Differential Equation - Proof - Feedback.

[PFStudent]

Hey,

1. Homework Statement .
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,
<br /> {{f} + {f^{\prime\prime}}} = {0}<br />
<br /> {f(0)} = {0}<br />
<br /> {f^{\prime\prime}(0)}} = {0}<br />

Then,
<br /> {f(x)} = {0}<br />

2. Homework Equations .
Knowledge of Calculus.

3. The Attempt at a Solution .
Let,
<br /> {{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} = {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}<br />

Prove that,
<br /> {f} = {0}<br />

Proof,
<br /> {{f}+{f^{\prime\prime}}} = {0}<br />

<br /> {{\left({{f}^{\prime}}\right)}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {\left({0}\right)}{\left({{f}^{\prime}}\right)}<br />

<br /> {{{f}^{\prime}}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {0}<br />

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

<br /> {{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}<br />

Let,
<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {u} = {f}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {v} = {{f}^{\prime}}<br />

<br /> {du} = {{{f}^{\prime}}{dx}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {dv} = {{{f}^{\prime\prime}}{dx}}<br />

<br /> {{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}<br />

<br /> {{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{C}}\right)}<br />

Let,
<br /> {{{C}_{1}}+{{C}_{2}}} = {C}<br />

<br /> {{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{\left({C}\right)}} = {{{\left({0}\right)}{x}}+{C}}<br />

<br /> {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}<br />

<br /> {{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}<br />

<br /> {{{\left({f(x)}\right)}^{2}}{+}{{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {0}<br />
, for all {x}.

This implies that {{f(x)} = {0}} for all {x}.

<br /> {\therefore}<br />

<br /> {f} = {0}<br />

Thanks,

-PFStudent
EDIT: Thanks for the edit HallsofIvy.

 

[HallsofIvy]

Hey,

1. Homework Statement .
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,
<br /> {{f} + {f^{\prime\prime}}} = {0}<br />
<br /> {f(0)} = {0}<br />
<br /> {f^{\prime\prime}(0)}} = {0}<br />

Then,
<br /> {f(0)} = {0}<br />

You mean f(x)= 0 for all x.

2. Homework Equations .
Knowledge of Calculus.

3. The Attempt at a Solution .
Let,
<br /> {{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} = {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}<br />

Prove that,
<br /> {f} = {0}<br />

Proof,
<br /> {{f}+{f^{\prime\prime}}} = {0}<br />

<br /> {{\left({{f}^{\prime}}\right)}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {\left({0}\right)}{\left({{f}^{\prime}}\right)}<br />

<br /> {{{f}^{\prime}}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {0}<br />

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

<br /> {{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}<br />

Let,
<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {u} = {f}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {v} = {{f}^{\prime}}<br />

<br /> {du} = {{{f}^{\prime}}{dx}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {dv} = {{{f}^{\prime\prime}}{dx}}<br />

<br /> {{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}<br />

<br /> {{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{C}}\right)}<br />

Let,
<br /> {{{C}_{1}}+{{C}_{2}}} = {C}<br />

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.

<br /> {{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{\left({C}\right)}} = {{{\left({0}\right)}{x}}+{C}}<br />

<br /> {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}<br />

<br /> {{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}<br />

<br /> {{{\left({f(x)}\right)}^{2}}{+}{{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {0}<br />
, for all {x}.

This implies that {{f(x)} = {0}} for all {x}.

<br /> {\therefore}<br />

<br /> {f} = {0}<br />

Thanks,

-PFStudent

 

[PFStudent]

Hey,

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.

Thanks, ok I understand I need to use the given facts,
<br /> {f(0)} = {0}<br />
<br /> {f^{\prime\prime}(0)}} = {0}<br />

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,

<br /> {{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}<br />

Let,
<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {u} = {f}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {v} = {{f}^{\prime}}<br />

<br /> {du} = {{{f}^{\prime}}{dx}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {dv} = {{{f}^{\prime\prime}}{dx}}<br />

<br /> {{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}<br />

<br /> {{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{{C}_{3}}}\right)}<br />

<br /> {{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{{\left({0}\right)}{x}}+{{C}_{3}}}<br />

<br /> {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{C}_{3}}<br />

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

Thanks,

-PFStudent

 

[HallsofIvy]

Hey,


Thanks, ok I understand I need to use the given facts,
<br /> {f(0)} = {0}<br />
<br /> {f^{\prime\prime}(0)}} = {0}<br />

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,

<br /> {{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}<br />

Let,
<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {u} = {f}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {v} = {{f}^{\prime}}<br />

<br /> {du} = {{{f}^{\prime}}{dx}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {\textcolor{white}{.}}<br /> {dv} = {{{f}^{\prime\prime}}{dx}}<br />

<br /> {{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}<br />

<br /> {{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{{C}_{3}}}\right)}<br />

<br /> {{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{{\left({0}\right)}{x}}+{{C}_{3}}}<br />

<br /> {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{C}_{3}}<br />

Which is the same as saying
{{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}} = C
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.

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

Thanks,

-PFStudent

 

[PFStudent]

Hey,

Which is the same as saying
<br /> {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}} = C<br />
Since, while you cannot assume the constants cancel, you can combine them into a single constant.

Thanks, ok so combining the constants I end up with the following,
<br /> {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}+{{C}_{1}}+{{C}_{2}}} = {{C}_{3}}<br />

Let,
<br /> {C} = {{{C}_{3}}-{\left({{{C}_{1}}+{{C}_{2}}}\right)}}<br />

<br /> {{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {\left({C}\right)}<br />

<br /> {{\frac{1}{2}}{{\left({f(x)}\right)}^{2}}+{{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {C}<br />

Now use the fact that f(0)= f"(0)= 0 to determine C.

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

It should be,
<br /> {{{f}^{\prime}}{(0)}} = {0}<br />

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

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 {C}. Maybe I would integrate the {{f}^{\prime}} function but that doesn't make since, because I only know the value of the function when {{x} = {0}}. A little more elaboration would be greatly appreciated, thanks for the help.

Thanks,

-PFStudent