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Differential Equation - Proof - Feedback.

  1. Oct 29, 2008 #1
    Hey,

    1. The problem statement, all variables and given/known data.
    I wrote my own proof for the below, I was wondering if you guys could take a look at it and give me some feedback please. Particularly, I would like to know if this proof is rigorous enough.

    Lemma 2.4 - Suppose f has a second derivative everywhere, and that,
    [tex]
    {{f} + {f^{\prime\prime}}} = {0}
    [/tex]
    [tex]
    {f(0)} = {0}
    [/tex]
    [tex]
    {f^{\prime\prime}(0)}} = {0}
    [/tex]

    Then,
    [tex]
    {f(x)} = {0}
    [/tex]

    2. Relevant equations.
    Knowledge of Calculus.

    3. The attempt at a solution.
    Let,
    [tex]
    {{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} = {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}
    [/tex]

    Prove that,
    [tex]
    {f} = {0}
    [/tex]

    Proof,
    [tex]
    {{f}+{f^{\prime\prime}}} = {0}
    [/tex]

    [tex]
    {{\left({{f}^{\prime}}\right)}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {\left({0}\right)}{\left({{f}^{\prime}}\right)}
    [/tex]

    [tex]
    {{{f}^{\prime}}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {0}
    [/tex]

    [tex]
    {{\int_{}^{}}{\left[{{{f}^{\prime}}{\left({{f}{+{{f}^{\prime\prime}}}}\right)}}\right]}{dx}} = {{\int_{}^{}}{\left[{0}\right]}{dx}}
    {\textcolor{white}{.}}
    ,
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    Let
    {\textcolor{white}{.}}
    {0} = {a}
    [/tex]

    [tex]
    {{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}
    [/tex]

    Let,
    [tex]
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {u} = {f}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {v} = {{f}^{\prime}}
    [/tex]

    [tex]
    {du} = {{{f}^{\prime}}{dx}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {dv} = {{{f}^{\prime\prime}}{dx}}
    [/tex]

    [tex]
    {{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}
    [/tex]

    [tex]
    {{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{C}}\right)}
    [/tex]

    Let,
    [tex]
    {{{C}_{1}}+{{C}_{2}}} = {C}
    [/tex]

    [tex]
    {{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{\left({C}\right)}} = {{{\left({0}\right)}{x}}+{C}}
    [/tex]

    [tex]
    {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}
    [/tex]

    [tex]
    {{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}
    [/tex]

    [tex]
    {{{\left({f(x)}\right)}^{2}}{+}{{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {0}
    [/tex]
    , for all [itex]{x}[/itex].

    This implies that [tex]{{f(x)} = {0}}[/tex] for all [itex]{x}[/itex].

    [tex]
    {\therefore}
    [/tex]

    [tex]
    {f} = {0}
    [/tex]

    Thanks,

    -PFStudent
    EDIT: Thanks for the edit HallsofIvy.
     
    Last edited: Oct 30, 2008
  2. jcsd
  3. Oct 29, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You mean [tex]f(x)= 0[/tex] for all x.

    No. C1, C2, and C are arbitrary constants. You can't just assume they cancel. You can, of course, use the given facts that f(0)=0 and f'(0)= 0.

     
  4. Nov 2, 2008 #3
    Hey,
    Thanks, ok I understand I need to use the given facts,
    [tex]
    {f(0)} = {0}
    [/tex]
    [tex]
    {f^{\prime\prime}(0)}} = {0}
    [/tex]

    However, I'm just not sure how to use those facts. In addition, addressing the point about the constants I reworked the problem to where I am still stuck,

    [tex]
    {{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}
    [/tex]

    Let,
    [tex]
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {u} = {f}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {v} = {{f}^{\prime}}
    [/tex]

    [tex]
    {du} = {{{f}^{\prime}}{dx}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {\textcolor{white}{.}}
    {dv} = {{{f}^{\prime\prime}}{dx}}
    [/tex]

    [tex]
    {{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}
    [/tex]

    [tex]
    {{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{{C}_{3}}}\right)}
    [/tex]

    [tex]
    {{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{{\left({0}\right)}{x}}+{{C}_{3}}}
    [/tex]

    [tex]
    {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{C}_{3}}
    [/tex]

    I'm not sure from here how to use the given facts to finish the proof, so any help is appreciated.

    Thanks,

    -PFStudent
     
    Last edited: Nov 2, 2008
  5. Nov 2, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Which is the same as saying
    [tex]{{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}} = C[/tex]
    Since, while you cannot assume the constants cancel, you can combine them into a single constant. Now use the fact that f(0)= f"(0)= 0 to determine C.

     
  6. Nov 2, 2008 #5
    Hey,
    Thanks, ok so combining the constants I end up with the following,
    [tex]
    {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{C}_{3}}
    [/tex]

    Let,
    [tex]
    {C} = {{{C}_{3}}-{\left({{{C}_{1}}+{{C}_{2}}}\right)}}
    [/tex]

    [tex]
    {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {\left({C}\right)}
    [/tex]

    [tex]
    {{\frac{1}{2}}{{\left({f(x)}\right)}^{2}}+{{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {C}
    [/tex]

    Err, it seems I made a typo in the beginning rather than,
    [tex]
    {{{f}^{\prime\prime}}{(0)}} = {0}
    [/tex]

    It should be,
    [tex]
    {{{f}^{\prime}}{(0)}} = {0}
    [/tex]

    That being said, I do not think that changes the problem significantly. So, going back to using the facts given,
    [tex]
    {f(0)} = {0}
    [/tex]
    [tex]
    {{{f}^{\prime}}{(0)}} = {0}
    [/tex]

    While I am familiar with solving differential equations using initial conditions and I think that is what is being hinted at in this step. I'm not quite sure how I would use the two equations above to solve for the constant [tex]{C}[/tex]. Maybe I would integrate the [tex]{{f}^{\prime}}[/tex] function but that doesn't make since, because I only know the value of the function when [tex]{{x} = {0}}[/tex]. A little more elaboration would be greatly appreciated, thanks for the help. :smile:

    Thanks,

    -PFStudent
     
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