Example Sheet 2 Solutions
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1 Exmple Sheet Solutions. i L f, g f, L g efinition of joint L g, f property of inner prouct g, Lf efinition of joint Lf, g property of inner prouct ii L L f, g Lf, g L f, g liner opertor property f, L g f, L g efinition of joint f, L g f, Lg from Qu. i f, L Lg liner opertor property b Therefore L L is self-joint. iii We cn write L L f, g Lf, g L f, g Therefore L L is skew-joint. liner opertor f, L g f, L g efinition of joint f, L g f, Lg from Qu. i f, L L g liner opertor L L L L L. Then from the properties shown in Qu ii this is the prouct of self-joint n skew-joint opertor.. i We first pply the opertor n tke the inner prouct s follows v, Lu vxu u u vu vu vu }{{}}{{}}{{} b c
2 Using integrtion by prts on n b gives ] vu vu u v therefore b vu b c vu ] v u ] vu v u ] vu ] u v ] vu v u vu v u vu vu u v v v vu v v vu where the cncelltion is ue to pplying the bounry conitions. This expression implies n in orer to stisfy the reltion L, v, Lu L v, u we require the following joint bounry conitions B {v, v }. ii If we use the prtition foun in Qu iii we hve L L L L L ] ]. 3. i In orer to prove tht this is necessry n sufficient conition we nee to prove the impliction in both irections, i.e. we must prove L is self-joint f, Lf R Let us first prove the impliction. Suppose L is self-joint so L L n so f, Lf L f, f Lf, f f, Lf.
3 Therefore f, Lf is rel vlue since z z z R. To prove the impliction, suppose f, Lf R f S. Then m C n some f, g S where f g, we hve f mg, Lf mg R. Then I f mg, Lf mg ] I f, Lf ] I f, Lmg ] I mg, Lf ] I mg, Lmg ] I f, Lmg ] I mg, Lf ] I m f, Lg ] I m g, Lf ] ] I m f, Lg ] I m Lf, g therfore I m f, Lg ] I I ] m Lf, g ] m Lf, g I m Lf, g ] since I z Iz. Let m, then I f, Lg ] I Lf, g ]. Let m i, then I i f, Lg ] I i Lf, g ], R f, Lg ] I Lf, g ]. Therefore both rel n imginry prts re in self-joint form so L is selfjoint. ii An opertor is positive if the inner prouct rel n positive vlue, n from prt i if the inner prouct is rel then L is self-joint. 4. Let Lu px u qxu, with p >, q n u ub. In orer for L to be positive the inner prouct f, Lf must be rel n positive. Tking the inner prouct, we hve u, Lu ū ū ū px u p u p u 3 ] qxu qūu q u
4 Using integrtion by prts on first term of RHS gives ] b b u, Lu ūpu pu ū q u pplying BCs since p > n q. p u q u 5. iv Using integrtion by prts for the inner prouct we hve v, Lu v u vu ] vu u v vu ] ] vu u v u v vu u v ] vu u v ] L v, u. v v u vu Therefore L L. In orer to stisfy this conition we require ] vu u v vu uv vu uv vu uv vu uv u v v u v v. Therefore B {v v, v v }. Then, since L L n B B the opertor is self-joint. v Using integrtion by prts for the inner prouct we hve v, Lu v u iu u vu i vu u v ] vu ] L v, u. vu vu ] v u i vu v iv vu u v vu 4
5 Therefore L L with B {v, v } n so the opertor is selfjoint. vi Integrting by prts over the given intervl gives v, Lu u x v Therefore with vx u vu vu ] u x v ] u x v u vx vx u] vx u x vu] ] x vu u x v... u x v ] x vu... u v x v u v 4x v x v u v x v x v x u v L x 3x, so the opertor is not self-joint. B {v v, v }, vii Consier the limit s n b. Then v, Lu vu k u vu u v ] b vu u v ] b L v, u. v k vu u v x v vu vu vu u x v 3x v v Therefore L L. In orer to to stisfy this expression we require lim vbu b ubv b lim vu uv b lim vbikub ubv b lim viku uv b lim ubv b ikvb lim uv ikv b 5
6 N.B. Here cre must be tken with signs since complex conjugte over the whole brcket chnges the sign of the imginry prt. We then hve lim x v x ikvx, lim x v x ikvx. So L L but B B so the opertor is only formlly self-joint. 6. Following the usul process we hve v, Lu vpu ru qu p vu ] b p vu p vu ] b u u p v r vu q vu ] b p v r vu ] b u p v... u r v ] b r vu We now choose B so tht the bounry terms re. We then hve v, Lu p v r v q v u p v p v p v p v p v p v r v r v r v then v, Lu L v, u p v p r v p r q v u pv p rv p r qvu... q vu u r v q vu therefore s require. L p p r p r q 7. Let L vlu L vu. Then L vpu ru qu up v p r v p r q v p vu u v r vu u p r v p r v] 6
7 so pulling out erivtives we hve L p vu u v p vu u v p v u u v r vu r vu r v u p v u r v u }{{} leve lone for now thus p r vu p r v u p r vu p vu u v r vu p r vu p vu u v r vu r v u p v u r v u p v u r v u p vu r vu } {{ } r vur v ur vu r vu s require. L p vu u v r p u v 8. i For L to be formlly self-joint we require L L. Compring the coefficients of ech term we hve : : p p p is rel p r r p r r p r r p Rr Rr iir Rr iir constnt : but p Rr so p r q q q p r q Rr Rr iir q q i Ir q q Ir q q i iiq i Iq ii If L is rel then p, q, r re rel functions. From prt i we lrey hve tht p is rel. Then from the constnt term we h Ir Iq, 7
8 but since q n r re rel we hve Ir Iq so the bove expression is stisfie. Tht leves us with the expression p Rr but since r is rel this implies p r s require. 9. i Since u, v re rel we hve ū u n v v. Then vlu ulv v 4 u u 4 v 4 4, v u u v, ii which is n exct ifferentil s require. vlu ulv vu uv ] vu uv ] v u vu u v uv v u vu u v uv... iii First cncel terms in u:... v u vu u v uv vlu ulv v u vu u v uv... then in v:... v u vu u v uv vlu ulv vu u v v u vu. iv In orer to give the require result, we choose the other set of bounry conitions, i.e. φ, φ, φ, φ. 8
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