2xt dz/dt (x,t) + dz/dt (x,t) + z (x,t) = 0 z(x,0) = 0 -infinity < x < +infinity

Question

2xt dz/dt (x,t) + dz/dt (x,t) + z (x,t) = 0 z(x,0) = 0 -infinity < x < +infinity Use method of characteristic curves. I'm stuck on this one, any help is appreciated.

Explanation / Answer

I think your first derivative should be dz/dx and in that case you will have a famous advection equation which comes up in chemical engineering. I also change the IC to z(x, 0) = x because otherwise your solution will be trivially zero. The idea of method of characteristics is to transform to a new coordinate system where the PDE can be solved as an ODE. The following link gives you a good run-down of the process: http://www.scottsarra.org/shock/shock.ht… The characteristic equations are dt/ds = 1 which gives t = s using the IC s(0) = 0 dx/ds = 2xt => dx/dt = 2xt (by t = s) and integration and applying the IC gives x = x0 e^(t^2) so that x0 = x e^(-t^2) dz/ds = -z which gives z = A e^(-s) and using the IC, A = x0 so with s = t we get z = x0 e^(-t). Now use the value for x0 we obtained above... You are supposed to get the answer z(x, t) = x e^(-t - t^2). Here is a check: dz/dx = e^(-t - t^2), and dz/dt = x (-1 - 2t) e^(-t - t^2) = - x e^(-t - t^2) - 2xt e^(-t - t^2) after distribution So 2xt dz/dx + dz/dt + z = 2 xt e^(-t - t^2) - x e^(-t - t^2) - 2xt e^(-t - t^2) + x e^(-t - t^2) = 0 //

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