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I have difficulties understanding how to solve a PDE in $\mathbb{R}^{4}$ using the method of characteristics. I have a limited background in solving PDEs. I have seen only 2-dim examples and none for higher dimensions, and really don't understand how to generalize the method to the 4-dim case.

Let $f$ be a $C^{1}$ function on $\mathbb{R}^{4}$. The goal is to solve in general

$Xf=uf$, with $X=\sum_{i=1}^{4}X_{i}\frac{\partial}{\partial x_{i}}$ a smooth vector field and $u$ a continuous function on $\mathbb{R}^{4}$ (not explicitly given).

No initial data is given so I'm free to prescribe it myself.

Attempt at the solution:

I have four variables, so we consider the curve $(\gamma_{1}(t),\gamma_{2}(t),\gamma_{3}(t),\gamma_{4}(t),\gamma_{5}(t))$ in $\mathbb{R}^{5}$ such that we have five ODEs:

$\frac{d\gamma_{1}}{dt}=X_{1}(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})$

$\frac{d\gamma_{2}}{dt}=X_{2}(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})$

$\frac{d\gamma_{3}}{dt}=X_{3}(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})$

$\frac{d\gamma_{4}}{dt}=X_{4}(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})$

$\frac{d\gamma_{5}}{dt}=u$

(is this correct so far?)

Now, from this point, I don't understand how and where to prescribe the initial values. That is, according to the general theory (for an equation for $\mathbb{R}^{n}$) one has to pick any transversal hypersurface on which one has the initial values, and it has dimension $n-1$. So is it right that it's 3-dim in our case? But then one variable "falls out" since I have four variables. I really don't understand how, and exactly for what, to give initial values.

Could anyone clarify this please? And how do I proceed from there? I'd be very grateful not to receive references to look in books, since I've done so multiple times and the theory there doesn't help... Thank you.

This is not homework.

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closed as off-topic by Deane Yang, Stefan Kohl, José Figueroa-O'Farrill, Johannes Hahn, Yoav Kallus Jul 14 '15 at 1:58

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You get a separate ODE for every integral curve (which you denote by $\gamma$) of the vector field $X$. The space $\mathbb R^4$ is a disjoint union of these integral curves. Each ODE is of order one, so you need to fix the value of $f$ at exactly one point from every integral curve. (Alternatively, you could also take asymptotic behaviour on each integral curve as a boundary condition.) You can fix the values on a three dimensional surface that meets every integral curve exactly once, provided such a surface exists. A good choice for such a surface or a collection of such surfaces depends on the vector field $X$.

The curves $\gamma$ should be curves in $\mathbb R^4$. We want $\gamma$ to satisfy $\gamma'(t)=X(\gamma(t))$ for every $t\in\mathbb R$ (this is what an integral curve mean) and we want $f$ to satisfy the ODE $\frac{d}{dt}f(\gamma(t))=u(\gamma(t))f(\gamma(t))$ along $\gamma$ (this for all integral curves $\gamma$ is equivalent with $Xf=uf$ if $f\in C^1$). This ODE can be solved by standard techniques.

The exact choice of the hypersurface on which we want to fix the values of $f$ depends on the vector field $X$. If $X(x)=(0,1+x_3^2,0,0)$, then one possibility is to fix the values of $f$ on the hypersurface given by $x_2=0$. There is no choice of the hypersurface that works for all possible $X$.

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  • $\begingroup$ thank you for the response. Could you please specify/give an example of what you mean by "fix the values on a three dimensional surface that meets every integral curve exactly once"? I have seen such statement before in the textbooks, but i don't see what it means in my concrete example? Could you write down what you mean? @Joonas Ilmavirta $\endgroup$ – GregVoit Jul 13 '15 at 13:17
  • $\begingroup$ and second question: are my equations correct then? And how does the final solution $f=…$ then look like (in terms of my $\gamma$)? $\endgroup$ – GregVoit Jul 13 '15 at 13:18
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    $\begingroup$ @GregVoit, I added some details. Your curve had one dimension too many. If you write $F=f\circ\gamma$ and $U=u\circ\gamma$, the ODE I wrote becomes $F'=UF$, which you should be able to solve. $\endgroup$ – Joonas Ilmavirta Jul 13 '15 at 13:27
  • $\begingroup$ oh, ok now I see that my equations were wrong. So about the initial data: my vector field $X$ has non-vanishing components, i.e. $X_{i}\neq 0$, $i=1,2,3,4$. Does it mean then that I can choose myself on which hyper surface to pick the values? For example, if I take $x_{2}=0$, then I have to prescribe value $f(x_{1},0,x_{3},x_{4})$ right? But that's just one value… @Joonas Ilmavirta $\endgroup$ – GregVoit Jul 13 '15 at 13:30
  • $\begingroup$ @GregVoit, if all components of $X$ are non-vanishing (everywhere), then you can pick any $i=1,2,3,4$ and $c\in\mathbb R$ and fix the values on the hyperplane where $x_i=c$. Your example with $i=2$ and $c=0$ works. Then you have to prescribe the values of $f$ for all choices of $x_1,x_3,x_4$ to cover all integral curves. $\endgroup$ – Joonas Ilmavirta Jul 13 '15 at 13:35

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