Eigenfunction expansion of homogeneous pde

Question

Use eigenfunction expansion to solve

$$u_{t}=u_{xx}$$ $$u_{x}(0,t)=u_{x}(1,t)=0$$ $$u(x,0)=\pi$$ for $$0<x<\frac{1}{2}$$ and $$u(x,0)=0$$ for $$\frac{1}{2}<x<1$$

I'm not well versed in this method can anyone offer some help?

Answer 1

Let $u(x,t)=\sum\limits_{n=0}^\infty C(n,t)\cos2n\pi x$ so that it automatically satisfies $u_x(0,t)=u_x(1,t)=0$ and fit for the piecewise zones of $0<x<\dfrac{1}{2}$ and $\dfrac{1}{2}<x<1$ ,

Then $\sum\limits_{n=0}^\infty C_t(n,t)\cos2n\pi x=-\sum\limits_{n=0}^\infty4n^2\pi^2C(n,t)\cos2n\pi x$

$\therefore C_t(n,t)=-4n^2\pi^2C(n,t)$

$\dfrac{C_t(n,t)}{C(n,t)}=-4n^2\pi^2$

$\int\dfrac{C_t(n,t)}{C(n,t)}dt=\int-4n^2\pi^2$

$\ln C(n,t)=-4n^2\pi^2t+f(n)$

$C(n,t)=F(n)e^{-4n^2\pi^2t}$

$\therefore u(x,t)=\sum\limits_{n=0}^\infty F(n)e^{-4n^2\pi^2t}\cos2n\pi x$

$u(x,0)=\pi$ for $0<x<\dfrac{1}{2}$ :

$\sum\limits_{n=0}^\infty F(n)\cos2n\pi x=\pi$

$F(n)=\begin{cases}\pi&\text{when}~n=0\\0&\text{otherwise}\end{cases}$

$\therefore u(x,t)=\pi$ for $0<x<\dfrac{1}{2}$

$u(x,0)=0$ for $\dfrac{1}{2}<x<1$ :

$\sum\limits_{n=0}^\infty F(n)\cos2n\pi x=0$

$F(n)=0$

$\therefore u(x,t)=0$ for $\dfrac{1}{2}<x<1$