User:Egm6321.f10.team1.lang/HW1

To prove that $${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}$$ is a nonlinear equation with respect to $${{y}^{1}}$$, we will consider a specific case. In an effort to make obvious the nonlinear nature of the of the previous equation we choose the trigonometric function sine to be included in our definition of $${{c}_{3}}({{y}^{1}},t)$$.

Let

$${{c}_{3}}(y,t)=\sin (y)+{{t}^{2}}$$

Next we need to define a criterion to determine whether or not a given equation is linear or nonlinear. For L to be considered a linear operator, it must have the following properties:

$$L(\phi +\psi )=L(\phi )+L(\psi )$$

$$L(k\phi )=kL(\phi )$$

where k is any scalar.

In order to test our newly formed rules for linearity (and simplify the writing), lets express the equation

$${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}=[\sin ({{y}^{1}})+{{t}^{2}}]\frac{d{{t}^{2}}}$$

as an operator L.

Let

$$L(\cdot )=[\sin (\cdot )+{{t}^{2}}]\frac{{{d}^{2}}(\cdot )}{d{{t}^{2}}}$$

We may now apply the prior definitions for linearity as a test for whether L is linear or nonlinear. First we check if the distributive property holds by computing the LHS of the defining equation for linear operators.

$$L(\phi +\psi )=[\sin (\phi +\psi )+{{t}^{2}}]\frac{{{d}^{2}}(\phi +\psi )}{d{{t}^{2}}}$$

and using the trigonometric identity for sine:

$$\sin (\phi +\psi )=\sin (\phi )\cos (\psi )+\cos (\phi )\sin (\psi )$$

we arrive at the following:

$$=[\sin (\phi )\cos (\psi )+\cos (\phi )\sin (\psi )+{{t}^{2}}][\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]$$

which then becomes

$$=[\sin (\phi )\cos (\psi )[\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]+\cos (\phi )\sin (\psi )[\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]+{{t}^{2}}[\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\frac{{{d}^{2}}\psi }{d{{t}^{2}}}]$$

and expanding into the 6 term expression we have

$$=[\sin (\phi )\cos (\psi )\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\sin (\phi )\cos (\psi )\frac{{{d}^{2}}\psi }{d{{t}^{2}}}+\cos (\phi )\sin (\psi )\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\cos (\phi )\sin (\psi )\frac{{{d}^{2}}\psi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\psi }{d{{t}^{2}}}$$

We will see in a momment why I picked the trigonometric funcation sine to demonstrate a test for linearity of an operator. Now we compute the RHS of the defining equation for linear operators.

$$L(\phi )+L(\psi )=[\sin (\phi )+{{t}^{2}}]\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+[\sin (\psi )+{{t}^{2}}]\frac{{{d}^{2}}\psi }{d{{t}^{2}}}$$

distributing the derivatives over the sums we arrive at the following 4 term expression

$$=\sin (\phi )\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\phi }{d{{t}^{2}}}+\sin (\psi )\frac{{{d}^{2}}\psi }{d{{t}^{2}}}+{{t}^{2}}\frac{{{d}^{2}}\psi }{d{{t}^{2}}}$$

As we can see, the nonlinearities in L with respect to $${{y}^{1}}$$ are obvious in the sine term, and L as defined does not distribute over the given sum. If L did distribute, then we would have seen equivalent expressions between both the RHS and LHS of the defining equation for linearity (i.e. the RHS gave 6 terms which were not equivalent to the 4 terms the LHS gave). Therefore, $${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}$$ is a nonlinear equation with respect to $${{y}^{1}}$$.

Egm6321.fall10.team1.lang 07:22, 14 September 2010 (UTC)