PDE with complex roots of characterictic solution
$begingroup$
I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.
As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
$$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
pde linear-pde
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
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add a comment |
$begingroup$
I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.
As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
$$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
pde linear-pde
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
$endgroup$
add a comment |
$begingroup$
I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.
As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
$$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
pde linear-pde
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
$endgroup$
I have the following equation $$frac{partial^2 V}{partial x^2 }+frac{partial^2 V}{partial y^2}=-4pi(x^2+y^2)$$ Such that $V(x,y)$ is real and $V(x,y)=0$ at $y=0$.
As it is clear form the problem the characteristic equation has complex roots. I have following as the solution
$$z=phi_1(y+ix)+phi_2(y-ix)-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
pde linear-pde
pde linear-pde
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
asked Jan 19 at 5:27
Piyush DivyanakarPiyush Divyanakar
3,348327
3,348327
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.
I cannot see how you get a so complicated particular solution :
$$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?
Without having the details of your calculus it is not possible to find where a possible mistake occurred.
There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
$$z_p=f(x)+g(y)$$
$$f''(x)+g''(y)=-4pi(x^2+y^2)$$
$$begin{cases}
f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
end{cases}$$
No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
$$z_p=-frac{pi}{3} (x^4+y^4)$$
$$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
$endgroup$
$begingroup$
I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
$endgroup$
– Piyush Divyanakar
Jan 20 at 5:34
$begingroup$
It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
$endgroup$
– JJacquelin
Jan 20 at 7:19
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
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$begingroup$
$z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.
I cannot see how you get a so complicated particular solution :
$$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?
Without having the details of your calculus it is not possible to find where a possible mistake occurred.
There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
$$z_p=f(x)+g(y)$$
$$f''(x)+g''(y)=-4pi(x^2+y^2)$$
$$begin{cases}
f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
end{cases}$$
No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
$$z_p=-frac{pi}{3} (x^4+y^4)$$
$$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
$endgroup$
$begingroup$
I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
$endgroup$
– Piyush Divyanakar
Jan 20 at 5:34
$begingroup$
It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
$endgroup$
– JJacquelin
Jan 20 at 7:19
add a comment |
$begingroup$
$z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.
I cannot see how you get a so complicated particular solution :
$$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?
Without having the details of your calculus it is not possible to find where a possible mistake occurred.
There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
$$z_p=f(x)+g(y)$$
$$f''(x)+g''(y)=-4pi(x^2+y^2)$$
$$begin{cases}
f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
end{cases}$$
No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
$$z_p=-frac{pi}{3} (x^4+y^4)$$
$$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
$endgroup$
$begingroup$
I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
$endgroup$
– Piyush Divyanakar
Jan 20 at 5:34
$begingroup$
It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
$endgroup$
– JJacquelin
Jan 20 at 7:19
add a comment |
$begingroup$
$z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.
I cannot see how you get a so complicated particular solution :
$$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?
Without having the details of your calculus it is not possible to find where a possible mistake occurred.
There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
$$z_p=f(x)+g(y)$$
$$f''(x)+g''(y)=-4pi(x^2+y^2)$$
$$begin{cases}
f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
end{cases}$$
No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
$$z_p=-frac{pi}{3} (x^4+y^4)$$
$$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
$endgroup$
$z(x,y)$ is the sum of the solution of the associated homogenous PDE and any particular solution.
I cannot see how you get a so complicated particular solution :
$$z_p=-dfrac{pi x^2left(6left(y-mathrm{i}xright)^2+left(3mathrm{i}-3right)x^2+left(8mathrm{i}+4right)xleft(y-mathrm{i}xright)right)}{3}$$
Moreover this function seems not correct. Did you put it into the PDE to check if the equation is satisfied ?
Without having the details of your calculus it is not possible to find where a possible mistake occurred.
There is a much simpler way to find a particular solution. For example, look for a particular solution on the form
$$z_p=f(x)+g(y)$$
$$f''(x)+g''(y)=-4pi(x^2+y^2)$$
$$begin{cases}
f''(x)=-4pi(x^2) & ;quad f(x)=-frac{pi}{3} x^4 \
g''(y)=-4pi(y^2) & ;quad g(x)=-frac{pi}{3} y^4
end{cases}$$
No need for more terms in $f(x)$ and in $g(y)$ since we are not looking for many particular solutions but only for one. Doesn't matter which one.
$$z_p=-frac{pi}{3} (x^4+y^4)$$
$$z=phi_1(y+ix)+phi_2(y-ix)-frac{pi}{3} (x^4+y^4)$$
Of course, since any particular solution is convenient, other particular solutions could be used instead of the above one. So, the solution can be written on an infinity of equivalent forms. If you find a solution which at first sight looks different from the above, check if they are not equivalent , due to different form of functions $phi_1$ and $phi_2$ .
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
edited Jan 19 at 8:43
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
answered Jan 19 at 8:33
JJacquelinJJacquelin
45.8k21858
45.8k21858
$begingroup$
I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
$endgroup$
– Piyush Divyanakar
Jan 20 at 5:34
$begingroup$
It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
$endgroup$
– JJacquelin
Jan 20 at 7:19
add a comment |
$begingroup$
I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
$endgroup$
– Piyush Divyanakar
Jan 20 at 5:34
$begingroup$
It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
$endgroup$
– JJacquelin
Jan 20 at 7:19
$begingroup$
I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
$endgroup$
– Piyush Divyanakar
Jan 20 at 5:34
$begingroup$
I used the general method of getting particular integral which is like $frac{1}{D+iD'}f(x,y)=int f(x,c+ix)$ where $y-ix=c$. So if I get particular integral through other methods it is still valid>
$endgroup$
– Piyush Divyanakar
Jan 20 at 5:34
$begingroup$
It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
$endgroup$
– JJacquelin
Jan 20 at 7:19
$begingroup$
It is easy to check if it is valid or not : put the particular solution that you obtain into the PDE and see if it agrees or not.
$endgroup$
– JJacquelin
Jan 20 at 7:19
add a comment |
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