# To solve non-linear Integro-differential equation

I am just begin to study integral equations, in which i come with following problem regarding second kind Volterra non-linear integro-differential equation, $$u'(x)=-1+\int_{0}^{x}u^{2}(t)dt$$ with the initial condition $\displaystyle u_{0}(x)=-x$. I want to know the exact solution of above problem i don't know how to proceed.... Thanks in advance....

We derive $$u''(x)=u(x)^2$$

By trial and error I found that $u$ could be $$u(x)=\frac6{(x+k)^2}$$

where $k$ is a constant that can be easily computed from your integro-differential equation (I obtained $k=\sqrt 6$). I don't know if there are more solutions.

By the other hand, I don't understand that initial condition: $u_0(x)=x$. Isn't that a function definition? What is exactly $u_0$?.

$u'(x)=-1+\int_0^xu^2(t)~dt$

$u''(x)=u^2(x)$

According to http://www.wolframalpha.com/input/?i=u%22%3Du%5E2,

$u(x)=\sqrt6~\wp\left(\dfrac{x+C_1}{\sqrt6};0,C_2\right)$

But I don't know how to substitute $u'(0)=-1$ .

The equation implies $$u''(x)=u(x)^2$$ when differentiating. Multiply by $$6u'(x)$$ to obtain $$6u'(x)u''(x)=6u(x)^2u'(x)=3(u'^2)'(x)=2(u^3)'(x).$$ Therefore, $$3u'(x)^2-3u'(0)^2=2u(x)^3-2u(0)^3,$$ which is equivalent to $$u'(x)^2=\frac23[u(x)^3+C_0],$$ where $$C_0=\frac{u'(0)^2}2-\frac{u(0)^3}3.$$ This implies the restriction $$u(x)^3\geq-C_0,$$ which is equivalent to $$u(x)\geq-\sqrt{C_0}.$$ The case where $$u(x)=-\sqrt{C_0}$$ is excluded, unless $$C_0=0,$$ and the reason for this is because, after multiplying by $$u'(x),$$ the equation $$6u'(x)u''(x)=6u(x)^2u'(x)$$ is trivially satisfied whenever $$u'(x)=0,$$ but this is not so for $$u''(x)=u(x)^2$$ unless $$u(x)=0,$$ which satisfies it. Therefore, for the non-trivial solutions, we must have $$u(x)\gt-\sqrt{C_0}.$$ With this having been said, there are now two possible equations, $$\frac1{\sqrt{u(x)^3+C_0}}u'(x)=-\sqrt{\frac23}$$ $$\frac1{\sqrt{u(x)^3+C_0}}u'(x)=\sqrt{\frac23}$$ Consider a function $$W_{C_0}:\left(-\sqrt{C_0},\infty\right)\rightarrow\mathbb{R}$$ such that $$W_{C_0}'(t)=\frac1{\sqrt{t^3+C_0}}.$$ Therefore, $$W_{C_0}[u(x)]-W_{C_0}[u(0)]=\pm\sqrt{\frac23}x,$$ or simply, $$W_{C_0}[u(x)]=W_{C_0}[u(0)]\pm\sqrt{\frac23}x.$$ This is the implicit solution, and one cannot go any further without analyzing $$W_{C_0}.$$ In the special case that $$C_0=0,$$ we have that $$W_0'(t)=t^{-\frac32},$$ implying that $$W_0(t)=K-\frac2{\sqrt{t}}$$ is possible, meaning that $$-\frac2{\sqrt{u(x)}}=-\frac2{\sqrt{u(0)}}\pm\sqrt{\frac23}x,$$ and you can do the rest.

The interesting case is when $$C_0\neq0.$$ According to Wolfram Alpha, $$W_{C_0}(t)=K+\frac{t\sqrt{\frac{t^3+C_0}{C_0}}\,_2F_1\left(\frac13,\frac12;\frac43;-\frac{t^3}{C_0}\right)}{\sqrt{t^3+C_0}}.$$ Therefore, $$\frac{u(x)\sqrt{\frac{u(x)^3+C_0}{C_0}}\,_2F_1\left(\frac13,\frac12;\frac43;-\frac{u(x)^3}{C_0}\right)}{\sqrt{u(x)^3+C_0}}=\frac{u(0)\sqrt{\frac{u(0)^3+C_0}{C_0}}\,_2F_1\left(\frac13,\frac12;\frac43;-\frac{u(0)^3}{C_0}\right)}{\sqrt{u(0)^3+C_0}}\pm\sqrt{\frac23}x.$$ There is the implicit solution. The function $$\,_2F_1$$ is the hypergeometric function.