1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $$ x^{n-2}:n(n-1)b_n +(2(n-2)+1)b_{n-2}=0\\~\\ b_0=1,~~b_1=-1,\\~\\ b_{n}=-\frac{2n-1}{n(n-1)}b_{n-2}\implies \begin{aligned} b_{2m}&=(-1)^m\frac{(4m-1)(4m-5)\dots7\cdot 3}{(2m)!}b_0\\~\\ b_{2m+1}&=(-1)^m\frac{(4m+1)(4m-3)\dots9\cdot 5}{(2m+1)!}b_1 \end{aligned} $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.