x)=1$] and $u=f(x)=2x^2$But,this relation does not satisfy $xu_{x}+yu_{y}=2u$ and so the choice&quot;(a)$f(x)=2x,g(x)=1$ has no solution,&quot; is right. Now I am stuck here and could not progress further.Am i going in the right direction? Please help.Thanks in advance for your time.
48245
Solve the equation $xU_x+yU_y=0$.   From Partial Differential Equations: An Introduction, 2nd Ed. (Strauss, pg. 10). There are no boundary conditions.  Solution: The PDE can be written $(x,y)\cdot\nabla U(x,y)=0$, implying that the vector $(x,y)$ is tangent to the PDE's characteristic equation. Then I can write $dy/dx=y/x$, and to solve this ODE,  $$\frac{dy}{dx}=\frac yx \\ \ln y=\ln x+c_1 \\ y=cx \\ y/x=c$$  Hence, the PDE is dependent on $c$ and I can write $U(x,y)=f(y/x)$ for some function $f$, solving the problem.  Question: Is this solution correct? The hint for the question (supplied by the professor) said I should use a substitution of variables, but this seems reasonably valid.
117539
It is a problem in Evan's PDE. I want to prove the smooth solution of the following PDE is zero: $$u_{tt}-du_t-u_{xx}=0, (0,1)\times(0,T) $$$u|_{x=0}=u|_{x=1}=u|_{t=0}=u_t|_{t=0}=0$.    The hint is to use the energy $\frac{1}{2}(||\partial_tu||_{L^2[0,1]}+||\partial_xu||_{L^2[0,1]})$, but when I differentiate the energy, I can only get the first and third term of PDE
51137
The problem is from Evans PDE 2.5 problem 18 How to solve the equations?  Assume u solves the initial-value problem  $u_{tt}-\Delta u=0$ in $R^n \times (0,\infty)$  $u=0$, $u_t=h$ on $R^n\times \{t=0\}$  Show that $v:=u_t$ solves  $v_{tt}-\Delta v=0$ in $R^n \times (0,\infty)$  $v=h$, $\ v_t=0$ on $R^n\times \{t=0\}$  (This is Stokes' rule)
122611
Finding all differentiable $f(z) = u(x) + iv(y)$ in $\mathbb{C}$ where $u(x),v(y)$ are real valued functions.   I’m not sure what to do. Would $f$ be differentiable simply if and only if both $u$ and $v$ were differentiable?   My thinking is that if the limit $$ \lim_{h \to 0}\frac{f(z+h)-f(z)}{h} $$  exists then it would be equal to  $$ \lim_{h \to 0}\frac{u(x+h)-u(x)}{h} + i\lim_{h \to 0}\frac{v(y+h)-v(y)}{h} $$  and therefore would exist if they existed. But that doesn’t seem quite right.  I also thought about using Cauchy-Riemann equations, but seeing that I’m trying to find ones that are differentiable rather than those that aren’t, I thought they wouldn’t be of much help.
73393
For $u:\mathbb{R}\times [0,1]$ with boundary conditions $u(0,x)=\cos (2\pi x)$ and $u(t,0)=u(t,1)$.   Solve $u_t=u_{xx}+u_x$.  I had this on an exam and tried to write $u$ as a product of two single variate functions and convert to an ode using the usual methods but things got messy and I couldn't finish. I also tried to write $u$ as a Fourier series in $x$ with coefficients $a_k(t)$ depending on $t$, but this also didn't seem to yield anything.  Is there a more clever way to approach this, possibly with Fourier series? I heard from a friend there may be a Fourier transform method since the equation was "homogeneous in momentum space."   Thanks!
63081
What is the general solution to the differential equation:  $$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial x}$$  I'm a little stuck because all the techniques I know are unable to solve it.
120456
We have an open set $U$ in $\mathbb{R}^2$ and two functions $u, v \in C^1 (U)$. $u$ and $v$ satisfy the Cauchy-Riemann equations in $U$, i.e., $u_x = v_y$ and $u_y = -v_x$. Also, $u^2 + v^2 \neq 0$ for all $(x,y) \in U$. Then, we have to show that $$\frac{uu_x + vv_x}{u^2 + v^2}$$ is harmonic in $U$, i.e., the Laplacian of this function is zero.  On the face of it, it seems there is nothing to do but calculate the second partial derivatives and verify the result. But that would be very messy and complicated. Is there something easier that I am missing?  Note: $u_x$ means the partial derivative of $u$ with respect to $x$, etc.
35399
Let $\Omega $ be a smooth bounded subset of $\mathbb R^d$. Assume that $u=0$ if $u\in H^1(\Omega )$ and $u$ satisfies $$-\Delta u-u=0\ \ in\ \ \Omega \quad \text{and}\quad \partial _\nu u=u=0\ \ on \ \ \partial \Omega .$$  Let $f\in L^2(\Omega )$ and assume that $v$ is solution of $-\Delta v-v=f$ in $\Omega $. Prove that   $$\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{-1/2}(\partial \Omega )})),$$ where $C$ is independent of $v$ and $f$.  My attempts  My attempts are unfortunately very few... I'm stuck on this problem since yesterday. I tried to make the substitution $w=v-u$, then $w$ is a solution of $\nabla v-v=f$, but it doesn't look to help a lot.
76345
Let $f(x,y)$ be a function of two variables. Given that  $$\frac{\partial f}{\partial x}=0$$ for all $y$, and  $$\frac{\partial f}{\partial y}=0$$ for all $x$, is there a theorem that states $f(x,y)=$ constant for all $x$, $y$? Or how would I prove this rigorously?