Green's theorem proof

• May 15th 2011, 04:40 PM
morganfor
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!!
• May 15th 2011, 09:11 PM
TheEmptySet
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.

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

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