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Arguments That Seem True but Are Actually False in Differential Equations

  1. The author did not formulate correctly what he meant, and thus left a loophole.
        In [Ches, p.181, l.10-l.12], Chester says, "The fact that t is missing from the right-hand sides makes the family of solutions an n- instead of an (n+1)-parameter family." It is easy to find a counterexample: dx/dt = 1, but x = t+c. I think [Joh, p.9, l.-6-p.10, l.6] was what Chester meant. In fact, there is a better interpretation than John's. We can reduce the three equations of [Joh, p.9, (4.4)] to the two equations of [Joh, p.9, (4.3)]. Then we have only two integration constants.

  2. Handwaving arguments
    1.  
      1. The author's wording creates a facade that puts readers under the impression that they have finished the proof even though they have not.
      2. Author omits so many details that readers are left with no clue about how to start a rigorous proof.
      Example 1. [Kre, p.49, l.-2-l.-1]. Kreyszig should have said, "Since a1 = b1 and a2 = b2, for the first order all we need to prove is (da3/ds) |s0 = (db3/ds*) |s*0.
    2. Either the argument's loose ends need to be tied together or its confusing points need to be clarified.
      Example 1. In the proof of [Spi, vol.1, p.203, Theorem 5], Spivak leaves out the the part of the proof given in [Arn1, p.306, l.19-p.307, l.2]. By contrast, in Arnold's proof he points out what tasks need to be done in order to tie the loose ends together, and what analyses need to be refined in order to clarify the confusion.
      Example 2. [Spi, vol.1, p.214, Theorem 10] leaves its proof unfinished. In order to complete the proof, we must prove the following lemma: If Xf=Yf for every C f: M R, then X=Y.
      Proof: Let X = ai(/xi) and Y = bi(/xi). Then ai = bi locally X=Y locally.
    3. Analysis needs to be refined and the notation abuse needs to be analyzed.
      Example. The equality given in [Spi1, p.102, l.-1].
      Case ji: Let 1mk and mi. If mj, I(j,a)*(dxm) = dxm. If m=j, I(j,a)*(dxm) = 0. If ji, ($m)1mk : m=j.
      Case j=i: Strictly speaking, f should be defined on [0,1]k.  Spivak's notation is sloppy: f on [0,1]k-1 should have been written as f(j,a). For k=2, the integral given in [Spi1, p.102, l.-1] is actually a line integral. It can be considered an area integral only if we add a factor [0,1] dxi = 1.
    4. Information that should have been provided is not.
          In [Spi, vol. 1, p.290, l.13], Spivak says, "It is easy to prove that dw is alternating, .", but he fails to provide any hints to start the proof. It would be helpful if he could provide the following hints: Assume Xio = Xjo, where i0 < j0. For S1, consider the cancellation between  Xiow and Xjow. For S2, consider the cancellations among the four cases: i<io, io<j, i<jo, and j0<j.
    5. If a quoted theorem [Har, p.25, Lemma 2.1] has several versions whose proofs are similar, we sill have to present every version in detail even though we may prove one version and omit the proofs of other versions. If we omit any version, serious confusions may arise (compare [Har, p. 25, (2.2)] with the equality given in [Har, p.25, l.13]). A random guess only leads to an incorrect result (e.g., u0 v(t0) given in [Har, p.27, l.-10] should have been v(t0) u0). In addition, we must clarify the meaning of ambiguous statements (the large n in [Har, p.25, l.-14; l.-7] refers to n such that an = min (a, b/(M+n-1)) > a').
      Remark. The various versions of [Har, p.25, Lemma 2.1]
      (*)max     u' = U(t,u), u(t0) = u0
      (**)max    u(t) u0(t)
      (*)min     u' = U(t,u), u(t0) = u0
      (**)min    u(t) u0(t)
      1. (The left-half interval) Let U(t,u) be continuous on a rectangle R: t0-att0, |y-y0|b; let |U(t,u)|M and a=min (a, b/M). Then
        1. (The maximal solution) (*)max has a solution u=u0(t) on [t0-a,t0] with the property that every solution u=u(t) of u' = U(t,u), u(t0) u0 satisfies (**)max on [t0-a,t0].
        2. (The minimal solution) (*)min has a solution u=u0(t) on [t0-a,t0] with the property that every solution u=u(t) of u' = U(t,u), u(t0) u0 satisfies (**)min on [t0-a,t0].
      2. (The right-half interval) Let U(t,u) be continuous on a rectangle R: t0tt0+a, |y-y0|b; let |U(t,u)|M and a=min (a, b/M). Then
        1. (The maximal solution) (*)max has a solution u=u0(t) on [t0,t0+a] with the property that every solution u=u(t) of u' = U(t,u), u(t0) u0 satisfies (**)max on [t0,t0+a].
        2. (The minimal solution) (*)min has a solution u=u0(t) on [t0,t0+a] with the property that every solution u=u(t) of u' = U(t,u), u(t0) u0 satisfies (**)min on [t0,t0+a].
    6. Be specific. It is inappropriate to say, "It is impossible for Statement A to be true." Instead, we must pinpoint which previous statement it would contradict if Statement A were true. Otherwise, the readers will have no clue to follow.
      Example. This is impossible [Har, p.40, l.3]. Hartman should be more specific and say that
      [(V(y(t))c as t) ((d/dt)[V(y(t)]0 as t)].
    7. The proof that may apply to a particular case, but cannot apply to the general case.
      Example. We want to prove [z, z+2K] dn2 zdz = [0, 2K] dn2 zdz [Gon1, p.441, (5.24-5)]. The proof given in [Gon1, p.4441, l.18] applies only to the case where z is a real number. If z = x+iy, we should use [Gon1, (5.17-25), (5.18-4) and (5.18-6)].
    8. The argument is misleading even though the conclusion is correct.
          In [Gon1, p.476, l.4-l.5], Gonzlez says that e1 is the largest root of 4u3 -g2u -g3 = 0 because '(x)<0 for 0 <x <w1. However, the reason he provides is not fundamental. Assume e2 > e1. Then there exists an x0 such that 0 <x0 <w1 and (x0) = e2 [Gon1, p.475, l.-14-l.-13]. By [Gon1, p.376, Theorem 5.17], the existence of the root x0  contradicts [Gon1, p.449, Corollary 5.23a].
    9. The proof given in [Wat1, p.452, l.9-l.16] provides rough ideas, but fails to specify a method to execute them. In contrast, the proof of [Gon1, p.378, Theorem 5.19] provides specific details.
    10. Links {1}.

