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Problem: Solve the first order equation ${{u}_{x}}+{{u}_{y}}+z{{u}_{z}}=0$ with initial curve $x=t,\text{ }y=0,\text{ }z=\sin t$ .

Is my solution correct?

$\frac{dx}{ds}=1,\text{ }\frac{dy}{ds}=1,\text{ }\frac{ds}{ds}=z,\text{ }\frac{du}{ds}=0.$

$x=s+{{c}_{1}},\text{ }y=s+{{c}_{2}},\text{ }z={{c}_{3}}{{e}^{s}},\text{ }u={{c}_{4}}.$

From the initial curve, we get $x=s+t,\text{ }y=s,\text{ }z={{e}^{s}}\sin t.$

Eliminating $s$ and $t$ , we get $z={{e}^{y}}\sin \left( x-y \right).$

Finally, $u=f\left( z-{{e}^{y}}\sin \left( x-y \right) \right)$ for any smooth function $f$ .

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Corrections:

  1. $\frac{dz}{ds} = z$ not $\frac{ds}{ds} = z$

  2. $(c_1, c_2, c_3, c_4) = (c_1(t), c_2(t), c_3(t), c_4(t))$ (I mean, I think you should make it explicit)

  3. $(x,y,z,u) = (x(s,t), y(s,t), z(s,t), u(x,y,z))$ (I mean, I think you should make it explicit)


Questions on the solution:

  1. How did you get $u = f(z - e^y\sin(x-y))$?

  2. What does $$f(z - e^y\sin(x-y))$$ even mean?

$$f(z - e^y\sin(x-y)) = f(0) \ \text{right?}$$

If so, then $u$ is some constant $f'(0)$ s.t. $$f'(0) + f'(0) + zf'(0) = 0$$

$$\to u = f'(0) = 0$$ since $z = -2$ doesn't satisfy $z(s,0) = \sin(0)$?

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    $\begingroup$ 1. You are right. I should have made them explicit. $\endgroup$ – Patrick Windance Nov 18 '15 at 3:57
  • $\begingroup$ 2. $u$ is constant anlong $z={{e}^{y}}\sin \left( x-y \right).$ ? $\endgroup$ – Patrick Windance Nov 18 '15 at 4:00
  • $\begingroup$ @PatrickWindance Why? $\endgroup$ – BCLC Nov 18 '15 at 15:34
  • $\begingroup$ The first three terms give the characteristic curve. The forth term says $u$ is constant along the curve. $\endgroup$ – Patrick Windance Nov 19 '15 at 0:05

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