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 $\mathbb R$ ?
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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...
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...
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?
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...
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!