Alan,

You posted an answer before I did a final update, Ignore remark about u_t - u_x = 0.

- Cuchulainn
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Alan,

You posted an answer before I did a final update, Ignore remark about u_t - u_x = 0.

You posted an answer before I did a final update, Ignore remark about u_t - u_x = 0.

Yeah, I hate when that happens, too. Esp. when math is posted, it's always best to wait several minutes for edits.

- Cuchulainn
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Some of the questions just were in my original post. Alan has just started some feedback on them.2. The solution does not need any boundary conditions.

3. Plugging in, [$]u(0,t) = f(t/(1+t))[$]. Does this mean anything?

4. [$]u(1,t) = f(1))[$],

5. What happens in the empty quarter [$]y > 1[$]? for mean, it is beyond infinity.

No, I wasn't just reposting Alan's solution! It was my solution.

TBH I never actually looked at any of the details of any posts here...it was too painful!

I just assumed that whatever answer Alan had, that Cuch was referring too, was very unlikely to be wrong!!!

z=1/(y-1) ...

A) They are straight lines at 45 degrees. y between zero and one corresponds to z from - infinity to -1.

B) They are hyperbolae going "south east". E.g. the one from (0,0) zero goes to - infinity at t=1. The one from (0,1) is degenerate, it is y=1. There will be solutions for some y<0 even though data only given on [0,1]

C) Hyperbolae go the other way.

D) Too messy for now

E) No solution for all y in [0,1]. The bottom hyperbola defines region

F) If going backwards in time there will be a jump from just below y=1 to just above (depending on initial data on last characteristic)

G) Jump in solution at finite time. A "shock" and ...

H) one of the reasons why need to know this stuff before doing numerics

I) Because b is only a function of time this is relatively easy. Can assume it's one, except for any singularities.

Alan I think you meant

[$]y=\frac{t}{t-1}[$]

as the bottom characteristic(?)

TBH I never actually looked at any of the details of any posts here...it was too painful!

I just assumed that whatever answer Alan had, that Cuch was referring too, was very unlikely to be wrong!!!

z=1/(y-1) ...

A) They are straight lines at 45 degrees. y between zero and one corresponds to z from - infinity to -1.

B) They are hyperbolae going "south east". E.g. the one from (0,0) zero goes to - infinity at t=1. The one from (0,1) is degenerate, it is y=1. There will be solutions for some y<0 even though data only given on [0,1]

C) Hyperbolae go the other way.

D) Too messy for now

E) No solution for all y in [0,1]. The bottom hyperbola defines region

F) If going backwards in time there will be a jump from just below y=1 to just above (depending on initial data on last characteristic)

G) Jump in solution at finite time. A "shock" and ...

H) one of the reasons why need to know this stuff before doing numerics

I) Because b is only a function of time this is relatively easy. Can assume it's one, except for any singularities.

Alan I think you meant

[$]y=\frac{t}{t-1}[$]

as the bottom characteristic(?)

Alan, you say that the solution is 'undefined.' That's because of the 'flow' of information. It's like putting origami boats in a flowing river, you can predict the number of boats downstream of you but unless you know whether someone upstream has done the same you can't say how many boats will be passing you in the future!

Actually, my [$]\xi(t)[$] with [$]\xi(0) = y \in (0,1)[$] are increasing in [$]t[$]. After all [$]\dot{\xi}(t) = (1 - \xi(t))^2 > 0[$]. This may be a nomenclature issue on what we are calling the 'characteristics'??

Anyway, plot 'my' characteristics starting at [$]y>1[$] and you'll see the issue leading to my [$]y^*(t)[$].

At this point, I think a "full analysis" requires examining initial data (and the solution) on the whole [$]y[$]-axis. Then, nothing is undefined. But, I don't want to do it.

Anyway, plot 'my' characteristics starting at [$]y>1[$] and you'll see the issue leading to my [$]y^*(t)[$].

At this point, I think a "full analysis" requires examining initial data (and the solution) on the whole [$]y[$]-axis. Then, nothing is undefined. But, I don't want to do it.

Last edited by Alan on January 29th, 2020, 5:18 pm, edited 4 times in total.

I don't think it's any different. You'll just have to keep track of your signs. (But get of that stupid -1 first of all!!!)

But I could be wrong. I'll check.

But I could be wrong. I'll check.

V quick, might be wrong...

Chars are

[$]y=\frac{y_0}{1+y_0t}[$]

going through (y_0,0).

