Q31 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Consider the question of Example 5 $\mathbf{y}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{0}$

1. Does Theorem 1 imply the existence of a unique solution to (13) that satisfies $\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$?
2. Show that when ${\mathbf{x}}_{\mathbf{0}}\mathbf{\ne }\mathbf{0}$ equation (13) can’t possibly have a solution in a neighbourhood of $\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$ that satisfies $\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$.
3. Show two distinct solutions to (13) satisfying $\mathbf{y}\left(0\right)\mathbf{=}\mathbf{0}$ ( See Figure 1.4 on page 9).

Step-by-Step Solution

Answer

1. Theorem 1 does not imply the existence of a unique solution that satisfies $\mathrm{y}\left({\mathrm{x}}_0\right)=0$.
2. Equation (13) doesn’t have a solution in the neighbourhood of $\mathrm{x}={\mathrm{x}}_0$ that satisfies $\mathrm{y}\left({\mathrm{x}}_0\right)=0$.
3. There are two distinct solutions to (13) satisfying $\mathrm{y}\left(0\right)=0$.

Step 1: Finding the partial derivative of the given relation concerning y

Here, $f\left(x,y\right)=\frac{4x}{y}$ and $\frac{\partial f}{\partial y}=-\frac{4x}{y^2}$.

Step 2(a): Determining whether Theorem 1 implies the existence of a unique solution or not

From Step 1, we find that $\frac{\partial f}{\partial y}$ is not continuous or even defined when $\mathrm{y}=0$. Consequently, there is no rectangle containing the point $({\mathrm{x}}_0=0)$, in which both $\mathrm{f}(\mathrm{x},\mathrm{y})$ and $\frac{\partial f}{\partial y}$ are continuous.

Because the hypotheses of Theorem 1 do not hold, we cannot use Theorem 1 to determine whether the given initial value problem does or does not have a unique solution. It turns out that this initial value problem has more than one solution.

 

Hence, Theorem 1 does not imply the existence of a unique solution that satisfies $\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$.

Step 3(b): Showing that the equation (13) has a solution in a neighbourhood of x = x 0 that satisfies y ( x 0 ) = 0

Because y comes in the denominator, the function is not defined at $\mathrm{y}=0$.

Hence, when ${\mathit{x}}_{\mathbf{0}}\mathbf{\ne }\mathbf{0}$  equation (13) can’t possibly have a solution in a neighbourhood of $\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$ that satisfies $\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$.

Step 4(c): Showing that there are two distinct solutions to (13) satisfying y ( 0 ) = 0 .

In Figure 1.4, the implicit solutions for $\mathrm{c}=0,\pm 1,\pm 4$ are sketched.

The curves are hyperbolas with common asymptotes $y=\pm 2x$. Notice that the implicit solution curves (with c arbitrary) fill the entire plane and are non-intersecting for $\mathrm{c}\ne 0$. For $\mathrm{c}=0$, the implicit solution gives rise to the two explicit solutions $\mathrm{y}=2\mathrm{x}$ and $y=-2x$, both of which pass through the origin.

Thus, the two distinct solutions to (13) satisfying $\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ are $\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}$ and $\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}$.