MATH 308 Lecture 12

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

Lecture Notes

Theorem of Existence and Uniqueness

$y^{″}+p(t)y+q(t)y=g(t)y(t_0)=y_0{y}^{\prime }(t_0)=y{\text{'}}_0$

Only one solution

Exercise 8

Find the largest interval on which the solution exists.

$\begin{array}{rlrlrl}t(t-1)y^{″}+3t{y}^{\prime }+4y& =2& y(-1)& =0& y^{\prime }(-1)& =3\\ y^{″}+\frac{3}{t-1}{y}^{\prime }+\frac{4}{t(t-1)}y& =\frac{2}{t(t-1)}& t& \in ̸\{0,1\}\end{array}$

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. $y(0)=0$)

$\begin{array}{rl}y& =\sin t^2\\ y^{\prime }& =2t\cos t^2\\ y^{″}& =-4t^2\sin t^2+2\cos t^2{y}^{″}+p(t)y^{\prime }+q(t)y& =0\end{array}$

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.

$W(y_1,y_2,\dots ,y_n)=\left|\begin{array}{cccc}{y}_1& y_2& \dots & y_n\\ {y_1}^{\prime }& {y_2}^{\prime }& \dots & {y_n}^{\prime }\\ ⋮& ⋮& \ddots & ⋮\\ {y_1}^{(n-1)}& {y_2}^{(n-1)}& \dots & {y_n}^{(n-1)}\end{array}\right|$

Exercise 10

$y^{″}+y=0$

$y_1=\sin x$ is a solution, and $y_2=\cos (x+\frac{\pi }{2})$ is also a solution. Therefore, $y=c_1\sin x+c_2\cos (x+\frac{\pi }{2})$ is a general solution.

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

The theorem guarantees that $Y$ exists.

Notice that $\cos (x+\frac{\pi }{2})=-\sin x$, 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 $c_2=1$.

$\begin{array}{rl}W(\sin x,\cos (x+\frac{\pi }{2})& =\dots =-\cos (x-(x+\frac{\pi }{2}))=0\\ W(\sin x,\cos x)& =-1\end{array}$

Theorem

Suppose $y_1$ and $y_2$ are two solutions of the equation

$y^{″}+p(t)y^{\prime }+q(t)y=0$

For any initial conditions $y(t_0)=y_0$, $y^{\prime }(t_0)=y{\text{'}}_0$, 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 $W(y_1,y_2)\ne 0$ at $t_0$.

The solutions to the differential equation is the set of functions

$y=c_1{y}_1+c_2{y}_2$

iff $W(y_1,y_2)\ne 0$ at $t_0$.

$y_1$ and $y_2$ are said to form a fundamental set of solutions if $W(y_1,y_2)\ne 0$

Abel's Theorem

If $y_1$ and $y_2$ are two solutions to the differential equation $y^{″}+p(t)y^{\prime }+q(t)y=0$, then there exists a constant $C$ such that

$W(y_1,y_2)=C\exp (-\int p(t)\mathrm{d}t)$

Exercise 13

If $y_1$ and $y_2$ are two solutions of the differential equation $t^2{y}^{″}-2y^{\prime }+t{\mathrm{e}}^{t}y=0$, and if $W(y_1,y_2)(2)=3$, find $W(y_1,y_2)(4)$.

$\begin{array}{r}W(y_1,y_2)=C{\mathrm{e}}^{\int \frac{2}{t^2}\mathrm{d}t}=C{\mathrm{e}}^{-\frac{2}{t}}\\ W(y_1,y_2)(2)& =3=C{\mathrm{e}}^{\frac{2}{2}}=\frac{C}{\mathrm{e}}\\ C& =3\mathrm{e}\\ W(y_1,y_2)(4)& =C{\mathrm{e}}^{-\frac{2}{4}}=3\mathrm{e}{\mathrm{e}}^{-\frac{1}{2}}=3\sqrt{\mathrm{e}}\end{array}$