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Hello, I have the following 4 PDEs which I am trying to solve for $G(x,y,z)$ :

(1) $G_{xy}=0$

(2) $G_{xz}=0$

(3) $G_{yz}=0$

(4) $G_{xx}-G_{yy}=0$.

It is not hard to see that the general solution is:

$$G(x,y,z)=C_1+C_2x+C_3y+C_4(x^2+y^2)+F_1(z).$$

Here $C$ represents constants and $F_1$ a function.

I am trying to approach this solution as follows: Do $(1)_{x}$-$(4)_{y}$ to get $G_{yyy}(x,y,z)=0$. Now consider this along the $y$-axis, i.e. $G_{yyy}(0,y,0)=0$. This in a third order ODE so its solution $G(0,y,0)$ involves 3 constants. Next integrate (1) w.r.t $y$ and get an arbitrary function of $x$ and $z$. Consider this function along the $y$-axis $(x=z=0)$ and it becomes an arbitrary constant i.e. $G_x(0,y,0)$ is a constant. So we end up with Cauchy data along the $y$ axis in terms of 4 constants. This Cauchy data together with equation (4) allows us to solve for $G(x,y,0)$, that is, to get the solution on the $x$-$y$ plane.

My next step was to extend this to the $z$ axis to get $G(x,y,z)$. Now for Cauchy data I need $G(x,y,0)$ which I already have, and the normal derivative $G(x,y,0)_z$ . Using equations (2) and (3) we can get $G(x,y,0)_z$ in terms of one additional constant, to a total of 5 constants. Now (finally...) to my question!

Given $G(x,y,0)$ and $G(x,y,0)_z$ I have Cauchy data in the $x$-$y$ plane that can be used to solve a PDE for $G(x,y,z)$, but I guess non of equations (1)..(4) or their combinations are that (hyperbolic) PDE since the general solution contains an arbitrary function $F_1(z)$. Are equations (1)..(4) parabolic? is the problem that the Cauchy data is on a charateristic surface?? I need a little help explaining this. Finally, please note that if I add a fifth equation (5) $G_{xx}-G_{zz}=0$ we get:

$$G(x,y,z)=C_1+C_2x+C_3y+C_4z+C_5(x^2+y^2+z^2)$$

that is, no arbitrary functions, just constants.

Is it trivial to conclude that $z=0$ is a charateristic plane to equations (1)-(4) but not to the fifth one? Here is a procedure I found which does not seem so trivial:

(a) Consider a charateristic surface $P(x,y,z)$. we have:$A_{ij}P_iP_j=0$ where $A$ is a 3X3 matrix containing the coefficients of the second order derivatives. The subscripts of$P$ denote differentiation and the summation is implied.

(b) Consider eq. $G_{yz}=0$ and a characteristic surface $z=N(x,y)$. Using (a) we get $A_{23}=A_{32}=1$ yielding $2N_x=0$2N_y=0$. We get$N(x,y)=F(y)=z$N(x,y)=F(x)=z$ with $F$ arbitrary function. Take $F=0$ and get $z=0$ char surface.

(c) $G_{xz}$ gives $z=0$ char surface the same way.

(d) For $G_{xy}$ I got $F(z)=x$ with $F$ arbitrary. Choosing it large can take us as close as we want to $z=0$, e.g. take $F(z)=10^{10}z$. So, again $z=0$ is a charactristic.

(e) Similar arguments apply for $G_{xx}-G_{yy}$

(f) Finally regarding the fifth equation: Using this procedure, the equation $G_{xx}-G_{zz}$ does not give $z=0$ as characteristic surface, and therefore we can use the Cauchy data there to generate the full space solution.

Does this make sense or is this over complicating things?

Thanks

Uri

15 deleted 1 characters in body

Hello, I have the following 4 PDEs which I am trying to solve for $G(x,y,z)$ :

(1) $G_{xy}=0$

(2) $G_{xz}=0$

(3) $G_{yz}=0$

(4) $G_{xx}-G_{yy}=0$.

It is not hard to see that the general solution is:

$$G(x,y,z)=C_1+C_2x+C_3y+C_4(x^2+y^2)+F_1(z).$$

Here $C$ represents constants and $F_1$ a function.

I am trying to approach this solution as follows: Do $(1)_{x}$-$(4)_{y}$ to get $G_{yyy}(x,y,z)=0$. Now consider this along the $y$-axis, i.e. $G_{yyy}(0,y,0)=0$. This in a third order ODE so its solution $G(0,y,0)$ involves 3 constants. Next integrate (1) w.r.t $y$ and get an arbitrary function of $x$ and $z$. Consider this function along the $y$-axis $(x=z=0)$ and it becomes an arbitrary constant i.e. $G_x(0,y,0)$ is a constant. So we end up with Cauchy data along the $y$ axis in terms of 4 constants. This Cauchy data together with equation (4) allows us to solve for $G(x,y,0)$, that is, to get the solution on the $x$-$y$ plane.

My next step was to extend this to the $z$ axis to get $G(x,y,z)$. Now for Cauchy data I need $G(x,y,0)$ which I already have, and the normal derivative $G(x,y,0)_z$ . Using equations (2) and (3) we can get $G(x,y,0)_z$ in terms of one additional constant, to a total of 5 constants. Now (finally...) to my question!

