Prove if wronskian is zero then solutions of ODE are linearly dependant

calculusordinary differential equations

Suppose we have $$y'' + P(x)y' + Q(x)y = 0 $$

and let the solutions be $y_1, y_2$ in $[a,b]$. The wronskian is $y_1 y_2' – y_2 y_1'$, which is identically zero. Now lets consider 3 cases:

If $y_1$ is identically zero, then so is $W(x)$ and so $y_1 = 0(y_2)$ so they are linearly dependant.
If $y_1$ is never zero in $[a,b]$ then $\frac{W(x)}{y_1^2} = \left(\frac{y_2}{y_1}\right)' = 0$ or $y_2 = c y_1$ so they are L.D.
What if $y_1$ is zero only on some points in $[a,b]$ and say $y_1'$ is also zero at those points, how do we show $y_2, y_2'$ is also zero at those points

Best Answer

If the list $\{y_1, y_2 \} $ were independent, then we can find nontrivial $(c_1, c_2)$ so that

$$ c_1 y_1 + c_2 y_2 = 0$$

differentiation gives

$$ c_1 y_1' + c_2 y_2' = 0 $$

and so the system

$$ \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = 0 $$

has a nontrivial solution which means that $det $ of the coefficient matrix, that is $y_1 y_2' - y_1' y_2$ cannot be zero.

Related Solutions

[Math] Proof that ODE solutions with Wronskian identically zero are linearly dependent

Suppose that $y_1(t), \ldots, y_n(t)$ are solutions of $\frac{d^n y}{dt} + p_{n-1}(t) \frac{d^{n-1} y}{dt} + \cdots + p_1(t) \frac{dy}{dt} + p_0(t) y = 0$, and suppose that their Wronskian is zero for $t = t_0$, i.e. \begin{equation*} \left| \begin{array}{cccc} y_1(t_0) & y_2(t_0) & \cdots & y_n(t_0) \\ y_1'(t_0) & y_2'(t_0) & \cdots & y_n'(t_0) \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)}(t_0) & y_2^{(n-1)}(t_0) & \cdots & y_n^{(n-1)}(t_0) \end{array} \right| = 0. \end{equation*} Then the corresponding matrix is not invertible, and the system of equations \begin{array}{c} c_1 y_1(t_0) &+& c_2 y_2(t_0) &+& \cdots &+& c_n y_n(t_0) &=& 0 \\ c_1 y_1'(t_0) &+& c_2 y_2'(t_0) &+& \cdots &+& c_n y_n'(t_0) &=& 0 \\ \vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\ c_1 y_1^{(n-1)}(t_0) &+& c_2 y_2^{(n-1)}(t_0) &+& \cdots &+& c_n y_n^{(n-1)}(t_0) &=& 0 \\ \end{array} has a nontrivial solution for $c_1, c_2, \ldots, c_n$ not all zero.

Let $y(t) = c_1 y_1(t) + \cdots + c_n y_n(t)$. Because $y(t)$ is a linear combination of solutions of the differential equation, $y(t)$ is also a solution of the differential equation. Additionally, because the weights satisfy the above system of equations, we have $y(t_0) = y'(t_0) = \cdots = y^{(n-1)}(t_0) = 0$.

These initial conditions and the original differential equation define an initial-value problem, of which $y(t)$ is a solution. If $p_0(t), p_1(t), \ldots, p_{n-1}(t)$ are continuous, then any initial-value problem associated with the differential equation has a unique solution. Obviously $y^*(t) = 0$ is a solution of the initial-value problem; since we know that $y(t)$ is also a solution of the same initial-value problem, it follows that $y(t) = 0$ for all $t$, not just $t = t_0$.

We now have $c_1 y_1(t) + \cdots + c_n y_n(t) = 0$ for all $t$, where $c_1, \ldots, c_n$ are not all zero. Thus the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent.

Conversely, if the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent, then the system of equations \begin{array}{c} c_1 y_1(t) &+& c_2 y_2(t) &+& \cdots &+& c_n y_n(t) &=& 0 \\ c_1 y_1'(t) &+& c_2 y_2'(t) &+& \cdots &+& c_n y_n'(t) &=& 0 \\ \vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\ c_1 y_1^{(n-1)}(t) &+& c_2 y_2^{(n-1)}(t) &+& \cdots &+& c_n y_n^{(n-1)}(t) &=& 0 \\ \end{array} has a nontrivial solution for every $t$, the corresponding matrix is not invertible for any $t$, and $W[y_1, \ldots, y_n](t) = 0$.

Does the Wronskian of three or more linearly independent functions change its sign

Background :

$f, g $ are differentiable functions of an open interval $I$ and $f, g$ both are real valued.

Let $W(f,g)(x) =\begin{vmatrix}f(x) &g(x) \\f'(x)&g'(x)\end{vmatrix}$ denote the Wronskian of $f, g$ at $x\in I$

If two functions are solutions of a differential equation $y''+p(x) y'+q(x) y=0\tag 1$ on $I$ where $p, q\in C(I) $ then by Abel's identity we have

$$W(f, g) (x) =W(f, g) (x_o) e^{-\int_{x_0}^{x} p(t) dt}$$

Then $W(f, g) (x_0) \neq 0$ for some $x_0\in I$ implies $W(f, g) \neq 0$ on $I$

Moreover $W(f,g)$ different from zero with the same sign at every point ${\displaystyle x} \in {\displaystyle I}$

Back to the main problem:

$f, g$ are solution of $y''+p(x) y'+q(x) y=0$ on $I$ where $p, q\in C(I) $ then $f, g\in C^1(I) $ i.e $f, g$ both are continuously differentiable.

Then $W(f,g)(x) =\begin{vmatrix}f(x) &g(x) \\f'(x)&g'(x)\end{vmatrix}=f(x)g'(x)-f'(x)g(x) $

is continuous forall $x\in I$.

Hence $W(f, g) $ has Darboux property (or I.V.P) . Hence if $W(f, g) $ changes it's sign then $W(f, g)(x_0)=0$ for some $x_0\in I$.

Then by Abel's identity $W(f, g) =0$ on $I$.

Conclusion: If $f, g$ are two linearly independent solutions of $(1) $ then $W(f, g) $ can't change the sign on $I$.

Abel's identity can be generalized for $n$ solutions of a $n$ the order homogeneous linear equation. (See here).

Again $n$ solutions are $n$-th times differentiable.Hence $f_1, f_2, \ldots , f_n\in C^{n-1}(I) $.

Since $W(f_1, f_2, \ldots, f_n) $ are function of $f_1, f_2, \ldots, f_n$ and their derivatives of order upto $n-1$ , so $W(f_1, f_2, \ldots, f_n) $ is continuous on $I$ .

Now again by Darboux property we can conclude that if Wronskian changes sign then it must attains $0$.

Then by Generalized Abel's identity, it follows that $W(f_1, f_2, \ldots, f_n) =0$ on $I$.

Conclusion: If Wronskian of $n$ solutions of a $n$th order homogenous ode doesn't attin $0$ , it can't change the sign.

But more stronger result is true.

See here ,here and here.

Best Answer

Related Solutions

[Math] Proof that ODE solutions with Wronskian identically zero are linearly dependent

Does the Wronskian of three or more linearly independent functions change its sign

Related Question