Insulated endpoint conditions(Neumann BVP)

Now consider the BVP.

$$u_t = k u_xx~~0< x<L,$$ (3.21)

$$u_x(0,t) =u_x(L,t) = 0,$$ (3.22)

$$u(x,0) = f(x)$$ (3.23)

which corresponds to heat distribution in a thin (1D) rod of length L with initial temperature $f(x)$ and its endpoints insulated.

Figure:

As in the previous case,follow the method of separation of variables.

u(x,t)=X(x)T(t)
$\frac{X''}{x} = \frac{T'}{kt} = \lambda$

If $u(x,t)=X(x)T(t)$ is solution of (19), then it must satisfy B.C.s given in Eq. () and Eq. (), then

$$X'(0)T(t)=X'(L)T(t)=0, X'(0)=0$$    and $$X'(L)=0$$

Thus $X(x)$ satisfies

$$X''-\lambda X=0,X'(0)=0~~X'(L)=0.$$ (3.24)

Again we have to consider three possible $\lambda>0,\lambda=0,\lambda<0.$

1. If $\lambda >0$

   $$X''=\lambda X$$ (3.25)

   $$X(x)=\alpha exp( \sqrt{\lambda}x )+\beta exp(-\sqrt{\lambda}x)$$ (3.26)

   $$X(x)=\alpha cosh( \sqrt{\lambda}x )+\beta sinh(\sqrt{\lambda}x)$$ (3.27)

   $$X'(x)=\sqrt{\lambda}\alpha exp( \sqrt{\lambda}x )-\sqrt{\lambda}\beta exp(-\sqrt{\lambda}x)$$ (3.28)

   using B.C's
   

   $$X'(0) = \sqrt{\lambda}\alpha-\sqrt{\lambda}\beta$$ (3.29)

   $$X'(x) = \sqrt{\lambda}\alpha exp( \sqrt{\lambda} L)-\sqrt{\lambda}\beta exp(-\sqrt{\lambda}L) \alpha exp( \sqrt{\lambda}x )-\beta exp(-\sqrt{\lambda}x)=0$$ (3.30)

   
   from $\alpha=\beta$
   $$\alpha (exp( \sqrt{\lambda}x )- exp(-\sqrt{\lambda}x))=0$$
   This bracket term never be zero for $\lambda =0.$ Which implies that $\alpha=0$ and $\beta=0$ and we get a trivial solution.Thus we can have positive eigenvalues. Which means $\lambda \not>0.$

2. If $\lambda=0,$
   $$X''=0\implies X'= \alpha$$
   $$X=\alpha x+\beta$$
   $$X'(0)=0\implies \alpha =0$$
   $$\implies X=\beta,\beta \neq 0$$
   else we get, trivial solution. As a constant multiple of an eigen-functions is also function,we choose $\beta=1.$ Thus for zero eigenvalues $0,$ we have $\lambda_0=0$, $X_0 (x)=1$

3. If $\lambda < 0,$

   Replace $\lambda$ by $-\lambda$ in Eq. ()
   

   $$X''+\lambda X =$$ 0 (3.31)

   $$X = \alpha \cos(\sqrt{\lambda}x)+\beta \sin(\sqrt{\lambda}x)$$ (3.32)

   $$X' = \sqrt{\lambda}\alpha \sin(\sqrt{\lambda}x)+\sqrt{\lambda}\beta \cos(\sqrt{\lambda}x)$$ (3.33)

   
   
   $$X'(0)=0\implies -\lambda \alpha . 0+\sqrt{\lambda}\beta .1=0$$
   $$\beta =0$$
   $$X'(L)=0\implies-\sqrt{\lambda}sin(\sqrt{\lambda}L)=0$$
   $\alpha \neq 0$,  else we get the trivial solution
   $$\implies\sqrt{\lambda}L=n \pi,~~n=1,2,3,...$$
   $$\lambda_n=\frac{n^2 \pi^2}{L^2}$$
   are the eigenvalues and corresponding eigen-functions are
   $$X_n(x)=\alpha_n \cos(\frac{n\pi}{L}x)$$
   we can combine both the case as
   $$X_n(x)=\alpha_n \cos(\frac{n\pi}{L}x)~~n=0,1,2,...$$

Solving for $T(t)$: For $\lambda =0$

$$T'(t)=0$$

$$\implies T(t)=\gamma _n exp(\frac{-n^2\pi^2}{L^2}kt)~~n=1,2,...$$

we can combine the choose two cases as

$$T_n(t)=\gamma_n exp(-\frac{n^2\pi^2}{L^2}kt)~~n=0,1,2,...$$

we obtain sequence of separated solutions.Thus

$$u_n(x,t) = ((\alpha_n \cos(\frac{n\pi}{L}x)(\gamma_n \exp(-\frac{n^2\pi^2}{L^2}kt))~~n=0,1,2,...$$ (3.34)

$$= \chi_n cos(\frac{n\pi}{L}x)( \exp(-\frac{n^2\pi^2}{L^2}kt)$$ (3.35)

