Mixing
Transversality condition and cauchy problem
$begingroup$
Can anyone give me an intuition of the transversality condition for solving PDEs?
What happens when the initial characteristic curve is tangent to a characteristic at a point ?
pde
asked Jan 28 at 17:24
mat2mat2
84
$endgroup$
add a comment |
$begingroup$
Can anyone give me an intuition of the transversality condition for solving PDEs?
What happens when the initial characteristic curve is tangent to a characteristic at a point ?
pde
asked Jan 28 at 17:24
mat2mat2
84
$endgroup$
add a comment |
$begingroup$
Can anyone give me an intuition of the transversality condition for solving PDEs?
What happens when the initial characteristic curve is tangent to a characteristic at a point ?
pde
asked Jan 28 at 17:24
mat2mat2
84
$endgroup$
Can anyone give me an intuition of the transversality condition for solving PDEs?
What happens when the initial characteristic curve is tangent to a characteristic at a point ?
pde
pde
asked Jan 28 at 17:24
mat2mat2
84
asked Jan 28 at 17:24
mat2mat2
84
asked Jan 28 at 17:24
mat2mat2
84
asked Jan 28 at 17:24
mat2mat2
84
asked Jan 28 at 17:24
mat2mat2
84
84
add a comment |
add a comment |
1 Answer
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To fully determine the solution $u(x,y)$ we have to determine unambiguously every single characteristic curve. We can determine a single characteristic curve assigning to its projection on the plane $x;y$ the value for the function $u$ at one point. So, it have to be given some curve cutting every projection of the characteristics once and values assigned for $u$ along this curve. This is what is called boundary conditions and this determines every characteristic and hence, the solution. Formally $u(x,g(x))=f(x)$
Now, we can face to two problems: one, the curve doesn't cut every projection, two, the curve cuts some projections more than once. For the first case, there are many solution as some of the characteristics are free. For the second case we have two or more different conditions on some projections, so is, two or more values for a single value of $x$ and $y$, having then no solution.
Being the curve for the boundary conditions $g(x)$ tangent at some point to one of the projections amounts that $g$ "bounces" back, cutting again some projections and/or leaving untouched others. Draw a family of parallel lines as the projection of the characteristics and an arc with any shape representing the curve for the boundary conditions, you will see that it cuts twice some projections because is tangent to one projection at some point.
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
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1 Answer
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1 Answer
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active
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$begingroup$
To fully determine the solution $u(x,y)$ we have to determine unambiguously every single characteristic curve. We can determine a single characteristic curve assigning to its projection on the plane $x;y$ the value for the function $u$ at one point. So, it have to be given some curve cutting every projection of the characteristics once and values assigned for $u$ along this curve. This is what is called boundary conditions and this determines every characteristic and hence, the solution. Formally $u(x,g(x))=f(x)$
Now, we can face to two problems: one, the curve doesn't cut every projection, two, the curve cuts some projections more than once. For the first case, there are many solution as some of the characteristics are free. For the second case we have two or more different conditions on some projections, so is, two or more values for a single value of $x$ and $y$, having then no solution.
Being the curve for the boundary conditions $g(x)$ tangent at some point to one of the projections amounts that $g$ "bounces" back, cutting again some projections and/or leaving untouched others. Draw a family of parallel lines as the projection of the characteristics and an arc with any shape representing the curve for the boundary conditions, you will see that it cuts twice some projections because is tangent to one projection at some point.
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
$endgroup$
add a comment |
$begingroup$
To fully determine the solution $u(x,y)$ we have to determine unambiguously every single characteristic curve. We can determine a single characteristic curve assigning to its projection on the plane $x;y$ the value for the function $u$ at one point. So, it have to be given some curve cutting every projection of the characteristics once and values assigned for $u$ along this curve. This is what is called boundary conditions and this determines every characteristic and hence, the solution. Formally $u(x,g(x))=f(x)$
Now, we can face to two problems: one, the curve doesn't cut every projection, two, the curve cuts some projections more than once. For the first case, there are many solution as some of the characteristics are free. For the second case we have two or more different conditions on some projections, so is, two or more values for a single value of $x$ and $y$, having then no solution.
Being the curve for the boundary conditions $g(x)$ tangent at some point to one of the projections amounts that $g$ "bounces" back, cutting again some projections and/or leaving untouched others. Draw a family of parallel lines as the projection of the characteristics and an arc with any shape representing the curve for the boundary conditions, you will see that it cuts twice some projections because is tangent to one projection at some point.
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
$endgroup$
add a comment |
$begingroup$
To fully determine the solution $u(x,y)$ we have to determine unambiguously every single characteristic curve. We can determine a single characteristic curve assigning to its projection on the plane $x;y$ the value for the function $u$ at one point. So, it have to be given some curve cutting every projection of the characteristics once and values assigned for $u$ along this curve. This is what is called boundary conditions and this determines every characteristic and hence, the solution. Formally $u(x,g(x))=f(x)$
Now, we can face to two problems: one, the curve doesn't cut every projection, two, the curve cuts some projections more than once. For the first case, there are many solution as some of the characteristics are free. For the second case we have two or more different conditions on some projections, so is, two or more values for a single value of $x$ and $y$, having then no solution.
Being the curve for the boundary conditions $g(x)$ tangent at some point to one of the projections amounts that $g$ "bounces" back, cutting again some projections and/or leaving untouched others. Draw a family of parallel lines as the projection of the characteristics and an arc with any shape representing the curve for the boundary conditions, you will see that it cuts twice some projections because is tangent to one projection at some point.
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
$endgroup$
To fully determine the solution $u(x,y)$ we have to determine unambiguously every single characteristic curve. We can determine a single characteristic curve assigning to its projection on the plane $x;y$ the value for the function $u$ at one point. So, it have to be given some curve cutting every projection of the characteristics once and values assigned for $u$ along this curve. This is what is called boundary conditions and this determines every characteristic and hence, the solution. Formally $u(x,g(x))=f(x)$
Now, we can face to two problems: one, the curve doesn't cut every projection, two, the curve cuts some projections more than once. For the first case, there are many solution as some of the characteristics are free. For the second case we have two or more different conditions on some projections, so is, two or more values for a single value of $x$ and $y$, having then no solution.
Being the curve for the boundary conditions $g(x)$ tangent at some point to one of the projections amounts that $g$ "bounces" back, cutting again some projections and/or leaving untouched others. Draw a family of parallel lines as the projection of the characteristics and an arc with any shape representing the curve for the boundary conditions, you will see that it cuts twice some projections because is tangent to one projection at some point.
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
answered Jan 29 at 17:00
Rafa BudríaRafa Budría
5,9121825
5,9121825
add a comment |
add a comment |
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