 Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

1... $u_t+u_x=0$ .
2 ...$u_t+ sgn(x) u_x=0$
3...$u_t- sgn(x) u_x=0$
I suspect classification of characteristics. May need to do a bit of reading up.
1, unique solution
2. multiple solutions
3. non-unique solution (after a while)

Is $x = 0$ a pathological case?
Good question; we need more than $u(x,0)$ in domain of dependence. Paul
Posts: 10771
Joined: July 20th, 2001, 3:28 pm

### Re: About solving a transport equation

By George I think I she’s got it! Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

1519870931505.png

Had to test "place inline". This is great!
Is that St. Paul's London in the background? Alan
Posts: 10165
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: About solving a transport equation

1519870931505.png

Had to test "place inline". This is great!
Is that St. Paul's London in the background?
Yes, whenever Hope Hicks testifies, they have to find a secret location. Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

1519870931505.png

Had to test "place inline". This is great!
Is that St. Paul's London in the background?
Yes, whenever Hope Hicks testifies, they have to find a secret location.
Can we expect some shock waves? Alan
Posts: 10165
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: About solving a transport equation

Is that St. Paul's London in the background?
Yes, whenever Hope Hicks testifies, they have to find a secret location.
Can we expect some shock waves?
Certainly, given her unique characteristics. Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

Not to mention expansion waves, compression waves and contact discontinuities!
Not necessarily in that order. Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

Let me just address 1. for now. I just followed the prescription in my Vol I book, pg 33. That is, the characteristics $\xi$ are the solutions to the ODE:

$\frac{d \xi}{dt} = b(\xi) \equiv (1 - \xi)^2$ for this problem, with $\xi(t=0) = y$.

Since the solutions $\xi(t)$ also depend upon y, write them finally as $\xi(y,t)$.
I am now looking at equation A2.10 on page 33. Don't see how my solution and it are the same. Alan
Posts: 10165
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: About solving a transport equation

Really? A2.10 refers to P3. Take c=k=0, $\phi \rightarrow f$, $\tau \rightarrow t$, and $X_{\tau} \rightarrow \xi(t)$, suppressing the initial value x. That's the same, right, up to the various different labels for things?

p.s. Thanks for buying the book! Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

Really? A2.10 refers to P3. Take c=k=0, $\phi \rightarrow f$, $\tau \rightarrow t$, and $X_{\tau} \rightarrow \xi(t)$, suppressing the initial value x. That's the same, right, up to the various different labels for things?

p.s. Thanks for buying the book!
I can see that P3 is a 1st order hyperbolic PDE and a special case of P2 when diffusion coefficient = 0 (it reminds more of a singular (or regular?) perturbation problem). The analysis seems to switch between PDE and SDE which is alien to me,. The conclusion seems to be that the solution of P2 is via an expectation and this holds when $a(x)= 0$? In general, don't we get a boundary layer problem because only 1 BC is allowed for P3?

I am lost on the martingale discussion.

What I don't see from here is how the following solution was produced:

$u(y,t) = f \left( \xi(y,t) \right)$ where the characteristic  $\xi(y,t) = \frac{y + t - t y}{1 + t - t y}$.

My approach was a change of variables applied to 1st order hyperbolic PDE:

$\xi = \xi(x,y)$
$\eta = \eta(x.y)$

and this leads to the above solution. Alan
Posts: 10165
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: About solving a transport equation

Let me switch to (mostly) book notation for a moment. My main goal in Appendix 1.2 is to make the standard PDE-SDE probability connection for parabolic PDE's on the real line (so no bc). So, for each PDE treated, one finds the (formal) probabilistic solution: run an SDE and take an expectation. Effectively, this would be a Monte Carlo solution.

For example, to formally solve $u_t = \frac{1}{2} a^2(x) u_{xx} + b(x) u_x$ with $u(x,0) = f(x)$, you run the SDE

$dX_t = b(X_t) \, dt + a(X_t) \, dB_t$,  with $X_0 = x$.

Then the probabilistic solution is $u(x,t) = E[f(X_t)]$, where $E[\cdots]$ is a time-0 expectation, referencing the time-t value of all possible $\{X_t\}$ paths that start at $x$.

