# Solve the given partial differential equation

• chwala
In summary, In this problem, u(x,y) is a function of two variables x and y. The boundary condition states that u=sin(3x+y). Using the initial condition and substitution, we find that sin y = u(0,y) and that f(z) is the cosine of -z/6. Finally, u(x,y) is equal to u(\eta(x,y)) when \eta=3x+y.

#### chwala

Gold Member
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]##

Last edited:
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]##

anuttarasammyak said:
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.

Last edited:
BvU
Please find attached the sketch for explanation. Best.

chwala said:
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##...

Last edited:
SammyS