maximum interval of existence of a solution of differential equations

Question

I have these initial value problems:

1. $x⋅u'(x) = u(x) + x^3⋅e^{x^2}$; $u(1)=e/2$

2. $u'(x) = 2x\cdot u(x)^3$; $u(0) = a$; ($a \in \mathbb R$, hint: case differentiation)

3. $x^3⋅u'(x)-u(x)^3-x^2⋅u(x) = 0$; $u(1)=1$

I think, I already solved them correctly, with these solutions:

1. $u(x) = \dfrac{xe^{x^2}}{2}$

2. $u(x) = \dfrac{1}{\sqrt{-2x^2+\frac{1}{a^2}}}$

3. $u(x) = \dfrac{x}{\sqrt{-2\log(x)+1}}$

How can I find the maximum interval of existence now? I hope y'all understand me, I'm not a native english speaker.

Answer 1

The first DE equation is defined for $x\neq 0$ and hence the solution is defined either in $(-\infty, 0)$ or in $(0, \infty)$. The second interval contains the initial point $x=1$, so the maximal interval is $(0, \infty)$. Also the unique solution is $y=\frac{1}{2}x^2+\frac{e-1}{2}x$.

For the second one consider two cases $a=0$ and $a\neq 0$ separately. When $a=0$ the IVP has a unique solution $u=0$ defined for all $x$, so maximal interval is $(-\infty, \infty)$. For $a\neq 0$, it follows from your solution that you must determine an interval satisfying the inequality $-x^2+1/a^2>0$ and contain the initial point $x=0$. This interval would be your maxiamal interval of existence. I think you can do the last one by yourself. In this case observe that the DE is defined for all $x\neq 0$.

Answer 2

I will briefly go through each problem, but it's up to you to fill in the details.

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1. Rewrite the equation in standard form

$$ u' -\frac{1}{x}u = x^2e^{x^2} $$

This is a standard first-order linear equation. The integrating factor is $\mu = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = x^{-1}$. Multiplying through, we can simplify

$$ \left(\frac{u}{x}\right)' = xe^{x^2} \implies \frac{u}{x} = \frac{1}{2}e^{x^2} + c $$

The initial condition gives $c=0$, confirming your solution is correct.

As for the domain of existence, notice the coefficient of $u$ in the standard form is $1/x$, which is undefined at $x=0$, making it a singular point. Therefore, the domain can only be one of $(-\infty,0)$ or $(0,\infty)$. The one that contains the given initial point is the answer here.

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2. Upon separating and integrating, you end up with the solution

$$ \frac{1}{u^2} = \frac{1}{a^2} - 2x^2 $$

Since the left hand side cannot be negative, a required condition is $2x^2 \le \frac{1}{a^2}$ or $|x| \le \frac{1}{\sqrt{2}|a|}$. This is your domain.

Do note that $a$ can still be negative, so it's preferrable to rewrite the solution as

$$ u = \frac{a}{\sqrt{1-2a^2x^2}} $$

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3. This is a Bernoulli equation. The substitution $y=u^{-2}$ transforms the equation to one that's similar to problem 1, and you'll once again find that $x=0$ is a singular point. You'll also want to restrict the domain further so that $y = u^{-2} \ge 0$, just like in problem 2.