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Consider the ODE $$\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.$$ I have to find the interval of unique solution by using Picard method. As in my local book it is solved as $$|f(x,y)|=|x^{2}+y^{2}|\leq a^{2}+b^{2}=M. $$ $$h=\min\{a,\frac{b}{M}\}$$ where $M=a^{2}+b^{2}$. So $h=\min\{a,\frac{b}{a^{2}+b^{2}}\}$. Then my main problem is the author do like $$a=\frac{b}{a^{2}+b^{2}}$$ which gives a quadratic in $b$ as $ab^{2}-b+a^{3}=0$ with discriminant as $1-4a^{4}=0$ gives $a=\frac{1}{\sqrt{2}}$ so $|h|\leq\frac{1}{\sqrt{2}}.$ Please suggest me how he find $h=\min\{a,\frac{b}{a^{2}+b^{2}}\}$? Why he put $a=\frac{b}{a^{2}+b^{2}}$? Thanks in advance.

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  • $\begingroup$ Yes but there is no process to find $h$ $\endgroup$
    – neelkanth
    Jan 20, 2017 at 2:35
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    $\begingroup$ This equation can be solved exactly in terms of Bessel functions as $\frac{x J_{\frac{3}{4}}\left(\frac{x^2}{2}\right)}{J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)}$. Combining this with the fact that $y$ is an odd function, one can prove that the interval of existence is exactly $(-\sqrt{2 \alpha},\sqrt{2 \alpha})$ where $\alpha \approx 2.0063$ is the smallest (positive) zero of $J_{-1/4}$. I hope this helps. $\endgroup$
    – user1337
    Jan 21, 2017 at 15:49
  • $\begingroup$ Thanks....i am asking about the method discussed in question... $\endgroup$
    – neelkanth
    Jan 21, 2017 at 16:51

1 Answer 1

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We want to find the interval for a unique solution using the Picard–Lindelöf theorem given

$$\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.$$

In this case, $x_0 = y_0 = 0$. For some $a, b $, the function $f(x, y) = x^2 + y^2 $ is defined and Lipschitz continuous on the rectangle $R=\{|x|\leq a, |y|\leq b\}$.

The Picard existence theorem states that the IVP has a unique solution on the interval $[-h, h]$.

By Theorem 1 or the Wiki, the supremum is given by $|f(x,y)|=|x^{2}+y^{2}|\leq a^{2}+b^{2}=M$, and therefore $$h= \min\left\{a,\dfrac{b}{M}\right\} = \min\left\{a,\frac{b}{a^2 + b^2}\right\}$$

We now have to choose $a$ and $b$ such that $h$ becomes as large as possible, that is, we want $$ h = \max_{a, b}~ \min\left\{a,\frac{b}{a^2 + b^2}\right\}$$

Among all rectangles of a given perimeter the square has the largest area. Using the AM-GM inequality, we want to find when $\sqrt{l~ w} \le \dfrac{l + w}{2}$. This occurs if and only if $l = w$, that is, the rectangle is a square and the sides are equal (that is why the author equates them). As an aside, this is a consequence of Is there a "simple" proof of the isoperimetric theorem for squares?.

Thus in our case

$$a = \dfrac{b}{a^2 + b^2} \implies a(a^2 + b^2) - b = 0 \implies a \ne 0, b = \dfrac{1-\sqrt{1-4a^4}}{2 a}$$

The expression under the radical must be positive, hence

$$1 - 4a^4 \ge 0 \implies | a | \le \dfrac{1}{\sqrt{2}}$$

We could have stopped here, but lets use this result for $a$ and do an additional step to validate that we get the same result for $b$. Using $a$, we find

$$ h = \max_b~\min\left\{\pm~ \dfrac{1}{\sqrt{2}},\frac{b}{\frac{1}{2} + b^2}\right\} \implies |b| \le \dfrac{1}{\sqrt{2}}$$

Conclude that for any choice of $a, b$, such that $|a| \le \frac 12, a \ne 0, |b| \le \frac 12$, the maximum $h$ is given by

$$| h | \le \dfrac{1}{\sqrt{2}}$$

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  • $\begingroup$ can we apply the same method for any question of this type ? $\endgroup$
    – neelkanth
    Aug 25, 2021 at 3:55
  • $\begingroup$ Assuming the conditions of the theorem are met, that is, see Picard–Lindelöf theorem proof at people.math.wisc.edu/~seeger/522/picard.pdf, then yes. This Wiki even shows an optimization of the interval. $\endgroup$
    – Moo
    Aug 25, 2021 at 10:35

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