Solution for 1st order, homogenous PDE

  • #1
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##u_t + t \cdot u_x = 0##

The equation can be written as ##<1, t, 0> \cdot <d_t, d_x, -1>## where the second vector represents the perpendicular vector to the surface and since the dot product is zero, the first vector must necessarily represent the tangent to the surface. We parameterize this by s:

characteristic equations:

##\frac{dt}{ds} = 1##
##\frac{dx}{ds} = t##
##\frac{dz}{ds} = 0##

After switching around variables for a bit, we get a solution set of:

##t = s + c_1##
##x = s^2 + c_2 \cdot s + c_3##
##z = c_4##

Equations one and two can be combined to get:

##x = (t + c_1)^2+ c_2 \cdot (t + c_1) + c_3##

And we can manipulate that around a bit more, but in the end, we can't isolate any constant term. I've been finishing the problems here by equating the z = constant and another solution equals a constant. In my book, there's a problem like this, but boundary conditions are used to get rid of certain terms and make it work out.
 

Answers and Replies

  • #2
scottdave
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The constants are determined by the given conditions (boundary conditions and initial conditions).
 
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