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 $b\neq 0$?

  • $\begingroup$ 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

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


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.

  • $\begingroup$ 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

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