Upper function/bound for $y'(x)=x^3+y^3(x)$

Question

The function is $y'(x)=x^3+y^3(x)$ for $x>0$ and $y(0)=1$

I was asked to find an upper & lower function so that $y_1'(x)<y'(x)<y_2'(x)$

I've found a lower bound for $y'(x)$ by seeing that $y_1'(x)=y^3(x)<x^3+y^3(x)=y'(x)$ with $y_1(0)=1$

With that, I got $y_1(x)=\frac{1}{\sqrt{1-2x}}$ as solution and now I can deduce that $0<x<\frac{1}{2}$

But not I'm having problems constructing the upper function for that problem.

My initial guess was $y_2(x)=1+y^3(x)$ but that proofed hard to solve. Can someone give me a hint for a better estimate?

Answer 1

Let $I = [0, a)$ denote the maximal interval on which a solution of the initial value problem exists.

You already showed that $a \le 1/2$. It follows that $2x < 1 \le y(x)$ for all $x \in I$, and therefore $$ y'(x) < \left( \frac{y(x)}{2} \right)^3 + y(x)^3 = \frac 98 y(x)^3 \, . $$

That leads to $$ y(x) < y_2(x) = \frac{2}{\sqrt{4-9x}} \, . $$ In particular, $a \ge \frac 4 9 \approx 0.444$.

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For a better estimate you can use your lower bound $$ y(x) > \frac{1}{\sqrt{1-2x}} $$ and verify that $$ \frac{1}{\sqrt{1-2x}} > 5x $$ for $0 \le x <1/2$. It follows that $x < \frac 15 y(x)$, and therefore $$ y'(x) < \left( \frac{y(x)}{5} \right)^3 + y(x)^3 = \frac {126}{125} y(x)^3 \, . $$ That leads to the upper bound $$ y(x) < \frac{{{5}^{3/2}}}{\sqrt{125-252x}} $$ and $a \ge \frac{125}{252} \approx 0.496$.

Answer 2

If one adds more complications to the (first, for $x<\frac14$, now improved) strategy of the answer of Martin R, one could observe that $$ y'(x)\ge x^3+1 \implies y(x)\ge 1+x+\frac{x^4}4\ge 1+x+\frac{x^4}8+\frac{x^4}8 \ge 1+3\sqrt[3]{x·\frac{x^4}8·\frac{x^4}8}\ge\frac34 x^3 $$ (exact at $x=2$, factor can be improved to slightly above $0.8$) so that one obtains an upper bound via the differential inequality $$ y'(x)\le \frac43y(x)+y(x)^3. $$ This can be solved via the Bernoulli approach to get $$ y(x)^{-2}+\frac34\ge(1+\frac34)e^{-\frac83x} $$ which gives a singularity-free region for $x<-\frac38\ln\frac37=0.317736697...$

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Reducing the initial estimate to the interval $[0,\frac12]$ one has there $$ y(x)\ge x^3(x^{-3}+x^{-2}+\tfrac14x)>12x^3. $$ This gives the upper bound a differential inequality of the Bernoulli type $$ y'(x)\le \frac1{12}y(x)+y(x)^3 \implies y(x)^{-2}+12\ge(1+12)\exp(-\tfrac16x). $$ This bound prevents divergence to infinity for $x<6\ln(1+\frac1{12})=0.4802562460...$