- #1

#### member 428835

Hi PF!

Given operator ##B## defined as $$ B[u(s)] = c u(s) - u''(s) - \frac{1}{2 s_0}\int_{-s_0}^{s_0}(c u(s) - u''(s))\, ds$$ I'm trying to find it's inverse operator ##B^{-1}##. The journal I'm reading states ##B^{-1}## is an integral operator $$B^{-1}(u(s)) = \int_{-s_0}^{s_0}G(s,z)u(s)\,ds$$ whose kernal is the Green's function of ##B(u)=f : \pm u'(\pm s_0)+c \cot(\alpha) u(\pm s_0) = 0##. The paper I'm reading applies variation of parameters to conclude that

where evidently ##v_1## and ##v_2## solve $$cv(s)-v''(s) = 0;\\

v_1(0)=0,\,v_1'(0) = 1;\\

v_2(0)=1,\,v_1'(0) = 0.$$

Any idea how they determined the ##w## function? I've looked everywhere but can't find a link. I should possibly state that ##c = \cos \alpha## where ##\alpha## is constant.

Given operator ##B## defined as $$ B[u(s)] = c u(s) - u''(s) - \frac{1}{2 s_0}\int_{-s_0}^{s_0}(c u(s) - u''(s))\, ds$$ I'm trying to find it's inverse operator ##B^{-1}##. The journal I'm reading states ##B^{-1}## is an integral operator $$B^{-1}(u(s)) = \int_{-s_0}^{s_0}G(s,z)u(s)\,ds$$ whose kernal is the Green's function of ##B(u)=f : \pm u'(\pm s_0)+c \cot(\alpha) u(\pm s_0) = 0##. The paper I'm reading applies variation of parameters to conclude that

where evidently ##v_1## and ##v_2## solve $$cv(s)-v''(s) = 0;\\

v_1(0)=0,\,v_1'(0) = 1;\\

v_2(0)=1,\,v_1'(0) = 0.$$

Any idea how they determined the ##w## function? I've looked everywhere but can't find a link. I should possibly state that ##c = \cos \alpha## where ##\alpha## is constant.

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