I am trying to solve the problem of a insulated heated rod given by $$u_t=ku_{xx}$$ With $$u(0,t)=0$$ $$u(x,0)=f(x)$$ and finally for $0<x<L$ at $x=L$ the rod gives off heat into a medium of temperature 0.
Using newtons Law of cooling $$T(t)=t_s + (t_0 - t_s)e^{-kt}$$ and letting $t_s=0$ I get $$u(L,t)=f(x)e^{-kt}$$ Perhaps it would be best to solve the steadystate solutions?
which sort of makes sense to me as when $t=0$ my starting temperature is $f(x)$ and as t gets very large the rod end point approaches 0
Im not sure if this boundary is correct or not and if it is I am kind of lost on how to solve the equation, trying seperation of variables leads to a tricky solution as $X(x)=Asin(λx)$ but Im not really sure how to solve for my $λ$
If I have $$u(L,t)=0$$ my soultion would simply be $$u_n(x,t)=\sum_{n=0}^\infty bnsin((n*\pi/L)x)e^{-n^2\pi^2t/l^2}$$
Where $b_n$ is definied as $b_n= \int f(x)sin(n\pi x/l)$ from $0<x<l$
Your other boundary condition is $u(L,t) = 0$ in general this looks like
$$\frac{\partial u}{\partial t} =k\frac{\partial^{2}u}{\partial x^{2}}$$
with the boundary conditions
$$ BC1 : u(0,t) = 0\\ BC2 : u(L,t) = 0 \\ IC1 : u(x,0) = f(x) $$
from the boundary conditions and initials conditions we get
$$ \frac{d\phi^{2}}{dt^{2}} +\lambda \phi =0 $$
$$ \phi(t) = c_{1}\cos(\sqrt{\lambda} t) + c_{2} \sin(\sqrt{\lambda}t)$$ $$ \phi(0) = c_{1} = 0 \implies\phi(t) = c_{2}\sin(\sqrt{\lambda}t)$$ $$ \phi(L) = c_{2}\sin(\sqrt{\lambda}L) = 0 \implies \sin(\sqrt{\lambda}L) = 0$$ $$ \sqrt{\lambda}L = n\pi \implies \lambda = (\frac{n\pi}{L})^{2} , n\in \mathbb{N}$$
$$ b_{n} = \frac{2}{L} \int_{0}^{L} f(x)\sin(\sqrt{\lambda}t) dx $$
$$ u(x,t) = \sum_{n=1}^{\infty} b_{n}\sin(\sqrt{\lambda}t)e^{-\lambda k t} $$ using fouriers trick leads to this due to orthogonality of sines
$$ f(x) = \sum_{i=1}^{\infty} b_{n}\sin(\sqrt{\lambda}t)$$ your initial condition is satified by since if $t=0$ then the exponential is $1$. See $u(x,0)$ says at time $0$ we have $f(x)$ equal to...