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How can I use the following initial condition for a partial differential equation describing heat diffusion?

$$f(x) = \begin{cases} 0, & 0<x<0.45 \\ 1, & 0.45<x<0.55 \\ 0, & 0.55<x<1 \end{cases}$$

I have a heat equation $u_{t} = \beta u_{xx}$ with boundary conditions $u(0,t)=u(L,t)=0$, where $L$ denotes the right end of the space coordinates. The space coordinates range from $0$ to $L$. With separation of variables these conditions lead us to set $X(0)=X(L)=0$ in order to solve for $X(x)$. Putting together the solution for $T(t)$, the general solution for the heat equation is:

$$u(x,t) = e^{\lambda\beta t}c_2 sin\left(\dfrac{n \pi x}{L}\right) $$

where $\lambda=-(n\pi/L)^2$, where $n$ can be any positive integer.

Once again, my initial condition is the following:

$$f(x) = \begin{cases} 0, & 0<x<0.45 \\ 1, & 0.45<x<0.55 \\ 0, & 0.55<x<1 \end{cases}$$

The function $f(x)$ can be represented by the following Fourier series:

$$f(x) \simeq \dfrac{1}{20} + \sum_{n=1}^{\infty}\dfrac{1}{n\pi}a_n \text{cos}\left(\dfrac{n \pi x}{L}\right) + \sum_{n=1}^{\infty}\dfrac{1}{n\pi}b_n \text{sin}\left(\dfrac{n \pi x}{L}\right)$$

where $a_n = \dfrac{1}{n\pi}\left[\text{sin}\left(n\pi 0.55\right)-\text{sin}\left(n\pi 0.45\right) \right]$ and $b_n = \dfrac{1}{n\pi}\left[\text{cos}\left(n\pi 0.45\right)-\text{cos}\left(n\pi 0.55\right) \right]$ for all $n=1,2,3,...$. The constant is $a_0/2=1/20$. If I take this Fourier representation of $f(x)$ and plug it on the general solution I get the following expression.

$$u(x,t) = \dfrac{1}{20} + \sum_{n=1}^{\infty}\dfrac{e^{-(n\pi/L)^2\beta t}}{n\pi}a_n \text{cos}\left(\dfrac{n \pi x}{L}\right) + \sum_{n=1}^{\infty}\dfrac{e^{-(n\pi/L)^2\beta t}}{n\pi}b_n \text{sin}\left(\dfrac{n \pi x}{L}\right)$$

This is where I get confused. When $t=0$ the exponential vanishes and I get an approximation of my step function, which is my initial condition, so far everything is OK. However, as $t>0$ the temperature $u(x,t) \neq 0$ and as $\lim t \rightarrow \infty$ the temperature never reaches zero; it converges to the constant term $1/20$. Even more puzzling, although not noticeable in the picture here sometimes the temperature dips below zero at the ends. I am doubting if the equation representing the initial condition is meant to only be used at $t=0$ and not for $t>0$, thus using the general solution afterwards. I suspect this is wrong. So basically, how can I incorporate the initial condition $f(x)$ into the heat equation???

A bit of interactive graphs, where b represents time $t$ can be found here

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  • $\begingroup$ If you built the general solution only with a linear combination of terms $ e^{\lambda\beta t}c_2 sin\left(\dfrac{n \pi x}{L}\right) $ , the fonction is odd with respect to the variable $x$. The initial condition with a not odd Fourier series from $f(x)$ cannot be fitted. So, either you use an odd Fourier series (as in my answer below) or you have to consider a more general form for $u(x,t)$ not odd, but more complicated to deal with and to fit with the condition $u(0,t)=u(L,t)=0$. $\endgroup$ – JJacquelin Jul 23 '16 at 10:28
  • $\begingroup$ Thank you for your answer and comment. It makes sense that I should have used an odd function for the initial condition, extending the range of the periodicity did the trick! Just curious, do you have any reference on the other more complicated, general form of $u(x,t)$, which would include non-odd functions as well? $\endgroup$ – Sophie Jul 23 '16 at 16:28
  • $\begingroup$ Sorry, I don't have at hand a reference. But, probably one can find such examples, may-be on Stack Exchange. $\endgroup$ – JJacquelin Jul 23 '16 at 17:24
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Together with the general solution of the PDE, the conditions $u(0,t)=u(L,t)=0$ lead to : $$u(x,t)= \sum_{n=1}^{\infty}c_n \sin\left(\dfrac{n \pi x}{L}\right)\exp\left(-\left(\frac{n\pi}{L}\right)^2 \beta t \right) $$ The condition $u(x,0)=f(x)$ can be expressed in terms of Fourier series of various manner, depending on the chosen bounds. In the present case, in order to fit with the above form of $u(x,0)=\sum_{n=1}^{\infty}c_n \sin\left(\dfrac{n \pi x}{L}\right)$ which is an odd function, we extend the domain to $-L<x<L$

$$f(x) = \begin{cases} 0, & -1<x<-0.55 \\ 1, & -0.55<x<0.45 \\ 0, & -0.45<x<0.45 \\ 1, & 0.45<x<0.55 \\ 0, & 0.55<x<1 \end{cases}$$

This can be surprising because $f(x)\neq 0$ somewhere on $x<0$. It doesn't matter since latter, only the range $0<x<L$ is taken into account. This is the same when, instead of $-L<x<L$, we take the Fourier series on $0<x<L$ : also $f(x)\neq 0$ somewhere on $x<0$.

That way, we obtain : $$f(x)\simeq \sum_{n=1}^{\infty}c_n \sin\left(\dfrac{n \pi x}{L}\right)$$ $$c_n=\frac{1}{L}\int_{-0.55}^{-0.45}(-1)\sin\left(\frac{n\pi x}{L} \right)dx+\frac{1}{L}\int_{0.45}^{0.55}(+1)\sin\left(\frac{n\pi x}{L} \right)dx$$

$$c_n=\frac{2}{n\pi}\left(\cos\left(n\pi\frac{0.45}{L} \right)-\cos\left(n\pi\frac{0.55}{L} \right) \right)$$ You can draw $\sum_{n=1}\frac{2}{n\pi}\left(\cos\left(n\pi\frac{0.45}{L} \right)-\cos\left(n\pi\frac{0.55}{L} \right) \right) \sin\left(\dfrac{n \pi x}{L}\right)$ and verify that it fit to $f(x)$ on the range $0<x<L$. Doesn't matter outside this range.

Finally : $$u(x,t)= \sum_{n=1}^{\infty}\frac{2}{n\pi}\left(\cos\left(n\pi\frac{0.45}{L} \right)-\cos\left(n\pi\frac{0.55}{L} \right) \right) \sin\left(\dfrac{n \pi x}{L}\right)\exp\left(-\left(\frac{n\pi}{L}\right)^2 \beta t \right) $$ $u(x,t\to \infty)\to 0$

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