Question

8Uxx-8Uxy+2Uyy+17Ux-13Uy=0

Answer 1

Solution;

We reduce the equation into conical form;

From the equation;

a=8

b=-8

c=2

Hence;

$b^2-4ac$ = (-8)²-(4×8×2)=0

Hence the equation is parabolic.

By;

$2ar_x+br_y=0$

Gives;

$8r_x-4r_y=0$

Solving it obtains;

$r=2x+4y$ and $s=y$

The first derivatives will be ;

$U_x=U_rr_x+U_ss_x$

$U_x=2Ur$

$U_y=U_rr_y+U_ss_y$

$U_y=4U_r$

The second derivatives will be as follows;

$U_{xx}=U_{rr}r_x^2+2U_{rs}r_xs_x+U_{ss}s_x^2+U_rr_{xx}+U_ss_{x}$

$U_{xx}=4U_{rr}$

$U_{xy}=U_{rr}r_xr_y+U_{rs}(r_xs_y+r_ys_x)+U_{ss}s_ys_x+U_rr_{xy}+U_ss_{xy}$

$U_{xy}=8U_{rr}+2U_{rs}$

$U_{yy}=U_{rr}r_y^2+2U_{rs}r_ys_y+U_{ss}s_y^2+U_rr_{yy}+U_ss_{yy}$

$U_{yy}=16U_{rr}+8U_{rs}+U_{ss}$

Substitute into the given equation;

$32U_{rr}-64U_{rr}-16U_{rs}+32U_{rr}+16U_{rs}+2U_{ss}+34U_r-52U_r=0$

Simplifying;

$2U_{ss}-18U_r=0$

Which is a heat equation;

$u_{ss}=9u_r$