MATH 308 Lecture 12

« previous | Wednesday, February 13, 2013 | next »

Lecture Notes

Theorem of Existence and Uniqueness

$y''+p(t)\,y+q(t)\,y=g(t)\quad y(t_{0})=y_{0}\quad y'(t_{0})=y'_{0}$

Only one solution

Exercise 8

Find the largest interval on which the solution exists.

${\begin{aligned}t(t-1)y''+3t\,y'+4y&=2&y(-1)&=0&y'(-1)&=3\\y''+{\frac {3}{t-1}}\,y'+{\frac {4}{t(t-1)}}\,y&={\frac {2}{t(t-1)}}&t&\not \in \{0,1\}\end{aligned}}$

Since the function is not continuous at $t=t_{0}=0$, the theorem does not apply.

Exercise 9

Can $y=\sin {t^{2}}$ be a solution to a second order homogeneous linear equation on an interval containing 0?

(i.e. $y(0)=0$)

${\begin{aligned}y&=\sin {t^{2}}\\y'&=2t\,\cos {t^{2}}\\y''&=-4t^{2}\,\sin {t^{2}}+2\cos {t^{2}}y''+p(t)\,y'+q(t)\,y&=0\end{aligned}}$

Since the function is homogeneous, it has solution $y(t)=0$. Assuming $p$ and $q$ are continuous for all real numbers, the theorem applies and states that there will be only one unique solution.

Therefore $\sin {t^{2}}$ cannot be a solution.

Note: The constant 0 function is a solution to any homogeneous linear differential equation.

Wronskian

Applies to homogeneous equation; may have non-constant coefficients.

From the beginning of the section, we found that if $y_{1}$ and $y_{2}$ are solutions, then any linear combination $c_{1}\,y_{1}+c_{2}\,y_{2}$ is a solution.

$W(y_{1},y_{2},\ldots ,y_{n})={\begin{vmatrix}y_{1}&y_{2}&\ldots &y_{n}\\{y_{1}}'&{y_{2}}'&\ldots &{y_{n}}'\\\vdots &\vdots &\ddots &\vdots \\{y_{1}}^{(n-1)}&{y_{2}}^{(n-1)}&\ldots &{y_{n}}^{(n-1)}\end{vmatrix}}$

Exercise 10

$y''+y=0$

$y_{1}=\sin {x}$ is a solution, and $y_{2}=\cos {\left(x+{\frac {\pi }{2}}\right)}$ is also a solution. Therefore, $y=c_{1}\,\sin {x}+c_{2}\,\cos {\left(x+{\frac {\pi }{2}}\right)}$ is a general solution.

For the initial value problem $y(0)=1$, $y'(0)=0$, the theorem of existence and uniqueness implies that there exists a (unique) solution $Y$ to the initial value problem. However, $y(0)=0\neq 1$, which is impossible!

The theorem guarantees that $Y$ exists.

Notice that $\cos {\left(x+{\frac {\pi }{2}}\right)}=-\sin {x}$, so our $y$ is essentially the function $y=c_{1}\,\sin {x}-c_{2}\,\sin {x}=c_{3}\,\sin {x}$.

What about $y=c_{1}\,\sin {x}+c_{2}\cos {x}$? This satisfies the initial value problem with $c_{1}=0$ and $c_{2}=1$.

${\begin{aligned}W(\sin {x},\cos {\left(x+{\frac {\pi }{2}}\right)}&=\ldots =-\cos {\left(x-\left(x+{\frac {\pi }{2}}\right)\right)}=0\\W(\sin {x},\cos {x})&=-1\end{aligned}}$

Theorem

Suppose $y_{1}$ and $y_{2}$ are two solutions of the equation

$y''+p(t)\,y'+q(t)\,y=0$

For any initial conditions $y(t_{0})=y_{0}$, $y'(t_{0})=y'_{0}$, there exists two constants $c_{1}$ and $c_{2}$ such that $y=c_{1}\,y_{1}+c_{2}\,y_{2}$ is a solution to the initial value problem iff $W(y_{1},y_{2})\neq 0$ at $t_{0}$.

The solutions to the differential equation is the set of functions

$y=c_{1}\,y_{1}+c_{2}\,y_{2}$

iff $W(y_{1},y_{2})\neq 0$ at $t_{0}$.

$y_{1}$ and $y_{2}$ are said to form a fundamental set of solutions if $W(y_{1},y_{2})\neq 0$

Abel's Theorem

If $y_{1}$ and $y_{2}$ are two solutions to the differential equation $y''+p(t)\,y'+q(t)\,y=0$, then there exists a constant $C$ such that

$W(y_{1},y_{2})=C\,\exp {\left(-\int p(t)\,\mathrm {d} t\right)}$

Exercise 13

If $y_{1}$ and $y_{2}$ are two solutions of the differential equation $t^{2}\,y''-2y'+t\,\mathrm {e} ^{t}\,y=0$, and if $W(y_{1},y_{2})(2)=3$, find $W(y_{1},y_{2})(4)$.

${\begin{aligned}W(y_{1},y_{2})=C\,\mathrm {e} ^{\int {\frac {2}{t^{2}}}\,\mathrm {d} t}=C\,\mathrm {e} ^{-{\frac {2}{t}}}\\W(y_{1},y_{2})(2)&=3=C\,\mathrm {e} ^{\frac {2}{2}}={\frac {C}{\mathrm {e} }}\\C&=3\mathrm {e} \\W(y_{1},y_{2})(4)&=C\,\mathrm {e} ^{-{\frac {2}{4}}}=3\mathrm {e} \,\mathrm {e} ^{-{\frac {1}{2}}}=3{\sqrt {\mathrm {e} }}\end{aligned}}$