Are both answers for $xu_x + yu_y = 0$ valid?
$begingroup$
Solving this problem by the method of characteristic curves we have to solve the ODE
$$frac{mathrm{d}y}{mathrm{d}x} = frac{y}{x}$$
which gives us
$$C = ln(y/x)$$
where $C$ is constant.
Therefore, the solution should be
$$u(x,y) = f(ln(y/x))$$
We can go further by noting that $e^C$ is also a constant which we can name $C_2$ giving us
$$C_2 = y/x$$
and the solution would give
$$u(x,y) = f(y/x)$$
Are both solutions valid?
pde characteristics
edited Dec 13 '18 at 10:35
Harry49
7,48431340
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
$endgroup$
add a comment |
$begingroup$
Solving this problem by the method of characteristic curves we have to solve the ODE
$$frac{mathrm{d}y}{mathrm{d}x} = frac{y}{x}$$
which gives us
$$C = ln(y/x)$$
where $C$ is constant.
Therefore, the solution should be
$$u(x,y) = f(ln(y/x))$$
We can go further by noting that $e^C$ is also a constant which we can name $C_2$ giving us
$$C_2 = y/x$$
and the solution would give
$$u(x,y) = f(y/x)$$
Are both solutions valid?
pde characteristics
edited Dec 13 '18 at 10:35
Harry49
7,48431340
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
$endgroup$
$begingroup$
All you are doing is rescaling how "quickly" your solution travels along the characteristics. Both solutions are essentially the same. With both formulations you may run into problems with things blowing up, but that is part of the differential equation, not your formulation
$endgroup$
– whpowell96
Dec 13 '18 at 5:17
add a comment |
$begingroup$
Solving this problem by the method of characteristic curves we have to solve the ODE
$$frac{mathrm{d}y}{mathrm{d}x} = frac{y}{x}$$
which gives us
$$C = ln(y/x)$$
where $C$ is constant.
Therefore, the solution should be
$$u(x,y) = f(ln(y/x))$$
We can go further by noting that $e^C$ is also a constant which we can name $C_2$ giving us
$$C_2 = y/x$$
and the solution would give
$$u(x,y) = f(y/x)$$
Are both solutions valid?
pde characteristics
edited Dec 13 '18 at 10:35
Harry49
7,48431340
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
$endgroup$
Solving this problem by the method of characteristic curves we have to solve the ODE
$$frac{mathrm{d}y}{mathrm{d}x} = frac{y}{x}$$
which gives us
$$C = ln(y/x)$$
where $C$ is constant.
Therefore, the solution should be
$$u(x,y) = f(ln(y/x))$$
We can go further by noting that $e^C$ is also a constant which we can name $C_2$ giving us
$$C_2 = y/x$$
and the solution would give
$$u(x,y) = f(y/x)$$
Are both solutions valid?
pde characteristics
pde characteristics
edited Dec 13 '18 at 10:35
Harry49
7,48431340
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
edited Dec 13 '18 at 10:35
Harry49
7,48431340
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
edited Dec 13 '18 at 10:35
Harry49
7,48431340
edited Dec 13 '18 at 10:35
Harry49
7,48431340
edited Dec 13 '18 at 10:35
Harry49
7,48431340
7,48431340
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
asked Dec 13 '18 at 2:44
The BoscoThe Bosco
613212
613212
$begingroup$
All you are doing is rescaling how "quickly" your solution travels along the characteristics. Both solutions are essentially the same. With both formulations you may run into problems with things blowing up, but that is part of the differential equation, not your formulation
$endgroup$
– whpowell96
Dec 13 '18 at 5:17
add a comment |
$begingroup$
All you are doing is rescaling how "quickly" your solution travels along the characteristics. Both solutions are essentially the same. With both formulations you may run into problems with things blowing up, but that is part of the differential equation, not your formulation
$endgroup$
– whpowell96
Dec 13 '18 at 5:17
$begingroup$
All you are doing is rescaling how "quickly" your solution travels along the characteristics. Both solutions are essentially the same. With both formulations you may run into problems with things blowing up, but that is part of the differential equation, not your formulation
$endgroup$
– whpowell96
Dec 13 '18 at 5:17
$begingroup$
All you are doing is rescaling how "quickly" your solution travels along the characteristics. Both solutions are essentially the same. With both formulations you may run into problems with things blowing up, but that is part of the differential equation, not your formulation
$endgroup$
– whpowell96
Dec 13 '18 at 5:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
They are the same solution but not the same function to express it. Using the same name for both functions is confusing. Call the second $g$, so $f(ln(x/y))=g(x/y)$ For any chosen $f$ we have determined $g$ as $g=fcirc ln$
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
I like to think of it that when:
$$C = ln left ( frac{y}{x} right ) iff frac{y}{x} = e^{C} = C_1$$
So you can just think of it as unwrapping the independent variables until you can't go any further.
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
They are the same solution but not the same function to express it. Using the same name for both functions is confusing. Call the second $g$, so $f(ln(x/y))=g(x/y)$ For any chosen $f$ we have determined $g$ as $g=fcirc ln$
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
They are the same solution but not the same function to express it. Using the same name for both functions is confusing. Call the second $g$, so $f(ln(x/y))=g(x/y)$ For any chosen $f$ we have determined $g$ as $g=fcirc ln$
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
They are the same solution but not the same function to express it. Using the same name for both functions is confusing. Call the second $g$, so $f(ln(x/y))=g(x/y)$ For any chosen $f$ we have determined $g$ as $g=fcirc ln$
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
$endgroup$
They are the same solution but not the same function to express it. Using the same name for both functions is confusing. Call the second $g$, so $f(ln(x/y))=g(x/y)$ For any chosen $f$ we have determined $g$ as $g=fcirc ln$
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
edited Dec 14 '18 at 16:19
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
answered Dec 13 '18 at 5:39
Rafa BudríaRafa Budría
5,8151825
5,8151825
add a comment |
add a comment |
$begingroup$
I like to think of it that when:
$$C = ln left ( frac{y}{x} right ) iff frac{y}{x} = e^{C} = C_1$$
So you can just think of it as unwrapping the independent variables until you can't go any further.
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
$endgroup$
add a comment |
$begingroup$
I like to think of it that when:
$$C = ln left ( frac{y}{x} right ) iff frac{y}{x} = e^{C} = C_1$$
So you can just think of it as unwrapping the independent variables until you can't go any further.
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
$endgroup$
add a comment |
$begingroup$
I like to think of it that when:
$$C = ln left ( frac{y}{x} right ) iff frac{y}{x} = e^{C} = C_1$$
So you can just think of it as unwrapping the independent variables until you can't go any further.
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
$endgroup$
I like to think of it that when:
$$C = ln left ( frac{y}{x} right ) iff frac{y}{x} = e^{C} = C_1$$
So you can just think of it as unwrapping the independent variables until you can't go any further.
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
answered Dec 13 '18 at 19:38
gdepaulgdepaul
613
613
add a comment |
add a comment |
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$begingroup$
All you are doing is rescaling how "quickly" your solution travels along the characteristics. Both solutions are essentially the same. With both formulations you may run into problems with things blowing up, but that is part of the differential equation, not your formulation
$endgroup$
– whpowell96
Dec 13 '18 at 5:17