### Author Topic: Uniqueness/ continuity of solutions  (Read 1983 times)

#### Ioana Nedelcu

• Full Member
• Posts: 29
• Karma: 3
##### Uniqueness/ continuity of solutions
« on: January 24, 2018, 12:29:49 AM »
I'm still trying to understand how the initial and boundary conditions affect the existence and types of solutions.

For example, in one of the tutorials the solution to $$xu_{t} + u_{x} = 0$$ $$u(x, 0) = f(x)$$ is not uniquely determined. What exactly determines this?

(It's also possible I understood the question completely wrong but I'm generally confused about uniqueness)

#### Victor Ivrii

• Elder Member
• Posts: 2553
• Karma: 0
##### Re: Uniqueness/ continuity of solutions
« Reply #1 on: January 24, 2018, 03:27:49 AM »
Also existence problems at $(0,0)$.

Draw integral lines and consider their intersections with $\{(x,t)\colon t=0\}$.

#### Jingxuan Zhang

• Elder Member
• Posts: 106
• Karma: 20
##### Re: Uniqueness/ continuity of solutions
« Reply #2 on: January 24, 2018, 06:01:01 AM »
But should there be issue with existence? I think there at the origin we have this parabolic characteristic tangent to it and what's wrong with that?

#### Victor Ivrii

• Elder Member
• Posts: 2553
• Karma: 0
##### Re: Uniqueness/ continuity of solutions
« Reply #3 on: January 24, 2018, 06:48:43 AM »
Look, at the picture. In the "yellow" domain initial condition $u(x,0)=f(x)$ does not define solution (why?).
In  the white domain initial condition could be impossible to satisfy (why?)
« Last Edit: January 24, 2018, 06:50:34 AM by Victor Ivrii »

#### Jingxuan Zhang

• Elder Member
• Posts: 106
• Karma: 20
##### Re: Uniqueness/ continuity of solutions
« Reply #4 on: January 24, 2018, 07:13:23 AM »
Look, at the picture. In the "yellow" domain initial condition $u(x,0)=f(x)$ does not define solution (why?).
In  the white domain initial condition could be impossible to satisfy (why?)
But you mentioned in another post that existence problem is at $(0,0)$. For the white domain indeed you have to have even initial function to meet the symmetrical curves, but $t-x^{2}=0$ does not require this, and so shouldn't it rather be no issue at origin?

#### Victor Ivrii

• Elder Member
• Posts: 2553
• Karma: 0
##### Re: Uniqueness/ continuity of solutions
« Reply #5 on: January 24, 2018, 07:43:34 AM »
It means "in the arbitrarily small vicinity of the origin"

#### Ioana Nedelcu

• Full Member
• Posts: 29
• Karma: 3
##### Re: Uniqueness/ continuity of solutions
« Reply #6 on: January 24, 2018, 05:40:54 PM »
For the yellow domain, is it because when t=0 there are infinite possibilities for f(x)? So then there is no unique solution there

Impossible solution in the white domain because the characteristic curve doesn't cross the x-axis so there is no function of x, ie solution doesn't exist

Because the solution u(x, t) is constant on/ determined by the characteristic curves, do we essentially just look at how those curves behave at the initial and boundary conditions?

#### Victor Ivrii

1) In yellow domain characteristics do not intersect $\{t=0\}$ (non-unicity)
2) In white domain they intersect $\{t=0\}$ twice (possible non-existence)