1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial Differential Equations

  1. Sep 23, 2012 #1
    I am working on some problems for an assignment in my PDEs class and find myself either not understanding what I am supposed to do or being unsure of my answers or the next step.

    I am going to outline my understanding of the problems, provide my attempt at a solution and highlight where I am having difficultly. Please correct me anywhere that I am wrong and offer as much assistance and helpful prods in the right direction as you can. Thank you very much in advance!


    Constant Coefficient PDEs
    The general solution of a partial differential equation of the form ##au_{x} + bu_{y} = 0## where a and b are constants is of the form ##u(x,y) = f(bx -ay)## where f is any arbitrary function of one variable. You can not find a specific solution to this PDE unless provided with further information, such as an intitial condition (ie. when x or y is equal to 0).


    My homework question:

    "Consider the equation ##u_{x} - 3u_{y} = 0## in ##{x>0, y>0}## with the initial condition at ##x = 0, u = y, y >0##. Where is this solution defined? Is it defined everywhere in ##{x>0,y>0}## or do we need to impose an initial condition at y = 0? If we need to impose such a condition use at ##y = 0, u = x (x>0)## and solve this IVBP."

    Given ##u_{x} - 3u_{y} = 0## we find the general solution ##u(x,y) = f(-3x -y)##.
    Using the initial condition at ##x = 0, u = y, y >0##:

    ##u(x,y) = u(0,y) = f(-y) = y##

    Let ##w = -y \rightarrow y = -w##; then:

    ##f(w) = -w##

    ##u(x,y) = -(-3x-y)##

    ##u(x,y)= 3x + y##


    Now, if I understand the question, I must determine if ##u(x,y)= 3x + y## is a valid solution of ##u_{x} - 3u_{y} = 0## for all x, y such that x>0 and y>0, correct? I'm not sure how to do this exactly. Any help here would be greatly appreciated.

    Note: After this question there is a very similar question in which we are asked to do the same thing (given the same initial condition) and we are considering {x<0, y>0} instead.



    EDIT:

    Variable Coefficient PDEs

    My homework question:

    "Find the general solution of ##xu_{x} + 4yu_{y} = 0## in ##{(x,y)\neq(0,0}##; when is this solution continuous at (0,0)?"

    In the one example in my textbook (for ##u_{x} + yu_{y} = 0## they write ##\frac{dy}{dx} = \frac{y}{1}##. I do not understand how or why they can write this (they turn it into a separable ODE). Can someone please explain why/how they do this?

    If I accept the idea of writing it as a separable equation I can write:

    ##\frac{dx}{dy} = \frac{x}{4y}##

    ##\frac{dx}{x} = \frac{dy}{4y}##

    Integrating both sides, we find:

    ##lnx + c = \frac{1}{4} lny + d##

    ##lnx + c = lny^{\frac{1}{4}} + d##

    ##lnx + b= lny^{\frac{1}{4}}##, where ##b = c-d##

    ##e^(lnx + b) = e^(lny^{\frac{1}{4}})##

    ##x e^b = y^{\frac{1}{4}}##, but ##e^b## is a constant, so we say ##e^b = C##

    ##Cx = y^{\frac{1}{4}}##

    ##(Cx)^4 = y##

    So ##y = (Cx)^4## describes the characteristic curves of the ODE, but I'm unsure of what I must do next to solve the PDE.


    EDIT:

    After browsing around for a little bit I think I've managed to find the general solution for this problem.

    ##u(x,y) = u(x,[Cx]^4)##
    ##\frac{du(x, [Cx]^4)}{dx} = \frac{du}{dx} + \frac{du}{dy}\frac{dy}{dx} = \frac{du}{dx} + \frac{du}{dy}\frac{4y}{x} = xu_{x} + 4yu_{y} = 0##; this is the initial question, so we know that this works. Now we know:

    ##u(x,y) = u(x, [Cx]^4) = f(C)##, but we can solve for C in terms of x and y to find ##C = \frac{y^(\frac{1}{4})}{x}## so we have the general solution:

    ##u(x,y) = f(\frac{y^(\frac{1}{4})}{x})##
     
    Last edited: Sep 23, 2012
  2. jcsd
  3. Sep 23, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's easy to check. Just plug it into the equation and initial conditions and see if it works.
     
  4. Sep 23, 2012 #3
    Well the solution ##u(x,y) = 3x +y## does satisfy the given PDE, but I'm unsure how to consider the idea of whether or not its valid for all x and y greater than 0.

    For example, ##u_{x} = 3, u_{y} = 1## and these satisfy the PDE regardless of the values of x and y because they do not depend on either x or y. Does this mean that it holds true for all x and y greater than 0 or is there more that I have to show?
     
  5. Sep 23, 2012 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Like you say, it works for all x and y, and that includes x>0 and y>0. If it satisfies the boundary conditions how could anyone argue that it isn't a solution when it is?
     
  6. Sep 23, 2012 #5
    Then by the same reasoning it is also a valid solution provided that x<0, y>0 (for the next part of the problem). I felt like there was more I had to show before I could come to such a conclusion because it seemed so obvious.

    Thank you very much for your help.


    I've edited my initial post to include a question about Variable Coefficient PDEs; if you have the time would you mind providing me with a hint as to how to proceed? Again, thank you very much!
     
  7. Sep 23, 2012 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That generally isn't a good idea. You should post new questions in a separate post. With several replies already, many readers may well just skip the thread assuming it is finished. I will answer your first question. Suppose you have ##u(x,y) = C##, which defines y implicitly as a function of ##x## and you want to calcuate ##dy/dx##. You would differentiate implicitly: ##u_x + u_y\frac {dy}{dx} = 0##. This gives$$
    \frac {dy}{dx} = -\frac{u_x}{u_y}$$So for an equation like ##xu_x+4yu_y = 0## you would have$$
    \frac {dy}{dx}= -\frac{u_x}{u_y}=\frac{4y}{x}$$ I'm not a PDE expert so I will leave your question about the characteristics to someone else.
     
    Last edited: Sep 23, 2012
  8. Sep 23, 2012 #7
    Okay, thank you very much! I think I have the answer to my second question now (just not sure about the continuity aspect). Thanks again for your time!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partial Differential Equations
Loading...