As a by-product, you also get the solution to the hyperbolic problem $u_t = b(x) u_x$ with $u(x,0) = f(x)$ -- by taking $a = 0$. Doing so, the SDE reduces to the ODE problem for the characteristic, namely:

(*) $dX_t = b(X_t) \, dt$, with $X_0 = x$.

Then, the probabilistic solution reduces to the deterministic solution $u(x,t) = f(X_t)$, by simply dropping the expectation. There's no longer any randomness in the SDE. For each starting value $x$, there's a unique path solution. The ODE (*)  is easily solved: $X_t = \xi(x,t)$, where, as I've posted several times

(**) $\int_x^{\xi} \frac{dy}{b(y)} = t$.

You fix x and t, and solve (**)  for $\xi = \xi(x,t)$.

This is all for general problems. Finally, for the specific problem at hand, take $b(y) = (1 - y)^2$, and solve (**) for $\xi$, taking in this case $x \in (0,1)$ and so $\xi \in (0,1)$. This is a slight 'cheat' because I started out saying the problems were on the whole real axis and now restrict the domain. But, I think it is legitimate for this case.  Go ahead and do the integral in (**), which is a very easy one, and you'll get

$\xi(x,t) = \frac{x + t (1-x)}{1 + t (1-x)}$.

So, that's how I got the answer I posted. Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

That's very good and clear, Alan.

As I see it the main issues were:

1. Probabilistic representation of the solution of problem P1 (or P2).
2. The solution of problem P3, where it comes from (it works in practice and in theory).
3. Using method of characteristics of problem P3.

1.
Indeed, as you mention in your references Friedman (great book), chapter 6, Theorem 5.3 elaborates the problem. This is the Cauchy problem in "full" space. (actually, similar results hold for elliptic PDE and parabolic PDE with Dirichlet BC.)
2. One assumption in Theorem 5.3 is that it is a uniformly elliptic (A1 on page 141). Important question is if this is necessary or sufficient for further results to hold. Miltstein and Tretyakov 1991, 1996 address this by the small noise SDE (reminds me of singular perturbations).

$dX = bdt + \varepsilon^2 c dt + \varepsilon dW$

Their solution is the same as your ODE solution.

Regarding the PDE itself, no BCs are needed. This can be proved by Fichera and by the standard energy inequality (well-posed) approach.  See attachment. Numerical boundary conditions are however, needed. Not an issue.
Regarding energy approach; it is widespread in PDE/FEM and generalises well.

I get the same solution as your integral approach. BTW do you use some kind of Leibniz' rule for this? I might have missed something along the way.
Attachments
EnergyEstmates.pdf Alan
Posts: 10165
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: About solving a transport equation

That's very good and clear, Alan.

I get the same solution as your integral approach. BTW do you use some kind of Leibniz' rule for this? I might have missed something along the way.
Thanks.

No Leibniz rule needed for what I wrote; you solve $\frac{dX}{dt} = b(X)$ simply by writing it as $\frac{dX}{b(X)} = dt$ and integrating both sides. The result is (**).

However, Leibniz rule does get used here when you want to know: what is $\xi_x$, the partial of the characteristic trajectory with respect to x, holding t constant? To get that, apply $\frac{\partial}{\partial x}$ to both sides of my (**), and  Leibniz rule yields:

(*** )$\frac{\xi_x}{b(\xi)} - \frac{1}{b(x)} = 0 \Rightarrow \xi_x(x,t) = \frac{b \left( \xi(x,t) \right)}{b(x)}$.

I am forever forgetting how to derive (***) and have to puzzle it out each time. Paul
Posts: 10771
Joined: July 20th, 2001, 3:28 pm

### Re: About solving a transport equation

My guess is that this works as long as the coeff b(x) is benign. Cuchulainn
Posts: 62113
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: About solving a transport equation

On a related note of 'using it or losing it' I would guess that > 90% of people have great difficulty with the chain rule of differentiation and applying it to e.g front-fixing in Stefan problem/early exercise. They forget to see that it is a 2d problem $(S,t) \rightarrow (y,\tau)$

where

$y = S/B(t)$
$\tau = t$

where $B(t)$ is the moving boundary.

I am forever forgetting how to derive (***) and have to puzzle it out each time.
Similar to transforming an elliptic PDE to canonical form.

Once the individual steps in the process have been externalised (and are not implicit anymore) then it becomes easier.  