2
$\begingroup$

I am looking at the following theorem:

If $a_1, a_2 \in \mathbb{R}$ and $x_0,y_0,y_1 \in \mathbb{R}$

then the initial value problem $\left\{\begin{matrix} y''+a_1y'+a_2y=0, x \in \mathbb{R}\\ y(x_0)=y_0\\ y'(x_0)=y_1 \end{matrix}\right.$

has (at least ) one solution.

It is as follows:

We are looking for a solution of the form $\phi(x)=e^{rx}$, where $r \in \mathbb{R}$ or $\mathbb{C}$.

If $\phi$ is a solution of the initial value problem, then it has to hold:

$$\phi''+a_1 \phi'+a_2 \phi=0 (A)$$

and $\phi(x_0)=y_0, \phi'(x_0)=y_1$.

$\phi$ is a solution of the differential equation $A$ iff:

$$(r^2+a_1 r+a_2)e^{rx}=0, x \in \mathbb{R} \ \text{ iff }\\ r^2+a_1r+a_2=0 \to \text{the equation has always a solution in } \mathbb{C}. \text{So, there exists a }\\ r \in \mathbb{C} \text{ so that } \phi(x)=e^{rx} \text{is a solution of } A.$$

The equation $r^2+a_1r+a_2=0$ has:

  • two real roots $r_1, r_2$ with $r_1 \neq r_2$
  • a double root $r$
  • two complex roots: $r_1$ and $r_2=\overline{r_1}$

Then we have the solution of $A$:

  • $\phi_1(x)=e^{r_1 x}, \phi_2(x)=e^{r_2x}$
  • $\phi_1(x)=e^{rx}, \phi_2(x)=xe^{rx}$
  • $\phi_1(x)=e^{r_1x}, \phi_2(x)=e^{\overline{r_1}x}$

We are looking for a solution of the initial value problem of the form

$$\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x)$$

Then $\phi$ satifies the differential equation $A$ $\forall c_1, c_2 \in \mathbb{R}$ (or $\mathbb{C}$) and so it suffices that $\phi$ satisfies the following:

$$c_1 \phi_1(x_0)+c_2 \phi_2(x_0)=y_0\\ c_1 \phi_1'(x_0)+c_2 \phi_2'(x_0)=y_1$$

The above system has a unique solution as for $c_1, c_2$ if:

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix} \neq 0$$

  • If $r_1, r_2$ real roots with $r_1 \neq r_2$ then

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}=(r_2-r_1)e^{(r_1+r_2)x_0} \neq 0$$

  • If $r$ is a double root, then:

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}=e^{2rx} \neq 0$$

  • If $r_1, r_2$ complex solutions, then:

$$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}=(\overline{r_1}-r_1)e^{(r_1+\overline{r_1})x_0} \neq 0$$

Since in each case, the determinant $$\begin{vmatrix} \phi_1(x_0) & \phi_2(x_0) \\ \phi_1'(x_0) & \phi_2'(x_0) \end{vmatrix}$$ is $\neq 0$ doesn't this imply that the initial value problem $\left\{\begin{matrix} y''+a_1y'+a_2y=0, x \in \mathbb{R}\\ y(x_0)=y_0\\ y'(x_0)=y_1 \end{matrix}\right.$ has exactly one solution?

$\endgroup$
1
$\begingroup$

Yes, it is true that there is a exactly one solution. But it is also true that there is at least one solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.