Heat Equation: Homogeneous Dirichlet boundary conditions

Suppose that we want to find the temperature $u(x,t)$ in the thin (one dimensional) rod of finite length L extending form $x=0$ to $x=L$ extending from $x=0~$to$~x=L$. It is governed by the equation

$$\frac{\partial u}{\partial t} = k\frac{\partial u^2}{\partial x^2 },$$ (3.1)

where $k>0$ is thermal diffusing (Heat Conduction Coefficient) of the rod. Suppose that end points of the rod are kept at temperature zero. ( e.g. by damping against a large ice block). Thus we get the homogeneous boundary conditions

$$u(0,t) = u(L,t)=0 ~~ \forall~~ t\geq0,$$ (3.2)

Also, Assume that across the length the rod is heated at time $t=0$ as given by the initial condition

$$u(x,0)=f(x) ~~ 0 \leq x \leq L,$$ (3.3)

In order to make () consistent with (), we assume the compatibility condition

$$f(0)=f(L)=0$$ (3.4)

Figure:

Question: Any restriction on $f(x)?$

(If $f(x)$ is piece-wise continuous function, it can be showed that a formal series solution always satisfy the BC's and it is unique solution of the BVP. For part see,chap.6,Churchill and Brawn,Fainter series and solution of BVP.,3rd Ed.)

Solution by method of separation of variables:

Suppose that solution of model problem (),()and () can be expressed as

$$u(x,t)=X(x)T(t)$$ (3.5)

in which variables $x$ and $t$ are "separated". Substituting () in PDE (), we get

$$X(x)T'(t)=kX''(x)T(t)$$ (3.6)

where ' stands for derivative w.r.t respective variables.

$$\frac{X''}{X}=\frac{T'}{kT}$$ (3.7)

LHS of () is function of $x$ only ,where as RHS. is a function of t only. For a fixed t on RHS and LHS must remain constant as $x$ varies and it should be constant say $\lambda$

$$\frac{X''}{X}=\frac{T'}{kT}=\lambda$$ (3.8)

This $\lambda$ is also known as a separation constant. As a result we get two equations

$$X''(x)-\lambda X(x)=0 ~~ 0 < x < L$$ (3.9)

$$T''(x)-\lambda T(x)=0 ~~ 0 < x < L$$ (3.10)

$u(x,t)$ in () satisfy the BC's () iff

$$u(0,t)=X(0)T(t)=0 ~ \forall t\geq 0$$ (3.11)

$$u(L,t)=X(L)T(t)=0~ \forall t\geq 0$$ (3.12)

Since $u$ is not a trivial solution.

$$\implies X(0)=X(L)=0.$$

Thus the solution $X(x)$ is the solution of BVP.

$$X''(x)+\lambda X(x) = 0, ~~ 0 < x < L$$

$$X(0)=X(L)=0$$

We can interpret the above system as eigenvalue problem. A non-trivial solution of the system is called eigen-function of the problem with an eigenvalue $\lambda$. Depending on the sign of $\lambda$, we have different terms of solution of the above problem

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

1. If $\lambda > 0 : X(x)=\alpha \exp{(\sqrt{\lambda}x)}+\beta\exp{(-\sqrt{\lambda}x)},$
2. If $\lambda=0 : \alpha+\beta x,$
3. If $\lambda<0 : X(x)=\alpha \cos{(\lambda x)}+\beta \sin{(\lambda x)} =\alpha exp(-\iota \lambda x)+\beta exp(\iota \lambda x)$

Where $\alpha$ and $\beta$ are arbitrary constants. Here we have implicitly assumed that $\lambda$ is purely real Why? Answer is hidden in Strum-Liouville theory. In fact it can be shown that all eigenvalues of the problem are real constants.

If $\lambda >0$

$$X(0)=0 \implies \alpha \cosh(0)+\beta \sinh(0)=0.$$

$\cosh(x)$ is a strictly positive function $\implies \alpha =0.$

$$X(L)=0 \implies \beta \sinh(\sqrt{\lambda}L)=0.$$

$\sinh(x)$ has a unique root at $x=0 \implies \beta =0.$ Thus for $\lambda >0$, we have $X=0.$

$\implies$ This eigenvalue problem can not have positive eigenvalues.

