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Homework Help: Help integrating dx/dt = u(x)

  1. Nov 11, 2011 #1
    In my differential equations class I have been given a problem which involves solving dx/dt=u where u=u(x)

    I know that this is done by seperation such that ∫dx/u = ∫dt, and then the constant of integration found using initial conditions, however I am getting myself all worked up and confused as I dont know how to treat the integral on the left, i.e. ∫dx/u

    Any help would be greatly appreciated

    Thanks in advance,

    Last edited: Nov 11, 2011
  2. jcsd
  3. Nov 11, 2011 #2


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    What is the question? To solve for x(t)? The integral on the left is simply the integral (with respect to x) of 1/u And since you have u as a function of x, then you're integrating a function of x with respect to x. And on the right-hand side, you're integrating 1 with respect to t.
  4. Nov 11, 2011 #3
    Thanks for the quick reply :)

    The actual question is to use the method of characteristics to solve a PDE

    From this I have obtained that one of the characteristic equations is dx/dt = u, which I need to rearange to get an equation for x in terms of t.

    I have also been given the intial data : u(x,y=0) = 2, x<0 and 1, x>0 however I believe this is not needed until later in the question
  5. Nov 11, 2011 #4
    Am I being really stupid?!?

    All I know is that u=u(x), and so I dont know how to integrate 1/u wrt x. Surely its not ln(u) because u doesnt equal x?!?

    Ive been stuck on this all day and its starting to drive me mad so Im sorry if the answers staring me in the face
  6. Nov 11, 2011 #5


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    Its not ln(u). In fact, dx/dt = u is not enough information to get an equation for x in terms of t. I reckon you will have to use your initial data somehow.

    In your initial data, you have the x dependence of u at y=0. So this implies you could write u as a function of y and x. So this means you have:

    dx/dt = u(x,y)

    But you don't know how u depends on y (or how y depends on x or t), so I think you still don't have enough information.
  7. Nov 11, 2011 #6
    Ok, so im clearly missing something blindingly obvious:

    Initial data: t=0, u(x,y=0) = 2, x < 0 and 1, x > 0

    I have 3 characteristic equations given by:

    1. du/dt=0 therefore u is constant wrt t

    2. dy/dt=1 therefore y=t + constant

    Using that at t=0, y=0 we get that constant = 0 and so y=t

    3. dx/dt=u therefore ∫dx/u = ∫dt

    As far as I can see there is no way to reconcile the problem using the other two equations.
    It follows that the initial data must therefore be the key.

    Would putting in the value of u for x (being positive or negative) invalidate the DE

    i.e. for x positive: ∫dx/1=∫dt therefore x(+) = t + constant
    and for x negative: ∫dx/2=∫dt therefore x(-) = 2t + constant

    Your opinion on this would be greatly appreciated

    Last edited: Nov 11, 2011
  8. Nov 11, 2011 #7


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    Well, you've now told me there are more equations! OK, so y = t +constant. So this means that when t=0, y=constant. This doesn't mean the constant is zero.

    The problem does look pretty tricky actually. It might help to consider product rule, etc. I'm not sure. I'll come back to this tomorrow!
  9. Nov 11, 2011 #8
    Sorry, I didnt think the other equations would make a difference to the one in question, which is probably the reason why I've been sweating over this all day.

    I'm pretty sure that the constant does equal zero since the initial data gives at t=0, u(x,y=0)

    Thanks again for your help
  10. Nov 11, 2011 #9


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    Oh I see, yes you are right, y = t. Well from there, I think it is simpler to replace y by t in all the places where y appears, so then you just have 3 variables: u, x and t.

    So now you have 2 important equations: du/dt=0 and dx/dt=u. You could try writing out the full differential for u, and use these equations to get a relationship between the two partial differentiations. But I don't think this gives u(x)...
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