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Task: Solving the PDE $au_x+bu_y+cu=0$.
(Source: PDE, 2ndE by Walter A. Strauss, Exercise 1.2.19. Lots of books have it, though.)

Solution 1

The PDE can be transformed by the coordinate method via $$\begin{cases}x'=ax+by\\y'=bx-ay\\\end{cases}$$ and we shall obtain $(a^2+b^2)u_{x'}+cu=0$, which gives $u=f(bx-ay)\exp(-\frac{c}{a^2+b^2}x)$.

Solution 2

When I was googling I came across this second solution here (links to PDF) which states that we may do another transformation by letting $$\begin{cases}x'=-\frac{b}{a}x+y\\y'=x\\\end{cases}$$ and getting $u=f(-\frac{b}{a}x+y)\exp(-\frac{c}{a}x)$

By the link, I sense that by choosing $f(bx-ay)$ smartly, between the two generic solutions will shine equivalence.

(I'm aware of a third solution in which we divide the left side of the PDE by $u$ and then set $v=\ln(u)$ which will yield the same solution as solution II. My God, so many solutions??)

Thoughts and Questions

I think that the way we solve PDEs in Solution I is nice and clean because we chose another orthogonal coordinate system, and that the Solution II is harder to envisage because the coordinates are not orthogonal. I'm not so confident, so I would like to ask the following questions:

  1. How to show the equivalence of the two generic solutions?

  2. What's the goal of Solution II? And how is it achieving its desired goal (why does it transform like that by going non-orthogonal)?

  3. Are there any advantages in using one transform over another? If so, why?
    (I see for sure that the link's author likes this Solution II)
    I'd like some cool suggestions.

  4. (Thank you! And something extra on the other side of the fence.)
    I can't solve PDEs like:
    $au_x+bu_y+cu=g(x,y)=\exp(dx+ey)$ What should I do about it?

Thanks!

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1 Answer 1

up vote 1 down vote accepted

The equation involves the directional derivative of $u$. So we want to solve it by considering the lines with direction vector $(a,b)$, since on each such line the equation becomes an ODE. Specifically, it becomes a linear homogeneous ODE with constant coefficients: $u_t+ku=0$ and we know that the standard approach to such ODE is to look for solutions of the form $u=C\exp(\lambda t)$. The difference between solutions II and III is the order of operation: we could recognize the constant-coefficient structure first (hence take an exponential ansatz) or the directional structure first (hence introduce new coordinate axes).

Whether to use I or II depends on how initial values are given. Since we can solve the equation separately on each line in the direction $(a,b)$ , the solution is uniquely determined by its values on any line going in any other direction. If we know the values of $u$ on a line perpendicular to $(a,b)$, the orthogonal transformation of I fits the task perfectly. But if we are given $u(0,y)$ or $u(x,0)$ instead, then we better keep that axis as one of our coordinates, orthogonality be damned.

As for IV, it's a nonhomogeneous equation which reduces to a nonhomogeneous ODE of the form $u_t+cu=\alpha \exp(\beta t)$. The solution is obtained as a sum of the general solution of the homogeneous ODE and a particular solution, which you will find in the form of the right hand side (or, if you are unlucky to hit resonance $\beta=-c$, RHS times a linear function). This is standard ODE material, http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

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Thank you! The initial value "what-if" tip is what I had never really thought of before. I'm still thinking of what sol II is trying within the elegant exponential outfit. I think it's based on a shear + transpose matrix which shears the original coordinate system $xOy$ so that either one of $u_x$ or $u_y$ drops out of the PDE. It's obvious that Solution I is a rotation + zoom matrix. So after all, skewing and rotation can both transform $(a,b)$ so that the new vector parallels one of the new axes Also, you're right in that I need to go over ordinary ODE. –  FrenzY DT. Jul 30 '12 at 7:25
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