Solve the given partial differential equation

[chwala]

Homework Statement

Find the solutions satisfying ##2u_x-6u_y=0## given ##u(0,y)=\sin y##.

Relevant Equations

method of characteristics

Looking at pde today- your insight is welcome...

##η=-6x-2y##

therefore,

##u(x,y)=f(-6x-2y)##

applying the initial condition ##u(0,y)=\sin y##; we shall have

##\sin y = u(0,y)=f(-2y)##

##f(z)=\sin \left[\dfrac{-z}{2}\right]##

##u(x,y)=\sin \left[\dfrac{6x+2y}{2}\right]##

 

[anuttarasammyak]

Excellent. In short
[tex]u(x,y)=u(\eta(x,y))[/tex]
[tex]u_x=\frac{du}{d\eta}\eta_x[/tex]
[tex]u_y=\frac{du}{d\eta}\eta_y[/tex]
From the condition given
[tex]\eta_x=3\eta_y[/tex]
Thus we can make
[tex]\eta=3x+y[/tex]
[tex]\eta(0,y)=y[/tex]
From the boundary condition given
[tex]u=\sin\eta=\sin(3x+y)[/tex]

 

[chwala]

just a quick question, can we also approach this using;

##u(ξ,η )=f(ξ)##

then it follows that,

##u(x,y)=f(2x-6y)##

##\sin y=u(0,y)=f(-6y)##

##f(z)=\sin\left[\dfrac{-z}{6}\right]##

##u(x,y)=\sin\left[\dfrac{-2x+6y}{6}\right]##

 

[chwala]

Excellent. In short
[tex]u(x,y)=u(\eta(x,y))[/tex]
[tex]u_x=\frac{du}{d\eta}\eta_x[/tex]
[tex]u_y=\frac{du}{d\eta}\eta_y[/tex]
From the condition given
[tex]\eta_x=3\eta_y[/tex]
Thus we can make
[tex]\eta=3x+y[/tex]
[tex]\eta(0,y)=y[/tex]
From the boundary condition given
[tex]u=\sin\eta=\sin(3x+y)[/tex]

Thus we can make
[tex]\eta=3x+y[/tex]
[tex]\eta(0,y)=y[/tex]
From the boundary condition given
[tex]u=\sin\eta=\sin(3x+y)[/tex]

It is interesting on how you applied the boundary condition here... my understanding is that it applies to ##u(x,y)## looks like in your case it applies to ##η(x,y)##...the steps before this are quite clear.

 

[anuttarasammyak]

Please find attached the sketch for explanation. Best.

 

[anuttarasammyak]

just a quick question, can we also approach this using;

##u(ξ,η )=f(ξ)##

then it follows that,

##u(x,y)=f(2x-6y)##

##\sin y=u(0,y)=f(-6y)##

##f(z)=\sin\left[\dfrac{-z}{6}\right]##

##u(x,y)=\sin\left[\dfrac{-2x+6y}{6}\right]##

From this
[tex]u_x=-\frac{1}{3}\cos\{-\frac{x}{3}+y\}[/tex]
[tex]u_y=\cos\{-\frac{x}{3}+y\}[/tex]
So [tex]-3u_x=u_y[/tex]
which is diffent from the problem statement.

 

[chwala]

@anuttarasammyak can you kindly show how you applied the boundary conditions? As requested in post ##4##...

 

[anuttarasammyak]

@chwala I tried it in post #5 figure. Your thoughts on that will be appreciated.
Supplement: https://www.wolframalpha.com/input?i=graph+of+u=sin(3x+y)+&lang=en