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Two PDE's

  1. Dec 27, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Solve the following PDE's:
    [itex]\frac{\partial u }{\partial t }+c \frac{\partial u }{\partial x}[/itex] with [itex]u(x,0)=h(x)[/itex]. (1)
    [itex]\frac{\partial u }{\partial t }+u \frac{\partial u }{\partial x}[/itex] with [itex] u(x,0)=h(x)[/itex]. (2)
    Hints:
    Specify the characteristic field of directions associated to each equation.
    Consider the curve [itex]\gamma (s) =(s,0,h(s)) in \mathbb{R}^3[/itex]. Get the characteristic curve [itex]\gamma (s,t)[/itex] that passes through in each [itex]\gamma (s)[/itex] in [itex]t=0[/itex], solving the DE that determines the field of characteristic directions.


    2. Relevant equations
    No idea, self studying. Tried separation of variables method, assuming that [itex]u(x,t)=f(x)g(t)[/itex].


    3. The attempt at a solution
    Using separation of variables I reach [itex]\int _{g(t_0)}^{g(t_1)} \frac{dg}{g}=K(t_1-t_0)[/itex] and [itex]\int _{f(x_0)}^{f(x_1)} \frac{df}{f}=K(x_1-x_0)[/itex]. So that [itex]\ln g (t_1)- \ln g(t_0)=K(t_1-t_0)[/itex] and [itex]\ln f (x_1)- \ln f(x_0)=K(x_1-x_0)[/itex].
    I'm not confident in myself nor do I know how to proceed further if this is right.
     
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  3. Dec 28, 2011 #2

    HallsofIvy

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    I don't believe that separation of variables is intended here- the problem says specifically, "Specify the characteristic field of directions associated to each equation." Do you understand what the "characteristic field of directions" is?

    (By the way, in order to have "Partial Differential Equations", you have to have equations! Are we to assume that those are equal to 0?)
     
  4. Dec 28, 2011 #3
    May I suggest a text book which I've used for some time which I think is very accessible: "Basic Partial Differential Equations" by Bleecker and Csordas. There is a chapter on first-order PDE which I think you'd find accessible (easy to follow) and I bet you'd like it and it's in the library on most campuses. Start simple, first-order, constant coefficients. Get that straight, then first-order, variable coefficients, then attempt to tackle first-order non-linear (quasi-linear) like the second one above.
     
  5. Dec 28, 2011 #4
    Both first order PDE's (the standard diffusion equation and the inviscid Burgers equation) can be solved by the method of characteristics. This online handout from stanford explains the basic method:
    www.stanford.edu/class/math220a/handouts/firstorder.pdf

    Try to understand what a characteristic is. It is actually quite important, mathematically as well as physically.
     
  6. Dec 28, 2011 #5

    fluidistic

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    Ok thanks guys, I'm definitely looking to study more before attacking the first problem.
    No I don't know what it is (I'm going to spend some time now on it!). And yes I forgot to equal them to 0, sorry about that.

    Too bad my campus library is closed up till last week of January, because they have the book! (checked out in their internet search books option).
    Ok thanks a lot; this might be where I'll get most of my info for now.
    I will repost here later.
     
  7. Dec 29, 2011 #6

    fluidistic

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    To my understanding the characteristic lines are lines along which u(x,t) is constant.
    So that if I take the directional partial derivative of u in the direction of these particular lines, I should get 0.
    I think this means [itex]\vec \nabla u(x,t) \cdot \hat r=0[/itex]. The problem is that I don't know what u(x,t) is.
    Now I look back at the hint and see that I should consider [itex]t=0[/itex], so in a 3d dimensional graph of [itex]u(x,t)[/itex] sketched in function of [itex]x[/itex] and [itex]t[/itex], [itex]\gamma (s,0)[/itex] lies in the [itex]x-u(x,t)[/itex] plane.
    I think I understand what curve [itex]\gamma (s,0)[/itex] they ask me to find (I've skected it) but I'm not sure how to do find it. I must first find the characterstic lines of [itex]u(x,t)[/itex] and then [itex]\gamma (s,0)[/itex] will be the intersection of these lines with the [itex]x-u(x,t)[/itex] plane.
    I don't know how to solve the problem precisely because I don't know u(x,t) (although I know that if I know this function the problem is instantly solved).
     
