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I am looking at the proof of the following theorem.

Theorem(General solution of the homogeneous differential equation of second order with constant coefficients)

Let the differential equation $y''+a_1y'+a_2y=0, \ a_1, a_2 \in \mathbb{R}$.

We consider the characteristic polynomial of the above differential equation $p(r)=r^2+a_1r+a_2$.

Then, if there are two different solutions of the polynomial $p$, let $r_1, r_2 (r_1, r_2 \in \mathbb{R} \text{ or } r_1,r_2 \in \mathbb{C})$ then the functions $\phi_1(x)=e^{r_1 x}, \phi_2(x)=e^{r_2 x}$ are solutions of the differential equation in $\mathbb{R}$.

If there is a double root of the polynomial $p$, let $r$, then the functions $\phi_1(x)=e^{rx}, \phi_2(x)=xe^{rx}$ are solutions of the differential equation in $\mathbb{R}$.

It holds that : if $\phi$ is a solution of the differential equation in $\mathbb{R}$ then there are $c_1, c_2 \in \mathbb{R} (\text{ or } \mathbb{C})$ so that $\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x) \forall x \in \mathbb{R}$ and obviously for all $c_1, c_2 \in \mathbb{R} (\text{ or } \mathbb{C})$ the function $c_1 \phi_1(x)+c_2 \phi_2(x)$ is a solution of the differential equation.

Proof

We consider the solutions $\phi_1, \phi_2$ of the differential equation as at the formulation of the theorem.

Let $\phi$ be a random solution of the differential equation in $\mathbb{R}$.

We pick a random $x_0 \in \mathbb{R}$.

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

We consider the initial value problem $\left\{\begin{matrix} y''+a_1y'+a_2y=0\\ y(x_0)=y_0\\ y'(x_0)=y_1 \end{matrix}\right.$.

Then we know that there is a solution $\psi$ of the initial value problem of the form $\psi(x)=c_1 \phi_1(x)+c_2 \phi_2(x)$ for appropriate $c_1, c_2$.

Furthermore, $\phi$ is a solution of the same initial value problem.

From the uniqueness of the initial value problem we have $\phi(x)=\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x), \forall x \in \mathbb{R}$.

I haven't understood the following:

We pick a random $x_0$ such that $\phi(x_0)=y_0$ and $\phi'(x_0)=y_1$ and then we consider the initial value problem.

How do we deduce that $c_1 \phi_1(x_0)+ c_2 \phi_2(x_0)=y_0$ and $c_2 \phi_1'(x_0)+c_2 \phi_2'(x)=0$, i.e. that $c_1 \phi_1(x)+c_2 \phi_2(x)$ is a solution of the specific initial value problem?

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2 Answers 2

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How do we deduce that $c_1 \phi_1(x_0)+ c_2 \phi_2(x_0)=y_0$ and $c_2 \phi_1'(x_0)+c_2 \phi_2'(x)=0$, i.e. that $c_1 \phi_1(x)+c_2 \phi_2(x)$ is a solution of the specific initial value problem?

We do it the other way. Pick any $x_0\in R$ and calculate the value of $\phi(x_0)$ and $\phi'(x_0)$. Now choose $c_1$ and $c_2$ such that they satisfy the equations: $$c_1\phi_1(x_0)+c_2\phi_2(x_0)=\phi(x_0)$$ $$c_1\phi_1'(x_0)+c_2\phi_2'(x_0)=\phi'(x_0)$$

Now $\psi = c_1\phi_1+c_2\phi_2$ also satisfies the initial value problem satisfied by $\phi$. By uniqueness, $\phi=\psi$.

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    $\begingroup$ I see.. zed111 But could you also explain why we are sure that there are such $c_1, c_2$ ? $\endgroup$
    – evinda
    Commented Feb 26, 2015 at 23:13
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    $\begingroup$ Nice question. This is because $\phi_1$ and $\phi_2$ are linearly independent solutions so that Wronskian is non-zero. The Wronskian is also the determinant of the coefficient matrix for solving the system of 2 equations. And we know that if the determinant of the coefficient matrix is non-zero, we have a unique solution for the equation. $\endgroup$
    – zed111
    Commented Feb 27, 2015 at 4:12
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Let $\phi_1$ and $\phi_2$ be linearly independent solutions to the differential equation. Then $$ (c_1\phi_1 + c_2\phi_2)'' + a_1(c_1\phi_1 + c_2\phi_2)' + a_2(c_1\phi_1 + c_2\phi_2) $$ $$ = $$ $$ c_1(\phi_1'' + a_1\phi_1' + a_2\phi_1) + c_2(\phi_2'' + a_1\phi_2' + a_2\phi_2). $$ So, any linear combination of $\phi_1$ and $\phi_2$ is a solution to the differential equation. Now, we're given the boundary conditions $y(x_0) = y_0$ and $y'(x_0) = y_1$. We want to find $c_1$ and $c_2$ such that $$ c_1\phi_1(x_0) + c_2\phi_2(x_0) = y_0 $$ $$ c_1 \phi_1'(x_0) + c_2\phi_2'(x_0) = y_1. $$ But since $x_0$ is fixed, $\phi_i(x_0), \phi_i'(x_0)$ are just some fixed constants. So the above system of 2 equations is linear with two variables, $c_1$ and $c_2$. We can always solve such a system (since $\phi_1$ and $\phi_2$ are linearly independent), so we have the desired constants.

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  • $\begingroup$ We have suppose that $\phi_1, \phi_2$ are solutions of the differential equation and so we decude that any linear combination of them is also a solution. $$$$ But how do we know that $c_1 \phi_1(x_0)+c_2 \phi_2(x_0)=y_0$ and $c_1 \phi_1'(x_0)+c_2 \phi_2'(x_0)=y_1$? $$$$ Do we have an initial condition for $\phi_1$ and $\phi_2$? $\endgroup$
    – evinda
    Commented Feb 26, 2015 at 9:01
  • $\begingroup$ $c_1$ and $c_2$ are chosen to yield a solution to the boundary value problem. Since any solution to the DE is a linear combination of $\phi_1$ and $\phi_2$, there exist $c_1$ and $c_2$ such that $c_1\phi_1 + c_2\phi_2$ is a solution to the boundary value problem. $\endgroup$ Commented Feb 26, 2015 at 9:06
  • $\begingroup$ I haven't understood why we can find $c_1$, $c_2$ such that these initial conditions are satisfied. :/ Could you explain it further to me? $\endgroup$
    – evinda
    Commented Feb 26, 2015 at 9:10
  • $\begingroup$ Sure. Let my edit my answer. $\endgroup$ Commented Feb 26, 2015 at 9:12

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