pset_4sol

# pset_4sol - ACM 95/100b Problem Set 4 Solutions Faisal...

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Unformatted text preview: ACM 95/100b Problem Set 4 Solutions Faisal Amlani February 8, 2009 Problem 1 (10 points) Using integration by parts, verify the Lagrange identity. That is show that if we define the operator L [ y ( x )] = − d dx parenleftbigg p ( x ) d dx y ( x ) parenrightbigg + q ( x ) y ( x ) , and examine the expression integraldisplay 1 L [ u ( x )] v ( x ) dx = integraldisplay 1 bracketleftbigg − d dx parenleftbigg p ( x ) d dx u ( x ) parenrightbigg v ( x ) + q ( x ) u ( x ) v ( x ) bracketrightbigg dx, then integraldisplay 1 { L [ u ( x )] v ( x ) − u ( x ) L [( v ( x )] } dx = − braceleftbigg p ( x ) bracketleftbigg du dx v ( x ) − u dv dx bracketrightbiggbracerightbiggvextendsingle vextendsingle vextendsingle vextendsingle 1 . Show that the right hand side of the Lagrange identity does indeed vanish if one assumes the homogeneous separable boundary conditions at x = 0 , 1 discussed in class. SOLUTION : Integration by parts on the first term of the integrand yields integraldisplay 1 − d dx parenleftbigg p ( x ) d dx u ( x ) parenrightbigg v ( x ) dx = − p ( x ) du dx v ( x ) vextendsingle vextendsingle vextendsingle 1 + integraldisplay 1 p ( x ) du dx dv dx dx. But by parts integraldisplay 1 p ( x ) du dx dv dx dx = u ( x ) p ( x ) dv dx vextendsingle vextendsingle vextendsingle 1 − integraldisplay 1 u ( x ) d dx parenleftbigg p ( x ) d dx v ( x ) parenrightbigg dx and so integraldisplay 1 − d dx parenleftbigg p ( x ) d dx u ( x ) parenrightbigg v ( x ) dx = − p ( x ) du dx v vextendsingle vextendsingle vextendsingle 1 + up ( x ) dv dx vextendsingle vextendsingle vextendsingle 1 − integraldisplay 1 u ( x ) d dx parenleftbigg p ( x ) d dx v ( x ) parenrightbigg dx = − braceleftbigg p ( x ) bracketleftbigg du dx v − u dv dx bracketrightbiggbracerightbiggvextendsingle vextendsingle vextendsingle vextendsingle 1 − integraldisplay 1 u ( x ) d dx parenleftbigg p ( x ) d dx v ( x ) parenrightbigg dx. 1 Substituting this into the expression, we have integraldisplay 1 L [ u ( x )] v ( x ) dx = integraldisplay 1 bracketleftbigg − d dx parenleftbigg p ( x ) d dx u ( x ) parenrightbigg v ( x ) + q ( x ) u ( x ) v ( x ) bracketrightbigg dx = − braceleftbigg p bracketleftbigg du dx v − u dv dx bracketrightbiggbracerightbiggvextendsingle vextendsingle vextendsingle vextendsingle 1 − integraldisplay 1 u ( x ) d dx parenleftbigg p ( x ) dv dx parenrightbigg dx + integraldisplay 1 q ( x ) u ( x ) v ( x ) dx = − braceleftbigg p bracketleftbigg du dx v − u dv dx bracketrightbiggbracerightbiggvextendsingle vextendsingle vextendsingle vextendsingle 1 + integraldisplay 1 bracketleftbigg − u ( x ) d dx parenleftbigg p ( x ) dv dx parenrightbigg + q ( x ) v ( x ) u ( x ) bracketrightbigg dx = − braceleftbigg p ( x ) bracketleftbigg du dx v ( x ) − u ( x ) dv dx bracketrightbiggbracerightbiggvextendsingle vextendsingle vextendsingle vextendsingle 1 + integraldisplay 1 u ( x ) L [( v ( x )] dx which after rearranging gives integraldisplay...
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## This note was uploaded on 12/12/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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pset_4sol - ACM 95/100b Problem Set 4 Solutions Faisal...

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