# Partial integro-differential equation

I don't know if there is a method to solve this following integro - differential equation:

$$\partial_t{u(x,t)}\int_{0}^{x}u(\zeta,t)d\zeta=\partial_{xx}u(x,t)$$ Can someone give me some hint? Thanks.

• One solution is zero. What answer would you like to have on the same question say for the heat equation? – Andrew Apr 26 '12 at 14:06
• @Andrew: zero is a trivial solution. I would like to know if there is a method to solve this kind equations. – Riccardo.Alestra Apr 26 '12 at 15:05

Let $v(x,t)=\int_0^xu(\zeta,t)\,d\zeta$. Then $0=v_x$ and the original equation is equivalent to $$\tag{1}v\,v_{tx}=v_{xxx}.$$ This is a non-linear equation for which I doubt that a general solution can be found. However, you can find spcial solutions by reducing it to an ODE.
Solutions in separated variables. Looking for solutions of the form $v(x,t)=X(x)\,T(t)$ leads to $$\frac{X'''}{X\,X'}=T'=a,\quad\text{a constant.}$$ The solution of the equation $T'=\lambda$ is $T(t)=a\,t+b$. The equation for $X$ is $$X'''=a\,X\,X'=\frac{a}{2}\,(X^2)'\implies X''=\frac{a}{2}\,X^2+b.$$ Multiplying the last equation by $X'$ and integrating we get the first order equation $$(X')^2=3\,a\,X^3+2\,b\,X+c,$$ whose solution is $$\int\frac{dx}{\sqrt{3\,a\,X^3+2\,b\,X+c}}\,dx=\pm\,x+d.$$ Taking $a=0$ gives $u(x,t)=A\,x+B$, $A,B\in\mathbb{R}$. Taking $a\ne0$ and $b=c=0$ gives the family of solutions $$u(x,t)=\frac{a\,(a\,t+B)}{12\,(C\pm\,x)^{3}}.$$
Self-similar solutions. For any $\beta\in\mathbb{R}$, the function $v(x,t)=t^{1+2\beta}w(x\,t^\beta)$ is a solution of (1) if $w(\xi)$ is a solution of the ODE $$(1+2\,\beta)\,w\,w'+\beta\,\xi\,w''=w'''.$$
Travelling wave solutions. $v(x,t)=\phi(x+c\,t)$ leads to the ODE $$c\,\phi\,\phi''=\phi'''.$$
• I think there are not any traveling wave solutions because $u_{xx}(0,t)=0$, which implies $\phi'''=0$, etc. But I am curious what is the purpose for this equation? – Bob Terrell Apr 27 '12 at 19:07