Show that $\dot x=x^2-t^2$ has a solution on $(-1,1)\subset\mathbb R$

Question

I want to show that $\dot x=x^2-t^2,\ x(0)=1$ has a solution on $(-1,1)$ using some kind of power series argument.

So let's assume $x(t)=\sum_{n=0}^\infty a_n t^n$. The derivative is given by $\dot x(t)=\sum_{n=0}^\infty a_{n+1}(n+1)t^n$ and by the Cauchy-formula we have $x^2=\sum_{n=0}^\infty\sum_{k=0}^na_ka_{n-k}t^n$. So, by plugging those into the equation and comparing the coefficients, I have that $$a_{n+1}\cdot(n+1)=\sum_{k=0}^na_ka_{n-k}$$ for all $n\neq 2$. In the special case $n=2$ we have $3a_3=-1+\sum_{k=0}^2a_ka_{2-k}$.

Now, since $x(0)=a_0=1$, I can say that $a_1=1$ and $a_2=1$ by the recursion computed above. Then $a_3=2/3$ and so on. How do I move on from here? And is it correct so far?

Answer 1

If $0<a_k<1$ for all $0\le k\le n$ then

$$0<a_{n+1}=\frac1{n+1}\sum_{k=0}^n a_ka_{n-k}<\frac{n+1}{n+1}=1$$

and the radius of convergence around $0$ is at least $1$.

Answer 2

For Riccati-equations you get easier coefficient recursions if you parametrize $x$ as $$ x(t)=-\frac{\dot u(t)}{u(t)} $$ Then the poles of $x$ are at the roots of $u$ and every root of $u$ causes a pole of $x$.

The differential equation for $x$ transforms into a linear DE of second order for $u$, $$ -\frac{\ddot u(t)}{u(t)}+\frac{\dot u(t)^2}{u(t)^2}=\frac{\dot u(t)^2}{u(t)^2}-t^2\implies\ddot u(t)=t^2u(t) $$

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Consequently, the power series for $u(t)=\sum_{n=0}^\infty a_nt^n$ has the general coefficient recursion (set $a_n=0$ for $n<0$) $$ n(n-1)a_n=a_{n-4}. $$ together with $a_0=1$, $a_1=-a_0x(0)=-1$ we get $$ u(t)=1-t+\frac1{3\cdot 4}t^4-\frac1{4⋅5}t^5+\frac1{3⋅4⋅7⋅8}t^8-\frac1{4⋅5⋅8⋅9}t^9+\dots $$

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For the claim of the task, one has to test that $u(t)>0$ for $t\in(-1,1)$, as $u(0)=1>0$. As one can see, the terms are all positive for $t<0$. The pairs of coefficients $$1-t, \frac{t^4}{4}\left(\frac13-\frac t5\right),\frac{t^8}{4⋅8}\left(\frac1{3⋅7}-\frac{t}{5⋅9}\right),...$$ have positive sums for $t<1$, so we get also $u(t)>0$ for $t\in(0,1)$ which establishes the claim.

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The root of $u$ and thus the pole of $x$ can be found close to $1.037489135$.

Answer 3

Not a solution to the question, but too long for a comment:

One way to show that there is a solution valid on $(-1,1)$ is to note that $f(x,t) = x^2-t^2$ is smooth (in fact analytic) and hence a unique solution exists locally around $0$.

This can be done with differential inequalities. (A reference is Petrovitsch, 1901, I don't have a more accessible contemporary statement at the moment.)

The next step is to show that the solution is defined on (at least) the interval $(-1,1)$. Let $I$ be the maximal interval of definition of the solution, we know that it contains an open interval containing the origin.

One way to do this is to find continuous functions $x_1,x_2$ that are bounds for $x$ on $(-1,1)$. This allows the domain of definition of $x$ to be extended to $(-1,1)$.

Let $x_1,x_2$ satisfy $\dot{x_1} = x_1^2-1$ and $\dot{x_2} = x_2^2$ subject to $x_1(0) = x_2(0)=1$. It is straightforward to verify that $x_1(t) = 1$ and $x_2(t) = {1 \over 1-t}$ for $t \in (-1,1)$.

Note that $\dot{x_1} \le f(x_1,t)$ and $\dot{x_2} \ge f(x_2,t)$ for $t \in (-1,1)$, hence it follows that for $t \in [0,1) \cap I$ we have $x_1(t) \le x(t) \le x_2(t)$ and for $t \in (-1,0] \cap I$ we have $x_2(t) \le x(t) \le x_1(t)$.

Since $x_1,x_2$ are bounded on compact intervals in $(-1,1)$ it follows that $(-1,1) \subset I$.

Note that this is a weaker conclusion that the existence of a power series with radius of convergence one. Cauchy Kovalevskaya shows that there is a locally real analytic solution, however this is not enough to conclude, a priori, that there is a single series with radius one.

Yves Daoust's result shows that the $a_n$ are bounded, and hence the radius of convergence is at least one.