Characteristic curves for second-order Tricomi equation
Consider the Tricomi equation
$$yu_{xx} + u_{yy} = 0$$
Find ordinary differential equations describing the real characteristic
curves and solve these ODEs to obtain equations for the characteristic
curves. Sketch the characteristic curves where they exist.
So Tricomi equation is hyperbolic if $y<0$, parabolic if $y=0$, elliptic if $y>0$. We need to find the real characteristics, so we need to look at the hyperbolic case $y<0$. From the Wikipedia article, we can deduce that the characteristics for this equation are the curves $x pm frac{2}{3}(-y)^{3/2}=C$. How to derive the ODEs for characteristics?
pde characteristics hyperbolic-equations linear-pde
edited Dec 12 '18 at 13:27
Harry49
5,99121031
asked Dec 10 '18 at 21:11
dxdydz
1949
add a comment |
Consider the Tricomi equation
$$yu_{xx} + u_{yy} = 0$$
Find ordinary differential equations describing the real characteristic
curves and solve these ODEs to obtain equations for the characteristic
curves. Sketch the characteristic curves where they exist.
So Tricomi equation is hyperbolic if $y<0$, parabolic if $y=0$, elliptic if $y>0$. We need to find the real characteristics, so we need to look at the hyperbolic case $y<0$. From the Wikipedia article, we can deduce that the characteristics for this equation are the curves $x pm frac{2}{3}(-y)^{3/2}=C$. How to derive the ODEs for characteristics?
pde characteristics hyperbolic-equations linear-pde
edited Dec 12 '18 at 13:27
Harry49
5,99121031
asked Dec 10 '18 at 21:11
dxdydz
1949
add a comment |
Consider the Tricomi equation
$$yu_{xx} + u_{yy} = 0$$
Find ordinary differential equations describing the real characteristic
curves and solve these ODEs to obtain equations for the characteristic
curves. Sketch the characteristic curves where they exist.
So Tricomi equation is hyperbolic if $y<0$, parabolic if $y=0$, elliptic if $y>0$. We need to find the real characteristics, so we need to look at the hyperbolic case $y<0$. From the Wikipedia article, we can deduce that the characteristics for this equation are the curves $x pm frac{2}{3}(-y)^{3/2}=C$. How to derive the ODEs for characteristics?
pde characteristics hyperbolic-equations linear-pde
edited Dec 12 '18 at 13:27
Harry49
5,99121031
asked Dec 10 '18 at 21:11
dxdydz
1949
Consider the Tricomi equation
$$yu_{xx} + u_{yy} = 0$$
Find ordinary differential equations describing the real characteristic
curves and solve these ODEs to obtain equations for the characteristic
curves. Sketch the characteristic curves where they exist.
So Tricomi equation is hyperbolic if $y<0$, parabolic if $y=0$, elliptic if $y>0$. We need to find the real characteristics, so we need to look at the hyperbolic case $y<0$. From the Wikipedia article, we can deduce that the characteristics for this equation are the curves $x pm frac{2}{3}(-y)^{3/2}=C$. How to derive the ODEs for characteristics?
pde characteristics hyperbolic-equations linear-pde
pde characteristics hyperbolic-equations linear-pde
edited Dec 12 '18 at 13:27
Harry49
5,99121031
asked Dec 10 '18 at 21:11
dxdydz
1949
edited Dec 12 '18 at 13:27
Harry49
5,99121031
asked Dec 10 '18 at 21:11
dxdydz
1949
edited Dec 12 '18 at 13:27
Harry49
5,99121031
edited Dec 12 '18 at 13:27
Harry49
5,99121031
edited Dec 12 '18 at 13:27
Harry49
5,99121031
5,99121031
asked Dec 10 '18 at 21:11
dxdydz
1949
asked Dec 10 '18 at 21:11
dxdydz
1949
asked Dec 10 '18 at 21:11
dxdydz
1949
1949
add a comment |
add a comment |
1 Answer
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This linear second-order equation rewrites as
$$
L[u] = a u_{xx} + 2bu_{xy} + c u_{yy} = 0
$$
where $a = y$, $b = 0$ and $c = 1$. Computing the discriminant $Delta = b^2 - ac$ tells that the equation is hyperbolic if $Delta = -y > 0$. We introduce the change of coordinates $(x,y) mapsto (xi(x,y),eta(x,y))$. To obtain the coordinates $xi$, $eta$ which reduce the PDE to its canonical form $w_{xieta} = ell[w]$, we solve the polynomial equation
$a lambda^2 + 2b lambda + c = 0$,
which roots are $lambda = pm (-y)^{-1/2}$. The characteristic equations are
$$
frac{text d x}{text d t} = 1,
qquad
frac{text d y}{text d t} = -lambda,
qquad
frac{text d z}{text d t} = 0 .
$$
Therefore, $frac{text d y}{text d x} = frac{text d y/text d t}{text d x/text d t} = mp(-y)^{-1/2}$ gives the expected expression of characteristic curves, along which $xi$ or $eta$ is constant.
