Q29 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

(a) For the initial value problem (12) of Example 9. Show that ${\mathbf{\varphi }}_{\mathbf{1}}\left(x\right)\mathbf{=}\mathbf{0}$ and ${\mathbf{\varphi }}_{\mathbf{2}}\left(x\right)\mathbf{=}{\left(x-2\right)}^{\mathbf{3}}$ are solutions. Hence, this initial value problem has multiple solutions. (See also Project G in Chapter 2.)

(b) Does the initial value problem $\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathit{y}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{,}$ $\mathbf{y}\left(0\right)\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}$ have a unique solution in a neighbourhood of $\mathbf{x}\mathbf{=}\mathbf{0}$?

Step-by-Step Solution

Answer

1. For the given initial value problem (12), ${\varphi }_1\left(\mathrm{x}\right)=0$ and ${\varphi }_2\left(\mathrm{x}\right)={\left(\mathrm{x}-2\right)}^3$ are solutions. Hence, this initial value problem has multiple solutions.
2.  The initial value problem $\mathrm{y}\text{'}=3{\mathrm{y}}^{\frac{2}{3}}, \mathrm{y}\left(0\right)={10}^{-7}$ has a unique solution in a neighbourhood of $\mathrm{x}=0$.

Step 1(a): Showing that the given initial value problem (12) has multiple solutions

Clearly, ${\varphi }_1\left(\mathrm{x}\right)=0$ as y is a solution to the given initial value problem.

Now taking $\mathrm{y}={\mathrm{\varphi }}_2\left(\mathrm{x}\right)={\left(\mathrm{x}-2\right)}^3$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=3{\left(x-2\right)}^2\\ \frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{y}}^{\frac{2}{3}}\end{array}$

(Because, $\mathrm{y}={\left(\mathrm{x}-2\right)}^3$,  i.e., ${\mathrm{y}}^{\frac{1}{3}}=\left(\mathrm{x}-2\right)$)

which is identical to the given differential equation. So, ${\mathrm{\varphi }}_2\left(\mathrm{x}\right)={\left(\mathrm{x}-2\right)}^3$ is a solution to the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{y}}^{\frac{2}{3}}$. 

Hence, this initial value problem has multiple solutions.

Step 2: Finding the partial derivative of the given relation with respect to y

Here, $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=3{\mathrm{y}}^{\frac{2}{3}}$

$\begin{array}{l}\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=2{\mathrm{y}}^{\frac{-1}{3}}\\ \frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\frac{2}{{\mathrm{y}}^{\frac{1}{3}}}\end{array}$     

which is continuous in any rectangle containing the point $\left(0,{10}^{-7}\right)$. 

Step 3(b): Verify whether the given initial value problem y ' = 3 y 2 3 ,   y ( 0 ) = 10 - 7 has a unique solution in a neighbourhood of x = 0 or not

Now from Step 1, we find that both of the functions $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ and $\frac{\partial f}{\partial y}$ are continuous in any rectangle containing the point $\left(0,{10}^{-7}\right)$, so the hypotheses of Theorem 1 are satisfied.

It then follows from the theorem that the given initial value problem has a unique solution in an interval about $x=0$ of the form $\left(-\delta ,\delta \right)$, where $\delta$ is some positive number. 

Therefore, the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathbf{y}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{,}\mathbf{ }\mathbf{y}\left(0\right)\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}$ have a unique solution in a neighbourhood of $\mathbf{x}\mathbf{=}\mathbf{0}$.