0

Transform the equation changing to new independent variables $(u ; v)$ :

(a) $z_{t t}=a^{2} z_{x x}, u=x-a t, v=x+a t$

(b) $x^{2} z_{x x}-y^{2} z_{y y}=0, u=x y, v=\frac{y}{x}$

I understand how to perform this by chain rule to $u_x$, but with second derivative I have no idea.

Prox
  • 77
  • 6
  • 1
    If we write $z(x,t)=Z(u,v)$ then the chain rule twice gives $z_{tt}=Z_{uu} u_t^2 +2Z_{uv}u_t v_t +Z_{vv} v_t^2 +Z_{u}u_{tt}+Z_v v_{tt}$,and you know all the partial derivatives of $u,v$, so just substitute these and deal with $z_{xx}$ in same way. – ancient mathematician Jan 25 '21 at 13:39
  • @ancientmathematician could you, please, write how you applied chain rule twise in more detailed way? I just need to understand how it works for second and higher derivatives – Prox Jan 25 '21 at 13:49
  • 1
    $z_t=Z_u u_t + Z_v v_t$, by Chain Rule. Now $(Z_u u_t)t=(Z_u)_t u_t + Z_u (u_t)_t$ by product Rule. Then $(Z_u)_t=Z{uu} u_t + Z_{uv} v_t$ by Chain Rule etc. OK? – ancient mathematician Jan 25 '21 at 17:59

0 Answers0