To solve non-linear Integro-differential equation

Question

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....

Answer 1

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 $u_0$?.

Answer 2

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

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

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

$u(x)=\sqrt[3]6~\wp\left(\dfrac{x+C_1}{\sqrt[3]6};0,C_2\right)$

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

Answer 3

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 $u(x)\geq-\sqrt[3]{C_0}.$ The case where $u(x)=-\sqrt[3]{C_0}$ 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 $u(x)\gt-\sqrt[3]{C_0}.$ 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 $W_{C_0}:\left(-\sqrt[3]{C_0},\infty\right)\rightarrow\mathbb{R}$ 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.