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Here is my approach to find a criterion to determine whether the equation $\alpha u_{tt}+\beta u_{tx}+\mu u_{xx}=0$ can be transformed into a wave equation $v_{tt}-c^2v_{yy}=0$.

Suppose there is a coordinate transformation $u(t,x) \mapsto v(s,y)$ such that $u(t,x)=v(s,y)$ and $v_{ss}-c^2v_{yy}=0$.
Then since $$v_{ss}=(u_{xx}x_s+u_{x}t_s)x_s+u_xx_{ss}+(u_{tx}x_s+u_{tt}t_s)t_s+u_tt_{ss}$$ and $$v_{yy}=(u_{xx}x_y+u_{xt}t_y)x_y+u_xx_{yy}+(u_{tx}x_y+u_{tt}t_y)t_y+u_tt_{yy}$$, we have $$v_{ss}-c^2v_{yy}=(t_s^2-c^2t_y^2)u_{tt}+2(t_sx_s-c^2t_yx_y)u_{xt}+(x_s^2-c^2x_y^2)u_{xx}+(x_{ss}-c^2x_{yy})u_x+(t_{ss}-c^2t_{yy})u_t=0$$. So we get $$\alpha=t_s^2-c^2t_y^2$$ $$\beta=2(t_sx_s-c^2t_yx_y)$$ $$\mu=x_s^2-c^2x_y^2$$ $$x_{ss}-c^2x_{yy}=t_{ss}-c^2t_{yy}=0$$. Then $\frac{1}{4}\beta^2-\alpha\mu=c^2(t_sx_y-t_yx_s)^2\geq 0$.

Here, I am curious about the case of $t_s x_y=t_y x_s$ so that $\beta^2-4\alpha\mu=0$. Intuitively, it may reach to contradiction if $\beta^2-4\alpha\mu=0$, since it means $\alpha u_{tt}+\beta u_{tx}+\mu u_{xx}=\left(\partial_t +\frac{\beta}{2\alpha}\partial_x\right)^2u=0$ which is not a wave equation. But I cannot find any clue. Also, I have no information for the condition $x_{ss}-c^2x_{yy}=t_{ss}-c^2t_{yy}=0$. To solve this, I have to bring out some relations between $s$ and $y$ from $t_s x_y=t_y x_s$ first. Would you give me any help?

okw1124
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The short answer to your title question is that the Inverse Function Theorem does not apply to the mapping $(x,y)\to(u(x,y),v(x,y))$. Such a mapping has derivative matrix $$ \begin{bmatrix} u_x & u_y \cr v_x & v_y \end{bmatrix}. $$ The IFT says that if this matrix is continuous and is invertible at one point then the mapping is invertible in a small neighborhood of that point. This is certainly what you want as a change of variables $(x,y)$ to $(u,v)$.

So I think what you have shown so far is that to convert your PDE to the wave equation you need to have $\tfrac{1}{4}\beta^2-\alpha\mu>0$. Beyond that you need express, for example, $$ \mu = x_s^2-c^2x_y^2 $$ for some solution $x$ of the wave equation $x_{ss}-c^2x_{yy} = 0$. But all solutions of that are expressible as $$ x(y,s) = f(y-cs)+g(y+cs) $$ for some functions $f$ and $g$. Similar requirements for $\alpha$ and $\beta$, pretty involved.

You might want to look at the discussion of normal forms for certain PDEs in Courant and Hilbert, Methods of Mathematical Physics Vol II, page 154.

Bob Terrell
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  • Thank you! But I have one more question: Is there any contradiction if IFT does not apply to the mapping $(x,y)\mapsto (u,v)$? I understood that if the equation is elliptic, i.e. is a wave equation, then $\beta^2-4\alpha\mu>0$, but not the converse. Could you give me gsome intuitive or direct evidences? – okw1124 Mar 29 '22 at 01:23
  • A standard example is the Tricomi equation $u_{xx}+xu_{yy}=0$ with $x$ negative. It is hyperbolic (you meant hyperbolic above) but is a well-studied equation that I do not believe is reducible to the standard wave equation. – Bob Terrell Mar 29 '22 at 12:01
  • A simpler example is $u_{xx}+u_{xy}=0$. The solutions are $f(x-y)+g(y)$, not all waves. – Bob Terrell Mar 29 '22 at 17:54