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
u(x,y)=u(\eta(x,y))
u_x=\frac{du}{d\eta}\eta_x
u_y=\frac{du}{d\eta}\eta_y
From the condition given
\eta_x=3\eta_y
Thus we can make
\eta=3x+y
\eta(0,y)=y
From the boundary condition given
u=\sin\eta=\sin(3x+y)

 

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

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

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
u_x=-\frac{1}{3}\cos\{-\frac{x}{3}+y\}
u_y=\cos\{-\frac{x}{3}+y\}
So -3u_x=u_y
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