Solve Conduction of Heat | General Formula for Temperature u(x,t)

[viciado123]

Find a general formula for the temperature u(x,t) in the form of a series with general formula for the coefficients of the series.

\alpha^2 \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t} with 0 < x < L; t>0

u(0,t) = 0
\frac{\partial u}{\partial x} (L,t) = 0 with t>0

u(x,0) = f(x)

How can I resolve ?

 

[elibj123]

You assume that your basic functions (the ones spanning the space of solution) are of form u(x,t)=X(x)T(t). These functions are called separated variable functions.

You plug this form into the equation. Note that derivation by t affects only T(t), and derivation by x affects only X(x). So:

\alpha^{2}X''T=X\dot{T} (Here a prime ' is used to specify spatial derivative and a dot specifies temporal derivative)

Separating this:

\frac{\dot{T(t)}}{\alpha^{2}T(t)}=\frac{X''(x)}{X(x)}

Leads to the conclusions that both expression must equal some constant number. This is because on one side you only have a function of t, and on the other side a function of x.
Usually (out of convention) we name this constant -\lambda

And the problem is reduced into 2 ODE's:

1)X''+\lambda X=0; X(0)=0, X'(L)=0 (Derivation of the boundary conditions are not hard)
2)\dot{T}+\alpha^{2}\lambda T=0

Let me concentrate on the first problem, also known as a Sturm-Liouville problem.
Here, you must find special numbers \lambda for which there exists a function X(x) that both solves the equation, and satisfies the boundary condition. As you will see, there is an infinite number of them, but they form a discrete set of numbers. These lambda's are named the eigenvalues of the Sturm-Liouville problem (or the eigenvalue of the Sturm-Liouville operator) and corresponding solutions X are named the eigenfunctions (in fields like quantum physics they are name eigenstates) of the problem, and the set of eigenfunctions forms an orthonormal basis for all functions (this might remind you of Fourier Series theory, if you've learned it).

So now let's solve it.

First we assume that \lambda<0. Then the solutions are:

X(x)=Asinh(\sqrt{\lambda}x)+Bcosh(\sqrt{\lambda}x)

Plugging in the boundary conditions:

X(0)=0=B
X'(L)=A\sqrt{\lambda}cosh(\sqrt{\lambda}L)+B\sqrt{\lambda}sinh(\sqrt{\lambda}L)=0

So A=B=0 (because the function cosh doesn't have zeros), and what we have is X(x)=0. But {X} is supposed to be a basis to a space, so we are not interested in trivial functions which are identically zero. So we say there are no negative eigenvalues.

Now let's assume lambda is zero. So the solutions are
X(x)=Ax+B
Plugging in the boundary conditions reveals again that X must be the zero function.

Disappointed we proceed to positive lambda's. Now, the solutions are trigonometric:

X(x)=Acos(\sqrt{\lambda}x)+Bsin(\sqrt{\lambda}x)

Now we plug in the boundary conditions and we get that:

X(0)=A=0
X'(L)=-\sqrt{\lambda}Asin(\sqrt{\lambda}L)+\sqrt{\lambda}Bcos(\sqrt{\lambda}L)=0

or

Bcos(\sqrt{\lambda}L)=0

Now we don't want B=0, because we will have X(x)=0 again and that's it for us. Instead we can require

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

so:

\lambda_{n}=\frac{n^{2}\pi^{2}}{4L^{2}}

These are our eigenvalues. Our eigenfunctions will be:

X_{n}(x)=sin(\frac{n\pi x}{2L})
(Here the coefficient in front of the sin is not important since eigenfunction of the same eigenvalues are linearly dependent)

Another very important thing about the SL problem, is the inner product which is defined on the space. In this case (and to go into more general cases please read in the internet) the inner product is defined via:

<f,g>=\int^{L}_{0}f(x)g(x)dx

You can notice that in respect to this inner product our set of eigenfunctions are orthogonal, but are not normalized. This will harden our calculations in which we will try to expend an arbitrary function as a series of these functions. One thing you can do is to work with normalized functions from the beginning (calculate the norm of the current functions and divide by it).

So now we have our solutions to the first ODE, recall that we had another ODE to solve:

\dot{T}+\alpha^{2}\lambda T=0

Now that we have the lambds we can write:
\dot{T_{n}}=-\frac{\alpha^{2} n^{2} \pi^{2}}{4L^{2}}T_{n}

Solving this gives us:

T_{n}(t)=a_{n}e^{-\frac{\alpha^{2} n^{2} \pi^{2}}{4L^{2}}t}

With coefficient An to be determined.
Now that we have our T's and X's, let's go back to the heat equation:

every basic separated variable solution will look like this:

u_{n}(x,t)=X_{n}(x)T_{n}(t)=a_{n}sin(\frac{n\pi x}{2L})e^{-\frac{\alpha^{2} n^{2} \pi^{2}}{4L^{2}}t}

And if we want to express a general solution, we will do this with a series:

u(x,t)=\sum^{\infty}_{n=1}a_{n}sin(\frac{n\pi x}{2L})e^{-\frac{\alpha^{2} n^{2} \pi^{2}}{4L^{2}}t}

We still have an initial condition u(x,0)=f(x), which means that:

f(x)=\sum^{\infty}_{n=1}a_{n}sin(\frac{n\pi x}{2L})

This series is named the Generalized Fourier Series of f(x) with respect to our orthogonal basis. From orthogonality follows that:

a_{n}=\frac{<f,X_{n}>}{<X_{n},X_{n}>}=\frac{\int^{L}_{0}f(x)sin(\frac{n\pi x}{2L})dx}{\int^{L}_{0}sin^{2}(\frac{n\pi x}{2L})dx}

Usually these integrals are complicated to compute and depend on the form of f(x).
Note that while a series is defined for every f(x), it almost never uniformly converges to the function itself. You will say then that the series is identified with f(x), usually using the sign ~ instead of =.

But, if f(x) is continuous on the domain of the problem (in this case [0,L]) and its derivative f'(x) is piecewise continuous there, and also f(x) satisfies the boundary conditions of the SL problem, then the series will uniformly converge to f(x). In terms of a PDE problem, you will say that the solution u(x,t) is classic.

Calculating the above integrals and plugging An back here:

u(x,t)=\sum^{\infty}_{n=1}a_{n}sin(\frac{n\pi x}{2L})e^{-\frac{\alpha^{2} n^{2} \pi^{2}}{4L^{2}}t}

Gives you the entire solution to your problem.

I highly encourage you to read more about the theory of SL problems.
But in general, you handle the problem in a very similar way to the way I did (depends on the specific problem)