Sturm separation theorem

In mathematics, in the field of ordinary differential equations, Sturm separation theorem, named after Jacques Charles François Sturm, describes the location of roots of homogeneous second order linear differential equations. Basically the theorem states that given two linear independent solution of such an equations the roots of the two solutions are alternating.

Sturm separation theorem

Given a homogeneous second order linear differential equation and two continuous linear independent solutions "u"("x") and "v"("x") with "x"₀ and "x"₁ successive roots of "u"("x"), then "v"("x") has exactly one root in the open interval ] "x"₀, "x"₁ [.

Proof

The proof is by contradiction. Assume that "v" has no zeros in ] "x"₀, "x"₁ [. Since "u" and "v" are linearly independent, "v" cannot vanish at either "x"₀ or "x"₁, so the quotient "u" / "v" is well-defined on the closed interval ["x"₀, "x"₁] , and it is zero at "x"₀ and "x"₁. Hence, by Rolle's theorem, there is a point ξ between "x"₀ and "x"₁ where :$frac\{d\}\{dx\}\; left(\; frac\{u(x)\}\{v(x)\}\; ight)\; =\; frac\{u\text{'}(x)\; v(x)\; -\; u(x)\; v\text{'}(x)\}\{(v(x))^2\}$vanishes. Hence, $u\text{'}(xi)\; v(xi)\; =\; u(xi)\; v\text{'}(xi)$, which implies that "u" and "v" are linearly dependent. This contradicts our assumption, and thus, "v" has to have at least one zero between "x"₀ and "x"₁.

On the other hand, there can be only one zero between "x"₀ and "x"₁, because otherwise "v" would have two zeros and there would be no zeros of "u" in between, and it was just proved that this is impossible harv|Bhamra|Ratna Bala|2003|p=262.

An Alternative Proof

Since $displaystyle\; u$ and $displaystyle\; v$ are linearly independent it follows that the Wronskian $displaystyle\; W\; [u,v]$ must satisfy $W\; [u,v]\; (x)equiv\; W(x)\; eq\; 0$ for all $displaystyle\; x$ where the differential equation is defined, say $displaystyle\; I$. Without loss of generality, suppose that . Then:$u(x)v\text{'}(x)-u\text{'}(x)v(x)\; eq\; 0$So at $displaystyle\; x=x\_0$:$W(x)=-u\text{'}left(x\_0\; ight)vleft(x\_0\; ight)$and either $u\text{'}left(x\_0\; ight)$ and $vleft(x\_0\; ight)$ are both positive or both negative. Without loss of generality, suppose that they are both positive. Now, at $displaystyle\; x=x\_1$ :$W(x)=-u\text{'}left(x\_1\; ight)vleft(x\_1\; ight)$and since $displaystyle\; x=x\_0$ and $displaystyle\; x=x\_1$ are successive zeros of $displaystyle\; u(x)$ it causes . Thus, to keep we must have . We see this by observing that if $displaystyle\; u\text{'}(x)>0mbox\{\; \}forallmbox\{\; \}xin\; left(x\_0,x\_1\; ight]$ then $displaystyle\; u(x)$ would be increasing (away from the $displaystyle\; x$-axis), which would never lead to a zero at $displaystyle\; x=x\_1$. So for a zero to occur at $displaystyle\; x=x\_1$ at most $u\text{'}left(x\_1\; ight)=0$ (i.e., $u\text{'}left(x\_1\; ight)leq\; 0$ and it turns out, by our result from the Wronskian that $u\text{'}left(x\_1\; ight)leq\; 0$). So somewhere in the interval $left(x\_0,x\_1\; ight)$ $displaystyle\; v(x)$ changed signs. By the Intermediate Value Theorem there exists $x^*inleft(x\_0,x\_1\; ight)$ such that $vleft(x^*\; ight)=0$.

By the same reasoning as in the first proof, $displaystyle\; v(x)$ can have at most one zero for $xin\; left(x\_0,x\_1\; ight)$.

References

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