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 $0<x<l$

  • $\begingroup$ I don't understand your statement of the problem. Doesn't "at $x=L$ the rod gives off heat into a medium of temperature 0" imply the boundary condition $u(L,t)=0$? $\endgroup$ – Keith McClary Aug 13 '18 at 3:51
  • $\begingroup$ the problem states who's ends are $x=0$, $x=L$ has the end $x=0$ at temperature zero, while at the end $x=L$ radiation takes place into a medium of temperature 0, Does that imply that $u(L,t)=0$? If it does I think I can solve this problem quite easily, however the "hint" to this solution was Hint: Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings) but maybe that relates to the solution? $\endgroup$ – yipz Aug 13 '18 at 3:54
  • $\begingroup$ I added my solution for $u(L,t)=0$, perhaps what the hint was talking about is the fact the solution looks like the newtons law of cooling and when t=0 it is f(x) $\endgroup$ – yipz Aug 13 '18 at 4:05
  • $\begingroup$ I was assuming it meant that the medium is maintained at $0$ and it was conduction, not radiation. I can't think what it might mean. $\endgroup$ – Keith McClary Aug 13 '18 at 4:13

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...

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