User:Egm6321.f10.team4.petralanda.n

Problem Statement
Show that the second order differential equation given bellow is non-linear with respect to $$y^1$$.

$$\begin{align} {{c}_{3}}({{y}^{1}},t)\frac{\partial {{t}^{2}}} \\ \end{align}$$ where $$ {{c}_{3}}=M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right]$$

Solution
For an equation to be linear it has to satisfy both of the following properties.

Property 1:

$$ H(\alpha )=\alpha H $$

Property 2:

$$ \text{H(}\alpha \text{+}\beta \text{)=H(}\alpha \text{)+H(}\beta \text{)} $$

Based on the first property, the given equation will take the following form

$$ M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u(\alpha {{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}(\alpha {{y}^{1}})}{\partial {{t}^{2}}}=\alpha M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}({{y}^{1}})}{\partial {{t}^{2}}} $$

By definition, α is an arbitrary real number which would allow for the following expansion.

$$\alpha M\left[ 1-\alpha \bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}({{y}^{1}})}{\partial {{t}^{2}}}=\alpha M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}({{y}^{1}})}{\partial {{t}^{2}}}$$

It is now clear that the left and right hand sides of the equation are not equal. Therefore, this particular equation is non-linear with respect to $$y^1 $$