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Assume that $p$ and $q$ are continuous and that the functions $y_1$ and $y_2$ are solutions of the differential equation $$y''+p(t) y'+q(t)y=0$$ an open interval $I$. Prove that if $y_1$ and $y_2$ are zero at the same point in $I$, then they cannot be a fundamental set of solutions on that interval.

If $y_1$ and $y_2$ are zero at the same point does this cause the determinant of Wronskian matrix is equal zero?

If so is this the reason why there cannot be a fundamental set of solutions on that interval.

Thanks in advance for any help.

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  • $\begingroup$ You answered your own question: the Wronskian determinant is zero. $\endgroup$ – Giuseppe Negro Mar 13 '19 at 19:14
  • $\begingroup$ Recent similar question: math.stackexchange.com/q/3143942/115115 If you look for it, you will most probably find many more. $\endgroup$ – Lutz Lehmann Mar 13 '19 at 19:25
  • $\begingroup$ To your last question: it is not that "there cannot be a fundamental set of solutions on that interval" (since there always is such a system), but "the solutions $y_1$ and $y_2$ do not form a fundamental system of solutions." $\endgroup$ – user539887 Mar 13 '19 at 19:26
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The Wronskian matrix $W(y_1, y_2)$ of two solutions $y_1$ and $y_2$ of

$y'' + p(t)y'(t) + q(t)y(t) = 0 \tag 1$

may be defined as

$W(y_1, y_2) = \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix}, \tag 2$

with determinant

$\Delta_W = \vert W(y_1, y_2) \vert = \det \left (\begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \right ) = y_1y_2' - y_2 y_1'; \tag 3$

we calculate

$\Delta_W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_2y_1''; \tag 4$

we may now use (1) in the form

$y_i'' = -py_i' - qy_i, \; i = 1, 2, \tag 5$

to transform (4) to

$\Delta_W' = y_1(-py_2' -qy_2) - y_2(-py_1' - qy_1) = -py_1y_2' - qy_1y_2 + py_1'y_2 + qy_1y_2 = -p(y_1y_2' - y_1'y_2) = -p \Delta_W, \tag 6$

a simple first order, linear ordinary differential equation for $\Delta_W$; the solutions of this equation are

$\Delta_W(t) = \Delta_W(t_0) e^{-\int_{t_0}^t p(s)\; ds}; \; t_0, t \in I, \tag 7$

which the reader may easily check. It follows from this equation and the uniquenness of solutions that if

$\Delta_W(t_0) = 0, \tag 8$

then

$\Delta_W(t) = 0, \; \forall t \in I; \tag 9$

thus if

$y_1(t_0) = y_2(t_0) = 0, \tag{10}$

it follows that

$\Delta_W(t_0) = 0, \tag{11}$

and (9) binds as well. Thus $y_1$ and $y_2$ cannot be a fundamental solution system for (1) on $I$ since fundamental systems are characterized by the non-vanishing of $\Delta_W$ everywhere, that is, by the linear independence of the columns of (2); but of course when the columns are linearly dependent then $W(y_1, y_2) = 0$.

Finally, as pointed out by user539887 in his comment to the question itself, it's not that a fundamental system doesn't exist but that $y_1$, $y_2$ does not form one. A fundamental solution system always exists, as may be seen by taking

$W(t_0) = I = \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}, \tag{12}$

for then

$\Delta_W(t) \ne 0, \; \forall t \in I, \tag{13}$

as follows from (7) since

$\Delta_W(t_0) = 1 \tag{14}$

in this case.

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