existence of solutions for Cauchy problems
$begingroup$
Consider the equation
$$(1-cos x)u_{tt} - u_{tx} - u_{xx} = 0$$
with Cauchy data
$$u(x,0) = f(x), u_t(x,0) = g(x),text{ for } f,ginmathcal{C}^2$$
What compatibility condition do $f$ and $g$ have to satisfy for this problem to have a $mathcal{C}^2$ solution in a neighborhood of $t = 0$?
My idea is that rewrite the problem into a system of equations:
$$left( begin{array}{ccc}0 &1 &0\ 0 &0 &1\ 1-cos x &-1 &-1end{array} right) left( begin{array}{ccc} u_{tt}\ u_{tx}\ u_{xx} end{array} right) = left( begin{array}{ccc} g'\ f'' \ 0 end{array} right)$$
And the determinant is $0$: $cos x - 1 = 0$, when $x=2kpi$. So the determinant is not identically $0$. I'm stuck here and don't know what to do when the determinant is not identically $0$. Any ideas?
pde characteristics cauchy-problem
asked Dec 1 '18 at 15:19
QD666QD666
1276
$endgroup$
add a comment |
$begingroup$
Consider the equation
$$(1-cos x)u_{tt} - u_{tx} - u_{xx} = 0$$
with Cauchy data
$$u(x,0) = f(x), u_t(x,0) = g(x),text{ for } f,ginmathcal{C}^2$$
What compatibility condition do $f$ and $g$ have to satisfy for this problem to have a $mathcal{C}^2$ solution in a neighborhood of $t = 0$?
My idea is that rewrite the problem into a system of equations:
$$left( begin{array}{ccc}0 &1 &0\ 0 &0 &1\ 1-cos x &-1 &-1end{array} right) left( begin{array}{ccc} u_{tt}\ u_{tx}\ u_{xx} end{array} right) = left( begin{array}{ccc} g'\ f'' \ 0 end{array} right)$$
And the determinant is $0$: $cos x - 1 = 0$, when $x=2kpi$. So the determinant is not identically $0$. I'm stuck here and don't know what to do when the determinant is not identically $0$. Any ideas?
pde characteristics cauchy-problem
asked Dec 1 '18 at 15:19
QD666QD666
1276
$endgroup$
add a comment |
$begingroup$
Consider the equation
$$(1-cos x)u_{tt} - u_{tx} - u_{xx} = 0$$
with Cauchy data
$$u(x,0) = f(x), u_t(x,0) = g(x),text{ for } f,ginmathcal{C}^2$$
What compatibility condition do $f$ and $g$ have to satisfy for this problem to have a $mathcal{C}^2$ solution in a neighborhood of $t = 0$?
My idea is that rewrite the problem into a system of equations:
$$left( begin{array}{ccc}0 &1 &0\ 0 &0 &1\ 1-cos x &-1 &-1end{array} right) left( begin{array}{ccc} u_{tt}\ u_{tx}\ u_{xx} end{array} right) = left( begin{array}{ccc} g'\ f'' \ 0 end{array} right)$$
And the determinant is $0$: $cos x - 1 = 0$, when $x=2kpi$. So the determinant is not identically $0$. I'm stuck here and don't know what to do when the determinant is not identically $0$. Any ideas?
pde characteristics cauchy-problem
asked Dec 1 '18 at 15:19
QD666QD666
1276
$endgroup$
Consider the equation
$$(1-cos x)u_{tt} - u_{tx} - u_{xx} = 0$$
with Cauchy data
$$u(x,0) = f(x), u_t(x,0) = g(x),text{ for } f,ginmathcal{C}^2$$
What compatibility condition do $f$ and $g$ have to satisfy for this problem to have a $mathcal{C}^2$ solution in a neighborhood of $t = 0$?
My idea is that rewrite the problem into a system of equations:
$$left( begin{array}{ccc}0 &1 &0\ 0 &0 &1\ 1-cos x &-1 &-1end{array} right) left( begin{array}{ccc} u_{tt}\ u_{tx}\ u_{xx} end{array} right) = left( begin{array}{ccc} g'\ f'' \ 0 end{array} right)$$
And the determinant is $0$: $cos x - 1 = 0$, when $x=2kpi$. So the determinant is not identically $0$. I'm stuck here and don't know what to do when the determinant is not identically $0$. Any ideas?
pde characteristics cauchy-problem
pde characteristics cauchy-problem
asked Dec 1 '18 at 15:19
QD666QD666
1276
asked Dec 1 '18 at 15:19
QD666QD666
1276
edited Dec 1 '18 at 15:50
QD666
asked Dec 1 '18 at 15:19
QD666QD666
1276
asked Dec 1 '18 at 15:19
QD666QD666
1276
asked Dec 1 '18 at 15:19
QD666QD666
1276
1276
add a comment |
add a comment |
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