Solve the partial differential equation $u_{t} = \alpha^{2}u_{xx}$ with conditions: \begin{cases} u(x,0) = f(x)\\ u(0,t) = u_{x}(L,t) = 0 \end{cases} where $$f(x) = \begin{cases} \frac{2}{L}x,& x \in \left[0,\frac{L}{2}\right) \\ 1,& x \in \left[\frac{L}{2},L\right)\end{cases}.$$ You must justify the convergence of the series, and justify aspects of odd and even extensions if necessary.
My attempt.
Step 1. Separation of variables:
Write $u(x,t) = F(x)G(t)$. Thus, $F(x)G'(t) = \alpha^{2}F''(x)G(t)$, i.e, $$\frac{F''(x)}{F(x)} = y = \frac{1}{\alpha^{2}}\frac{G'(t)}{G(t)}$$ where the parameter $y$ no depends of $x$ and $t$. We have two ODE's: $$F''(x) - yF(x) = 0\tag{1}$$ and $$G'(t) - y\alpha^{2}G(t) = 0\tag{2}$$
Step 2: Bound conditions.
we have
- If $y = 0$, $F \equiv 0$,
- If $y > 0$, $F \equiv 0$,
- If $y < 0$, take $y = -\lambda^{2}$ then $F_{n}(x) = c_{2}\sin\left(\frac{n\pi x}{2L}\right)$
where the solution for $y > 0$ is $F(x) = c_{1}\cos(\lambda x) + c_{2}\sin(\lambda x)$. The solution for $G'(t) - y\alpha^{2}G(t) = 0$ is $G(t) = ce^{-y\alpha^{2}t}$. So, for $n = \pm 1, \pm 2,\dots$, we have $$u_{n}(x,t) = Ke^{-\frac{n\pi \alpha}{2L}t}\sin\left(\frac{n\pi x}{2L}\right).$$
Step 3. Find Fourier Series of $f$.
I tried to get an odd or even extension of $f$ and so, to find the Fourier Series of the extension (maybe is not possible to get a $\sin$ or $\cos$ Fourier Series). But I could not find a periodical extension of $f$. Precisely, I want to get a Fourier Series of the form $$\sum_{1}^{\infty}c_{n}\sin\left(\frac{n\pi x}{2L}\right).$$ With this, $$u(x,t) = K\sum_{1}^{\infty}c_{n}e^{-\frac{n\pi \alpha}{2L}t}\sin\left(\frac{n\pi x}{2L}\right)$$ is a solution.
Can someone help me?
Your eigenfunction is almost, but not quite correct. You need
$$ F_n'(L) = \frac{n\pi}{2L}\cos\left(\frac{n\pi}{2}\right) = 0 $$
This is only valid if $n$ is odd, so $n=1,3,5,\dots$
Furthermore, since $\sin$ is odd, you'll want to have an odd periodic extension of $f(x)$, i.e. $f(-x)_{x\in(-L,0)} = -f(x)_{x\in(0,L)}$ and so on. This gives a period of $2L$.
Using the definition, we have
$$ c_n = \frac{\int_{-L}^L f(x)\sin \left(\frac{n\pi}{2L}x\right) dx}{\int_{-L}^L\sin^2 \left(\frac{n\pi}{2L}x\right)dx} = \frac{2}{L}\int_0^L f(x) \sin \left(\frac{n\pi}{2L}x\right) dx $$
You can integrate piecewise, i.e.
$$ c_n = \frac{2}{L} \left(\int_0^{L/2} \frac{2}{L}x \sin \left(\frac{n\pi}{2L}x\right) \ dx + \int_{L/2}^L \sin \left(\frac{n\pi}{2L}x\right) dx \right) $$
for odd $n$. I'll leave this to you.