Differential equation involving evaluation of solution at a point

Question

I'm working on the differential equation: $$u_t(t,x,y) = u_y(t,x,y) + u_{xy}(t,1,1) \cdot u_x(t,x,y) $$

I haven't seen anything like this before in my courses, any ideas of what we could do? Would characteristics be appropriate here?

Thanks!

Answer 1

In the first step, the pde can be easily "solved" by the method of characteristics to achieve

$$ u(t,x,y) = C(x+\phi(t),y+t)\\ \phi'(t)=u_{xy}(t,1,1) $$ Assuming you have inital conditions at $t=0$, namely $u(0,x,y)=C(x+\phi(0),y)=u_0(x,y)$ we can eliminate $C$ and end up with $$ u(t,x,y)=u_0(x-\phi(0)+\phi(t),t+y)\\ \phi'(t)=(u_0)_{xy}(1-\phi(0)+\phi(t),1+t) $$ So, if you have a specific sufficiently smooth initial condition $u_0$ you can solve the second equation, get $\phi(t)$ and subsequently get an (explicit) solution $u$.

Example for $u_0=x^2y$: $$ \Rightarrow \phi'(t)=2(1-\phi(0)+\phi(t))\\ \Rightarrow\phi(t)=\phi(0)+e^{2t}-1\\ \Rightarrow u(t,x,y)=(x+e^{2t}-1)^2(t+y) $$

Note: this relatively easy approach works only for $\mathbb{R}^+\times\mathbb{R}^2$. Otherwise the first part will remain the same but the function $C$ will look different.