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June 21st, 2017, 01:11 PM   #21
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Since you asked in the first version of your post, and to spell it out for anyone else that's interested:

The initial value problem $$xy' = 2y \quad y(-2)=4$$
is satisfied by
1. $\displaystyle y=x^2$ as Zylo noted. This can be verified directly because $y(-2) = (-2)^2 = 4$ and $y'=2x$ so $xy' = 2x^2 = 2y$.
2. $$y = \begin{cases}x^2 & (x \lt 0) \\ -x^2 & (x \ge 0)\end{cases}$$
The initial condition is satisfied exactly as for the previous solution. Similarly for $x \ne 0$ we can differentiate $y$ and substitute into the original equation. At $x=0$ we have $y=0$ and so the differential equation is satisfied as long as $y'$ is defined. We can show that $y'$ is defined at $x=0$ by using the limit definition of the derivative $\displaystyle y'(0)= \lim_{h \to 0} \frac{y(h) - y(0)}{h}$. For $h \lt 0$ we have $\displaystyle y'(0) = \lim_{h \to 0} \frac{h^2 - 0}{h} = \lim_{h \to 0} h = 0$ and for $h \gt 0$ we have $\displaystyle y'(0) = \lim_{h \to 0} \frac{-h^2 - 0}{h} = \lim_{h \to 0} -h = 0$. These limits exist and are equal so we have $y'(0)=0$. More to the point, it exists and the differential equation is thus satisfied.
So, we have two solutions to the differential equation and both are valid for all $x$. Thus, either the theorem that Zylo cited (http://www2.fiu.edu/~aladrog/BasicTheoryLinear.pdf) is false, or the conditions for it are not met.

The theorem states:
Quote:
 Originally Posted by http://www2.fiu.edu/~aladrog/BasicTheoryLinear.pdf Let $$a_0 (x)y^{(n)} + a_1(x)y^{(n-1)} + \cdots + a_{n-1}(x)y" + a_n (x)y = b(x)$$ with $a_0(x), a_1(x), \ldots, a_n(x)$ and $b(x)$ are continuous real-valued functions on the interval $[a, b]$ and $a_0(x) ≠ 0$ for at least one $x$ in $[a, b]$. Let $x_0$ be a point on the interval $[a, b]$ and let $c_0, c_1, \ldots, c_{n-1}$ be $n$ arbitrary constants. Then there exists a unique function $f(x)$ defined on $[a, b]$ such that $y=f(x)$ is a solution of the linear differential equation and satisfies $f(x_0) = c_0, f’(x_0) = c_1, f’’(x_0) = c_2,\ldots , f^{(n-1)}(x_0) = c_{n-1}$.
In relation to our IVP $xy'=2y,\, y(-2)=4$ this means that $n=1$, $a_0(x)=x$, $a_1(x)=2$, $b(x)=0$ ,$x_0=-2$, $c_0=4$. Since we have found that there is no unique function that is a solution of the IVP, we can conclude that the theorem is false.

However, I believe that the theorem is intended to refer to analytic functions $y=f(x)$. In other words, functions for which $y^{(n)}=f^{(n)}(x)$ exist for all $x$ in $[a,b]$ for every natural number $n$. Solution 2 above is not analytic because $f''(0)$ does not exist ($f''(x) = 2$ for all $x \lt 0$ and $f''(x) = -2$ for all $x \gt 0$). And the theorem, I believe, therefore becomes true. (That's not a proof, only a demonstration that the counterexample doesn't apply when talking about analytic functions).

This is borne out by the corollary in the referenced document which says that if the initial conditions are such that $f(x_0)=f'(x)=f''(x)=\cdots=f^{(n-1)}(x_0)=0$ then the function $f(x)=0$ is the only solution.

Again, if we analyse this for $xy'=2y, y(0)=0$ we find that it is false (both solutions 1 and 2 remain solutions of the new problem). However, if we stipulate that every derivative of $y=f(x)$ should be zero at $x=0$, then neither solution 1 nor solution 2 fit the bill. Solution 1 has $y''(0)=2$ and we have seen that $y''(0)$ does not exist for solution 2. But the solution $y=0$ does work.

Thus, if we take the initial condition in the theorem as having been written incorrectly and that, instead of referring to only $(n-1)$ derivatives, it should refer to every derivative, we have a theorem that could work.

Last edited by skipjack; June 21st, 2017 at 02:38 PM.