By the principle of superposition of solutions,we get the general solution

$$u(x,t) = \sum_{n=0}^{\infty}\chi_n u_n(x,t)$$ (3.36)

$$= \sum_{n=0}^{\infty}\chi_n \cos(\frac{n\pi}{L}x)( exp(-\frac{n^2\pi^2}{L^2}kt)$$ (3.37)

$$u(x,t)=\chi_0 +\sum_{n=1}^{\infty}\chi_n \cos(\frac{n\pi}{L}x)( \exp(-\frac{n^2\pi^2}{L^2}kt)$$ (3.38)

The unknown coefficients ${\chi_n}$ are found by using the initial condition

$$u(x,0)=\chi_0 +\sum_{n=1}^{\infty}\chi_n \cos(\frac{n\pi}{L}x)=f(x)$$ (3.39)

multiply both sides by $\cos(\frac{n\pi}{L})x$ and integrate both sides $[0,L]$

$$\int_{0}^{L}\chi_0 \cos(\frac{n\pi}{L}x dx +\int_{0}^{L}\sum_{n=1... ...frac{m\pi}{L}x)\cos(\frac{n\pi}{L})xdx=\int_{0}^{L}f(x)\cos(\frac{n\pi}{L})x dx$$ (3.40)

Now we use the orthogonality property

$\int_{0}^{L}\cos(\frac{m\pi}{L}x)\cos(\frac{n\pi}{L}x dx=0~~m\not =n$ $\frac{L}{2}~~m=n \not =0$ L   m=n=0

$$L\chi_0=\int_{0}^{L}f(x) dx$$

$$\chi_0=\frac{\int_{0}^{L}f(x) dx}{L}$$

For m=1 

$$\chi_1\frac{L}{2}=\int_{0}^{L}f(x)\cos(\frac{2\pi}{L}x dx$$

$$\chi_1=\frac{2}{L}\int_{0}^{L}f(x)\cos(\frac{2\pi}{L}x dx$$

$$\chi_n = \frac{2}{L} \int_{0}^{L} f(x)\cos( \frac{n\pi}{L} x dx$$ (3.41)

Thus ${\chi_n}$ are the Fourier-cosine series coefficients of $f(x)$ on $[0,L]$ and the formal series solution of the model problem Eq. (),Eq. () and Eq. () is

$$u(x,t)=\chi_0 +\sum_{n=1}^{\infty} \chi_n \cos( \frac{n\pi}{L} x)\exp(- \frac{n^2\pi^2}{L^2} kt)$$ (3.42)

where ${\chi}$ are coefficients of Fourier cosine expansion of initial condition $f(x)$ on the initial $[0,L]$.

Remark Note that $\lim_{n\to\infty}$u(x,t)=$\chi_0$= $\frac{1}{L}$ $\int_{0}^{L}f(x)dx=$ average values of the initial temp. Thus with lateral surfaces and end points insulated,the initial heat content ultimately distributes itself uniformly throughout the rod and attains the average values $\chi_0.$

Illustration

We consider the same $50cm$ rod as in the previous example with the following modifications

1. At time $t=0$, the rod's lateral surfaces and its two ends are insulated.
2. The initial temp. profile is given in angular function.

The model equation for this setting for $~0< x<L$ is

$$u_t = k u_{xx},$$ (3.43)

$$u_x(0,t) =u_x(50,t) = 0,$$ (3.44)

$$u(x,0) = f(x)$$ (3.45)

where

$$f(x) = \begin{cases} 4x &0\leq x \leq 25\\ 100-4x& 25\leq x \leq 50 \end{cases}$$ (3.46)

Solution For this Neumann problem, the solution is given by

$$u(x,t)=\frac{\chi_0}{2} +\sum_{n=1}^{\infty}\chi_n cos(\frac{n\pi}{L}x)exp(-\frac{n^2\pi^2}{L^2}kt)$$ (3.47)

where ${\chi_n}$ are Fourier coefficients of Fourier-cosine expansion of the initial temp. profile $f(x)$ in the interval $[0,L=50].$ These coefficients are given by

$\frac{\chi_n}{2}=\frac{1}{L}\int_{0}^{L}f(x)cos(\frac{n\pi}{L}x dx~n=0,1,2,...$

Thus to find the required solution,all we are do is find Fourier-cosine expansion of $f(x).$

$$\frac{\chi_n}{2} =\frac{1}{L}\int_{0}^{50}f(x)\cos(\frac{n\pi}{L}... ...)\cos(\frac{n\pi}{L}x dx+\int_{25}^{50}(200-4x)\cos\big(\frac{n\pi}{L}x dx\big)$$

$$f(x)=50-\frac{1600}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{(n)^2} \cos... ...{\pi^2}\sum_{n=1}^{\infty}\frac{1}{(4n-2)^2}\cos\big(\frac{(4n-2)\pi}{50}x\big)$$

$$u(x,t)=50-\frac{1600}{\pi^2}\sum_{n=2,6,10}^{}\frac{1}{(n)^2} cos\big(\frac{n\pi}{50}\big)x\exp\big(\frac{-n^2\pi^2}{2500}kt\big)$$