  8. Dec 29, 2011 #7

    Dick

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    You don't need u(x,t) to find the characteristics. That's the whole point. Work through Example 1 in the Stanford pdf file. It's the same as your first equation and they are very thorough. You'll see they find the characteristics and use them to solve the equation, they don't use the solution to find the characteristics.
     
  9. Dec 30, 2011 #8

    fluidistic

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    Ok thanks.
    I'm stuck at understanding when they say
    I have [itex]x(s)=cs+x_0[/itex], [itex]t(s)=s+t_0[/itex] and [itex]z=z_0[/itex].
    Now I have to "eliminate the parameter s and reach [itex]x_0=x-ct[/itex]. How did they get that?
    My attempt: [itex]x(s)=x=cs+x_0[/itex]... So this is not the way to eliminate s.
    Edit: the closest thing I reach is [itex]x_0=x-c(t-t_0)[/itex].
     
  10. Dec 30, 2011 #9

    Dick

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    The constant t_0 in t(x)=s+t_0 is just an arbitrary constant. They just set it to zero. If you want to keep it around like you did in [itex]x_0=x-c(t-t_0)[/itex] write that as [itex]x_0-c*t_0=x-ct[/itex] putting all of the constants on one side. The message is still the same. The value of u(x,t) is constant along the characteristic lines x-ct=constant.
     
  11. Dec 30, 2011 #10

    fluidistic

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    Ah I see. I'm going to keep the constants of integration. Ok about that the message doesn't change, though it's not obvious to me that u(x,t) is constant for these lines.
    In the PDE file they say
    Perfect, this looks like my [itex]\gamma[/itex]. So [itex]\gamma = (x(s),t(s),z(s))=(vs+x_0,s+t_0,z_0)[/itex].
    They go on to say
    But I don't get it. When I look at [itex]z(x,t)[/itex], it's worth [itex]z_0[/itex] which is constant for any .... oh wait, [itex]x[/itex] and [itex]t[/itex]. But [itex]x=vs-x_0[/itex] which lead to [itex]x-vt=constant[/itex] as we saw. (sorry for changing letters, my c into v which is their a.) So I guess now I understand this point.
    Hmm not really. For any x means [itex]x=v(t-t_0)+x_0[/itex] ah but for any t means... well I'm really confused here. Why only the lines x-vt, I don't know. It looks like for any x and any t so... any line?
     
  12. Dec 30, 2011 #11

    Dick

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    You are maybe getting caught up in irrelevant details. Let's skip to the end. The idea is that u(x,t) is constant along curves where x-ct=constant. That means u(x,t) is really only a function of x-ct. Since if x1-c*t1=x2-c*t2 then you must have u(x1,t1)=u(x2,t2). Now what is that value? You have u(x,0)=h(x) as a boundary condition. That means u(x,t)=h(x-ct). Try substituting u(x,t)=h(x-ct) into your original PDE. Does it work?
     
  13. Dec 31, 2011 #12

    fluidistic

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    Yes it does satisfies the transport equation, but I'm not understanding anything. :/
    I'm currently searching for a free ebook on PDE's on the Internet, if someone has any suggestions, feel free to post.
     
  14. Jan 1, 2012 #13

    fluidistic

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  15. Jan 1, 2012 #14

    fluidistic

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    I'm confused on the second PDE.
    [itex]\frac{dx}{ds}=u(x,t)[/itex], [itex]\frac{dt}{ds}=1[/itex] and [itex]\frac{du}{ds}=0[/itex]. The latter equation tells me that [itex]u(x,t)[/itex] is constant along [itex]s[/itex].
    However [itex]t=s[/itex] from integration and choosing [itex]t_0=0[/itex], so that [itex]u(x,t)[/itex] is also a constant with respect to time, for a given [itex]x[/itex].
    Since [itex]u(x,t)[/itex] is constant with respect to [itex]s[/itex] and time [itex]t[/itex], [itex]\frac{dx}{ds}=u(x,t) \Rightarrow x_0=x-\underbrace {u(x,t)}_{\text{constant}}t[/itex]. This is similar to the previous equation.
    Now [itex]u(x,t)=constant=h(x,0)[/itex], I take [itex]x=x_0[/itex], so that [itex]u(x_0,t)=h(x_0)=h(x-ut)[/itex].
    When I plug back this function into the original PDE, I indeed reach [itex]0[/itex] so that [itex]h(x-ut)[/itex] (with u being a constant with respect to time) is a solution to the PDE.
    In the original problem I forgot to mention that they ask about this particular PDE, until what value of t can we extend the solution?
    In Stanford's PDF they say that there's a singularity from a certain t but I don't understand how to calculate this time t. Any help on this is welcome.
    Also, is what I've done correct?
     
