Question

[Partial Differential Equation]

solve the following boundary value problems

\[ \LARGE \frac{\partial^{2} u}{\partial x \partial y} \left( x,y \right) = 0 , u(x,0) = \sin x , u(0,y) = y \]

Answer 1

[Partial Differential Equation]

solve the following boundary value problems

\[ \LARGE \frac{\partial^{2} u}{\partial x \partial y} \left( x,y \right) = 0 \]
\[ \LARGE \frac{\partial}{\partial x } \left( \frac{\partial u}{\partial y } \left( x,y \right) \right)= 0 \]

integrate with respect to \(x\) yields

\[ \LARGE \frac{\partial u}{\partial y } \left( x,y \right)= f(y) \]
integrate with respect to \(y\) yields

\( \LARGE u( x,y )= F(y) + g(x) \)

with
\[ \LARGE \frac{\partial }{\partial y } F(y) = f(y) \]

is this correct?

what's next?

Answer 2

Now put your boundary conditions into the expression for \(u(x,y)\).

Answer 3

$$u(x,0) = \sin{x} \implies g(x) + F(0) = \sin{x} \implies g(x) = \sin{x} - F(0) $$

$$u(0,y) = y \implies g(0) + F(y) = y \implies F(y) = y- g(0) $$

$$u(x,y) = F(y) + g(x) = y - g(0) + \sin{x} - F(0) $$

like this?

what should I do with \(F(0) \) and \(g(0) \) ?

Answer 4

\(u(x,0)=\sin x=\sin x-g(0)-F(0)\)
\(u(0,y)=y=y-g(0)-F(0)\)
\(g(0)=-F(0)\)
If you put this into the expression for \(u(x,y)\) you will get
\(u(x,y)=\sin x+y\)

Answer 5

the third line..,
you get \( g(0) = -F(0) \) from \( -g(0) - F(0) = 0\) right?

don't we need to change \( -g(0) - F(0) \) become \(C\) (constant) ?
There usually the 'C' part in the general solution..

Answer 6

As you can see, \(C\) will not do. If we let \(-g(0)-F(0)=C\), then \(u(x,y)=\sin x+y+C\). Now check your boundary conditions again:
\(u(0,y)=y+C\)
It will satisfy only if \(C=0\).

Answer 7

You have enough conditions to find the definite solution.

Answer 8

ah.., I see.. :)

my other question,
do you get \( g(0) = -F(0)\) from \( -g(0) - F(0) = 0 \) ?

Answer 9

Yes.

Answer 10

ok..., thank you... :)