Question about characteristics and classification of second-order PDEs











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I am currently reading through the book 'Computational Techniques for Fluid Dynamics', by C.A.J. Fletcher. Chapter 2 discusses classification of PDEs by finding the number and nature of their characteristics. However, there is a section about finding characteristics of second-order PDEs (2.1.3), which I am a little confused about.



They give a generalized second-order PDE as:



$$Afrac{partial^2u}{partial x^2}+Bfrac{partial^2u}{partial xpartial y}+Cfrac{partial^2u}{partial y^2}+H=0tag{1}$$



where $A$, $B$ and $C$ are functions of $x,y$ and $H$ contains all the first-derivative terms. They then introduce some new variables to simplify the notation:



$$P=frac{partial u}{partial x}, Q=frac{partial u}{partial y}, R=frac{partial^2 u}{partial x^2}, S=frac{partial^2 u}{partial xpartial y}, T=frac{partial^2 u}{partial y^2}$$



They then state that a curve K is introduced and along a tangent to K, the differentials for $P$ and $Q$ satisfy:



$$dP=Rdx+Sdytag{2}$$



$$dQ=Sdx+Tdytag{3}$$



Using the substitutions above, the original PDE can be written:



$$AR+BS+CT+H=0$$



Equations (2) and (3) are then used to eliminate $R$ and $T$ in the above equation, which is then re-arranged to give the following:



$$SBigl[ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+CBigr]-Bigl{Bigl[ABigl(frac{dP}{dx}Bigr)+HBigr]frac{dy}{dx}+Cfrac{dQ}{dx}Bigr}=0tag{4}$$



It then states that if:



$$ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+C=0tag{5}$$



then that eliminates the left-hand term in equation (4), which yields a simpler relationship between $frac{dP}{dx}$ and $frac{dQ}{dx}$. The solutions of equation (5) define the characteristic curves for the PDE.



So, here is where I am confused: why did they choose to split up equation (4) in that manner? As I understand it, the goal of finding characteristic curves is to reduce a PDE to a total differential, so that it can be more easily solved. However, how has this goal been met, when the terms $P$, $Q$ and $H$ on the right-hand side of equation (4) still contain partial differentials? Given that it still contains partial differentials, how is the reduced form of the equation more useful and why does equation (5) provide the characteristic curves?



Thanks in advance!










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    The third coefficient in (1) should have been $C$, I believe. For a linear PDO, it is the principal (highest order homogeneous) part that defines the nature of the operator; you refer to lower order terms only if the principal symbol is semidefinite. Therefore you shouldn't be surprized that $H$ has no role in characteristics. I am not familiar with the book you mention, but I suggest to think about diagonalizing your principal symbol in the first place. If you are a physicist, try to set up a Hamiltonian dynamics with Hamiltonian being the principal symbol. Trajectories are characteristics.
    – Bedovlat
    yesterday










  • @Bedovlat thanks for catching that typo, I have now corrected it. Yes, I know that $H$ contains the first-order and lower terms and so is not relevant for the classification. I think I would have understood, if their substitutions had reduced it to a first-order pde; however, surely the $frac{dP}{dx}$ and $frac{dQ}{dx}$ are second-order terms (in $u$), so it doesn't seem to have even done that. I have to admit I'm not familiar with Hamiltonian Mechanics; however, I will read up on it, if you think it may help my understanding.
    – Time4Tea
    yesterday






  • 1




    Crossposted to physics.stackexchange.com/q/443899/2451
    – Qmechanic
    5 hours ago















up vote
23
down vote

favorite
5












I am currently reading through the book 'Computational Techniques for Fluid Dynamics', by C.A.J. Fletcher. Chapter 2 discusses classification of PDEs by finding the number and nature of their characteristics. However, there is a section about finding characteristics of second-order PDEs (2.1.3), which I am a little confused about.