Why is this bad?

Chars are

[$]y=\frac{y_0}{1+y_0t}[$]

going through (y_0,0).

Why is this bad?

Sorry, that’s after taking out the stupid 1!

- Cuchulainn
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Regarding the boundary condition, the exact formula gives [$]u(1,t) = f(1)[$] which is also the value predicted by special case of Fichera theory (no diffusion, and convection = 0 when y = 1). And corroborated numerically. Voila! we have a numerical BC at [$]y = 1[$]. No BC ay y = 0.

*At this point, I think a "full analysis" requires examining initial data (and the solution) on the whole y **axis. Then, nothing is undefined. But, I don't want to do it.*

ah, go on.

[ltr]y = 0 in the original is not 100% correct; y = -1 (minus infinity) is better and then we are back OP's pde:[/ltr]

[ltr][$]\frac{\partial u}{\partial t} + (1+y)(1-y)\frac{\partial u}{\partial y} = 0 [$] on the interval [$](-1,1)[$] AFAIR[/ltr]

You can take [$]f(y) = e^{-y^{2}/2}[$] on the real line as initial condition.

Q: do ye exact solutions still work?

[ltr]This is now a singular pde.[/ltr]

ah, go on.

[ltr]y = 0 in the original is not 100% correct; y = -1 (minus infinity) is better and then we are back OP's pde:[/ltr]

[ltr][$]\frac{\partial u}{\partial t} + (1+y)(1-y)\frac{\partial u}{\partial y} = 0 [$] on the interval [$](-1,1)[$] AFAIR[/ltr]

You can take [$]f(y) = e^{-y^{2}/2}[$] on the real line as initial condition.

Q: do ye exact solutions still work?

[ltr]This is now a singular pde.[/ltr]

Last edited by Cuchulainn on January 29th, 2020, 5:55 pm, edited 2 times in total.

If you are worried about initial/boundary conditions you need to know more about characteristics and what they mean. Then you will know what sort of conditions make sense, which don’t, where you will find solutions, where you won’t, where there will be a conflict, etc.

It’s such an interesting topic that I think you should get a basic book on the subject. The way you are going about this is too random. You are going to miss all the best bits and, more worryingly, get basic things wrong. I’m usually all in favor of reinventing the wheel. But this subject is fun and I don’t want you to miss that!!

And it has a big impact on numerics. These pdes are much harder than the diffusion equation!

It’s such an interesting topic that I think you should get a basic book on the subject. The way you are going about this is too random. You are going to miss all the best bits and, more worryingly, get basic things wrong. I’m usually all in favor of reinventing the wheel. But this subject is fun and I don’t want you to miss that!!

And it has a big impact on numerics. These pdes are much harder than the diffusion equation!

Original question has solution something like:

[$]u(y,t)=f\left(\frac{(1+y)e^{-t}-(1-y)e^t}{(1+y)e^{-t}+(1-y)e^t}\right)[$]

[$]u(y,t)=f\left(\frac{(1+y)e^{-t}-(1-y)e^t}{(1+y)e^{-t}+(1-y)e^t}\right)[$]

- Cuchulainn
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You don't say. Most of the famous schemes (Lax-Wendroff, Friedrichs, 1st order upwinding Vol III PW page 1224) don't use characteristics at all for 1st order pde. They are useful for motivation, How do you solve Asian PDEs using characteristics. They are very niche, but do come in handy.If you are worried about initial/boundary conditions you need to know more about characteristics and what they mean. Then you will know what sort of conditions make sense, which don’t, where you will find solutions, where you won’t, where there will be a conflict, etc.

It’s such an interesting topic that I think you should get a basic book on the subject. The way you are going about this is too random. You are going to miss all the best bits and, more worryingly, get basic things wrong. I’m usually all in favor of reinventing the wheel. But this subject is fun and I don’t want you to miss that!!

And it has a big impact on numerics. These pdes are much harder than the diffusion equation!

FYI I wrote a simulation system for a nonlinear Waterhammer flow for central heating of the Hague running on a CDC 6600 supercomputer in Fortran.

I think you have missed the point of this thread completely. The real pde is embedded in the Anchor article...and at the end of the day you need numeric boundary conditions.

The real challenge is numerical, especially convectIon+diffusion. That's not news.

Last edited by Cuchulainn on January 29th, 2020, 8:53 pm, edited 2 times in total.

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