forked from eggy/eifueo
ece204: add odes
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@@ -225,6 +225,13 @@ Two boundary conditions are requred to solve the problem for all $t>0$ — that
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- $u(x,0)=f(x),0\leq x\leq L$
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- e.g., $u(0,t)=u(L,t)=0,t>0$
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Thus the general solution is:
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$$
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\boxed{u(x,t)=\sum^\infty_{n=1}a_ne^{-\left(\frac{n\pi a}{L}\right)^2t}\sin(\frac{n\pi x}{L})} \\
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f(x)=\sum^\infty_{n=1}a_n\sin(\frac{n\pi x}{L})
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$$
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### Periodicity
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The **period** of a function is an increment that always returns the same value: $f(x+T)=f(x)$, and its **fundamental period** of a function is the smallest possible period.
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