They give a generalized second-order PDE as:



$$Afrac{partial^2u}{partial x^2}+Bfrac{partial^2u}{partial xpartial y}+Cfrac{partial^2u}{partial y^2}+H=0tag{1}$$



where $A$, $B$ and $C$ are functions of $x,y$ and $H$ contains all the first-derivative terms. They then introduce some new variables to simplify the notation:



$$P=frac{partial u}{partial x}, Q=frac{partial u}{partial y}, R=frac{partial^2 u}{partial x^2}, S=frac{partial^2 u}{partial xpartial y}, T=frac{partial^2 u}{partial y^2}$$



They then state that a curve K is introduced and along a tangent to K, the differentials for $P$ and $Q$ satisfy:



$$dP=Rdx+Sdytag{2}$$



$$dQ=Sdx+Tdytag{3}$$



Using the substitutions above, the original PDE can be written:



$$AR+BS+CT+H=0$$



Equations (2) and (3) are then used to eliminate $R$ and $T$ in the above equation, which is then re-arranged to give the following:



$$SBigl[ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+CBigr]-Bigl{Bigl[ABigl(frac{dP}{dx}Bigr)+HBigr]frac{dy}{dx}+Cfrac{dQ}{dx}Bigr}=0tag{4}$$



It then states that if:



$$ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+C=0tag{5}$$



then that eliminates the left-hand term in equation (4), which yields a simpler relationship between $frac{dP}{dx}$ and $frac{dQ}{dx}$. The solutions of equation (5) define the characteristic curves for the PDE.



So, here is where I am confused: why did they choose to split up equation (4) in that manner? As I understand it, the goal of finding characteristic curves is to reduce a PDE to a total differential, so that it can be more easily solved. However, how has this goal been met, when the terms $P$, $Q$ and $H$ on the right-hand side of equation (4) still contain partial differentials? Given that it still contains partial differentials, how is the reduced form of the equation more useful and why does equation (5) provide the characteristic curves?



Thanks in advance!










share|cite|improve this question

















This question has an open bounty worth +100
reputation from amWhy ending in 6 days.


This question has not received enough attention.












  • 1




    The third coefficient in (1) should have been $C$, I believe. For a linear PDO, it is the principal (highest order homogeneous) part that defines the nature of the operator; you refer to lower order terms only if the principal symbol is semidefinite. Therefore you shouldn't be surprized that $H$ has no role in characteristics. I am not familiar with the book you mention, but I suggest to think about diagonalizing your principal symbol in the first place. If you are a physicist, try to set up a Hamiltonian dynamics with Hamiltonian being the principal symbol. Trajectories are characteristics.
    – Bedovlat
    yesterday










  • @Bedovlat thanks for catching that typo, I have now corrected it. Yes, I know that $H$ contains the first-order and lower terms and so is not relevant for the classification. I think I would have understood, if their substitutions had reduced it to a first-order pde; however, surely the $frac{dP}{dx}$ and $frac{dQ}{dx}$ are second-order terms (in $u$), so it doesn't seem to have even done that. I have to admit I'm not familiar with Hamiltonian Mechanics; however, I will read up on it, if you think it may help my understanding.
    – Time4Tea
    yesterday






  • 1




    Crossposted to physics.stackexchange.com/q/443899/2451
    – Qmechanic
    5 hours ago













up vote
23
down vote

favorite
5









up vote
23
down vote

favorite
5






5





I am currently reading through the book 'Computational Techniques for Fluid Dynamics', by C.A.J. Fletcher. Chapter 2 discusses classification of PDEs by finding the number and nature of their characteristics. However, there is a section about finding characteristics of second-order PDEs (2.1.3), which I am a little confused about.



They give a generalized second-order PDE as:



$$Afrac{partial^2u}{partial x^2}+Bfrac{partial^2u}{partial xpartial y}+Cfrac{partial^2u}{partial y^2}+H=0tag{1}$$



where $A$, $B$ and $C$ are functions of $x,y$ and $H$ contains all the first-derivative terms. They then introduce some new variables to simplify the notation:



$$P=frac{partial u}{partial x}, Q=frac{partial u}{partial y}, R=frac{partial^2 u}{partial x^2}, S=frac{partial^2 u}{partial xpartial y}, T=frac{partial^2 u}{partial y^2}$$



They then state that a curve K is introduced and along a tangent to K, the differentials for $P$ and $Q$ satisfy:



$$dP=Rdx+Sdytag{2}$$



$$dQ=Sdx+Tdytag{3}$$



Using the substitutions above, the original PDE can be written:



$$AR+BS+CT+H=0$$



Equations (2) and (3) are then used to eliminate $R$ and $T$ in the above equation, which is then re-arranged to give the following:



$$SBigl[ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+CBigr]-Bigl{Bigl[ABigl(frac{dP}{dx}Bigr)+HBigr]frac{dy}{dx}+Cfrac{dQ}{dx}Bigr}=0tag{4}$$



It then states that if:



$$ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+C=0tag{5}$$



then that eliminates the left-hand term in equation (4), which yields a simpler relationship between $frac{dP}{dx}$ and $frac{dQ}{dx}$. The solutions of equation (5) define the characteristic curves for the PDE.