Given $G(x,y,0)$ and $G(x,y,0)_z$ I have Cauchy data in the $x$-$y$ plane that can be used to solve a PDE for $G(x,y,z)$, but I guess non of equations (1)..(4) or their combinations are that (hyperbolic) PDE since the general solution contains an arbitrary function $F_1(z)$. Are equations (1)..(4) parabolic? is the problem that the Cauchy data is on a charateristic surface?? I need a little help explaining this. Finally, please note that if I add a fifth equation (5) $G_{xx}-G_{zz}=0$ we get:

$$G(x,y,z)=C_1+C_2x+C_3y+C_4z+C_5(x^2+y^2+z^2)$$

that is, no arbitrary functions, just constants.

Is it trivial to conclude that $z=0$ is a charateristic plane to equations (1)-(4) but not to the fifth one? Here is a procedure I found which does not seem so trivial:

(a) Consider a charateristic surface $P(x,y,z)$. we have:$A_{ij}P_iP_j=0$ where $A$ is a 3X3 matrix containing the coefficients of the second order derivatives. The subscripts of$P$ denote differentiation and the summation is implied.

(b) Consider eq. $G_{yz}=0$ and a characteristic surface $z=N(x,y)$. Using (a) we get $A_{23}=$A_{32}=1$A_{23}=A_{32}=1$ yielding $2N_x=0$. We get $N(x,y)=F(y)=z$ with $F$ arbitrary function. Take $F=0$ and get $z=0$ char surface.

(c) $G_{xz}$ gives $z=0$ char surface the same way.

(d) For $G_{xy}$ I got $F(z)=x$ with $F$ arbitrary. Choosing it large can take us as close as we want to $z=0$, e.g. take $F(z)=10^{10}z$. So, again $z=0$ is a charactristic.

(e) Similar arguments apply for $G_{xx}-G_{yy}$

(f) Finally regarding the fifth equation: Using this procedure, the equation $G_{xx}-G_{zz}$ does not give $z=0$ as characteristic surface, and therefore we can use the Cauchy data there to generate the full space solution.

Does this make sense or is this over complicating things?

Thanks

Uri

14 added 107 characters in body; deleted 19 characters in body

Hello, I have the following 4 PDEs which I am trying to solve for $G(x,y,z)$ :

(1) $G_{xy}=0$

(2) $G_{xz}=0$

(3) $G_{yz}=0$

(4) $G_{xx}-G_{yy}=0$.

It is not hard to see that the general solution is:

$$G(x,y,z)=C_1+C_2x+C_3y+C_4(x^2+y^2)+F_1(z).$$

Here $C$ represents constants and $F_1$ a function.

I am trying to approach this solution as follows: Do $(1)_{x}$-$(4)_{y}$ to get $G_{yyy}(x,y,z)=0$. Now consider this along the $y$-axis, i.e. $G_{yyy}(0,y,0)=0$. This in a third order ODE so its solution $G(0,y,0)$ involves 3 constants. Next integrate (1) w.r.t $y$ and get an arbitrary function of $x$ and $z$. Consider this function along the $y$-axis $(x=z=0)$ and it becomes an arbitrary constant i.e. $G_x(0,y,0)$ is a constant. So we end up with Cauchy data along the $y$ axis in terms of 4 constants. This Cauchy data together with equation (4) allows us to solve for $G(x,y,0)$, that is, to get the solution on the $x$-$y$ plane.

My next step was to extend this to the $z$ axis to get $G(x,y,z)$. Now for Cauchy data I need $G(x,y,0)$ which I already have, and the normal derivative $G(x,y,0)_z$ . Using equations (2) and (3) we can get $G(x,y,0)_z$ in terms of one additional constant, to a total of 5 constants. Now (finally...) to my question!

Given $G(x,y,0)$ and $G(x,y,0)_z$ I have Cauchy data in the $x$-$y$ plane that can be used to solve a PDE for $G(x,y,z)$, but I guess non of equations (1)..(4) or their combinations are that (hyperbolic) PDE since the general solution contains an arbitrary function $F_1(z)$. Are equations (1)..(4) parabolic? is the problem that the Cauchy data is on a charateristic surface?? I need a little help explaining this. Finally, please note that if I add a fifth equation (5) $G_{xx}-G_{zz}=0$ we get:

$$G(x,y,z)=C_1+C_2x+C_3y+C_4z+C_5(x^2+y^2+z^2)$$

that is, no arbitrary functions, just constants. Thanks!!

Uri

Is it trivial to conclude that $z=0$ is a charateristic plane to equations (1)-(4) but not to the fifth one? Here is a procedure I found which does not seem so trivial:

(a) Consider a charateristic surface $P(x,y,z)$. we have:$A_{ij}P_iP_j=0$ where $A$ is a 3X3 matrix , $P$ containing the coefficients of the second order derivatives. The subscripts of$P$ denote differentiation and the summation is implied.

(b) Consider eq. $G_{yz}=0$ and a char characteristic surface $z=N(x,y)$. Using (a) we get $A_{23}=$A_{32}=1$ yielding $2N_x=0$ so we have . We get$N(x,y)=F(y)=z$with$F$arbitrary function. Take$F=0$and get$z=0$char surface. (c) $G_{xz}$ gives$z=0$char surface the same way. (d) For $G_{xy}$ I got $F(z)=x$ with$F$arbitrary. Choosing it large can take us as close as we want to$z=0$, e.g. take $F(z)=10^{10}z$. So, again$z=0$is a charactristic. (e) Similar arguments apply for $G_{xx}-G_{yy}$ (f) Finally regarding the fifth equation: Using this procedure, the equation $G_{xx}-G_{zz}$ does not give$z=0\$ as characteristic surface, and therefore we can use the Cauchy data there to generate the full space solution.

Does this make sense or is this over complicating things?

Thanks

Uri

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