If $\lambda =0$

$$X(0)=0~ \implies~\alpha=0,$$

$$X(L)=0~ \implies~\beta=0,$$

$\implies\lambda =0$ is not allowed. $X=0$ trivial solution.

If $\lambda < 0,$

$$X(0)=0 \implies \alpha \cos{(0)} + \beta \sin{(x)}=0$$

$$\implies \alpha=0$$

$$X(L)=0 \implies \beta \sin{(L)}=0$$

For the non-trivial solution

$$sin(L)=0$$

$$\sqrt{\lambda}L=n\pi ~~n=1,2,3,...$$

(why not $n=0$)

$$\lambda_{n}=\frac{n^2\pi^2}{L^2}$$

and the corresponding eigen-function is

$$X_{n}(x)=\sin{\left (\frac{n \pi x}{L} \right )}.$$

(They are unique up-to a multiplicative constant.)

Now, Let us look at ODE ()

$$T'(t)-(- \lambda )k T(t) =$$ 0 (3.13)

$$T'(t)+\frac{n^2\pi^2}{L^2}k T(t) =$$ 0

$$T'(t) = \lambda_{n}\exp{(-k\frac{n^2\pi^2}{L^2})t},$$ (3.14)

as $t \rightarrow{ \infty}$ $T(t)\rightarrow{0}$: consistent with physical expectation c $\implies$ choice of sign of $\lambda$ was right. Thus we obtain sequence of separated solutions

$$u_n(x,t) =\Bigg[\beta_n\sin{\big(\frac{n\pi x}{L}\big)}\Bigg]\Bigg[\gamma_n\exp(-k\frac{n^2\pi^2}{L^2}t)\Bigg],~n=1,2,3,...,$$

$$=\xi_n\sin(\frac{n\pi x}{L})\exp(-k\frac{n^2\pi^2}{L^2}t).$$

By the principle of the superposition of solution,the general solution is given by

$$u(x,t) = \sum_{n=1}^{+\infty} \xi_n u_n(x,t)$$ (3.15)

$$= \sum_{n=1}^{+\infty} \xi_n sin(\frac{n \pi x}{L})exp(\frac{-kn^2 \pi^2}{L^2t})$$ (3.16)

But we shall need to find the unknown coefficients $\{{\xi_n}_{1}^{\infty}\}.$ They can be found by using 'initial condition'

$$u(x,t)=\sum_{n=1}^{+\infty} \xi_n sin(\frac{n \pi x}{L})=f(x)$$ (3.17)

Multiply both sides by $sin(\frac{n \pi x}{L})x$ and integrate over 0 to L

$$\int_{0}^{L}\sum_{n=1}^{+\infty} \xi_n sin(\frac{n \pi x}{L})sin(\frac{m \pi x}{L})x dx=\int_{0}^{L} f(x)sin(\frac{n \pi x}{L})x$$ (3.18)

$$\sum_{n=1}^{+\infty} \xi_n\int_{0}^{L} sin(\frac{n \pi x}{L})sin(\frac{m \pi x}{L})x dx=\int_{0}^{L} f(x)sin(\frac{n \pi x}{L})x$$ (3.19)

Using the orthogonality of sin functions

$$\int_{-\pi}^{\pi} sin(\frac{n \pi x}{L})sin(\frac{m \pi x}{L})x dx=\pi \delta_{m.n}$$

For a function f(x)periodic on interval $[-L,L]$ instead of $[\pi ,\pi]$ a change of variable gives

$$\int_{-L}^{L} sin(\frac{n \pi x}{L})sin(\frac{m \pi x}{L})x dx=L \delta_{m.n}$$

$$\int_{0}^{L} sin(\frac{n \pi x}{L})sin(\frac{m \pi x}{L})x dx=2L \delta_{m.n}$$

$$\xi_n \frac{L}{2}=int_{0}^{L}f(x) sin(\frac{n \pi x}{L})dx$$

$$\xi_n =\frac{2}{L}int_{0}^{L}f(x) sin(\frac{n \pi x}{L})dx$$

These ${\xi_n}$ are the Fourier sine coefficients putting this in eq.( ).We get Fourier expansion of f(x) on [0,L].These Fourier coefficients and the Fourier expansion of f(x) on [0,L] is uniquely determined. Thus the formal series solution of the model problem (1)-(3)

