Change of variables Laplace Equation

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Homework Statement



Write the Laplace equation [tex]\dfrac {\partial ^{2}F} {\partial x^{2}}+\dfrac {\partial ^{2}F} {\partial y^{2}}=0[/tex] in terms of polar coordinates.



Homework Equations


[tex]
r=\sqrt {x^{2}+y^{2}}
[/tex]
[tex]
\theta =\tan ^{-1}(\frac{y}{x})
[/tex]
[tex]
\dfrac {\partial r} {\partial x}=\cos \theta
[/tex]
[tex]
\dfrac {\partial \theta } {\partial x}=-\dfrac {\sin \theta } {r}
[/tex]
[tex]
\dfrac {\partial r} {\partial y}=\sin \theta
[/tex]
[tex]
\dfrac {\partial \theta } {\partial y}=\dfrac {\cos \theta } {r}
[/tex]

The Attempt at a Solution



My goal was to find the second derivative of F with respect to x and y in terms of derivatives in terms of r and theta and then substitute into Laplace's equation. So I started by taking the first derivative with respect to x and y to get:
[tex]

\dfrac {\partial F} {\partial x}=\dfrac {\partial F} {\partial r}\cdot \dfrac {\partial r} {\partial x}+\dfrac {\partial F} {\partial \theta }\cdot \dfrac {\partial \theta } {\partial x} =\cos \theta \dfrac {\partial F} {\partial r}-\dfrac {\sin \theta } {r}\dfrac {\partial F} {\partial \theta }
[/tex]
[tex]
\dfrac {\partial F} {\partial y}=\dfrac {\partial F} {\partial r}\cdot \dfrac {\partial r} {\partial y}+\dfrac {\partial F} {\partial \theta }-\dfrac {\partial \theta } {\partial y} =\sin \theta \dfrac {\partial F} {\partial r}+\dfrac {\cos \theta } {r}\dfrac {\partial F} {\partial \theta }
[/tex]


Based on [itex] \dfrac {\partial F} {\partial x} [/itex], I found the operator [itex]\dfrac {\partial } {\partial x}[/itex] to be

[tex]\dfrac {\partial } {\partial x}=\left( \cos \theta \dfrac {\partial } {\partial r}-\dfrac {\sin \theta } {r}\dfrac {\partial } {\partial \theta }\right) [/tex]

and [itex]\dfrac {\partial } {\partial y}[/itex] to be

[tex]\dfrac {\partial } {\partial y}=\left( \sin \theta \dfrac {\partial } {\partial r}+\dfrac {\cos \theta } {r}\dfrac {\partial } {\partial \theta }\right) [/tex]

Now I need to find [itex]\dfrac {\partial ^{2}F} {\partial x^{2}}=\dfrac {\partial } {\partial x}\left( \dfrac {\partial F} {\partial x}\right) [/itex] and I get:

[tex]\dfrac {\partial } {\partial x}\left( \dfrac {\partial F} {\partial x}\right) =\left( \cos \theta \dfrac {\partial } {\partial r}-\dfrac {\sin \theta } {r}\dfrac {d} {\partial \theta }\right) \left( \cos \theta \dfrac {\partial F} {\partial r}-\dfrac {\sin \theta } {r}\dfrac {\partial F} {\partial \theta }\right) [/tex]

I distribute to and simplify to get:
[tex]\dfrac {\partial } {\partial x}\left( \dfrac {\partial F} {\partial x}\right) =\cos ^{2}\theta \dfrac {\partial ^{2}F} {\partial r^2}-\dfrac {2} {r}\sin \theta \cos \theta \dfrac {\partial f^{2}} {\partial rd\theta }+\dfrac {\sin ^{2}\theta } {r^{2}}\dfrac {d^{2}F} {d\theta ^{2}}[/tex]

Now I need the same for [itex]\dfrac {\partial ^{2}F} {\partial y^{2}}=\dfrac {\partial } {\partial y}\left( \dfrac {\partial F} {\partial y}\right) [/itex] and I get:

[tex]\dfrac {\partial } {\partial y}\left( \dfrac {\partial F} {\partial y}\right) =\left( \sin \theta \dfrac {\partial } {\partial r}+\dfrac {\cos \theta } {r}\dfrac {\partial } {\partial \theta }\right) \left( sin\theta \dfrac {\partial F} {\partial r}+\dfrac {\cos \theta } {r}\dfrac {\partial F} {\partial \theta }\right) [/tex]

I distribute and simplify to get:
[tex]\dfrac {\partial } {\partial y}\left( \dfrac {\partial F} {dy}\right) =\sin ^{2}\theta \dfrac {\partial ^{2}F} {\partial r^{2}}+\dfrac {2\sin \theta \cos \theta } {r}\dfrac {\partial ^{2}F} {\partial rd\theta }+\dfrac {\cos ^{2}\theta } {r^{2}}\dfrac {\partial ^{2}F} {\partial \theta ^{2}}[/tex]

When I add these two second derivatives (wrt to x and y) and simplify, I get:
[tex]\dfrac {\partial ^{2}F} {\partial x^{2}}+\dfrac {\partial ^{2}F} {\partial y^{2}}=\dfrac {d^{2}F} {\partial r^{2}}+\dfrac {1} {r^{2}}\dfrac {\partial ^{2}F} {\partial \theta ^{2}}[/tex]

But this is not completely right. The second term is correct, but the answer should be:
[tex]\dfrac {1} {r}\dfrac {\partial } {\partial r}\left( r\dfrac {\partial F} {\partial r}\right) +\dfrac {1} {r^{2}}\dfrac {\partial ^{2}F} {\partial \theta ^{2}}[/tex]

I know this is a long post, but any ideas on where I went wrong?

