Recent content by Markov2

Solve $u_x+2u_y+2u=0,$ $x,y\in\mathbb R$ where $u(x,y)=F(x,y)$ in the curve $y=x.$ I don't know what does mean with the $y=x.$ Well I set up the following $\dfrac{dx}{1}=\dfrac{dy}{2}=\dfrac{du}{-2} ,$ is that correct? but I don't know what's next. Thanks for the help!

Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$ a) Study the pointwise convergence of the Fourier series for $f.$ b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$ c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ? ------------- First I need to...

Let $f:\mathbb R\to\mathbb R$ be a continuous function of period $2\pi.$ Prove that if $\displaystyle\int_0^{2\pi}f(x)\cos(nx)\,dx=0$ for $n=0,1,\ldots$ and $\displaystyle\int_0^{2\pi}f(x)\sin(nx)\,dx=0$ for $n=1,2,\ldots,$ then $f(x)=0$ for all $x\in\mathbb R.$ I know this has to do with the...

Oh yes, that now makes sense!

Sorry for the delay of the reply, but those books are online? Can you give the links if so? Thanks!

I need to use the Fourier transform to solve the wave equation: $\begin{aligned} & {{u}_{tt}}={{c}^{2}}{{u}_{xx}},\text{ }x\in \mathbb{R},\text{ }t>0, \\ & u(x,0)=f(x), \\ & {{u}_{t}}(x,0)=g(x). \end{aligned} $ So I have $\dfrac{{{\partial }^{2}}F(u)}{\partial...

If I use $u(x,s)$ at $s=0,$ then (2) will give me problems with the third term. :(

Okay but, do I need to find the inverse now for (2) and that's all?

Given $\begin{aligned} & {{u}_{t}}={{u}_{xx}},\text{ }x>0,\text{ }t>0 \\ & u(x,0)={{u}_{0}}, \\ & {{u}_{x}}(0,t)=u(0,t). \end{aligned} $ I need to apply the Laplace transform to solve it. I'll denote $u(x,s)=\mathcal L(u(x,\cdot))(s),$ so for the first line I have $s\cdot...

Okay I get that, since I have $u(x,t)=\dfrac12\displaystyle\int_{x-c}^{x+c}g(s)\,ds,$ so the solution equals $u(x,t)=\displaystyle\frac{1}{2}\int_{x-t}^{x+t}{\left( H(s+1)-H(s-1) \right)\,ds},$ is that what you mean?

I still don't get it very well, how to do it with $H(x+1)$ for example? Thanks a lot!

Jester, we can actually use the direct results by using separation of variables. I know the solution is a series where the coefficients must be found, but I actually want to know is if I need to pick my $f$ on one interval and then pick it from the other interval, which means that I'd have to...

I managed to get it. I was drowning myself on a glass of water, I hadn't checked my notes. Thanks for the help anyway!

Okay, it looks a bit messy from there. I homogenized first the equation, so I let $v(x,t)=u(x,t)+\dfrac{A{{e}^{-x}}}{{{c}^{2}}}$ so that $u(x,t)=v(x,t)-\dfrac A{c^2}e^{-x}$ then $u_{tt}=v_{tt}$ and $u_x=v_x+\dfrac A{c^2}e^{-x}$ and $u_{xx}=v_{xx}-\dfrac A{c^2}e^{-x}$ so replacing this to the ODE...

Thank you Jester! I'm sorry but I'm a bit lost on how applying D'lembert's formula, do I need to apply it for $H(x-1)$ and $H(x-2)$ ? How to do so? Much appreciated!