MHB Calculating the coefficients A_n

[mathmari]

Hey!

Having the problem:
$$v_t=v_{xx}, 0<x<1, t>0$$
$$v(0,t)=v_x(1,t)=0, t>0$$
$$v(x,0)=e^x-x$$

I have found that the solution of the problem is of the form
$$v(x,t)=\sum_{n=0}^{\infty}{A_n \sin{(\frac{(2n+1) \pi x}{2})} e^{-(\frac{(2n +1) \pi}{2})^2t}}$$

Using the boundary condition $v(x,0)=e^x-x$, we get the following:
$$\sum_{n=0}^{\infty}{A_n \sin{(\frac{(2n+1) \pi x}{2})}}=e^x-x$$

Can I calculate the coefficients $A_n$ by using the Fourier series of $e^x-x$ although there is 2n+1 in the $\sin$? (Wondering)

 

[Ackbach]

The "Fourier Trick" will still "work" formally.

Spoiler

\begin{align*}
\sum_{n=0}^{\infty}A_n \sin\left(\frac{(2n+1) \pi x}{2}\right)&=e^x-x \\
\sum_{n=0}^{\infty}A_n \sin\left(\frac{(2n+1) \pi x}{2}\right)\sin\left(\frac{(2m+1) \pi x}{2}\right)&=(e^x-x)\sin\left(\frac{(2m+1) \pi x}{2}\right) \\
\sum_{n=0}^{\infty}A_n \int_{0}^{1}\sin\left(\frac{(2n+1) \pi x}{2}\right)\sin\left(\frac{(2m+1) \pi x}{2}\right) dx&=\int_{0}^{1}(e^x-x)\sin\left(\frac{(2m+1) \pi x}{2}\right)dx=:C_m \\
\sum_{n=0}^{\infty}A_n \delta_{mn} \int_{0}^{1}\sin^{2}\left(\frac{(2m+1) \pi x}{2}\right) dx&=C_m \\
A_m \frac{1}{2} &=C_m \\
A_m&=2C_m,
\end{align*}
where
$$C_m=\frac{2\,\left( {\left( \pi + 2\,m\,\pi \right) }^3 +
\left( -4 + \left( -1 + e \right) \,
{\left( \pi + 2\,m\,\pi \right) }^2 \right) \,
\left( 2\,(-1)^m \right)
\right) }{{\left( \pi + 2\,m\,\pi \right) }^2\,
\left( 4 + {\left( \pi + 2\,m\,\pi \right) }^2 \right) }.$$

The only question is, will the $X_n$ functions be a complete set? That is, can you actually write
$$\sum_{n}A_n \sin\left(\frac{(2n+1)\pi x}{2}\right)=e^{x}-x?$$
The answer is not clear to me. The $x$ of $e^{x}-x$ can definitely be taken care of by the sin functions, but the exponential function is neither even nor odd. It's not clear to me how purely odd functions can sum to anything other than a purely odd function...

What it comes down to is this: does the operator defined by the DE and the boundary conditions have a complete set of eigenfunctions? I could be wrong, but I don't think the operator is self-adjoint, so there's no guarantee that the eigenfunctions are complete.

Using Mathematica to plot the sum up versus the function reveals that you have a serious convergence issue when $x\to 0$.