Htykuut

Consider one of the solutions to the Laplace equation
$$u(x,t)=sum_{n=1}^{infty} C_{n}e^{-alpha _{n}^2kt}left [ frac{-alpha_{n} }{h}cos(alpha _{n}x) + B_{n}sin(alpha _{n}x) right ],$$
where
$alpha _{n}^2$ is a positive eigenvalue. An initial condition is given as
$$u(x,0)=sum_{n=1}^{infty}C_{n}left [ frac{-alpha_{n} }{h}cos(alpha _{n}x) + B_{n}sin(alpha _{n}x) right ].$$

I am not showing my work that got me to this solution, because that would be a bit of time waster, both for me and whoever is reading.

My question is:

From this information, what would be the constants(usually given as an expression with an integral) of $C_{n}$ and $B_{n}$?

I usually do not have trouble with this, but the $C_{n}$ seems to be where I am confused. I understand that this would be a full Fourier seires, but I am not too sure how to continue. Could anyone provide me some feedback?

Thanks in advance!

Edit: The boundary conditions are:
$$u_{x}(0,t) +hu(0,t)=0 text{and} u_{x}(1,t)=0, t>0.$$
Also, I am just showing the solution for the positive eigenvalues, since I do not need help finding the constants for the negative one.

Let's start off with the eigenfunction in $x$

$$ X_n(x) = -frac{alpha_n}{h}cos(alpha_n x) + B_nsin(alpha_n x) $$

The first B.C. gives

$$ X_n'(0) + hX_n(0) = alpha_n B_n - alpha_n = 0 implies B_n = 1 $$

With that out of the way, the second B.C. gives

$$ X_n'(1) = alpha_nleft[frac{alpha_n}{h}sin alpha_n + cosalpha_nright] = 0 $$

$$ implies alpha_ntanalpha_n = -h $$

This is a necessary condition for $alpha_n$. There is no closed form, but you can solve it numerically given any $h$.

To show the eigenfunctions are orthogonal, note that for $nne m$ we have

$$ int_0^1 {X_m}''X_n = {X_m}'X_nBigg|_0^1 - X_n{X_m}'Bigg|_0^1 + int_0^1 X_m{X_n}'' $$

However

begin{align}
{X_m}'X_nBigg|_0^1 - X_n{X_m}'Bigg|_0^1 &= -{X_m}'(0)X_n(0) + X_m(0){X_n}'(0) \
&= hX_m(0)cdot X_n(0) - X_m(0)cdot hX_n(0) = 0
end{align}

Therefore

$$ int_0^1 {X_m}''X_n - int_0^1 X_m{X_n}'' = 0 implies -(alpha_m^2-alpha_n^2)int_0^1 X_mX_n = 0 $$

But $alpha_m ne alpha_n$, so the integral must be $0$

The remaining condition is

$$ u(x,0) = f(x) = sum_n C_n X_n(x) $$

where $f(x)$ is a known function. Using orthogonality

$$ int_0^1 f(x)X_m(x) dx = sum_n C_n int_0^1 X_n(x) X_m(x) dx = C_m int_0^1 big[X_m(x)big]^2 dx $$

$$ implies C_n = frac{int_0^1 f(x) X_n(x) dx}{int_0^1 big[X_n(x)big]^2 dx} $$

where the integrals depend on $alpha_n$. Again, numerical computation is useful here

If you follow the same method for the other case, i.e. $X'' = mu^2 X$, you'll find

$$ X_n(x) = frac{mu_n}{h}cosh(mu_n x) + sinh(mu_n x) $$

and $ mu_ntanhmu_n = h $. There is one solution if $h > 0$.

Note that the formula for $C_n$ works the same whether the eigenvalue is positive, negative, or zero.