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I understand that the definition of spectrum of an operator(where operator means to be in functional analysis).
The point spectrum(eigenvlue) is used for solving some partial equations(e.g. heat equation, wave equation).

But I don't know what the continuous spectrum means.
Why do we need to consider the continuous spectrum?

I'd appreciate it if you could answer this question.

Consider the differential operator $L=frac{1}{i}frac{d}{dx}$ on $L^2[-pi n,pi n]$, where $n=1,2,3,cdots$ with a periodic condition $f(-pi n)=f(pi n)$ on the domain. The normalized eigenfunctions of $L$ are
$$
f_k(x)=frac{e^{ik/n}}{sqrt{2pi n}},;;; k=0,pm 1,pm 2,cdots.
$$
A function $fin L^2[-npi,npi]$ can be expanded in these eigenfunctions as
$$
f = sum_{k=-infty}^{infty}frac{1}{2pi n}langle f,e^{i kx/n}rangle e^{ikx/n}.
$$
As $nrightarrowinfty$, the eigenvalues are closer and closer to each other, and one may intuitively expect, based on Riemann integral ideas, an expansion of the form

$$
f = frac{1}{2pi}int_{-infty}^{infty}left(int_{-infty}^{infty}f(t)e^{-ist}dtright)e^{isx}ds
$$
In the limit, there is a "continuous" spectrum $(-infty,infty)$ and a continuum of "eigenfunctions" indexed by $sin(-infty,infty)$. One also has the generalized Parseval relation:
$$ |f|^2 = frac{1}{2pi}int_{-infty}^{infty}left|int_{-infty}^{infty}f(t)e^{-ist}dtright|^2ds.
$$
This is basically the idea that Fourier proposed in order to come up with the Fourier transform, though he used cosine and sine transforms.

As intuitive and compelling as these ideas may be, they are extremely difficult to justify in a rigorous way using these arguments. Though Fourier did not manage a rigorous argument, the final result is correct. This sort of thing happens when transitioning from finite to infinite domains for differential operators.

For a selfadjoint operator $A$, a point $s$ is in the continuous spectrum if, for every $epsilon > 0$, there is a unit vector $f_{s,epsilon}$ such that $|Af_{s,epsilon}-sf_{s,epsilon}| < epsilon$, even though $Af_{s,epsilon} ne s f_{s,epsilon}$. Every $sinmathbb{R}$ is in the continuous spectrum of $L=frac{1}{i}frac{d}{dx}$ on $L^2(mathbb{R})$. The term "continuous spectrum" is confusing unless you know this origin of the term, where you have a continuum of approximate eigenvalues in the limit of the discrete. Multiplication by $x$ on $L^2[a,b]$ has continuous spectrum only, and every $sin [a,b]$ is in the continuous spectrum. The Fourier transform turns $frac{1}{i}frac{d}{dx}$ into such a multiplication operator with continuous spectrum $(infty,infty)$. Multiplication is a model for selfadjoint operators.