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

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SUMMARY

This discussion focuses on solving a system of partial differential equations (PDEs) defined by the equations {u_x} + 4{v_y} = 0 and {v_x} + 9{u_y} = 0, with initial conditions u(x,0) = 2x and v(x,0) = 3x. The general solutions for u and v are expressed as u(x,y) = h(x + 6y) + g(y - 6x) and v(x,y) = h(x + 6y) + g(y - 6x). The conversation highlights the need to differentiate the functions u and v to satisfy the initial conditions and suggests substituting u into the system to derive v.

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arrow27
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\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?
 
Last edited by a moderator:
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arrow27 said:
\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.
 
What can i do?
 
arrow27 said:
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.
 
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?
 
Last edited:
arrow27 said:
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.
 

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