Q26E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Boundary Value Problems. When the values of a solution to a differential equation are specified at two different points, these conditions are called boundary conditions. (In contrast, initial conditions specify the values of a function and its derivative at the same point.) The purpose of this exercise is to show that for boundary value problems there is no existence–uniqueness theorem that is analogous to Theorem 1. Given that every solution to (17)y" + y = 0 is of the form y(t) = c₁cost + c₂sint, where c₁ and c₂ are arbitrary constants, show that

 

(a) There is a unique solution to (17) that satisfies the boundary conditionsand $\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{y}\left(\frac{\pi }{2}\right)\mathbf{=}\mathbf{0}$.

(b) There is no solution to (17) that satisfies and $\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$.

(c) There are infinitely many solutions to (17) that satisfy y()) = 2 and $\mathit{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$.

Step-by-Step Solution

Answer

1. The equation has a unique solution $\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathit{t}$ that satisfies the boundary conditions.
2. The equation does not have a solution that satisfies the boundary conditions.
3. The equation has infinitely many solutions that satisfy the boundary conditions.

Step 1: Find the unique solution for boundary conditions.

The given equation is y"+y = 0 and the solution for the equation is y(t)=c₁cost + c₂sint.

 

The given initial conditions are y(0)= 2 and $y\left(\frac{\mathrm{\pi }}{2}\right)=0$

Use the given initial condition.

$\begin{array}{rcl}y\left(0\right)& =& c_{1}cos0+c_2\sin 0\\ 2& =& c_1\\ y\left(\frac{\mathrm{\pi }}{2}\right)& =& c_1\cos \left(\frac{\mathrm{\pi }}{2}\right)+c_2\sin \left(\frac{\mathrm{\pi }}{2}\right)\\ c_2& =& 0\\ y\left(t\right)& =& 2\cos t\end{array}$

Therefore the equation has a unique solution that satisfies the boundary conditions.

Step 2: There is no solution.

The given initial conditions are y(0) = 2 and $y\left(\mathrm{\pi }\right)=0$.

$$\begin{gathered} y\left(0\right)=c_1\cos 0+c_2\sin 0 \\ 2=c_1 \\ y\left(\mathrm{\pi }\right)={\mathrm{c}}_1\cos \left(\mathrm{\pi }\right)+{\mathrm{c}}_2\mathrm{sin\pi } \\ {\mathrm{c}}_1=0 \end{gathered}$$

Since there can’t be two values of c₁.

Therefore, the equation does not have a solution that satisfies the boundary conditions.

Step 3: Show that there are infinitely many solutions.

The given initial conditions are y(0) = 2 and $y\left(\mathrm{\pi }\right)=-2$.

$\begin{array}{rcl}y\left(0\right)& =& c_1\cos 0+c_2\sin 0\\ 2& =& c_1\\ y\left(\mathrm{\pi }\right)& =& {\mathrm{c}}_1\cos \left(\mathrm{\pi }\right)+{\mathrm{c}}_2\mathrm{sin\pi }\\ -{\mathrm{c}}_1& =& -2\\ {\mathrm{c}}_1& =& 2\end{array}$

Therefore the equation has infinitely many solutions that satisfy the boundary conditions.