Solve the given partial differential equation

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.
  • #1
chwala
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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]##
 
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  • #2
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]
 
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  • #3
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]##
 
  • #4
anuttarasammyak said:
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.
 
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  • #5
Please find attached the sketch for explanation. Best.
img20230109_22334057.jpg
 
  • #6
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
[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.
 
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  • #7
@anuttarasammyak can you kindly show how you applied the boundary conditions? As requested in post ##4##...
 

FAQ: Solve the given partial differential equation

What is a partial differential equation (PDE)?

A partial differential equation (PDE) is a mathematical equation that involves functions of several variables and their partial derivatives. PDEs are used to formulate problems involving functions of several variables and are either solved by hand or used to create a computer model.

What are the common methods to solve a PDE?

Common methods to solve PDEs include separation of variables, method of characteristics, Fourier transform, Laplace transform, and numerical methods such as finite difference, finite element, and finite volume methods.

What is the difference between an ordinary differential equation (ODE) and a partial differential equation (PDE)?

An ordinary differential equation (ODE) involves functions of a single variable and their derivatives, whereas a partial differential equation (PDE) involves functions of multiple variables and their partial derivatives. ODEs are generally simpler to solve than PDEs.

Can all PDEs be solved analytically?

No, not all PDEs can be solved analytically. Many PDEs, especially those modeling complex physical phenomena, do not have closed-form solutions and must be solved using numerical methods.

What are boundary and initial conditions in the context of PDEs?

Boundary conditions specify the behavior of the solution at the boundaries of the domain, while initial conditions specify the state of the system at the beginning of the observation (usually time t=0). Both types of conditions are essential to uniquely determine the solution of a PDE.

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