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

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The discussion focuses on solving a system of partial differential equations (PDEs) involving functions u(x,y) and v(x,y) with specific initial conditions. The general solutions for both functions are expressed in terms of arbitrary functions h and g, leading to the equations u_x + 4v_y = 0 and v_x + 9u_y = 0. A key point raised is the inconsistency in assuming u and v are identical, which contradicts the initial conditions. Participants suggest substituting one function into the system and integrating separately to derive the other function, emphasizing the importance of correctly applying boundary conditions. The conversation highlights the need for careful treatment of the relationships between the variables and their derivatives in the context of the PDEs.
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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?
 
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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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