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Math Help - Green's theorem proof

  1. #1
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    Green's theorem proof

    How does one show that v(x) solves the equation Lv=f

    where L is the differential operator: Lu= -(pu')'+qu=f

    and v(x)= -(int from a to x) f(y)*u1(y)*u2(x)/(p(y)*W(y)) dy - (int from x to b) f(y)*u1(x)*u2(y)/(p(y)*W(y)) dy

    where W(y) is the Wronskian


    Thank you!!
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  2. #2
    Behold, the power of SARDINES!
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    This is just an exercise is plug and chug.

    v(x)=-u_2(x)\underbrace{\int_{a}^{x}\frac{f(y)u_1(y)}{p(  y)W(y)}dy}_{I_1}-u_1(x)\underbrace{\int_{x}^{b}\frac{f(y)u_2(y)}{p(  y)W(y)}dy}_{I_2}

    Now lets compute some derivatives!

    v'(x)=-u_2'(x)I_1-u_2(x)I_1' -u_1'(x)I_2-u_1(x)I_2' =-u_2'(x)I_1 -u_1'(x)I_2

    v''(x)=-u_2''(x)I_1-u_2(x)'I_1'-u_1''(x)I_2-u_1(x)'I_2'

    -\frac{d}{dx}\left(p(x)v'(x) \right)=-p'(x)v'(x)-p(x)v''(x)=

    I_1(-p(x)u_2''(x)-p'(x)u_2'(x))+I_2(-p(x)u_1''(x)-p'(x)u_2'(x))-p(x)[u_2'(x)I_1'+u_1'(x)I_2'])

    W(x)=u_1(x)u_2'(x)-u_2(x)u_1'(x)

    Now just plug into the ODE to get

    I_1(-p(x)u_2''(x)-p'(x)u_2'(x))+I_2(-p(x)u_1''(x)-p'(x)u_2'(x))+p(x)[u_2'(x)I_1'+u_1'(x)I_2'])+q(x)u_2(x)I_1+q(x)u_1(x)I_2

    I_1(-p(x)u_2''(x)-p'(x)u_2'(x)+q(x)u_2(x))+I_2(-p(x)u_1''(x)-p'(x)u_2'(x)+q(x)u_1(x))+p(x)[u_2'(x)I_1'+u_1'(x)I_2'])

    Since u_1 and u_2 satisfy the homogenous equation the first two terms are zero.

    p(x)[u_2'(x)I_1'+u_1'(x)I_2'])=p(x)u_2'(x)\left( \frac{f(x)u_1(x)}{p(x)W(x)}\right)-p(x)u_1'(x)\left( \frac{f(x)u_2(x)}{p(x)W(x)}\right)

    \frac{f(x)}{W(x)}\underbrace{\left(u_2'(x)u_1(x)-u_1'(x)u_2(x)\right)}_{W(x)}=f(x)
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