How to Integrate and Compare Solutions for a Partial Differential System?

[arrow27]

\begin{array}{l}
u = u(x,y) \\
v = v(x,y) \\
and\\
{u_x} + 4{v_y} = 0 \\
{v_x} + 9{u_y} = 0 \\
with\ the\ initial\ conditions \\
u(x,0) = 2x _(3)\\
v(x,0) = 3x _(4)\\
\end{array}

Easy,
u_{xx}-36u_{yy}=0 and v_{xx}-36v_{yy}=0
General solution u\left ( x,y \right )=h\left ( x+6y \right )+g\left ( y-6x \right )
Similar,
v\left ( x,y \right )=h\left ( x+6y \right )+g\left ( y-6x \right )

From (3) : 2x=h\left ( 6x \right )+g\left ( -6x \right )
From (4) : 3x=h\left ( 6x \right )+g\left ( -6x \right )

How to continue?

 

[Mathwebster]

\begin{array}{l}
u = u(x,y) \\
v = v(x,y) \\
and\\
{u_x} + 4{v_y} = 0 \\
{v_x} + 9{u_y} = 0 \\
with\ the\ initial\ conditions \\
u(x,0) = 2x _(3)\\
v(x,0) = 3x _(4)\\
\end{array}

Easy,
u_{xx}-36u_{yy}=0 and v_{xx}-36v_{yy}=0
General solution u\left ( x,y \right )=h\left ( x+6y \right )+g\left ( y-6x \right )
Similar,
v\left ( x,y \right )=h\left ( x+6y \right )+g\left ( y-6x \right )

From (3) : 2x=h\left ( 6x \right )+g\left ( -6x \right )
From (4) : 3x=h\left ( 6x \right )+g\left ( -6x \right )

How to continue?

Why are you assuming the solutions $u$ and $v$ are exactly the same? As you're seeing this is inconsistent with the initial conditions.

 

[arrow27]

What can i do?

 

[Mathwebster]

What can i do?

Choose one of them, say $$u = g(x+6y) + g(y-6x)$$, then sub this back into your system. Then integrate each giving the solution for $$v(x,y)$$. Then use your BC's.

 

[arrow27]

But we have u_x and u_x in equations.
For example,

\[u = {g'(x)}(x + 6y) - 6{g'(x)}(x - 6y)\]
Sub this in the first equation?

 

[Mathwebster]

But we have u_x and u_x in equations.
For example,

\[u = {g'(x)}(x + 6y) - 6{g'(x)}(x - 6y)\]
Sub this in the first equation?

Sorry, that was a typo. If

$$u = h(6x+y) + g(6x-y)$$

then

$$u_x = 6h'(6x+y) + 6g'(6x-y)$$

$$u_y = h'(6x+y) - g'(6x-y)$$

So from the original set of PDEs we have

$$6h'(6x+y) + 6g'(6x - y) + 4 v_y = 0$$

$$v_x + 9\left(h'(6x+y) - g'(6x-y)\right) = 0$$

or

$$v_x = - 9h'(6x+y) + 9 g'(6x-y)$$

$$v_y = - \dfrac{3}{2} h'(6x+y) - \dfrac{3}{2} g'(6x - y). $$

Now integrate each separately and compare.