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Consider the equation $u_{xx}+2u_{xy}+u_{yy}=0.$ Write the equation in the coordinates $s=x$ and $t=x-y$ and find the general solution of the equation.

We have that $x=s$ and $y=s-t,$ thus $u(x,y)=u(s,s-t)=v(s,t)$. Now I know I must compute partial derivatives of $x$ and $y$ by using the chain rule to find a general solution, but I am not sure how to apply it here?

Robben
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1 Answers1

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$s=x$ and $t=x-y$ hense $s_x=1$ ; $s_y=0$ ; $t_x=1$ ; $t_y=-1$

$$u_x=v_s s_x+v_t t_x=v_s+v_t$$ $$u_{xx}=(v_{ss} s_x+v_{st} t_x)+(v_{ts} s_x+v_{tt} t_x)=v_{ss}+v_{st}+v_{ts}+v_{tt}=v_{ss}+2v_{st}+v_{tt}$$ $$u_{xy}=(v_{ss} s_y+v_{st} t_y)+(v_{ts} s_y+v_{tt} t_y)=-v_{st}-v_{tt}$$ $$u_y=v_s s_y+v_t t_y=-v_t$$ $$u_{yy}=-v_{ts} s_y-v_{tt} t_y=v_{tt}$$

$$u_{xx}+2u_{xy}+u_{yy}=v_{ss}+2v_{st}+v_{tt}+2(-v_{st}-v_{tt})+v_{tt}=v_{ss}$$ $$v_{ss}=0$$

There is no difficulty to continue...

JJacquelin
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