Chapter 8 The Wronskian
Recall the Sturm–Liouville differential expression \(Du:=\frac {(pu')'+qu}{w}\) (we assume that \(w\), \(q\) and \(p\) satisfy the Sturm–Liouville conditions: in particular the sign and smoothness conditions).
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Proof. This follows from integrating by parts twice:
\(\seteqnumber{0}{8.}{0}\)\begin{multline*} \int _\gamma ^\delta (Du)(x)v(x)\,w(x)\,dx =\int _\gamma ^\delta [(pu')'+qu]v(x)\,dx \\ =\left [pu'v\right ]_\gamma ^\delta -\int _\gamma ^\delta p(x)u'(x)v'(x)\,dx+\int _\gamma ^\delta q(x)u(x)v(x)\,dx \\ =\left [pu'v-puv'\right ]_\gamma ^\delta +\int _\gamma ^\delta u(x)[p(x)v'(x)]'\,dx+\int _\gamma ^\delta q(x)u(x)v(x)\,dx \\ =\left [pu'v-puv'\right ]_\gamma ^\delta +\int _\gamma ^\delta u(x)(Dv)(x)\,w(x)\,dx. \end{multline*} The condition that \(u,v\in D_{\max }\) ensures that the integrals in the above calculation exist and that \(u\) and \(v\) are continuously differentiable on \((\a ,\b )\) so that the function evaluations in \(\left [pu'v-puv'\right ]_\gamma ^\delta \) make sense (for the latter it is important that \([\gamma ,\delta ]\subset (\a ,\b )\)). □
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Definition 8.4. The Wronskian \(W:[\a ,\b ]\to \mR \) of the Sturm–Liouville problem is defined for \(u,v\in D_{\max }\) as follows. If \(x\in (\a ,\b )\) then
\[ W(u,v;p)(x)=p(x)[u'(x)v(x)-u(x)v'(x)], \]
and further
\[ W(u,v;p)(\a )=\lim _{x\downarrow \a }W(u,v;p)(x),\qquad W(u,v;p)(\b )=\lim _{x\uparrow \b }W(u,v;p)(x). \]
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Remark 8.5. Since \(u,v\in D_{\max }\) implies that \(u\) and \(v\) are continuously differentiable on \((\a ,\b )\), the Wronskian is well-defined on \((\a ,\b )\). That the (one-sided) limits at \(\a \) and \(\b \) exist follows from Lemma 8.3: picking \(\gamma \in (\a ,\b )\) arbitrary we have by Lemma 8.3
\[ W(u,v;p)(\delta )=W(u,v;p)(\gamma ) +\int _\gamma ^\delta (Du)(x)v(x)\,w(x)\,dx-\int _\gamma ^\delta u(x)(Dv)(x)\,w(x)\,dx, \]
and the limit of the right-hand side as \(\delta \uparrow \b \) exists since \(u,v\in D_{\max }\); therefore the limit of the left-hand side exists.
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Example 8.7. We reconsider the second derivative operator from Example 7.6 (i.e. \(w=1\), \(p=1\), \(q=0\)). The Wronskian is \(W(u,v;p)=u'v-uv'\). Since the boundary points are regular, we can omit limits when considering the Wronskian at the boundary points \(\a \) and \(\b \).
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Example 8.8. We reconsider Example 6.5 (i.e. \(\a =0\), \(\b =b\), \(w=x\), \(p=x\), \(q=\frac {-\sigma ^2}{x}\)). The Wronskian is \(W(u,v;p)=x(u'v-uv')\). The right boundary point is regular so that we can omit limits when considering the Wronskian at the boundary point \(\b \). The left boundary point is not regular, so we do need to consider a limit for \(W(u,v;p)(\a )\).
Consider \(\sigma =0\) and \(v(x)=\ln (x)\) (we have \(v\in D_{\max }\), in fact we have \(Dv=0\) by Example 7.8). We then have
\[ W(u,\ln (x);x)(\a )=\lim _{x\downarrow 0}x\left (u'(x)\ln (x)-u(x)\frac {1}{x}\right ) =\lim _{x\downarrow 0}\left (u'(x)x\ln (x)-u(x)\right ). \]
Choosing \(u=x\sqrt {x}\) (we have \(u\in D_{\max }\) as shown below) we obtain
\[ W(x\sqrt {x},\ln (x);x)(\a )=\lim _{x\downarrow 0}\left (\frac {3}{2}x\sqrt {x}\ln (x)-x\sqrt {x}\right )=0, \]
by standard limits. Note that due to the presence of \(\ln (x)\) (which is not defined at zero), we really do need to take limits.
We now verify that we indeed have \(u\in D_{\max }\) for \(u=x\sqrt {x}\). We have \(u\in L^2(0,b;x)\) since
\[ \int _0^b (x\sqrt {x})^2x\,dx<\infty , \]
as the integrand is continuous on \([0,b]\) and the integration interval \([0,b]\) is finite. We have \(Du=u''+\frac {1}{x}u'=\frac {3}{4}x^{-1/2}+\frac {3}{2}x^{-1/2}=\frac {9}{4}x^{-1/2}\). This belongs to \(L^2(0,b;x)\) since
\[ \int _0^b (x^{-1/2})^2x\,dx=\int _0^b 1\,dx=b, \]
which is finite.