Given :
$\frac{\partial{u}}{\partial{t}}=\alpha\frac{\partial^2{u}}{\partial{x^2}}$
Boundary conditions :
$\frac{\partial{u(0,t)}}{\partial{x}}=\delta(t)=-\frac{q}{\lambda}$
$\frac{\partial{u(L,t)}}{\partial{x}}=0$
$u(x,0)=T_0=298,15K$
$0<x<L$, $L=1mm$
$t>0$
I am looking for a response of the system at x = L to see a temperature rise at the end of the rod and plot T(x=L,t) to find how much time is needed to reach peak temperature. I've tried homogenizing the problem and used SoV by assuming u(x,t) = w(x,t) + v(x,t)
$\frac{\partial{w}}{\partial{t}}+\frac{\partial{v}}{\partial{t}}=\alpha\frac{\partial^2{w}}{\partial{x^2}}+\alpha\frac{\partial^2{v}}{\partial{x^2}}$
Where $w(x,t)=Ax^2+Bx+Ct$
Then
$\alpha\frac{\partial^2{w(x,t)}}{\partial{x^2}}=2A\alpha=\frac{\partial{w(x,t)}}{\partial{t}}=C$
$\frac{\partial{w(0,t)}}{\partial{x}}=B=-\frac{q}{\lambda}$, $\frac{\partial{w(L,t)}}{\partial{x}}=2AL+B=0$
I was left with
$\frac{\partial{v}}{\partial{t}}=\alpha\frac{\partial^2{v}}{\partial{x^2}}$
Boundary conditions :
$\frac{\partial{v(0,t)}}{\partial{x}}=0$
$\frac{\partial{v(L,t)}}{\partial{x}}=0$
$v(x,0)=T_0-w(x,0)=\phi(x)$
My solution is $u(x,t)=w(x,t)+\frac{a_0}{2}+\sum_{n=1}^\infty cos(\mu_nx)e^{-\mu_n^2\alpha t}$
$a_n=\frac{2}{L}\int_{0}^L\phi(x)cos(\mu_nx)dx$
But I coudn't slove it with $\delta(t)$, because then $a_n=0$, $a_0=T_0$, $v(x,t)=T_0$ and everything blows up, which is non-physical. By replacing $\delta(t)$ with $-\frac{q}{\lambda}$ I found the solution, but This is what happens at x = L= 1 mm
Evolution of temperature at x=L. I used constants such as $\alpha$,$\lambda$ etc. for iron
Can someone help me to compute it with $\delta(t)$ in B.Cs?