4. Consider the singular Sturm-Liouville eigenvalue problem for Legendre's equation
\begin{aligned}
& -\left[\left(1-x^2\right) u^{\prime}\right]^{\prime}=\lambda u \quad-1<x<1, \\
& \left(1-x^2\right) u^{\prime}(x) \rightarrow 0 \quad \text { as } x \rightarrow \pm 1 .
\end{aligned}(a) Solve the ODE for $\lambda=0$ and show that both endpoints $x= \pm 1$ are in the limit circle case.SolutionIf $\lambda=0$ then $\left[\left(1-x^2\right) u^{\prime}\right]^{\prime}=0$. Integrating this once, we get
$$
u^{\prime}=\frac{c_2}{1-x^2}
$$
where $c_2$ is a constant. Integrating again, we get
$$
u(x)=c_1+\frac{1}{2} c_2 \log \left|\frac{1+x}{1-x}\right| \text {. }
$$
Thus, a fundamental pair of solutions of the ODE is
$$
u_1(x)=1, \quad u_2(x)=\log \left|\frac{1+x}{1-x}\right| .
$$
A function with a logarithmic singularity is square-integrable $e . g$.
$$
\int_0^1(\log x)^2 d x<\infty
$$
so these functions are square-integrable on $(-1,1)$. Both endpoints are therefore in the limit circle case.
(b) For $n=0,1,2, \ldots$, define the Legendre polynomials $P_n(x)$ by
$$
P_n(x)=\frac{1}{2^n n !} \frac{d^n}{d x^n}\left[\left(x^2-1\right)^n\right]
$$
(Note that $P_n$ is a polynomial of degree $n$.) Show that the Legendre polynomials are eigenfunctions of the Legendre equation with eigenvalues
$$
\lambda_n=n(n+1)
$$
Hint. Let $v(x)=\left(x^2-1\right)^n$ and differentiate the equation $\left(x^2-1\right) v^{\prime}=2 n x v$ $n+1$ times.
SolutionIf $v(x)=\left(x^2-1\right)^n$, then
$$
\left(x^2-1\right) v^{\prime}=\left(x^2-1\right) \cdot 2 n x\left(x^2-1\right)^{n-1}=2 n x v .
$$
According to the Leibniz rule,
$$
\frac{d^{n+1}}{d x^{n+1}}(f g)=f \frac{d^{n+1} g}{d x^{n+1}}+(n+1) \frac{d f}{d x} \frac{d^n g}{d x^n}+\frac{1}{2} n(n+1) \frac{d^2 f}{d x^2} \frac{d^{n-1} g}{d x^{n-1}}+\cdots+\frac{d^{n+1} f}{d x^{n+1}} g
$$
Since all derivatives of $\left(x^2-1\right)$ of order greater than or equal to three are zero, we have
$$
\frac{d^{n+1}}{d x^{n+1}}\left[\left(x^2-1\right) v^{\prime}\right]=\left(x^2-1\right) \frac{d^{n+2} v}{d x^{n+2}}+2(n+1) x \frac{d^{n+1} v}{d x^{n+1}}+n(n+1) \frac{d^n v}{d x^n}
$$
Similarly, since all derivatives of $x$ of order greater than or equal to two are zero, we have
$$
\frac{d^{n+1}}{d x^{n+1}}(x v)=x \frac{d^{n+1} v}{d x^{n+1}}+(n+1) \frac{d^n v}{d x^n} .
$$
Hence, differentiating (1) $n+1$ times, we get
\begin{aligned}
& \left(x^2-1\right) \frac{d^{n+2} v}{d x^{n+2}}+2(n+1) x \frac{d^{n+1} v}{d x^{n+1}}+n(n+1) \frac{d^n v}{d x^n} \\
& =2 n x \frac{d^{n+1} v}{d x^{n+1}}+2 n(n+1) \frac{d^n v}{d x^n} .
\end{aligned}Dividing this equation by $2^n n$ ! and using the definition of $P_n(x)$, we find that
$$
\left(x^2-1\right) P_n^{\prime \prime}+2(n+1) x P_n^{\prime}+n(n+1) P_n=2 n x P_n^{\prime}+2 n(n+1) P_n,
$$
which simplifies to
$$
-\left[\left(1-x^2\right) P_n^{\prime}\right]^{\prime}=n(n+1) P_n
$$
This shows that $P_n(x)$ is an eigenfunction of the Legendre equation with eigenvalue $n(n+1)$.
It follows, for example, from the Weierstrass approximation theorem that the Legendre polynomials $\left\{P_n: n=0,1,2, \ldots\right\}$ form a complete set in $L^2(-1,1)$, so there are no other eigenvalues or eigenfunctions.
(c) With $v$ as in (b), show that
$$
\int_{-1}^1\left[\frac{d^n v}{d x^n}\right]^2=(2 n) ! \int_{-1}^1(1-x)^n(1+x)^n d x=(n !)^2 \int_{-1}^1(1+x)^{2 n} d x
$$
and deduce that
$$
\int_{-1}^1 P_n(x)^2 d x=\frac{2}{2 n+1}
$$
(d) Write out the orthogonality relations for the Legendre polynomials and the eigenfunction expansion of a function $f \in L^2(-1,1)$ with respect to the Legendre polynomials.
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Czhang271828
发表于 2023-4-16 19:35
