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Partial differential equation, characteristic equations.

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Given the initial value problem:
    [itex]\frac{(u)}{(1-e^-(2x))}[/itex]u[itex]_{x}[/itex]+ [itex]\frac{\sqrt{t}}{u}[/itex]u[itex]_{t}[/itex]=1, with x, t, u > 0

    Subject to condition u(x,1)=[itex]e^{-x}[/itex]

    2. Relevant equations

    a) Classify given partial differential equation.

    b) Write the characteristic equations. By solving them, show that the given problem admits ilpmicit solution* for u(x,t) which can be written as

    [itex]u^2-2ln(u)=2x+e^{-2x}-4\sqrt{t}+4[/itex] where u(x,t)

    (Hint: solve the characteristic equation for u first)

    * To to obtain the explicit solution for u wuld require the use of inverse functions and is NOT required, doing so may result in loss of generality in the solution since u may be multi valued, as you are asked to checfk in question 2(d)

    c) Verify that the implicit solution given in part b) satisfies the assigned partiasl differential equation and the initial condition)

    d) Determinate analytically the regions in the plane (x,t) such that the solution u(x,t) to the assigned problem takes respectively
    -one value
    -two values
    -no value

    3. The attempt at a solution

    a) First order linear partial differential equation.

    c) To check the initiaql condition I insert u(x,1)=[itex]e^-x[/itex], and t=1 to the equation from the point b), simplify and get 0=0.

    b) 1. [itex]\frac{dx}{ds}[/itex]=[itex]\frac{u}{1-e^(-2x)}[/itex]

    2. [itex]\frac{dt}{ds}[/itex]=[itex]\frac{\sqrt{t}}{u}[/itex]

    3. [itex]\frac{du}{ds}[/itex]=1


    Than: From the 3. du=ds => u = s + c3

    taking the initial condition sets s=1 so [itex]e^{-x}[/itex]=1+c3


    From 2.

    [itex]\frac{dt}{ds}[/itex]=[itex]\frac{\sqrt{t}}{s+e^{-x}-1}[/itex]

    [itex]\frac{dt}{\sqrt{t}}[/itex]= [itex]\frac{ds}{s+e^{-x}-1}[/itex]

    [itex]\frac{1}{2}[/itex]*[itex]t^{1/2}*[/itex]= ln(s+e^{-x}-1) + c2


    From 1.

    [itex]\frac{dx}{ds}[/itex]

    (1-[itex]e^{-2x}[/itex]dx = ds u

    x-[itex]\frac{1}{2}[/itex]*[itex]e^{-2x}[/itex]= s*u + c1


    What shall I do next?
     
    Last edited: Dec 11, 2012
  2. jcsd
  3. Dec 11, 2012 #2
    I know this isn't much help, but you wrote linear when you meant quasilinear.
     
  4. Dec 11, 2012 #3

    pasmith

    User Avatar
    Homework Helper

    [itex]s[/itex] parametrizes distance along a characteristic, so you can choose to have [itex]s = 0[/itex] on the curve where the boundary condition is applied, which is [itex]t = 1[/itex]. This is equivalent to the initial conditions [itex]t(0) = 1[/itex], [itex]x(0) = x_0 \geq 0[/itex], and [itex]u(0) = e^{-x_0}[/itex].

    So [itex]u(s) = s + e^{-x_0}[/itex].

    Right idea, but the integral of [itex]1/\sqrt t[/itex] is [itex]2\sqrt t[/itex]. Set [itex]u = s + e^{-x_0}[/itex] in your equation (2) and integrate using the initial condition [itex]t(0) = 1[/itex]. Then substitute [itex]s = u - e^{-x_0}[/itex].

    The integral of [itex]e^{-2x}[/itex] is [itex]-\frac12 e^{-2x}[/itex]. Again, you need to set [itex]u(s) = s + e^{-x_0}[/itex] in your equation (1) and integrate using the initial condition [itex]x(0) = x_0[/itex]. Then substitute [itex]s = u - e^{-x_0}[/itex].

    If you have any [itex]x_0[/itex]'s left, eliminate them from one equation by using the other. You should get the given answer.
     
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