Find good substitution for PDE.












0












$begingroup$


Consider system of : $begin{cases}
u_{tt} = vartriangle{u}
\
u|_{t = 0} = frac{1}{x^{2}+y^2+z^2}
\
u_t|_{t = 0} = 0
end{cases}$



I've thought about $v = x^{2} + y^2+z^2$, then $v_{x} = 2x, v_{y} = 2y, v_{z} = 2z$. Hence we have : $vartriangle{u} = 6u_{v} + 4v u_{v} = u_{tt}$. And addititional property is : $u|_{t=0} = frac{1}{v}$.



But how should I finish it?



Edit :



After spherical substitution and deducing that $U(x,y,z,t) = U(r,t)$



We have : $begin{cases}
u_{tt} = u_{rr} + frac{2}{r}u_{r}
\
u|_{t = 0} = frac{1}{r^2}
\
u_t|_{t = 0} = 0
end{cases}$



After substitution: $xi = t-r$ and $eta = t + r$ we have
$u_{xi eta} = frac{u_{eta} - u_{xi}}{eta - xi}$. The solution for this part is :
$U(xi,eta) = frac{f(eta)+ g(xi)}{eta - xi}$. Hence we have :



$U(t,r) = frac{f(t+r) +g(t-r)}{2r}$ with initial data : $U(0,r) = frac{1}{r^2}$ and $U_t(0,r) = 0$.



But now how can we find exact form of $f$ and $g$?






ordinary-differential-equations proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have a look at the Laplacian in spherical coordinates here. The solution is radial: depends only on $r$, not on $theta,phi$.
    $endgroup$
    – Federico
    Dec 20 '18 at 20:12










  • $begingroup$
    @Federico I've thought about spherical coordinates. But Laplacian looks terrible. Why should solution depends only of $r$? Suppose function $f = t r theta phi + frac{t}{r^{2}}$ depends on all variables , but $f_t = frac{1}{r^2}$
    $endgroup$
    – openspace
    Dec 20 '18 at 20:16










  • $begingroup$
    initial datum is radially simmetric. you need also to specify the initial value of $partial_t u$. i assume it is $0$
    $endgroup$
    – Federico
    Dec 20 '18 at 20:18










  • $begingroup$
    @Federico there is only one initial property. So the task assumes to find all solutions.
    $endgroup$
    – openspace
    Dec 20 '18 at 20:19








  • 1




    $begingroup$
    $f = dfrac{1}{t+r}$ and $g = -dfrac{1}{t-r}$ satisfy the conditions
    $endgroup$
    – Dylan
    Dec 22 '18 at 10:06


















0












$begingroup$


Consider system of : $begin{cases}
u_{tt} = vartriangle{u}
\
u|_{t = 0} = frac{1}{x^{2}+y^2+z^2}
\
u_t|_{t = 0} = 0
end{cases}$



I've thought about $v = x^{2} + y^2+z^2$, then $v_{x} = 2x, v_{y} = 2y, v_{z} = 2z$. Hence we have : $vartriangle{u} = 6u_{v} + 4v u_{v} = u_{tt}$. And addititional property is : $u|_{t=0} = frac{1}{v}$.



But how should I finish it?



Edit :



After spherical substitution and deducing that $U(x,y,z,t) = U(r,t)$



We have : $begin{cases}
u_{tt} = u_{rr} + frac{2}{r}u_{r}
\
u|_{t = 0} = frac{1}{r^2}
\
u_t|_{t = 0} = 0
end{cases}$



After substitution: $xi = t-r$ and $eta = t + r$ we have
$u_{xi eta} = frac{u_{eta} - u_{xi}}{eta - xi}$. The solution for this part is :
$U(xi,eta) = frac{f(eta)+ g(xi)}{eta - xi}$. Hence we have :



$U(t,r) = frac{f(t+r) +g(t-r)}{2r}$ with initial data : $U(0,r) = frac{1}{r^2}$ and $U_t(0,r) = 0$.