 June 21st, 2017, 01:24 PM #22 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,399 Thanks: 100 Jogging along and suddenly occured to me what the point of confusion might be. If y=x$\displaystyle ^{2}$, y(x)=y(-x). But so what. Consider SHO with m=1 and k=1: y''+y=0 By trial, y=acosx is a solution. Unique sol? No because the Theorem requires 2 BC to be satisfied for 2nd order eq. Google link above. So you need another independent sol. y=bsinx works by trial and is independent by another theorem. Assume: y(0)=A y'(0)=B, then y=Acosx+Bsinx is a unique periodic solution And now to the point. Assume B=0. Then Acos(t)=Acos(-t). So what? Do you, for example, want to restrict solution for a pendulum to a quarter period? Last edited by zylo; June 21st, 2017 at 01:37 PM. Reason: y'(0) =B
 June 21st, 2017, 01:39 PM #23 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra Yet again your poor notation makes it hard to talk about the specifics of your post. I have repeatedly talked about applied situations. I also in the previous posts talked about analytic solutions. Furthermore, in previous posts I explained in some detail about why the domain should be restricted in the abstract case. The fact that your solution $y=x^2$ is even is neither here nor there. Note that a third solution to $xy'=2y, \, y(-2)=4$ is $$y = \begin{cases} x^2 & (x \lt 0) \\ 6x^2 & (x \ge 0) \end{cases}$$ There are infinitely many, only one of which is analytic, but the problem gives no indication that an analytic solution is required, only that $y$ and $y'$ should exist everywhere. Your SHO example has no discontinuities, so the question of restricting the domain doesn't apply. Any two linearly independent solutions to the homogeneous equation are sufficient to describe every possible solution. It doesn't matter whether you take $\cos(t)$ or $\cos(-t)$. But that is completely different to the suggestion that there is only one solution to the IVP. Whatever your LI solutions are, the combination will result in the same solution to the IVP once you've worked out the appropriate coefficients. Note that $y=A\cos(t) + B\cos(t)$ is equivalent to $y=C_1e^{it} + C_2e^{-it}$, we have just used a different basis for the solution space. Last edited by v8archie; June 21st, 2017 at 01:45 PM.
June 21st, 2017, 03:10 PM   #24
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Quote:
 Originally Posted by v8archie Your theorem (implicitly) assumes that $f(x)$ is analytic everywhere (in the given interval).
The author is giving basic theory and hasn't made any use of the term "analytic". He is listed under "Retired Faculty", but is still contactable if you want to find out what he had in mind.

 June 21st, 2017, 03:14 PM #25 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra Yes. Although simply assuming that the function is analytic doesn't save the corollary.
 June 22nd, 2017, 08:19 AM #26 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,399 Thanks: 100 "Theorem: Consider the initial value problem y″ + p(t) y′ + q(t) y = g(t), y(t0) = y0, y′(t0) = y′0. If the functions p, q, and g are continuous on the interval I: α < t < β containing the point t = t0. Then there exists a unique solution y = φ(t) of the problem, and that this solution exists throughout the interval I. That is, the theorem guarantees that the given initial value problem will always have (existence of) exactly one (uniqueness) twice-differentiable solution, on any interval containing t0 as long as all three functions p(t), q(t), and g(t) are continuous on the same interval. Conversely, neither existence nor uniqueness of a solution is guaranteed at a discontinuity of p(t), q(t), or g(t)." From http://www.math.ucsd.edu/~y1zhao/201...Section3.2.pdf page 1 "Theorem 1: Let p and g be continuous functions on the open interval I=(α,β), and let t0∈(α,β). Then for each t∈I there exists a unique solution y=ϕ(t) to the differential equation dy/dt+p(t)y=g(t) that also satisfies the initial value condition that y(t0)=y0." From Existence/Uniqueness of Solutions to First Order Linear Differential Eqs. - Mathonline "Theorem 1 (Existence and Uniqueness). [1, NSS, Section 6.1, Theorem 1]1 Suppose p1(x), . . . , pn(x) and g(x) are continuous real-valued functions on an interval (a, b) that contains the point x0. Then, for any choice of (initial values) γ0, . . . , γn−1, there exists a unique solution y(x) on the whole interval (a, b) to the nonhomogeneous differential equation y^(n)(x) + p1(x)y^(n−1)(x) + · · · + pn(x)y(x) = g(x) for all x ∈ (a, b) and y^(i)(x0) = γi for i = 0, . . . , n − 1." From https://math.berkeley.edu/~shinms/SP...iffeq-thms.pdf Get the idea yet? Last edited by zylo; June 22nd, 2017 at 08:40 AM.
 June 22nd, 2017, 09:19 AM #27 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra First theorem: irrelevant to $xy'=2y$ as it speaks of second order equations. Second and third theorems: irrelevant to $xy'=2y$ as it speaks of a continuous function $p(t)$. What is it about the counterexample that you don't understand?
 June 22nd, 2017, 11:11 AM #28 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra The last two theorems are the one you were thinking of, but they prove my point. With the equation arranged so that the coefficient of the highest derivative is unity, that is $$y'' -\frac2x y=0 \quad y(-2)=4$$ there is a unique solution on any interval containing the initial condition and over which the remaining coefficients are continuous. In this case, that means that the interval contains $x=-2$ and $-\frac2x$ is continuous. The maximal such interval is $(-\infty,0)$ which aligns exactly with my comments about the interval of validity with regard to the original equation $y'=\frac y x$. It also explains how I can generate infinitely many solutions to the equation $xy'=2y$ over the domain $(-\infty,+\infty)$. This occurs because in the integration step we integrate across a discontinuity and thus can select different coefficients of integration on each side of the discontinuity without introducing a new discontinuity which would not satisfy the original equation. The coefficients of integration are determined on one side of the discontinuity by the initial condition, but there is no condition to be satisfied on the other, so any value will work.

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