The two faces of the slab 0 ≦ x ≦ L are kept at temperature zero, and the initial temperature of

1. We know that the solution to the heat equation with given initial and boundary conditions can

The two faces of the slab 0 ≦ x ≦ L are kept at temperature...

The two faces of the slab $0 \leqq x \leqq L$ are kept at temperature zero, and the initial temperature of the slab is given by $u(x, 0)=A$ (a constant) for $0<x<L / 2, u(x, 0)=0$ for $L / 2<x<L$. Derive the formal series solution $u(x, t)=$ $$ \frac{4 A}{\pi} \sum_{n=1}^{\infty} \frac{\sin ^{2}(n \pi / 4)}{n} \exp \left(-n^{2} \pi^{2} k t / L^{2}\right) \sin \frac{n \pi x}{L} $$

Transcript

00:01 Hi, here for the given question, we are going to use a non -theorem which tells us about the road with zero end point temperature.

00:13 So here we know that equation can be written as del u by del t equals to k times del square u upon del x square.

00:22 Here we have zero less than x less than l and t greater than zero.

00:27 So here in our case, u of zero comma t will be equals to u of l comma t, which is equal to zero and u of x comma zero is equal to f of x.

00:39 Now here in our case, we know that this given formula has a series solution.

00:44 So the series solution can be written as u of x comma t equals to summation and running from one to infinity bn e to the power minus n square pi square kt upon l square.

00:59 Here instead of e, we will write exp multiplied with sine n pi x upon l where bn is equal to two by l integration over zero to l f of x sine n pi x upon l dx.

01:18 Now here in our case, further from the given differential equation, here we can say that we have del u upon del t is equal to k del square u upon dx square.

01:31 We have zero less than x less than l and t greater than zero.

01:34 Here we are given initial condition u of zero comma t is equal to u of l comma t, which is equal to zero.

01:41 So u of x comma zero will be equal to a and zero.

01:46 A is when if zero less than x less than l by two and l by two less than x less than l.

01:54 So here we have zero value for this condition and a value for this condition.

01:59 So here in our case, now further we know that the solution which is u of x comma t can be written as summation n running from one to infinity bn exp minus n square pi square kt upon l square multiplied with sine n pi x upon l...

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