  3. We must pinpoint which theorem we use in a proof. If the proofs of Theorems A, B, and C are similar and we use Theorem A in a proof, we should say that we use Theorem A rather than Theorem B or Theorem C in the proof.
    Example. In [Har, p.32, l.15-l.16, proof of Theorem 6.1], Hartman claims that he uses [Har, p.27, Corollary 4.3] and the remark following [Har, p.26, Theorem 4.1]. In view of all the versions of [Har, p.26, Theorem 4.1] listed below, Hartman actually uses its (the minimal solution; the left-half interval)-version rather than its other versions or [Har, p.27, Corollary 4.3].
    Remark. The various versions of [Har, p.26, Theorem 4.1]
    (*)max     u' = U(t,u), u(t0) = u0
     (**)max    v(t) u0(t)
     (*)min     u' = U(t,u), u(t0) = u0
     (**)min    v(t) u0(t)
    1. (The maximal solution) Let U(t,u) be continuous function on an open (t,u)-set E and u = u0(t) be the maximal solution of (*)max.
      1. (The left-half interval) Let v(t) be a continuous function on [t0-a,t0] satisfying the condition v(t0)u0, (t, v(t))E and v(t) has a left derivative DLv(t) on (t0-a,t0] such that DLv(t)U(t, v(t)). Then on a common interval of existence of u0(t) and v(t), (**)max holds.
      2. (The right-half interval) Let v(t) be a continuous function on [t0,t0+a] satisfying the condition v(t0)u0, (t, v(t))E and v(t) has a right derivative DRv(t) on [t0,t0+a) such that DRv(t)U(t, v(t)). Then on a common interval of existence of u0(t) and v(t), (**)max holds.
    2. (The minimal solution) Let U(t,u) be continuous function on an open (t,u)-set E and u = u0(t) be the minimal solution of (*)min.
      1. (The left-half interval) Let v(t) be a continuous function on [t0-a,t0] satisfying the condition v(t0)u0, (t, v(t))E and v(t) has a left derivative DLv(t) on (t0-a,t0] such that DLv(t)U(t, v(t)). Then on a common interval of existence of u0(t) and v(t), (**)min holds.
      2. (The right-half interval) Let v(t) be a continuous function on [t0,t0+a] satisfying the condition v(t0)u0, (t, v(t))E and v(t) has a right derivative DRv(t) on [t0,t0+a) such that DRv(t)U(t, v(t)). Then on a common interval of existence of u0(t) and v(t), (**)min holds.

