MHB Interval for the defined solution

[mathmari]

Hey!

I have to find the solution of the problem $$u_t(x,t)+u_x(x,t)=\frac{2x}{1+(x-t)^2}u^2(x,t), x \in \mathbb{R}, t>0 \\ u(x,0)=1, x \in \mathbb{R}$$

I found that the solution is $$u(x, t)=\frac{1+x^2-2xt+t^2}{x^2-4xt+2t^2+1}$$

Now we have to look for which values this solution is defined.

$$2t^2-4xt+(x^2+1)=0 \\ \Delta=8(x^2-1) \\ t_{1,2}=x\pm \sqrt{\frac{x^2-1}{2}}$$

• When $\Delta<0 \Rightarrow |x|<1$, $x^2-4xt+2t^2+1=0$ has no roots. So, the solution is defined for $t \geq 0$.
  
• When $\Delta=0 \Rightarrow x=\pm 1$ the solution is defined $\forall t \geq 0$ with $t \neq 1$.
  
• When $\Delta>0 \Rightarrow |x| >1 \Rightarrow x>1 \text{ or } x <-1$. For $x<-1$ the solution is defined $\forall t \geq 0$ and for $x>1$ the solution is defined $\forall t \geq 0$ with $t \neq x\pm \sqrt{\frac{x^2-1}{2}}$.

Is this correct?? (Wondering)

Because in my notes there is the following solution:

For $-1<x<1$ the solution is defined $\forall t>0$, for $x \leq -1$ the solution is defined $\forall t \geq 0$, but for $x \geq 1$ the solution is defined for $0 \leq t \leq x-\sqrt{\frac{x^2-1}{2}}$.

 

[I like Serena]

Hi! (Smile)

I agree with your analysis. (Nod)

Then, perhaps they want the domain of the solution where it is continuous, which would exclude the part at the other side of the hyperbola.
Still, in that case we should also have the part for $x \geq 1$ with $t > x+\sqrt{\frac{x^2-1}{2}}$. (Thinking)