1.8 KiB
Let's revisit the heat equation:
The first time the heat equation was introduced, we figured out that it's solution was of the form u(t,x)=\sum_{n=1}^\infty c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right)
Now that we learned about fourier series and eigen value problems, we can finally solve it! (for a given specific case.)
IBVP of heat eq:
\frac{ \partial u }{ \partial t }=D\frac{ \partial^{2} u }{ \partial x^{2} } for 0\leq x\leq L for t>0
u(t,0)=u(t,L)=0, \quad t>0
u(0,x)=f(x), \quad 0\leq x\leq L
lets choose L=\pi
f(x)=\begin{cases}-x & 0\leq x\leq \frac{\pi}{2} \\1-x & \frac{\pi}{2}<x\leq \pi\end{cases}
!Lec 30 2023-11-24 13.42.29.excalidraw
So we have a non-uniformly heated rod with both ends insulated. What happens to the temperature inside the rod over time? "[...]. Very interesting problem." -Prof (I agree.)
If we made this a series, where would it converge? Well it's continuous from 0 to pi and its windowed form when repeated will be convergent everywhere, this is good news for us.
Separation of variables: u(t,x)=T(t)X(x)
theres a theorem that this will give a unique solution.
T'X=DTX''
\frac{T'}{DT}=\frac{X''}{X}
LHS is a function of t only , RHS is function of x only.
"I don't know what is time, I know space, I can take a step and see the step I take, but can you see time? Can you see the future? Some can but I can't." -prof. Very philosophical.
X''+\lambda X=0 where X(0)=X(L=\pi)=0
u(t,0)=T(t)X(0)=0
case 1) \lambda<0, r_{1,2}=\pm \sqrt{ -\lambda }
X(x)=c_{1}e^{ \sqrt{ -\lambda }x }+c_{2}e^{ -\sqrt{ -\lambda }x }
X(0)=c_{1}+c_{2}=0
X(\pi)=c_{1}e^{ \sqrt{ -\lambda }\pi }+c_{2}e^{ -\sqrt{-\lambda }\pi }=0
c_{1}=c_{2}=0
we will continue the problem in the next lecture.
#end of lec 30