$$u(x,t)=\sum_{n=1}^{+\infty} \xi_n sin(\frac{n \pi x}{L})exp(\frac{-kn^2 \pi^2}{L^2}t),$$ (3.20)

where $\{{\xi_n}\}$ are the Fourier coefficients of $f(x)$ on $[0,L].$

1. Notice the power all method.For a given initial condition for f, we only have to compute its Fourier sine coefficients in order to obtain the explicit solution.

2. The series solution Eq. () due to converges rapidly(unless $t$ is very small), due to exponentially decrease solution. Therefore infinite sum can be truncated by finite sum taking sufficiently large $N.$ This is used for comparison with numerical solution.

(C.Henny Edwars,Darrd E.Penny ,Differential equations and BVP.s computing and modeling 3rd Ed.)

Example 3.1.1   Suppose that a thin rod of length L=50cm is immersed in steam until its temperature is $u_0=100C$ throughout. At time $t=0$, its lateral surface is insulated and its two ends are embedded in ice at 0'c. Calculate the rod's temperature at its midpoint after half an hour if it is made of

1. with $k=0.15$ for iron,
2. with $k=0.005$ concrete

Solution The BVP for the temperature $u(x,t)$ of rod is

$$u_t = k u_xx~~0< x<50cm$$

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

$$u(x,0) = u_0=100^o C$$

Figure:

For this Dirichlet problem, the solution is given by

$$u(x,t)=\sum_{n=0}^{\infty}b_n\sin(\frac{n\pi}{L}x) exp(-\frac{n^2\pi^2}{L^2}kt),$$

where ${\{b_n\}_n}^\infty$ are the Fourier sine coefficients pf the Fourier sine expansion of $u_0=100$ on $[0,50]$ given by

$$b_n=\frac{2}{L}\int_{0}^{L}u_0\sin(\frac{n\pi x}{L})dx.$$

To compute $b_n$

$$b_n =\frac{2u_0}{L}\int_{0}^{L}u_0\sin(\frac{n\pi x}{L})dx$$

$$=\frac{2u_0}{L}{\Bigg[-\cos\big(\frac{n\pi x}{L}\frac{L}{n\pi}\big)\Bigg]_0}^{L}$$

$$=\frac{2u_0}{n\pi}[-(-1)^n+1]$$

This can be written as depend on $n$ as

$$b_n=\begin{cases} 0 & \text{$n$\ is odd},\\ \frac{4u_0}{n\pi} & \text{$n$\ is even}, \end{cases}$$

Thus the rod's temperature is given by

$$u(x,t)=\frac{4u_0}{\pi}\sum_{n=odd}^{}\frac{1}{n}\sin(\frac{n\pi}{L}x) exp(-\frac{n^2\pi^2}{L^2}kt),$$

or

$$u(x,t)=\frac{4u_0}{\pi}\sum_{n=1}^{\infty}\frac{1}{2n-1}\sin(\frac{2n-1\pi}{L}x) exp(-\frac{(2n-1)^2\pi^2}{L^2}kt),$$

Given $L=50$ and $u_0=100$

$$u(x,t)=\frac{400}{\pi}\sum_{n=1}^{\infty}\frac{1}{2n-1}\sin(\frac{2n-1\pi}{50}x) exp(-\frac{(2n-1)^2\pi^2}{2500}kt),$$

We want $u$ at $25$cm, $t=1800s$

$$u(25,1800)=\frac{400}{\pi}\sum_{n=1}^{\infty}\frac{1}{2n-1}\sin(\frac{2n-1\pi}{2}x) exp(-\frac{(2n-1)^2\pi^218}{25}k),$$

1. With $k=0.15$ for iron, we get
   $$u(25,1800)=43.851871,$$
   
   
2. With $k=0.005$ for concrete, we get
   for $5$ times
   $$u=100.19$$
   and for $10$ times
   $$u=100.00,$$

which implies that concrete is very effective insulator.