  4. A statement is false because of the use of improper notations.
    Example. [Spi, vol.1, p.214, l.2]. Correction: g(t,p) should have been 01(/y)f(st,p)ds, where y=st.

  5. The final answer is correct, but the involved calculations are not.
    Remark. The statement given in [Inc1, p.120, l.-17-l.-13] is correct, but the coefficient of u(n)  is
    (-1)nD(u1,…,un) rather than D(u1,…,un) [Inc1, p.120, l.19]. Each step of a proof must be carefully examined. Any guess or wishful thinking should not be allowed.

  6. In order to prove the equality of two series, we have to consider their domains of convergence in addition to their algebraic sums [Wat1, p.16, l.-16-l.-10].

  7. A complex variable's argument assigned in the proof of a theorem should be consistent with the variable's argument assigned in the theorem's hypothesis [1].

  8. The statement that seems true but is actually false: If an 0
  9. for n1, then [Perr, p.289, (1)][Perr, p.290, (3)].
    From the above statement we should learn the following lessons:
    1. How does this guess arise?
      Ans. If for some positive integer n an+1= 0, then [Perr, p.289, Satz 44] holds.
    2. Formally, the statement seems true [Perr, p.13, l.15]. If it were true, what contradiction would we obtain?
      Ans. [Perr, p.290, l.5-l.16]
    3. How do we correct the statement?
      Ans. For the strong case given in [Perr, p.290, l.21], we can find the minimum requirement (i.e., necessary and sufficient conditions) [Perr, p.290, Satz 45]. For the weak case given in [Perr, p.291, l.-9], we can only find sufficient conditions [Perr, pp.291-292, Satz 46, B, C, or D].

  10. (A theorem's proof should be guided by its physical theme)
        Without being guided by a theorem's physical theme, one may easily get lost in the maze of its proof . [Cod, p.319, l.-15-l.-10] uses the following argument:
    if [((A and C)B) and (BC)], then (AB)   (*),
    where
    A = [Cod, p.318, (1.16) & (1.17)];
    C = [|j(t)|d and Cod, p.319, (1.22)] (see [Cod, p.319, l.15-l.16]);
    B = [Cod, p.319, (1.23)].
    If in (*) we substitute C into B, we see that the conclusion (AC) is false. Thus, Levinson's argument is incorrect. However, the hypothesis [((A and C)B) and (BC)] ensures that under the condition A, B and C are equivalent. We can correct Levinson's mistake by the following method:
    Even though the estimate given in [Pon, p.211, l.-11] is less effective than that given in [Cod, p.319, (1.23)], we may use the former estimate to prove C. Thereby, we obtain the better estimate B.


  11. [Inc1, p.196, l.7-l.8] says that limits of integration are to be determined so that the term [te(1-t)c-e(u/t)]ab vanishes identically. Actually, it leaves out another term -u(0)(p0v)(1)|ab, where p0 = t(1-t). See [Inc1, p.186, l.-8] and [Cod, p.86, (6.12)]. If u=F(a,b;e;xt), both terms are 0 when a=0, b=1 provided that e-1>0, c-e-1>0.
    Remark. The equality given in [Inc1, p.196, l.-9] holds when b>1, c>b+1. By analytic continuation [Guo, p.153, l.-10-l.-7], it also holds when b>0, c>b.


  12. Prove y(m+1)=1-1+2-1+3-1+…m-1-g [Wat, p.60, l.-4].
    Incorrect proof. y(m+1)=-g+Sn=1(n-1-(m+n)-1) [Guo, p.108, (10)]
    =-g+Sn=1n-1-Sn=m+1n-1 (incorrect step).
    Correct proof. G(z+1)=zG(z)
    G '(z+1)=G(z)+zG '(z)
    G '(z+1)[G(z)]-1=1+zG '(z)[G(z)]-1
    G '(z+1)[G(z+1)]-1=z-1+G '(z)[G(z)]-1.


  13. Links {1(the symbol O)}.