Tricomi equation canonical form and solution
$begingroup$
Consider the Tricomi equation
$$u_{xx}+xu_{yy}=0$$
With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:
$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$
pde linear-pde
$endgroup$
add a comment |
$begingroup$
Consider the Tricomi equation
$$u_{xx}+xu_{yy}=0$$
With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:
$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$
pde linear-pde
$endgroup$
$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52
add a comment |
$begingroup$
Consider the Tricomi equation
$$u_{xx}+xu_{yy}=0$$
With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:
$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$
pde linear-pde
$endgroup$
Consider the Tricomi equation
$$u_{xx}+xu_{yy}=0$$
With the transformation $(r,q)=left(-frac{2}{3}x^{3/2},yright)$, I found the canonical form but I could not solve it:
$$v_{qq}+v_{rr}+dfrac1{3r}v_r=0.$$
pde linear-pde
pde linear-pde
edited Dec 12 '18 at 18:26
Harry49
6,21331132
6,21331132
asked Nov 18 '14 at 14:42
Joemak BoblinkeJoemak Boblinke
895
895
$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52
add a comment |
$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52
$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52
$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
$$
u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
$$
The functions $r,q$ satisfy the Beltrami differential equations
$$
r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
$$
Particular solutions of the Tricomi equation can be found here.
Further reading: p. 162-163 of
R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234
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add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
$$
u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
$$
The functions $r,q$ satisfy the Beltrami differential equations
$$
r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
$$
Particular solutions of the Tricomi equation can be found here.
Further reading: p. 162-163 of
R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234
$endgroup$
add a comment |
$begingroup$
In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
$$
u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
$$
The functions $r,q$ satisfy the Beltrami differential equations
$$
r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
$$
Particular solutions of the Tricomi equation can be found here.
Further reading: p. 162-163 of
R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234
$endgroup$
add a comment |
$begingroup$
In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
$$
u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
$$
The functions $r,q$ satisfy the Beltrami differential equations
$$
r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
$$
Particular solutions of the Tricomi equation can be found here.
Further reading: p. 162-163 of
R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234
$endgroup$
In the elliptic case $x>0$, the change of coordinates $(r,q) = left(-x^{3/2},frac{3}{2}yright)$ in the Tricomi equation gives the Euler-Poisson-Darboux equation
$$
u_{xx} + xu_{yy} = tfrac{9}{4}xleft( u_{qq} + u_{rr} + frac{1}{3r}u_rright) = 0 .
$$
The functions $r,q$ satisfy the Beltrami differential equations
$$
r_x = -sqrt{x}, q_y,qquad r_y = frac{1}{sqrt{x}}, q_x .
$$
Particular solutions of the Tricomi equation can be found here.
Further reading: p. 162-163 of
R. Courant, D. Hilbert: Methods of Mathematical Physics vol. II: "Partial differential equations". Wiley-VCH, 1962. doi:10.1002/9783527617234
edited Dec 12 '18 at 18:32
answered Dec 12 '18 at 13:10
Harry49Harry49
6,21331132
6,21331132
add a comment |
add a comment |
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$begingroup$
This note from the arxiv will help you arxiv.org/pdf/1311.3338v2.pdf
$endgroup$
– Autolatry
Nov 18 '14 at 14:52