\[u_t=a^2u_{xx}+bu_x^2+cu_y\] \[(x,y)\in\mathbb R,t>0\]

的初值问题

考试没有写出来TAT

许轶臣的解法如下：

令

\[u = f(v)\]

有

\[u_t=f'(v)v_t\] \[u_y=f'(v)v_y\] \[u_x=f'(v)v_x\] \[u_{xx}=f'(v)v_{xx}+f''(v)v_x^2\]

所以

\[f'(v)v_t=a^2(f'(v)v_{xx}+f''(v)v_x^2)+b(f'(v)v_x)^2+cf'(v)v_y\] \[f'(v)v_t=a^2f'(v)v_{xx}+a^2f''(v)v_x^2+bf'(v)^2v_x^2+cf'(v)v_y\]

令

\[a^2f''(v)v_x^2+bf'(v)^2v_x^2=0\]

即

\[a^2f''(v)+bf'(v)^2=0\]

解得

\[f(v)=\frac{a^2}{b}\ln\frac{bv}{a^2}\] \[f'(v)=\frac{a^2}{bv}\]

带入原方程

得到

\[v_t=a^2v_{xx}+cv_y\]