Prove that there is at most one continuous function on $[0,2]$ that satisfies:
$$v(x)=f(x)+\int_0^2 e^{-(x-y)^2} \cos(0.3v(y)) \, dy$$
I don't know how to estimate this integral...
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So $f$ is known and the existence of $v$ is to be shown? – Michael Hardy Mar 25 '14 at 02:17
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This is exactly as my assignment is written. There is also a hint to "estimate and apply the mean value theorem" – kiwifruit Mar 25 '14 at 02:17
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@MichaelHardy: at most one. So existence is not to be shown, but uniqueness (assuming existence) is. – Robert Israel Mar 25 '14 at 02:20
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I'd have written "at most one continuous function $v$ on $[0,2]$", making it clear that $v$, not $f$ was the function whose uniqueness is to be shown. – Michael Hardy Mar 25 '14 at 02:28
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Yes, that would be good, but existence and uniqueness of $f$ given $v$ is too trivial. – Robert Israel Mar 25 '14 at 02:32
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If $v_1$ and $v_2$ are two solutions, $$v_1(x) - v_2(x) = \int_0^2 e^{-(x-y)^2} (\cos(0.3 v_1(y)) - \cos(0.3 v_2(y)))\ dy$$
Now $\cos(.3 v_1(y)) - \cos(.3 v_2(y) = -.3 \sin(.3 s) (v_1(y) - v_2(y))$ for some $s$ between $v_1(y)$ and $v_2(y)$. Only quite obvious estimates of $\exp(-(x-y)^2)$ and $|\sin(s)|$ are needed ... you should get $|v_1(x) - v_2(x)| \le c \max_y |v_1(y) - v_2(y)|$ with $c < 1$.
Robert Israel
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So we select anything we want for the exp part to come up with a bound? – kiwifruit Mar 25 '14 at 02:57
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And how does the final result prove that there is only one function? – kiwifruit Mar 25 '14 at 03:04
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Not "anything we want" - something that is obviously $\ge e^{-(x-y)^2}$. – Robert Israel Mar 25 '14 at 04:12
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If $t \le ct$ where $0 < c < 1$ and $t \ge 0$, what can you conclude? – Robert Israel Mar 25 '14 at 04:13