Question

8. A rod 20 cm long, has both its ends maintained at 0*C at all times. At time t = 0 the
temperature distribution in the rod is

T (x,0) = { 50*C for 0<x<10cm
{0*C for 10cm<x<20cm

Obtain the temperature distribution u(x,t) in the rod if u(x,t) satisfies the diffusion
equation:

d^2 u(x,t) / dx^2 = 0.5 ( du(x,t)/dt )

Answer 1

Answer on Question #58205/ Physics – Molecular Physics | Thermodynamics

A rod 20 cm long, has both its ends maintained at 0°C at all times. At time t = 0 the temperature distribution in the rod is

T(x,0) = {50°C for 0<x<10cm <x<20cm="" for="" 10cm<x<20cm="" obtain="" temperature="" the="" to="">

Obtain the temperature distribution $u(x,t)$ in the rod if $u(x,t)$ satisfies the diffusion equation:

$$\frac{d^2 u(x,t)}{dx^2} = 0.5 \left( \frac{du(x,t)}{dt} \right)$$

Solution

Let us write down whole differential equation with boundary and initial conditions

$u$ – is temperature of rod and function of $x,t$ $u=u(x,t)$, length of rod is $l=20$ cm

$$\frac{\partial^2 u}{\partial x^2} = \frac{1}{2} \frac{\partial u}{\partial t}, \quad 0 < x < l, \quad t > 0$$ $$u(0,t) = 0$$ $$u(l,t) = 0$$ $$u(x,0) = \begin{cases} 50 = T_0, & 0 < x < 10 \\ 0, & 10 < x < 20 \end{cases}$$

So, we start from separation of variables.

Let $u(x,t) = X(x) \cdot T(t)$, where $X(x)$ – function that depends only on $x$ variable, $T(t)$ – function that depends only on $t$ variable.

$$\frac{\partial^2 u}{\partial x^2} = T \cdot \frac{d^2 X}{dx^2}$$ $$\frac{\partial u}{\partial t} = X \cdot \frac{dT}{dt}$$

And

$$X(0) = 0$$ $$X(l) = 0$$

From initial equation

$$\frac{\partial^2 u}{\partial x^2} = \frac{1}{2} \frac{\partial u}{\partial t}$$ $$T \cdot \frac{d^2 X}{dx^2} = \frac{1}{2} X \cdot \frac{dT}{dt}$$

Or

$$\frac{\frac{d^2 X}{dx^2}}{X} = \frac{1}{2} \frac{\frac{dT}{dt}}{T} = \lambda$$

And our task now is to find $\lambda$</x<10cm>

$$\frac {d ^ {2} X}{\frac {d x ^ {2}}{X}} = \lambda$$

Here we see spectral problem with boundary conditions.

$$\frac {d ^ {2} X}{d x ^ {2}} = \lambda X$$ $$X (0) = 0$$ $$X (l) = 0$$

1) Put $\lambda = 0$

$$X = C x + D$$ $$X (0) = 0 = D$$ $$X (l) = 0 = C l \rightarrow C = 0$$

So, $\lambda = 0$ can't give us a solution

2) Put $\lambda > 0, \lambda = w^2$

$$\frac {d ^ {2} X}{d x ^ {2}} - w ^ {2} X = 0$$

Solution for this equation can be found in form of:

$$X = C s h (w x) + D c h (w x)$$ $$X (0) = 0 = C 0 + D \rightarrow D = 0$$ $$\begin{array}{l} X (l) = 0 = C s h (w l) \rightarrow o r C = 0 o r s h (w l) = 0. I f s h (w l) = 0 \rightarrow \lambda \\ = 0, \text{ and initial assumption was } \lambda > 0. I f C = 0, \text{ again solution is trivial} \\ \end{array}$$

3) Put $\lambda < 0, \lambda = -w^2$

$$\frac {d ^ {2} X}{d x ^ {2}} + w ^ {2} X = 0$$

Solution for this equation can be found in form of:

$$X = C \sin (w x) + D \cos (w x)$$ $$X (0) = 0 = C 0 + D \rightarrow D = 0$$ $$X (l) = 0 = C \sin (w l), \text{ so if } C \neq 0, \text{ then } \sin (w l) = 0$$ $$w l = n \pi , n = 1, 2, 3, \dots$$ $$w _ {n} = \frac {n \pi}{l}$$

And here we have spectrum of eigenvalues.

Eigenfunctions are $X_{n} = C_{n}\sin (w_{n}x)$

$$\lambda = - w ^ {2} = \frac {1}{2} \frac {\frac {d T}{d t}}{T}$$

From previous row: $T_{n} = D_{n}e^{-2w^{2}t}$

So,

$$u _ {n} (x, t) = \widetilde {D _ {n}} e ^ {- 2 w _ {n} ^ {2} t} \sin (w _ {n} x)$$ $$\mathrm{u}(\mathrm{x}, \mathrm{t}) = \sum_{\mathrm{n} = 1}^{+\infty} u_n(x, t)$$

Then, we put initial condition

$$u(x, 0) = \sum_{\mathrm{n} = 1}^{+\infty} u_n(x, 0) = \sum_{\mathrm{n} = 1}^{+\infty} \widetilde{D_n} \sin(w_n x)$$

And $\widetilde{D_n}$ are coefficients of Fourier transform

$$\widetilde{D_n} = \frac{2}{l} \int_0^l u(x, 0) \sin(w_n x) \, dx$$

From

$$u(x, 0) = \begin{cases} 50, & 0 < x < 10 \\ 0, & 10 < x < 20 \end{cases}$$ $$\begin{aligned} \widetilde{D_n} &= \frac{2}{l} \int_0^{10} T_0 \sin(w_n x) \, dx = \frac{2T_0}{l} \left( -\left. \frac{\cos(w_n x)}{w_n} \right|_0^{10} \right) = \frac{2T_0 l}{l} \left( \left. \frac{\cos \left(\frac{n\pi x}{l}\right)}{n\pi} \right|_{10 = \frac{l}{2}}^0 \right) \\ &= \frac{2T_0}{n\pi} \left(1 - \cos \left(\frac{n\pi}{2}\right)\right) = \frac{2T_0}{n\pi} 2 \sin^2 \left(\frac{n\pi}{4}\right) = \frac{4T_0}{n\pi} \sin^2 \left(\frac{n\pi}{4}\right) \end{aligned}$$

Result

$$\mathrm{u}(\mathrm{x}, \mathrm{t}) = \sum_{\mathrm{n} = 1}^{+\infty} u_n(x, t) = \sum_{\mathrm{n} = 1}^{+\infty} \frac{4T_0}{n\pi} \sin^2 \left(\frac{n\pi}{4}\right) e^{-2 \left(\frac{n\pi}{l}\right)^2 t} \sin \left(\frac{n\pi x}{l}\right)$$

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