Mixing
Solving a PDE first order in three variables
$begingroup$
Solve the following PDE
$$ u_{x_1} + u_{x_2} + x_3 u_{x_3} = u^3 $$
with cauchy data $u(x_1,x_2,1) = phi(x_1,x_2) $. Also, determine the values of $x_1,x_2$ and $x_3$ for which the IVP exists.
Try:
The characteristics are given by
$$ begin{align*}
frac{dx_1}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_1 \
frac{dx_2}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_2 \
frac{dx_3}{ds} &= x_3, ; ; ; x(r_1,r_2,0) = 1 \
frac{d z}{ds} &= z^3, ; ; ; x(r_1,r_2,0) = h(r_1,r_2) \
end{align*}$$
We see easily that $x_1(r_1,r_2,s) = s+r_1$ and similarly $x_2 = s + r_2$ and $x_3 = e^s $. Next, we have
$$ frac{dz}{z^3} = ds implies - frac{1}{2 z^2} = s + C implies -frac{1}{2h^2} = C $$
Therefore,
$$ z(r_1,r_2,s) = frac{ 2 |h(r_1,r_2)| }{sqrt{1 - 2h(r_1,r_2)} } $$
also, notice that $s = log x_3 $ and so $r_1 = x_1 - log x_3$ and $r_2 = x_2 - log x_3$. Therefore, our solution is
$$ u(x_1,x_2,x_3) = frac{ 2 | h( x_1 - log x_3, x_2 - log x_3) |}{sqrt{1-2h( x_1 - log x_3, x_2 - log x_3)}} $$
and the solution exists for all $(x_1,x_2,x_3)$ since
$$ Jac(x_1,x_2,x_3) = 1 neq 0 $$
Is this a correct solution?
pde
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
$endgroup$
add a comment |
$begingroup$
Solve the following PDE
$$ u_{x_1} + u_{x_2} + x_3 u_{x_3} = u^3 $$
with cauchy data $u(x_1,x_2,1) = phi(x_1,x_2) $. Also, determine the values of $x_1,x_2$ and $x_3$ for which the IVP exists.
Try:
The characteristics are given by
$$ begin{align*}
frac{dx_1}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_1 \
frac{dx_2}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_2 \
frac{dx_3}{ds} &= x_3, ; ; ; x(r_1,r_2,0) = 1 \
frac{d z}{ds} &= z^3, ; ; ; x(r_1,r_2,0) = h(r_1,r_2) \
end{align*}$$
We see easily that $x_1(r_1,r_2,s) = s+r_1$ and similarly $x_2 = s + r_2$ and $x_3 = e^s $. Next, we have
$$ frac{dz}{z^3} = ds implies - frac{1}{2 z^2} = s + C implies -frac{1}{2h^2} = C $$
Therefore,
$$ z(r_1,r_2,s) = frac{ 2 |h(r_1,r_2)| }{sqrt{1 - 2h(r_1,r_2)} } $$
also, notice that $s = log x_3 $ and so $r_1 = x_1 - log x_3$ and $r_2 = x_2 - log x_3$. Therefore, our solution is
$$ u(x_1,x_2,x_3) = frac{ 2 | h( x_1 - log x_3, x_2 - log x_3) |}{sqrt{1-2h( x_1 - log x_3, x_2 - log x_3)}} $$
and the solution exists for all $(x_1,x_2,x_3)$ since
$$ Jac(x_1,x_2,x_3) = 1 neq 0 $$
Is this a correct solution?
pde
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
$endgroup$
add a comment |
$begingroup$
Solve the following PDE
$$ u_{x_1} + u_{x_2} + x_3 u_{x_3} = u^3 $$
with cauchy data $u(x_1,x_2,1) = phi(x_1,x_2) $. Also, determine the values of $x_1,x_2$ and $x_3$ for which the IVP exists.
Try:
The characteristics are given by
$$ begin{align*}
frac{dx_1}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_1 \
frac{dx_2}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_2 \
frac{dx_3}{ds} &= x_3, ; ; ; x(r_1,r_2,0) = 1 \
frac{d z}{ds} &= z^3, ; ; ; x(r_1,r_2,0) = h(r_1,r_2) \
end{align*}$$
We see easily that $x_1(r_1,r_2,s) = s+r_1$ and similarly $x_2 = s + r_2$ and $x_3 = e^s $. Next, we have
$$ frac{dz}{z^3} = ds implies - frac{1}{2 z^2} = s + C implies -frac{1}{2h^2} = C $$
Therefore,
$$ z(r_1,r_2,s) = frac{ 2 |h(r_1,r_2)| }{sqrt{1 - 2h(r_1,r_2)} } $$
also, notice that $s = log x_3 $ and so $r_1 = x_1 - log x_3$ and $r_2 = x_2 - log x_3$. Therefore, our solution is
$$ u(x_1,x_2,x_3) = frac{ 2 | h( x_1 - log x_3, x_2 - log x_3) |}{sqrt{1-2h( x_1 - log x_3, x_2 - log x_3)}} $$
and the solution exists for all $(x_1,x_2,x_3)$ since
$$ Jac(x_1,x_2,x_3) = 1 neq 0 $$
Is this a correct solution?