Thanks so much in advance.
 
Last edited:

Answers and Replies

  • #2
pasmith
Homework Helper
2,019
651

Homework Statement



Write the Laplace equation [tex]\dfrac {\partial ^{2}F} {\partial x^{2}}+\dfrac {\partial ^{2}F} {\partial y^{2}}=0[/tex] in terms of polar coordinates.



Homework Equations


[tex]
r=\sqrt {x^{2}+y^{2}}
[/tex]
[tex]
\theta =\tan ^{-1}(\frac{y}{x})
[/tex]
[tex]
\dfrac {\partial r} {\partial x}=\cos \theta
[/tex]
[tex]
\dfrac {\partial \theta } {\partial x}=-\dfrac {\sin \theta } {r}
[/tex]
[tex]
\dfrac {\partial r} {\partial y}=\sin \theta
[/tex]
[tex]
\dfrac {\partial \theta } {\partial y}=\dfrac {\cos \theta } {r}
[/tex]

The Attempt at a Solution



My goal was to find the second derivative of F with respect to x and y in terms of derivatives in terms of r and theta and then substitute into Laplace's equation. So I started by taking the first derivative with respect to x and y to get:
[tex]

\dfrac {\partial F} {\partial x}=\dfrac {\partial F} {\partial r}\cdot \dfrac {\partial r} {\partial x}+\dfrac {\partial F} {\partial \theta }\cdot \dfrac {\partial \theta } {\partial x} =\cos \theta \dfrac {\partial F} {\partial r}-\dfrac {\sin \theta } {r}\dfrac {\partial F} {\partial \theta }
[/tex]
[tex]
\dfrac {\partial F} {\partial y}=\dfrac {\partial F} {\partial r}\cdot \dfrac {\partial r} {\partial y}+\dfrac {\partial F} {\partial \theta }-\dfrac {\partial \theta } {\partial y} =\sin \theta \dfrac {\partial F} {\partial r}+\dfrac {\cos \theta } {r}\dfrac {\partial F} {\partial \theta }
[/tex]


Based on [itex] \dfrac {\partial F} {\partial x} [/itex], I found the operator [itex]\dfrac {\partial } {\partial x}[/itex] to be

[tex]\dfrac {\partial } {\partial x}=\left( \cos \theta \dfrac {\partial } {\partial r}-\dfrac {\sin \theta } {r}\dfrac {\partial } {\partial \theta }\right) [/tex]

and [itex]\dfrac {\partial } {\partial y}[/itex] to be

[tex]\dfrac {\partial } {\partial y}=\left( \sin \theta \dfrac {\partial } {\partial r}+\dfrac {\cos \theta } {r}\dfrac {\partial } {\partial \theta }\right) [/tex]

Correct so far.

Now I need to find [itex]\dfrac {\partial ^{2}F} {\partial x^{2}}=\dfrac {\partial } {\partial x}\left( \dfrac {\partial F} {\partial x}\right) [/itex] and I get:

[tex]\dfrac {\partial } {\partial x}\left( \dfrac {\partial F} {\partial x}\right) =\left( \cos \theta \dfrac {\partial } {\partial r}-\dfrac {\sin \theta } {r}\dfrac {d} {\partial \theta }\right) \left( \cos \theta \dfrac {\partial F} {\partial r}-\dfrac {\sin \theta } {r}\dfrac {\partial F} {\partial \theta }\right) [/tex]

I distribute to and simplify to get:
[tex]\dfrac {\partial } {\partial x}\left( \dfrac {\partial F} {\partial x}\right) =\cos ^{2}\theta \dfrac {\partial ^{2}F} {\partial r^2}-\dfrac {2} {r}\sin \theta \cos \theta \dfrac {\partial f^{2}} {\partial rd\theta }+\dfrac {\sin ^{2}\theta } {r^{2}}\dfrac {d^{2}F} {d\theta ^{2}}[/tex]

Here is your error. You should find that [tex]
\frac{\partial}{\partial r}\left( \cos \theta \frac{\partial F}{\partial r} - \frac{\sin \theta}r \frac{\partial F}{\partial \theta} \right) =
\cos \theta \frac{\partial^2 F}{\partial r^2} - \frac{\sin\theta}{r} \frac{\partial^2 F}{\partial r\,\partial \theta} + \frac{\sin\theta}{r^2} \frac{\partial F}{\partial \theta}
[/tex] and [tex]
\frac{\partial}{\partial \theta} \left( \cos \theta \frac{\partial F}{\partial r} - \frac{\sin \theta}r \frac{\partial F}{\partial \theta} \right) =
\cos \theta \frac{\partial^2 F}{\partial r\,\partial\theta} - \sin \theta \frac{\partial F}{\partial r} - \frac{\sin \theta}r \frac{\partial^2 F}{\partial \theta^2} - \frac{\cos\theta}r \frac{\partial F}{\partial \theta}[/tex] and hence that [tex]
\frac{\partial^2 F}{\partial x^2} = \cos^2 \theta \frac{\partial^2 F}{\partial r^2} + \frac{\sin^2 \theta}{r} \frac{\partial F}{\partial r} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2 F}{\partial \theta^2}
- 2\frac{\cos \theta \sin \theta}r \frac{\partial^2 F}{\partial r\,\partial\theta} + 2\frac{\cos\theta \sin \theta}{r^2} \frac{\partial F}{\partial \theta}.[/tex]

I suspect you have made a similar error in your calculation of [itex]\frac{\partial^2 F}{\partial y^2}[/itex].
 

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