But now how can we find exact form of $f$ and $g$?






ordinary-differential-equations proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have a look at the Laplacian in spherical coordinates here. The solution is radial: depends only on $r$, not on $theta,phi$.
    $endgroup$
    – Federico
    Dec 20 '18 at 20:12










  • $begingroup$
    @Federico I've thought about spherical coordinates. But Laplacian looks terrible. Why should solution depends only of $r$? Suppose function $f = t r theta phi + frac{t}{r^{2}}$ depends on all variables , but $f_t = frac{1}{r^2}$
    $endgroup$
    – openspace
    Dec 20 '18 at 20:16










  • $begingroup$
    initial datum is radially simmetric. you need also to specify the initial value of $partial_t u$. i assume it is $0$
    $endgroup$
    – Federico
    Dec 20 '18 at 20:18










  • $begingroup$
    @Federico there is only one initial property. So the task assumes to find all solutions.
    $endgroup$
    – openspace
    Dec 20 '18 at 20:19








  • 1




    $begingroup$
    $f = dfrac{1}{t+r}$ and $g = -dfrac{1}{t-r}$ satisfy the conditions
    $endgroup$
    – Dylan
    Dec 22 '18 at 10:06
















0












0








0





$begingroup$


Consider system of : $begin{cases}
u_{tt} = vartriangle{u}
\
u|_{t = 0} = frac{1}{x^{2}+y^2+z^2}
\
u_t|_{t = 0} = 0
end{cases}$



I've thought about $v = x^{2} + y^2+z^2$, then $v_{x} = 2x, v_{y} = 2y, v_{z} = 2z$. Hence we have : $vartriangle{u} = 6u_{v} + 4v u_{v} = u_{tt}$. And addititional property is : $u|_{t=0} = frac{1}{v}$.



But how should I finish it?



Edit :



After spherical substitution and deducing that $U(x,y,z,t) = U(r,t)$



We have : $begin{cases}
u_{tt} = u_{rr} + frac{2}{r}u_{r}
\
u|_{t = 0} = frac{1}{r^2}
\
u_t|_{t = 0} = 0
end{cases}$



After substitution: $xi = t-r$ and $eta = t + r$ we have
$u_{xi eta} = frac{u_{eta} - u_{xi}}{eta - xi}$. The solution for this part is :
$U(xi,eta) = frac{f(eta)+ g(xi)}{eta - xi}$. Hence we have :



$U(t,r) = frac{f(t+r) +g(t-r)}{2r}$ with initial data : $U(0,r) = frac{1}{r^2}$ and $U_t(0,r) = 0$.



But now how can we find exact form of $f$ and $g$?






ordinary-differential-equations proof-verification







share|cite|improve this question











$endgroup$




Consider system of : $begin{cases}
u_{tt} = vartriangle{u}
\
u|_{t = 0} = frac{1}{x^{2}+y^2+z^2}
\
u_t|_{t = 0} = 0
end{cases}$



I've thought about $v = x^{2} + y^2+z^2$, then $v_{x} = 2x, v_{y} = 2y, v_{z} = 2z$. Hence we have : $vartriangle{u} = 6u_{v} + 4v u_{v} = u_{tt}$. And addititional property is : $u|_{t=0} = frac{1}{v}$.



But how should I finish it?



Edit :



After spherical substitution and deducing that $U(x,y,z,t) = U(r,t)$



We have : $begin{cases}
u_{tt} = u_{rr} + frac{2}{r}u_{r}
\
u|_{t = 0} = frac{1}{r^2}
\
u_t|_{t = 0} = 0
end{cases}$



After substitution: $xi = t-r$ and $eta = t + r$ we have
$u_{xi eta} = frac{u_{eta} - u_{xi}}{eta - xi}$. The solution for this part is :
$U(xi,eta) = frac{f(eta)+ g(xi)}{eta - xi}$. Hence we have :



$U(t,r) = frac{f(t+r) +g(t-r)}{2r}$ with initial data : $U(0,r) = frac{1}{r^2}$ and $U_t(0,r) = 0$.