So, here is where I am confused: why did they choose to split up equation (4) in that manner? As I understand it, the goal of finding characteristic curves is to reduce a PDE to a total differential, so that it can be more easily solved. However, how has this goal been met, when the terms $P$, $Q$ and $H$ on the right-hand side of equation (4) still contain partial differentials? Given that it still contains partial differentials, how is the reduced form of the equation more useful and why does equation (5) provide the characteristic curves?



Thanks in advance!










share|cite|improve this question















I am currently reading through the book 'Computational Techniques for Fluid Dynamics', by C.A.J. Fletcher. Chapter 2 discusses classification of PDEs by finding the number and nature of their characteristics. However, there is a section about finding characteristics of second-order PDEs (2.1.3), which I am a little confused about.



They give a generalized second-order PDE as:



$$Afrac{partial^2u}{partial x^2}+Bfrac{partial^2u}{partial xpartial y}+Cfrac{partial^2u}{partial y^2}+H=0tag{1}$$



where $A$, $B$ and $C$ are functions of $x,y$ and $H$ contains all the first-derivative terms. They then introduce some new variables to simplify the notation:



$$P=frac{partial u}{partial x}, Q=frac{partial u}{partial y}, R=frac{partial^2 u}{partial x^2}, S=frac{partial^2 u}{partial xpartial y}, T=frac{partial^2 u}{partial y^2}$$



They then state that a curve K is introduced and along a tangent to K, the differentials for $P$ and $Q$ satisfy:



$$dP=Rdx+Sdytag{2}$$



$$dQ=Sdx+Tdytag{3}$$



Using the substitutions above, the original PDE can be written:



$$AR+BS+CT+H=0$$



Equations (2) and (3) are then used to eliminate $R$ and $T$ in the above equation, which is then re-arranged to give the following:



$$SBigl[ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+CBigr]-Bigl{Bigl[ABigl(frac{dP}{dx}Bigr)+HBigr]frac{dy}{dx}+Cfrac{dQ}{dx}Bigr}=0tag{4}$$



It then states that if:



$$ABigl(frac{dy}{dx}Bigr)^2-BBigl(frac{dy}{dx}Bigr)+C=0tag{5}$$



then that eliminates the left-hand term in equation (4), which yields a simpler relationship between $frac{dP}{dx}$ and $frac{dQ}{dx}$. The solutions of equation (5) define the characteristic curves for the PDE.



So, here is where I am confused: why did they choose to split up equation (4) in that manner? As I understand it, the goal of finding characteristic curves is to reduce a PDE to a total differential, so that it can be more easily solved. However, how has this goal been met, when the terms $P$, $Q$ and $H$ on the right-hand side of equation (4) still contain partial differentials? Given that it still contains partial differentials, how is the reduced form of the equation more useful and why does equation (5) provide the characteristic curves?



Thanks in advance!







pde characteristics






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share|cite|improve this question













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edited yesterday

























asked Nov 16 at 22:05









Time4Tea

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This question has an open bounty worth +100
reputation from amWhy ending in 6 days.


This question has not received enough attention.








This question has an open bounty worth +100
reputation from amWhy ending in 6 days.


This question has not received enough attention.










