MHB Initial and boundary value problem

[mathmari]

Hey!

I have to solve the following initial and boundary value problem:

$$u_t=u_{xx}, 0<x<L, t>0 (1)$$
$$u_x(0,t)=u_x(L,t)=0, t>0$$
$$u(x,0)=H(x - \frac{L}{2} ), 0<x<L, \text{ where } H(x)=1 \text{ for } x>0 \text{ and } H(x)=0 \text{ for } x<0$$

I have done the following:
Using the method of Separation of Variables, the solution is of the form: $u(x,t)=X(x) \cdot T(t)$
Replacing this at $(1)$, we get the following two problems:
$$\left.\begin{matrix}
X''+\lambda X=0, 0<x<L\\
X'(0)=X'(L)=0
\end{matrix}\right\}(2)$$
and
$$\left.\begin{matrix}
T'+ \lambda T=0, t>0
\end{matrix}\right\}(3)$$
$$u(x,0)=X(x)T(0)=H(x - \frac{L}{2} )$$

Solving the problem $(2)$ we get that the eigenfunctions are:
$$X_n(x)= \left\{\begin{matrix}
1 & , n=0 \\
\cos{(\frac{n \pi x}{L})} & , n=1,2,3, \dots
\end{matrix}\right.$$

and solving the problem $(3)$ we get:
$$T_n(t)= \left\{\begin{matrix}
A_0 & , n=0 \\
e^{-(\frac{n \pi}{L})^2t} & , n=1,2,3, \dots
\end{matrix}\right.$$

So the solution of the initial problem is of the form
$$u(x,t)=A_0 + \sum_{n=1}^{\infty}{ A_n \cos{(\frac{n \pi x}{L})} e^{-(\frac{n \pi}{L})^2t}}$$

$$u(x,0)=H(x - \frac{L}{2} ) \Rightarrow H(x - \frac{L}{2} )=A_0+ \sum_{n=1}^{\infty}{A_n \cos{(\frac{n \pi x}{L})}}$$

We can write the function $H(x - \frac{L}{2} )$ as a Fourier series to calculate the coefficients $A_0$ and $A_n$.

$$H(x - \frac{L}{2} )=\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n \cos{(\frac{n \pi x}{L})}}$$

$$A_0=\frac{a_0}{2}=\frac{1}{2}$$

$$A_n=a_n=\frac{2}{L} \int_0^L{H(x - \frac{L}{2} ) \cos{(\frac{n \pi x}{L})}}dx=\frac{2}{L} \int_{\frac{L}{2}}^L{\cos{(\frac{n \pi x}{L})}}dx=-\frac{2}{n \pi} \sin{(\frac{n \pi}{2})}$$

Is this correct so far?? (Wondering)

Is there a general formula for $\sin{(\frac{n \pi}{2})}$? For even $n$ it's equal to $0$, but for odd $n$ it's equal to $1$ or $-1$.

 

[mathmari]

Is there a general formula for $\sin{(\frac{n \pi}{2})}$? For even $n$ it's equal to $0$, but for odd $n$ it's equal to $1$ or $-1$.

Or should I just let it $\sin{(\frac{n \pi}{2})}$ in the sum? (Wondering)

 

[Ackbach]

Or should I just let it $\sin{(\frac{n \pi}{2})}$ in the sum? (Wondering)

That's probably what I would do. Yes, you could split the sum up into even and odd parts, and then simplify, but I think the final result would be more complicated than what you suggest here.

 

[mathmari]

That's probably what I would do. Yes, you could split the sum up into even and odd parts, and then simplify, but I think the final result would be more complicated than what you suggest here.

Ahaa..Ok! (Smile)
So is the solution of the problem:
$$u(x,t)=\frac{1}{2}-\sum_{n=1}^{\infty}{\frac{2}{n \pi} \sin{(\frac{n \pi}{2})} \cos{(\frac{n \pi x}{L})} e^{-(\frac{n \pi }{L})^2t}}$$ (Wondering)

 

[Ackbach]

It looks right to me. If you plot partial sums, and increase the number of terms included, it starts to look more and more like the step function.

 

[mathmari]

It looks right to me. If you plot partial sums, and increase the number of terms included, it starts to look more and more like the step function.

Ok! Thank you for your answer! (Mmm)