Does a Unique Solution Exist for a PDE with Specific Boundary Conditions?

  • Thread starter Thread starter t.t.h8701
  • Start date Start date
  • Tags Tags
    Uniqueness
t.t.h8701
Messages
2
Reaction score
0
If I have a PDE like Ux-Uy=0 and U(x,0)=f(x) when x in [0,1]. Then is there an uniqueness solution exist at point (5,1)?
How can I explain it using characteristics lines?

Thanks
 
Physics news on Phys.org
t.t.h8701 said:
If I have a PDE like Ux-Uy=0 and U(x,0)=f(x) when x in [0,1]. Then is there an uniqueness solution exist at point (5,1)?
How can I explain it using characteristics lines?

Thanks
If you only know f(x) between 0 and 1, you are going to have a problem extending the solution to x= 5!

The "characteristic lines" are of the form x+ y= C and any solution to this equation is of the form F(x+y) where F is an arbitrary function of one variable. Since you require that U(x,0)= F(x+0)= F(x)= f(x) for x between 0 and 1, you will need to take F(x) to be f(x) between 0 and 1 but that does not define it for x+ y= 5+ 1= 6. Consider any number of functions "f(x)" which are identical between 0 and 1 but differ outside that interval.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top