User:EGM6321.f12.team7.Zhou/R1

=Problem 6: Proof of nonlinearity of $$\displaystyle C_3(Y^1,t)\ddot Y^1$$=

From lecture note {}

Given
From the lecture note[], general expression of $$\displaystyle C_3(Y^1,t)\ddot Y^1$$ is given by,
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$$\displaystyle c_3(Y^1,t)\ddot Y^1=[1-\bar{R}u_{,ss}(Y^1,t)]\ddot Y^1$$ (6.1) From the lecture note[}, we can Consider $$\displaystyle F(\cdot) $$ as an operator, $$\displaystyle F(\cdot)$$ is linear if and only if,
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$$\displaystyle F(\alpha f+\beta g)=\alpha F(f)+\beta F(g)$$ (6.2) Where$$\displaystyle f$$ and $$\displaystyle g$$ are two possible arguments for $$\displaystyle F(\cdot)$$, and $$\displaystyle \forall \alpha, \beta \in R.$$
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The derivative operator is a linear operator, which is given by,
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$$\displaystyle F(\cdot)=\frac{(\cdot)}{dx}$$ (6.3)
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Solution
Suppose we have,
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$$\displaystyle Y^1:[t_0,+\infty) \rightarrow R$$
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$$\displaystyle \forall u, v:[t_0,+\infty) \rightarrow R$$

$$\displaystyle \forall \alpha, \beta\in R$$ (6.4) For the right-hand side, we have,
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$$\displaystyle RHS:=\alpha C_3(u,t)\frac{\alpha d^2u}{dt^2}+\beta C_3(v,t)\frac{\beta d^2v}{dt^2}$$ (6.5) The left-hand side is given by,
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$$\displaystyle LHS:=C_3(\alpha u+\beta v,t)\frac{d^2 (\alpha u+\beta v)}{dt^2}$$ (6.6) Since the derivative operator is linear operator, we can get,
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$$\displaystyle \frac{d^2 (\alpha u+\beta v)}{dt^2} = \frac{\alpha d^2u}{dt^2} +\frac{\beta d^2v}{dt^2}$$ (6.7) So (6.6) turns into,
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$$\displaystyle C_3(\alpha u+\beta v,t)\frac{d^2 (\alpha u+\beta v)}{dt^2}=C_3(\alpha u+\beta v,t)\frac{\alpha d^2u}{dt^2} +C_3(\alpha u+\beta v,t)\frac{\beta d^2v}{dt^2}$$ (6.8) If $$\displaystyle C_3(Y^1,t)$$ is nonlinear with respect to $$\displaystyle Y^1$$, by comparision between (6.5) and (6.8) we can find that,
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$$\displaystyle LHS\neq RHS $$ (6.9) Even if $$\displaystyle C_3(Y^1,t)$$ is linear with respect to $$\displaystyle Y^1$$, on the left-hsnd side we have,
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$$\displaystyle LHS:=[\alpha C_3(u,t)+\beta C_3(v,t)]\frac{\alpha d^2u}{dt^2}+[\alpha C_3(u,t)+\beta C_3(v,t)]\frac{\beta d^2v}{dt^2}$$ (6.10) Obviously,
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$$\displaystyle LHS\neq RHS $$ (6.11)
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