# Show $au_x+bu_y=f(x,y)$ gives $u(x,y)=(a^2+b^2)^{-\frac{1}{2}}\int_{L}fds +g(bx-ay)$ if $a\neq 0$.

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 $b\neq 0$?

• Are you going to mark your other questions as solved? – Kaster Feb 15 '13 at 3:24
• $Kaster I always do, I just have not had the time yet to read the answer thoroughly. – Slugger Feb 15 '13 at 11:15 • @Teun Verstraaten:how get$\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$– M.H Apr 20 '13 at 6:42 ## 1 Answer 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. • 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! – Slugger Feb 16 '13 at 14:18
• @TeunVerstraaten: There were a mistake, but of a different sense. Fixed it and added new stuff. – Kaster Feb 16 '13 at 21:56
• 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 – clarkson Mar 23 '14 at 15:15