Nonlinear problems

Abstract

Influence functions, as the influence function for the longitudinal displacement u(x)of a rod

$${{u}_{1}}\left( x \right)=\underset{0}{\overset{1}{\mathop{\int }}}\,{{G}_{0}}\left( y,x \right){{p}_{1}}\left( y \right)dy,$$

(5.1)

are L₂-scalar products between the Green’s function and the exterior load p₁ and because the scalar product is distributive

$$underset{0}{\overset{1}{\mathop{\int }}}\,{{G}_{0}}\left( {{p}_{1}}+{{p}_{2}} \right)dy=\underset{0}{\overset{1}{\mathop{\int }}}\,{{G}_{0}}{{p}_{1}}dy+\underset{0}{\overset{1}{\mathop{\int }}}\,{{G}_{0}}{{p}_{2}}dy,$$

the influence function for a nonlinear equation cannot be of the form (5.1). If Eq.(5.1) were the solution of the problem

$${{D}^{NL}}u={{p}_{1}},$$

and

$${{u}_{2}}\left( x \right)=\underset{0}{\overset{1}{\mathop{\int }}}\,{{G}_{0}}\left( y,x \right){{p}_{2}}\left( y \right)dy$$

the solution of the problem

$${{D}^{NL}}u={{P}_{2}},$$

then the functionld

$$u\left( x \right)={{u}_{1}}\left( x \right)+{{u}_{2}}\left( x \right)=\underset{0}{\overset{1}{\mathop{\int }}}\,{{G}_{0}}\left( y,x \right)\left( {{p}_{1}}\left( y \right)+{{p}_{2}}\left( y \right) \right)dy$$

Would be the solution of the problem

$${{D}^{NL}}\left( {{u}_{1}}+{{u}_{2}} \right)={{p}_{1}}+{{p}_{2}}.$$

But this is a contradiction.