pde
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
$endgroup$
Solve the following PDE
$$ u_{x_1} + u_{x_2} + x_3 u_{x_3} = u^3 $$
with cauchy data $u(x_1,x_2,1) = phi(x_1,x_2) $. Also, determine the values of $x_1,x_2$ and $x_3$ for which the IVP exists.
Try:
The characteristics are given by
$$ begin{align*}
frac{dx_1}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_1 \
frac{dx_2}{ds} &= 1, ; ; ; x(r_1,r_2,0) = r_2 \
frac{dx_3}{ds} &= x_3, ; ; ; x(r_1,r_2,0) = 1 \
frac{d z}{ds} &= z^3, ; ; ; x(r_1,r_2,0) = h(r_1,r_2) \
end{align*}$$
We see easily that $x_1(r_1,r_2,s) = s+r_1$ and similarly $x_2 = s + r_2$ and $x_3 = e^s $. Next, we have
$$ frac{dz}{z^3} = ds implies - frac{1}{2 z^2} = s + C implies -frac{1}{2h^2} = C $$
Therefore,
$$ z(r_1,r_2,s) = frac{ 2 |h(r_1,r_2)| }{sqrt{1 - 2h(r_1,r_2)} } $$
also, notice that $s = log x_3 $ and so $r_1 = x_1 - log x_3$ and $r_2 = x_2 - log x_3$. Therefore, our solution is
$$ u(x_1,x_2,x_3) = frac{ 2 | h( x_1 - log x_3, x_2 - log x_3) |}{sqrt{1-2h( x_1 - log x_3, x_2 - log x_3)}} $$
and the solution exists for all $(x_1,x_2,x_3)$ since
$$ Jac(x_1,x_2,x_3) = 1 neq 0 $$
Is this a correct solution?
pde
pde
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
asked Jan 19 at 6:42
Jimmy SabaterJimmy Sabater
2,791324
2,791324
add a comment |
add a comment |
1 Answer
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$begingroup$
You have the right idea; you've just wrong somewhere along the way in forgetting a square and dropping an $s$ it looks like. You can always check if your result is a solution or not by directly applying the PDE to it and checking the condition(s).
begin{cases}
u_x + u_y + z u_z = u^3 \
u(x, y, 1) = phi(x, y).
end{cases}
We have the characteristic
begin{cases}
begin{align}
displaystyle &frac{dx}{dt} = 1 &x(0) &= s_1 &implies x(t) &= s_1 + t \
displaystyle &frac{dy}{dt} = 1 &y(0) &= s_2 &implies y(t) &= s_2 + t \
displaystyle &frac{dz}{dt} = z &z(0) &= 1 &implies z(t) &= e^t \
displaystyle &frac{dtilde{u}}{dt} = tilde{u}^3 &tilde{u}(0) &= phibig( s_1, s_2 big) &implies tilde{u}(t) &= sqrt{frac{phi^2(s_1, s_2)}{1-2phi^2(s_1, s_2) t}}.
end{align}
end{cases}
The system is readily solvable for ${t, s_1, s_2}.$
$$u(x,y,z) = sqrt{frac{phi^2(x - log z, y - log z)}{1 - 2phi^2(x - log z, y - log z)log z}}.$$
For the Jacobian, I get
$$J = z.$$
However, unless using the complex-valued logarithm, I would restrict $z$ to be positive.
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have the right idea; you've just wrong somewhere along the way in forgetting a square and dropping an $s$ it looks like. You can always check if your result is a solution or not by directly applying the PDE to it and checking the condition(s).
begin{cases}
u_x + u_y + z u_z = u^3 \
u(x, y, 1) = phi(x, y).
end{cases}
We have the characteristic
begin{cases}
begin{align}
displaystyle &frac{dx}{dt} = 1 &x(0) &= s_1 &implies x(t) &= s_1 + t \
displaystyle &frac{dy}{dt} = 1 &y(0) &= s_2 &implies y(t) &= s_2 + t \
displaystyle &frac{dz}{dt} = z &z(0) &= 1 &implies z(t) &= e^t \
displaystyle &frac{dtilde{u}}{dt} = tilde{u}^3 &tilde{u}(0) &= phibig( s_1, s_2 big) &implies tilde{u}(t) &= sqrt{frac{phi^2(s_1, s_2)}{1-2phi^2(s_1, s_2) t}}.
end{align}
end{cases}
The system is readily solvable for ${t, s_1, s_2}.$
$$u(x,y,z) = sqrt{frac{phi^2(x - log z, y - log z)}{1 - 2phi^2(x - log z, y - log z)log z}}.$$
For the Jacobian, I get
$$J = z.$$
However, unless using the complex-valued logarithm, I would restrict $z$ to be positive.
