0
$\begingroup$

For my homework I am asked to do the following:

Solve $au_x+bu_y=f(x,y)$, where $f(x,y)$ is a given function. If $a\neq 0$ write the solution in the form $$u(x,y)=(a^2+b^2)^{-\frac{1}{2}}\int_{L}fds +g(bx-ay)$$ where the integral is a line integral and $L$ is the characteristic line segment from the $y$-axis to the point $(x,y)$ and $g$ is an arbitrary function of one variable. A hint to use the coordinata method (change of coordinates) is given.

For the $g(bx-ay)$ part we have $g_x(bx-ay)=bg'$ and $g_y=-ag'$ so this satisfies $ag_x+bg_y=0$ and therefore is the homogeneous solution. For the rest I realized that $au_x+bu_y$ is the directional derivative of $u$ along the characteristic line $c=bx-ay$ and therefore integrating along this line to solve seems reasonable. However I am unclear about the particulars. If anyone could help me out I would be very thankful. Also, isn't it important that besides specifying $a\neq 0$ we also have $b\neq 0$?

$\endgroup$
  • $\begingroup$ Are you going to mark your other questions as solved? $\endgroup$ – Kaster Feb 15 '13 at 3:24
  • $\begingroup$ $Kaster I always do, I just have not had the time yet to read the answer thoroughly. $\endgroup$ – Slugger Feb 15 '13 at 11:15
  • $\begingroup$ @Teun Verstraaten:how get $\frac 1{a^2+b^2} \int f (a^2+b^2)^{\frac 12} ds + g(bx - ay)$ from $u = \frac 1{a^2+b^2} \int f dt + g(p)$ i dont know how change $\int f dt$ to $\int f ds$ $\endgroup$ – M.H Apr 20 '13 at 6:42
2
$\begingroup$

Again, according to my this answer, if you do transformation $$ t = bx + ay \\ p = bx - ay $$ then your equation will be reduced to $$ 2ab\ u_t = f(t,p) $$ which can be integrated $$ u(t,p) = \frac 1{2ab} \int f(t,p) dt + g(p) $$

or alternatively $$ u(x,y) = \frac 1{2ab} \int f(x,y) (bdx + ady) + g(bx - ay) = \\ = \frac {(a^2+b^2)^{\frac 12}}{2ab} \int fds + g(bx - ay) $$

Update

I indeed made a mistake, but still it isn't the same as your, but it is correct answer though, I just chose different parametrization and consequently characteristics are different. I picked $$ t = bx + ay\\ p = bx - ay $$ To get te answer you have, you need to parametrize it as follows: $$ t = ax + by \\ p = bx - ay $$ So equations is reduced to $$ (a^2+b^2) u_t = f \\ u = \frac 1{a^2+b^2} \int f dt + g(p) = \\ = \frac 1{a^2+b^2} \int f (a^2+b^2)^{\frac 12} ds + g(bx - ay) = (a^2+b^2)^{-\frac 12} \int fds + g(bx - ay) $$ Same parametrization can also be used for this equations. Edits there were made as well.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I am guessing there is a mistake somewhere as the book I am using does not have the extra factor of $2ab$. I will have to go over all the steps in detail before I can understand this answer and hopefully spot the error. Thanks! $\endgroup$ – Slugger Feb 16 '13 at 14:18
  • $\begingroup$ @TeunVerstraaten: There were a mistake, but of a different sense. Fixed it and added new stuff. $\endgroup$ – Kaster Feb 16 '13 at 21:56
  • $\begingroup$ I understand upto the place before using ds in last line.But I don't understand how ds is obtained. It was obtained through line integral right?But I don't understand how that line integral was connected to this.I don't have a good knowledge on line integrals. So can someone please explain how ds was connected $\endgroup$ – clarkson Mar 23 '14 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.