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
answered Dec 12 '18 at 11:58
Harry49
5,99121031
add a comment |
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1 Answer
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1 Answer
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This linear second-order equation rewrites as
$$
L[u] = a u_{xx} + 2bu_{xy} + c u_{yy} = 0
$$
where $a = y$, $b = 0$ and $c = 1$. Computing the discriminant $Delta = b^2 - ac$ tells that the equation is hyperbolic if $Delta = -y > 0$. We introduce the change of coordinates $(x,y) mapsto (xi(x,y),eta(x,y))$. To obtain the coordinates $xi$, $eta$ which reduce the PDE to its canonical form $w_{xieta} = ell[w]$, we solve the polynomial equation
$a lambda^2 + 2b lambda + c = 0$,
which roots are $lambda = pm (-y)^{-1/2}$. The characteristic equations are
$$
frac{text d x}{text d t} = 1,
qquad
frac{text d y}{text d t} = -lambda,
qquad
frac{text d z}{text d t} = 0 .
$$
Therefore, $frac{text d y}{text d x} = frac{text d y/text d t}{text d x/text d t} = mp(-y)^{-1/2}$ gives the expected expression of characteristic curves, along which $xi$ or $eta$ is constant.
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
answered Dec 12 '18 at 11:58
Harry49
5,99121031
add a comment |
This linear second-order equation rewrites as
$$
L[u] = a u_{xx} + 2bu_{xy} + c u_{yy} = 0
$$
where $a = y$, $b = 0$ and $c = 1$. Computing the discriminant $Delta = b^2 - ac$ tells that the equation is hyperbolic if $Delta = -y > 0$. We introduce the change of coordinates $(x,y) mapsto (xi(x,y),eta(x,y))$. To obtain the coordinates $xi$, $eta$ which reduce the PDE to its canonical form $w_{xieta} = ell[w]$, we solve the polynomial equation
$a lambda^2 + 2b lambda + c = 0$,
which roots are $lambda = pm (-y)^{-1/2}$. The characteristic equations are
$$
frac{text d x}{text d t} = 1,
qquad
frac{text d y}{text d t} = -lambda,
qquad
frac{text d z}{text d t} = 0 .
$$
Therefore, $frac{text d y}{text d x} = frac{text d y/text d t}{text d x/text d t} = mp(-y)^{-1/2}$ gives the expected expression of characteristic curves, along which $xi$ or $eta$ is constant.
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
answered Dec 12 '18 at 11:58
Harry49
5,99121031
add a comment |
This linear second-order equation rewrites as
$$
L[u] = a u_{xx} + 2bu_{xy} + c u_{yy} = 0
$$
where $a = y$, $b = 0$ and $c = 1$. Computing the discriminant $Delta = b^2 - ac$ tells that the equation is hyperbolic if $Delta = -y > 0$. We introduce the change of coordinates $(x,y) mapsto (xi(x,y),eta(x,y))$. To obtain the coordinates $xi$, $eta$ which reduce the PDE to its canonical form $w_{xieta} = ell[w]$, we solve the polynomial equation
$a lambda^2 + 2b lambda + c = 0$,
which roots are $lambda = pm (-y)^{-1/2}$. The characteristic equations are
$$
frac{text d x}{text d t} = 1,
qquad
frac{text d y}{text d t} = -lambda,
qquad
frac{text d z}{text d t} = 0 .
$$
Therefore, $frac{text d y}{text d x} = frac{text d y/text d t}{text d x/text d t} = mp(-y)^{-1/2}$ gives the expected expression of characteristic curves, along which $xi$ or $eta$ is constant.
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
answered Dec 12 '18 at 11:58
Harry49
5,99121031
This linear second-order equation rewrites as
$$
L[u] = a u_{xx} + 2bu_{xy} + c u_{yy} = 0
$$
where $a = y$, $b = 0$ and $c = 1$. Computing the discriminant $Delta = b^2 - ac$ tells that the equation is hyperbolic if $Delta = -y > 0$. We introduce the change of coordinates $(x,y) mapsto (xi(x,y),eta(x,y))$. To obtain the coordinates $xi$, $eta$ which reduce the PDE to its canonical form $w_{xieta} = ell[w]$, we solve the polynomial equation
$a lambda^2 + 2b lambda + c = 0$,
which roots are $lambda = pm (-y)^{-1/2}$. The characteristic equations are
$$
frac{text d x}{text d t} = 1,
qquad
frac{text d y}{text d t} = -lambda,
qquad
frac{text d z}{text d t} = 0 .
$$
Therefore, $frac{text d y}{text d x} = frac{text d y/text d t}{text d x/text d t} = mp(-y)^{-1/2}$ gives the expected expression of characteristic curves, along which $xi$ or $eta$ is constant.
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
answered Dec 12 '18 at 11:58
Harry49
5,99121031
edited Dec 12 '18 at 17:52
answered Dec 12 '18 at 11:58
Harry49
5,99121031
answered Dec 12 '18 at 11:58
Harry49
5,99121031
answered Dec 12 '18 at 11:58
Harry49
5,99121031
5,99121031
add a comment |
add a comment |
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