Given curve and surface, proving uniqueness of w for given partial differential equation

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Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique

My attempt:

On boundary $Gamma$:
f is not defined. So taking f(x,y) = 0 on $Gamma$

Inside Surface S, w is not defined. So taking w(x,y) = 0 in S

Now Solution of give PDE is given by

$w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$

But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.

So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$

Now for uniqueness,

Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$

Now to choose that $f_1 , f_2$ which satisfy above?

complex-analysis pde boundary-value-problem laplacian

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asked Jul 16 at 15:55

Magneto

793213

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Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique

My attempt:

On boundary $Gamma$:
f is not defined. So taking f(x,y) = 0 on $Gamma$

Inside Surface S, w is not defined. So taking w(x,y) = 0 in S

Now Solution of give PDE is given by

$w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$

But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.

So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$

Now for uniqueness,

Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$

Now to choose that $f_1 , f_2$ which satisfy above?

complex-analysis pde boundary-value-problem laplacian

share|cite|improve this question

asked Jul 16 at 15:55

Magneto

793213

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique

My attempt:

On boundary $Gamma$:
f is not defined. So taking f(x,y) = 0 on $Gamma$

Inside Surface S, w is not defined. So taking w(x,y) = 0 in S

Now Solution of give PDE is given by

$w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$

But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.

So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$

Now for uniqueness,

Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$

Now to choose that $f_1 , f_2$ which satisfy above?

complex-analysis pde boundary-value-problem laplacian

share|cite|improve this question

asked Jul 16 at 15:55

Magneto

793213

Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique

My attempt:

On boundary $Gamma$:
f is not defined. So taking f(x,y) = 0 on $Gamma$

Inside Surface S, w is not defined. So taking w(x,y) = 0 in S

Now Solution of give PDE is given by

$w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$

But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.

So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$

Now for uniqueness,

Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$

Now to choose that $f_1 , f_2$ which satisfy above?

complex-analysis pde boundary-value-problem laplacian

share|cite|improve this question

asked Jul 16 at 15:55

Magneto

793213

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 15:55

Magneto

793213

asked Jul 16 at 15:55

Magneto

793213

asked Jul 16 at 15:55

Magneto

793213

793213

add a comment | 

add a comment | 

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