  16. Jan 1, 2012 #15

    Dick

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    Yes, what you have done is correct. h(x-ut) is a solution, but it's a solution only in implicit form. To actually find u you would need to solve the equation u=h(x-ut). You are only going to be able to pull this off if h(x) is a REALLY simple function. To get an intuition for what's going on pick a simple form of h(x), say h(x)=Kx for a K a constant. Now you can find the actual solutions. How does the singularity depend on K? Now can you picture what's going on in terms of the characteristics?
     
  17. Jan 1, 2012 #16

    fluidistic

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    Ok thank you. If I was a mathematician I'd have to probably use the implicit function theorem or inverse function theorem to show that I can't solve for u in [itex]u=h(x-ut)[/itex] for any function h. Unless I'm wrong.
    So I took your example, this lead to [itex]u(x,t)=\frac{Kx}{1+Kt}[/itex], which indeed satisifies the PDE. When [itex]t=-\frac{1}{K}[/itex], [itex]u[/itex] becomes undefinied. So I think this mean that for this particular [itex]h(x)=Kx[/itex], the solution would be valid for any [itex]t \neq -1/K[/itex].
    And I'm not sure what's going on with the characteristics... It looks like it means that [itex]\frac{dx}{ds}=\frac{dx}{dt}[/itex] becomes infinitely large for that particular value of t.
     
  18. Jan 1, 2012 #17

    Dick

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    You usually think of solutions to these kind of equations as starting at t=0 and integrating forward in time until the solution becomes singular. Then you stop. So for the time being, just think of the cases that have a singularity at positive t. Focus on a singular case like K=(-1). Then you have a singularity at t=1. Now draw some characteristics. What's the characteristic passing through (x,t)=(0,0), through (x,t)=(1/4,0), through (x,t)=(1/2,0) etc?
     
  19. Jan 1, 2012 #18

    fluidistic

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    Ok thank you, this makes sense.
    So for this particular case of [itex]K=-1[/itex], I'm having some troubles for the characteristic curves. If I'm not wrong I must find [itex](x(s),t(s))[/itex] such that this passes by [itex](0,0)[/itex].
    But the dependence on s bothers me somehow. [itex]x=x_0+tu[/itex], and [itex]t=s[/itex].
    So I've plotted a 3d graph x-y-u and drew [itex]x_0[/itex] along the x axis. To graph well, I think I must get u(x,t) which is worth [itex]u=-\frac{x_0}{t}[/itex]. But I thought u was worth [itex]-x/(1-t)[/itex]... I'm totally confused.
     
  20. Jan 1, 2012 #19
    Very good notes from an excellent professor of a class I took on PDEs: http://www.math.ucsb.edu/~grigoryan/124A/lecs/lec2.pdf

    Also, has anyone recommended Strauss' book?

    I printed all of these lecture notes in preparation of the final for the course. Reads much better than any book I've found.
     
  21. Jan 1, 2012 #20

    Dick

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    I'm not sure you getting the characteristic idea. u(x,t) is constant along curves where x-ut=constant, right? That means u is a constant. So the curve is a straight line. Let's figure out what the characteristic curve is through (x,t)=(1,0). Put x=1 and t=0. That gives 1-u(1,0)*0=constant. u(1,0)=h(1) and the constant=1. So x-h(1)*t=1 is the characteristic line through (1,0). If h(x)=(-x), then the characteristic line is x+t=1. Do the same exercise and find the characteristic through (2,0). If the characteristic lines intersect, you likely have a singularity there. The idea is that the value of u along a characteristic is determined by h, if two characteristics intersect then they may have conflicting values of h.
     
    Last edited: Jan 1, 2012
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