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It is fairly straight forward to solve linear 1st order PDEs by the method of characteristics. For example, if

$\partial_tf+a\partial_xf=bf$ ,

we have that $\dfrac{df}{dt}=bf$ on the characteristic curve of $\dfrac{dx}{dt}=a$ . From this we deduce that $f(t,x)=g(C)e^{bt}$ where $x=at+C$ .

Now, how does this work when $f$ is multidimensional. Can I solve equations on the following form by characteristics, or by any other means?

$\partial_tf_i(t,x)+\sum_jA_{ij}\partial_xf_j(t,x)=\sum_jB_{ij}f_j(t,x)$

where the components of $A$ and $B$ might be dependent on $x$ and $t$.

In particular, I am trying to solve the following,

$\begin{cases}\partial_tf+\dfrac{c}{t}\partial_xg=-\left(a+\dfrac{1}{t}\right)f\\\partial_tg+\dfrac{c}{t}\partial_xf=-\left(b+\dfrac{1}{t}\right)g\end{cases}$

where $f$ and $g$ are functions of $x$ and $t$ , where $t>t_0>0$ , $c\neq0$ . Any help is highly appreciated.

Edit: Never mind the specific equation. It tured out to be the result of a erroneous derivation. But I still wonder if there is some procedure to solve the general equation above.

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  • $\begingroup$ "$A$ and $B$ might be dependent on $x$ and $t$, making it quasilinear". Errr, no. You have variable coefficients, but not quasilinear. A quasilinear equation will have $A$ (and possibly also $B$) depending on $f$ as well as $t,x$. $\endgroup$ Feb 18, 2013 at 10:04
  • $\begingroup$ Your system of equations is degenerate/singular at $t = 0$. I assume you are wanting to solve it for either $t > 0$ or $t < 0$? $\endgroup$ Feb 18, 2013 at 10:05
  • $\begingroup$ Right, I'll edit and be a bit more specific. $\endgroup$
    – user62728
    Feb 18, 2013 at 16:55

1 Answer 1

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$\begin{cases}\partial_tf+\dfrac{c}{t}\partial_xg=-\left(a+\dfrac{1}{t}\right)f~......(1)\\\partial_tg+\dfrac{c}{t}\partial_xf=-\left(b+\dfrac{1}{t}\right)g~......(2)\end{cases}$ , $c\neq0$

From $(1)$ ,

$\partial_tf+\dfrac{c}{t}\partial_xg=-\left(a+\dfrac{1}{t}\right)f$

$\dfrac{c}{t}\partial_xg=-\partial_tf-\left(a+\dfrac{1}{t}\right)f$

$\partial_xg=-\dfrac{t}{c}\partial_tf-\left(\dfrac{at}{c}+\dfrac{1}{c}\right)f~......(3)$

$\partial_{xt}g=-\dfrac{t}{c}\partial_{tt}f-\dfrac{1}{c}\partial_tf-\left(\dfrac{at}{c}+\dfrac{1}{c}\right)\partial_tf-\dfrac{a}{c}f$

$\partial_{xt}g=-\dfrac{t}{c}\partial_{tt}f-\left(\dfrac{at}{c}+\dfrac{2}{c}\right)\partial_tf-\dfrac{a}{c}f~......(4)$

From $(2)$ ,

$\partial_tg+\dfrac{c}{t}\partial_xf=-\left(b+\dfrac{1}{t}\right)g$

$\partial_{xt}g+\dfrac{c}{t}\partial_{xx}f=-\left(b+\dfrac{1}{t}\right)\partial_xg~......(5)$

Put $(3)$ and $(4)$ into $(5)$ ,

$-\dfrac{t}{c}\partial_{tt}f-\left(\dfrac{at}{c}+\dfrac{2}{c}\right)\partial_tf-\dfrac{a}{c}f+\dfrac{c}{t}\partial_{xx}f=-\left(b+\dfrac{1}{t}\right)\left(-\dfrac{t}{c}\partial_tf-\left(\dfrac{at}{c}+\dfrac{1}{c}\right)f\right)$

$-\dfrac{t}{c}\partial_{tt}f-\left(\dfrac{at}{c}+\dfrac{2}{c}\right)\partial_tf-\dfrac{a}{c}f+\dfrac{c}{t}\partial_{xx}f=\left(\dfrac{bt}{c}+\dfrac{1}{c}\right)\partial_tf+\left(\dfrac{abt}{c}+\dfrac{a+b}{c}+\dfrac{1}{ct}\right)f$

$\dfrac{c}{t}\partial_{xx}f=\dfrac{t}{c}\partial_{tt}f+\biggl(\dfrac{(a+b)t}{c}+\dfrac{3}{c}\biggr)\partial_tf+\left(\dfrac{abt}{c}+\dfrac{2a+b}{c}+\dfrac{1}{ct}\right)f$

$\partial_{xx}f=\dfrac{t^2}{c^2}\partial_{tt}f+\dfrac{(a+b)t^2+3t}{c^2}\partial_tf+\dfrac{abt^2+(2a+b)t+1}{c^2}f$

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