But now how can we find exact form of $f$ and $g$?






ordinary-differential-equations proof-verification




ordinary-differential-equations proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 18:40







openspace

















asked Dec 20 '18 at 20:06









openspaceopenspace

3,4452822




3,4452822












  • $begingroup$
    Have a look at the Laplacian in spherical coordinates here. The solution is radial: depends only on $r$, not on $theta,phi$.
    $endgroup$
    – Federico
    Dec 20 '18 at 20:12










  • $begingroup$
    @Federico I've thought about spherical coordinates. But Laplacian looks terrible. Why should solution depends only of $r$? Suppose function $f = t r theta phi + frac{t}{r^{2}}$ depends on all variables , but $f_t = frac{1}{r^2}$
    $endgroup$
    – openspace
    Dec 20 '18 at 20:16










  • $begingroup$
    initial datum is radially simmetric. you need also to specify the initial value of $partial_t u$. i assume it is $0$
    $endgroup$
    – Federico
    Dec 20 '18 at 20:18










  • $begingroup$
    @Federico there is only one initial property. So the task assumes to find all solutions.
    $endgroup$
    – openspace
    Dec 20 '18 at 20:19








  • 1




    $begingroup$
    $f = dfrac{1}{t+r}$ and $g = -dfrac{1}{t-r}$ satisfy the conditions
    $endgroup$
    – Dylan
    Dec 22 '18 at 10:06




















  • $begingroup$
    Have a look at the Laplacian in spherical coordinates here. The solution is radial: depends only on $r$, not on $theta,phi$.
    $endgroup$
    – Federico
    Dec 20 '18 at 20:12










  • $begingroup$
    @Federico I've thought about spherical coordinates. But Laplacian looks terrible. Why should solution depends only of $r$? Suppose function $f = t r theta phi + frac{t}{r^{2}}$ depends on all variables , but $f_t = frac{1}{r^2}$
    $endgroup$
    – openspace
    Dec 20 '18 at 20:16










  • $begingroup$
    initial datum is radially simmetric. you need also to specify the initial value of $partial_t u$. i assume it is $0$
    $endgroup$
    – Federico
    Dec 20 '18 at 20:18










  • $begingroup$
    @Federico there is only one initial property. So the task assumes to find all solutions.
    $endgroup$
    – openspace
    Dec 20 '18 at 20:19








  • 1




    $begingroup$
    $f = dfrac{1}{t+r}$ and $g = -dfrac{1}{t-r}$ satisfy the conditions
    $endgroup$
    – Dylan
    Dec 22 '18 at 10:06


















$begingroup$
Have a look at the Laplacian in spherical coordinates here. The solution is radial: depends only on $r$, not on $theta,phi$.
$endgroup$
– Federico
Dec 20 '18 at 20:12




$begingroup$
Have a look at the Laplacian in spherical coordinates here. The solution is radial: depends only on $r$, not on $theta,phi$.
$endgroup$
– Federico
Dec 20 '18 at 20:12












$begingroup$
@Federico I've thought about spherical coordinates. But Laplacian looks terrible. Why should solution depends only of $r$? Suppose function $f = t r theta phi + frac{t}{r^{2}}$ depends on all variables , but $f_t = frac{1}{r^2}$
$endgroup$
– openspace
Dec 20 '18 at 20:16




$begingroup$
@Federico I've thought about spherical coordinates. But Laplacian looks terrible. Why should solution depends only of $r$? Suppose function $f = t r theta phi + frac{t}{r^{2}}$ depends on all variables , but $f_t = frac{1}{r^2}$
$endgroup$
– openspace
Dec 20 '18 at 20:16












$begingroup$
initial datum is radially simmetric. you need also to specify the initial value of $partial_t u$. i assume it is $0$
$endgroup$
– Federico
Dec 20 '18 at 20:18




$begingroup$
initial datum is radially simmetric. you need also to specify the initial value of $partial_t u$. i assume it is $0$
$endgroup$
– Federico
Dec 20 '18 at 20:18












$begingroup$
@Federico there is only one initial property. So the task assumes to find all solutions.
$endgroup$
– openspace
Dec 20 '18 at 20:19






$begingroup$
@Federico there is only one initial property. So the task assumes to find all solutions.
$endgroup$
– openspace
Dec 20 '18 at 20:19






1




1




$begingroup$
$f = dfrac{1}{t+r}$ and $g = -dfrac{1}{t-r}$ satisfy the conditions
$endgroup$
– Dylan
Dec 22 '18 at 10:06






$begingroup$
$f = dfrac{1}{t+r}$ and $g = -dfrac{1}{t-r}$ satisfy the conditions
$endgroup$
– Dylan
Dec 22 '18 at 10:06












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