  • 1




    The third coefficient in (1) should have been $C$, I believe. For a linear PDO, it is the principal (highest order homogeneous) part that defines the nature of the operator; you refer to lower order terms only if the principal symbol is semidefinite. Therefore you shouldn't be surprized that $H$ has no role in characteristics. I am not familiar with the book you mention, but I suggest to think about diagonalizing your principal symbol in the first place. If you are a physicist, try to set up a Hamiltonian dynamics with Hamiltonian being the principal symbol. Trajectories are characteristics.
    – Bedovlat
    yesterday










  • @Bedovlat thanks for catching that typo, I have now corrected it. Yes, I know that $H$ contains the first-order and lower terms and so is not relevant for the classification. I think I would have understood, if their substitutions had reduced it to a first-order pde; however, surely the $frac{dP}{dx}$ and $frac{dQ}{dx}$ are second-order terms (in $u$), so it doesn't seem to have even done that. I have to admit I'm not familiar with Hamiltonian Mechanics; however, I will read up on it, if you think it may help my understanding.
    – Time4Tea
    yesterday






  • 1




    Crossposted to physics.stackexchange.com/q/443899/2451
    – Qmechanic
    5 hours ago














  • 1




    The third coefficient in (1) should have been $C$, I believe. For a linear PDO, it is the principal (highest order homogeneous) part that defines the nature of the operator; you refer to lower order terms only if the principal symbol is semidefinite. Therefore you shouldn't be surprized that $H$ has no role in characteristics. I am not familiar with the book you mention, but I suggest to think about diagonalizing your principal symbol in the first place. If you are a physicist, try to set up a Hamiltonian dynamics with Hamiltonian being the principal symbol. Trajectories are characteristics.
    – Bedovlat
    yesterday










  • @Bedovlat thanks for catching that typo, I have now corrected it. Yes, I know that $H$ contains the first-order and lower terms and so is not relevant for the classification. I think I would have understood, if their substitutions had reduced it to a first-order pde; however, surely the $frac{dP}{dx}$ and $frac{dQ}{dx}$ are second-order terms (in $u$), so it doesn't seem to have even done that. I have to admit I'm not familiar with Hamiltonian Mechanics; however, I will read up on it, if you think it may help my understanding.
    – Time4Tea
    yesterday






  • 1




    Crossposted to physics.stackexchange.com/q/443899/2451
    – Qmechanic
    5 hours ago








1




1




The third coefficient in (1) should have been $C$, I believe. For a linear PDO, it is the principal (highest order homogeneous) part that defines the nature of the operator; you refer to lower order terms only if the principal symbol is semidefinite. Therefore you shouldn't be surprized that $H$ has no role in characteristics. I am not familiar with the book you mention, but I suggest to think about diagonalizing your principal symbol in the first place. If you are a physicist, try to set up a Hamiltonian dynamics with Hamiltonian being the principal symbol. Trajectories are characteristics.
– Bedovlat
yesterday




The third coefficient in (1) should have been $C$, I believe. For a linear PDO, it is the principal (highest order homogeneous) part that defines the nature of the operator; you refer to lower order terms only if the principal symbol is semidefinite. Therefore you shouldn't be surprized that $H$ has no role in characteristics. I am not familiar with the book you mention, but I suggest to think about diagonalizing your principal symbol in the first place. If you are a physicist, try to set up a Hamiltonian dynamics with Hamiltonian being the principal symbol. Trajectories are characteristics.
– Bedovlat
yesterday












@Bedovlat thanks for catching that typo, I have now corrected it. Yes, I know that $H$ contains the first-order and lower terms and so is not relevant for the classification. I think I would have understood, if their substitutions had reduced it to a first-order pde; however, surely the $frac{dP}{dx}$ and $frac{dQ}{dx}$ are second-order terms (in $u$), so it doesn't seem to have even done that. I have to admit I'm not familiar with Hamiltonian Mechanics; however, I will read up on it, if you think it may help my understanding.
– Time4Tea
yesterday




@Bedovlat thanks for catching that typo, I have now corrected it. Yes, I know that $H$ contains the first-order and lower terms and so is not relevant for the classification. I think I would have understood, if their substitutions had reduced it to a first-order pde; however, surely the $frac{dP}{dx}$ and $frac{dQ}{dx}$ are second-order terms (in $u$), so it doesn't seem to have even done that. I have to admit I'm not familiar with Hamiltonian Mechanics; however, I will read up on it, if you think it may help my understanding.
– Time4Tea
yesterday




1




1




Crossposted to physics.stackexchange.com/q/443899/2451
– Qmechanic
5 hours ago




Crossposted to physics.stackexchange.com/q/443899/2451
– Qmechanic
5 hours ago















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