I have the following Cauchy problem $$ \begin{cases} y'= y^2 - x^2= f(x,y)\\ y(0)=1 \end{cases} $$ in the rectangle $R=\{(x,y), 0 \leq x \leq 1, |y-1| \leq 1\}$.

The question is to prove the existence and uniqueness of solution of the problem and find the exact solution or an approximate solution.

Using the fixed point Theorem, I proved that the Cauchy problem admits a unique solution in the interval $\big[0,\frac{1}{5}\big]$.

For calculation of the solution, we have by the Fixed Point Theorem that the sequence $(y_n)= T(y_{n-1})$ converge vers the unique solution of the problem. We have: $$ T\big(y(x)\big)= y_0+ \int_0^x (y^2(s)-s^2)\ ds, \ y \in R, \ x \in \left[0,\dfrac{1}{5}\right] $$ Then by the relation $y_n= T\big(y_{n-1}(x)\big)$, we have \begin{align} y_1(x) &= 1+ \int_0^x (1-s^2) ds = 1 + x-\dfrac{x^3}{3} \\ y_2(x) &= 1+ x +x^2 -\dfrac{1}{6} x^4 - \dfrac{2}{16} x^5 + \dfrac{1}{63} x^7 \end{align}

How do we find the exact solution $y(x)$?

Thanks in advance for the help.


Your equation is a Riccati equation, you can not find a simple symbolic solution. However, you can insert the power series, assuming it exists and converges, either into the differential equation or into the integral version. $$ y=\sum_{n=0}^\infty c_nx^n=1-\frac{x^3}3+\sum_{n=1}^\infty \frac{x^n}{n}\sum_{k+m=n-1}c_kc_m $$ so that by comparing coefficients you get \begin{align} c_0&=1\\ c_1&=c_0^2=1\\ c_2&=\frac12(c_0c_1+c_1c_0)=1\\ c_3&=-\frac13+\frac13(c_1^2+2c_0c_2)=\frac23\\ c_4&=\frac12(c_0c_3+c_1c_2)=\frac56\\ &\vdots\\ c_{2m}&=\frac1m(c_0c_{n-1}+c_1c_{n-2}+\dots+c_{m-1}c_m)\\ c_{2m+1}&=\frac1{2m+1}(2c_0c_{2m}+2c_1c_{2m-1}+\dots-2c_{m-1}c_{m+1}+c_m^2) \end{align}

A systematic way to solve it is to set $y=-\frac{u'}u$ so that the second order linear ODE $$ u''=x^2u, ~ u(0)=1,~u'(0)=-1 $$ results. This is easier to solve via power series expansion as a simple coefficient recursion results, $$ a_{n+4}=\frac{a_n}{(n+4)(n+3)},~~ a_0=1,~ a_1=-1,~ a_2=0,~a_3=0 $$ You can express the solution in terms of named special functions like Bessel functions, see Convert $\frac{d^2y}{dx^2}+x^2y=0$ to Bessel equivalent and show that its solution is $\sqrt x(AJ_{1/4}+BJ_{-1/4})$ and links there for conversion methods.

You can split off the growth envelope of $u$ as $u(x)=e^{x^2/2}v(x)$ which is the same as setting $y(x)=x-\frac{v'(x)}{v(x)}$. The ODE for $v$ is again order 2 linear, $$ (x^2+1)v+2xv'+v''=x^2v\\ v''+2xv'+v=0,~~v(0)=1, v'(0)=u'(0)=-1 $$ and we get for the power series expansion $$ x^{n-2}:n(n-1)b_n +(2(n-2)+1)b_{n-2}=0\\~\\ b_0=1,~~b_1=-1,\\~\\ b_{n}=-\frac{2n-1}{n(n-1)}b_{n-2}\implies \begin{aligned} b_{2m}&=(-1)^m\frac{(4m-1)(4m-5)\dots7\cdot 3}{(2m)!}b_0\\~\\ b_{2m+1}&=(-1)^m\frac{(4m+1)(4m-3)\dots9\cdot 5}{(2m+1)!}b_1 \end{aligned} $$

  • $\begingroup$ Please but how we find the exact solution using the sequence $x_n = T(x_{n-1})$ defined in my post? $\endgroup$ – rosy May 19 '18 at 11:17
  • $\begingroup$ You do not, not really. You can formalize the insertion of the power series using the Cauchy product and get ever expanding formulas for the coefficients, added on top. $\endgroup$ – LutzL May 19 '18 at 11:37

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