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
$begingroup$
You have the right idea; you've just wrong somewhere along the way in forgetting a square and dropping an $s$ it looks like. You can always check if your result is a solution or not by directly applying the PDE to it and checking the condition(s).
begin{cases}
u_x + u_y + z u_z = u^3 \
u(x, y, 1) = phi(x, y).
end{cases}
We have the characteristic
begin{cases}
begin{align}
displaystyle &frac{dx}{dt} = 1 &x(0) &= s_1 &implies x(t) &= s_1 + t \
displaystyle &frac{dy}{dt} = 1 &y(0) &= s_2 &implies y(t) &= s_2 + t \
displaystyle &frac{dz}{dt} = z &z(0) &= 1 &implies z(t) &= e^t \
displaystyle &frac{dtilde{u}}{dt} = tilde{u}^3 &tilde{u}(0) &= phibig( s_1, s_2 big) &implies tilde{u}(t) &= sqrt{frac{phi^2(s_1, s_2)}{1-2phi^2(s_1, s_2) t}}.
end{align}
end{cases}
The system is readily solvable for ${t, s_1, s_2}.$
$$u(x,y,z) = sqrt{frac{phi^2(x - log z, y - log z)}{1 - 2phi^2(x - log z, y - log z)log z}}.$$
For the Jacobian, I get
$$J = z.$$
However, unless using the complex-valued logarithm, I would restrict $z$ to be positive.
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
$begingroup$
You have the right idea; you've just wrong somewhere along the way in forgetting a square and dropping an $s$ it looks like. You can always check if your result is a solution or not by directly applying the PDE to it and checking the condition(s).
begin{cases}
u_x + u_y + z u_z = u^3 \
u(x, y, 1) = phi(x, y).
end{cases}
We have the characteristic
begin{cases}
begin{align}
displaystyle &frac{dx}{dt} = 1 &x(0) &= s_1 &implies x(t) &= s_1 + t \
displaystyle &frac{dy}{dt} = 1 &y(0) &= s_2 &implies y(t) &= s_2 + t \
displaystyle &frac{dz}{dt} = z &z(0) &= 1 &implies z(t) &= e^t \
displaystyle &frac{dtilde{u}}{dt} = tilde{u}^3 &tilde{u}(0) &= phibig( s_1, s_2 big) &implies tilde{u}(t) &= sqrt{frac{phi^2(s_1, s_2)}{1-2phi^2(s_1, s_2) t}}.
end{align}
end{cases}
The system is readily solvable for ${t, s_1, s_2}.$
$$u(x,y,z) = sqrt{frac{phi^2(x - log z, y - log z)}{1 - 2phi^2(x - log z, y - log z)log z}}.$$
For the Jacobian, I get
$$J = z.$$
However, unless using the complex-valued logarithm, I would restrict $z$ to be positive.
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
$endgroup$
You have the right idea; you've just wrong somewhere along the way in forgetting a square and dropping an $s$ it looks like. You can always check if your result is a solution or not by directly applying the PDE to it and checking the condition(s).
begin{cases}
u_x + u_y + z u_z = u^3 \
u(x, y, 1) = phi(x, y).
end{cases}
We have the characteristic
begin{cases}
begin{align}
displaystyle &frac{dx}{dt} = 1 &x(0) &= s_1 &implies x(t) &= s_1 + t \
displaystyle &frac{dy}{dt} = 1 &y(0) &= s_2 &implies y(t) &= s_2 + t \
displaystyle &frac{dz}{dt} = z &z(0) &= 1 &implies z(t) &= e^t \
displaystyle &frac{dtilde{u}}{dt} = tilde{u}^3 &tilde{u}(0) &= phibig( s_1, s_2 big) &implies tilde{u}(t) &= sqrt{frac{phi^2(s_1, s_2)}{1-2phi^2(s_1, s_2) t}}.
end{align}
end{cases}
The system is readily solvable for ${t, s_1, s_2}.$
$$u(x,y,z) = sqrt{frac{phi^2(x - log z, y - log z)}{1 - 2phi^2(x - log z, y - log z)log z}}.$$
For the Jacobian, I get
$$J = z.$$
However, unless using the complex-valued logarithm, I would restrict $z$ to be positive.
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
answered Jan 19 at 8:44
AEngineerAEngineer
1,5441317
1,5441317
add a comment |
add a comment |
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