Question

U_xx+3U_xy-4U_yy+U_x+4U_y=0 find the canonical form and solve the equation. (general solution)

Answer 1

The characteristic quadratic form of this equation has the form

$Q( \lambda_1 , \lambda_2)=\lambda^2_1 +3 \lambda_1 \lambda_2-4 \lambda_2^2$

Let's bring it to its canonical form:

$\lambda^2_1 +3 \lambda_1 \lambda_2-4 \lambda_2^2=(\lambda_1+3/2 \lambda_2)^2-( \sqrt{11/2} \lambda_2)^2= k_1^2-k_2^2$

Where

$k1= \lambda_1+3/2 \lambda_2$

$k_2= \sqrt{11/2} \lambda_2$

So

$\begin{pmatrix} k_1 \\ k_2 \end{pmatrix} =\begin{pmatrix} 1 & 3/2\\ 0 & \sqrt{11/2} \end{pmatrix}\begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix}$

Find the number replacement matrix

$( \begin{pmatrix} 1 & 3/2\\ 0 & \sqrt{11/2} \end{pmatrix}^T)^{-1}=\begin{pmatrix} 1 & 0 \\ -30/47 & 20/47 \end{pmatrix}$

We replace the number:

$\begin{pmatrix} a \\ b \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ -30/47 & 20/47 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$

a=x

b=(-30x+20y)/47

To substitute new variables in the original equation, we put

V(a,b)=u(x,y)

u_x=v_a-(30/47)v_b

u_y=20/47 v_b

u_xx=v_aa-(60/47)v_ab+900/2209v_bb

u_xy=(60/47)v_ab-1800/2209v_bb

u_yy=400/2209v_bb

v_aa-(60/47)v_ab+900/2209v_bb+(180/47)v_ab-5400/2209v_bb -1600/2209v_bb +v_a-(30/47)v_b+800/47 v_b =v_aa+(120/47)v_ab-6100/2209v_bb + v_a+770/47v_b=0

Answer: v_aa+(120/47)v_ab-6100/2209v_bb + v_a+770/47v_b=0

hyperbolic type

General solution:

V(a,b)=f1(a)+f2(b)

u(x,y)=f1(x)+f2((-30x+20y)/47)