Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $tgt0,-inftylt xlt +infty$Parabolic linear PDE with continuous...
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Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $tgt0,-inftylt xlt +infty$
Parabolic linear PDE with continuous coefficients; how to solve and explanation of text neededUsing d'Alembert's solution to solve the 1-D wave equationSolve initial value problem (C.S.I.R)?Solve the boundary value problem $ Delta u=1$how to solve non-linear pde in two variablesSolve transport equations by using Laplace transformStuck trying to solve a PDE by method of characteristicsHow to solve a Partial Differential EquationCauchy problem for non-linear pdeSolve the initial value problem : $u_{xy}=6x^2y$ ,$u(x;0)=1-cos x$ , $u(0;y)=3y^2$ .
$begingroup$
Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $tgt0,-inftylt xlt +infty$
For $u_{tt}=u_{xx}$ ,we can get $u(x,t)=f(x-t)+g(x+t)$ and $u(x,0)=1-2x+3x^2=f(x)+g(x)$, what's the next? Now do we need more information to get an explict solution?
pde
asked yesterday
Jaqen ChouJaqen Chou
451110
$endgroup$
add a comment |
$begingroup$
Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $tgt0,-inftylt xlt +infty$
For $u_{tt}=u_{xx}$ ,we can get $u(x,t)=f(x-t)+g(x+t)$ and $u(x,0)=1-2x+3x^2=f(x)+g(x)$, what's the next? Now do we need more information to get an explict solution?
pde
asked yesterday
Jaqen ChouJaqen Chou
451110
$endgroup$
$begingroup$
$u_t(x,0)=phi(x)$ ???
$endgroup$
– Aleksas Domarkas
yesterday
$begingroup$
so we do need such a condition such that we could complete the left work?
$endgroup$
– Jaqen Chou
yesterday
add a comment |
$begingroup$
Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $tgt0,-inftylt xlt +infty$
For $u_{tt}=u_{xx}$ ,we can get $u(x,t)=f(x-t)+g(x+t)$ and $u(x,0)=1-2x+3x^2=f(x)+g(x)$, what's the next? Now do we need more information to get an explict solution?
pde
asked yesterday
Jaqen ChouJaqen Chou
451110
$endgroup$
Solve $u_{tt}=u_{xx}$ , $u(x,0)=1-2x+3x^2$, $tgt0,-inftylt xlt +infty$
For $u_{tt}=u_{xx}$ ,we can get $u(x,t)=f(x-t)+g(x+t)$ and $u(x,0)=1-2x+3x^2=f(x)+g(x)$, what's the next? Now do we need more information to get an explict solution?
pde
pde
asked yesterday
Jaqen ChouJaqen Chou
451110
asked yesterday
Jaqen ChouJaqen Chou
451110
asked yesterday
Jaqen ChouJaqen Chou
451110
asked yesterday
Jaqen ChouJaqen Chou
451110
asked yesterday
Jaqen ChouJaqen Chou
451110
451110
$begingroup$
$u_t(x,0)=phi(x)$ ???
$endgroup$
– Aleksas Domarkas
yesterday
$begingroup$
so we do need such a condition such that we could complete the left work?
$endgroup$
– Jaqen Chou
yesterday
add a comment |
$begingroup$
$u_t(x,0)=phi(x)$ ???
$endgroup$
– Aleksas Domarkas
yesterday
$begingroup$
so we do need such a condition such that we could complete the left work?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
$u_t(x,0)=phi(x)$ ???
$endgroup$
– Aleksas Domarkas
yesterday
$begingroup$
$u_t(x,0)=phi(x)$ ???
$endgroup$
– Aleksas Domarkas
yesterday
$begingroup$
so we do need such a condition such that we could complete the left work?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
so we do need such a condition such that we could complete the left work?
$endgroup$
– Jaqen Chou
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$
u(x,t) = f(x-t)+g(x+t)
$$
$$
u(x,0) = f(x)+g(x) = 1-2x+3x^2
$$
so making $f = g$ we have
$$
f(x) = frac 12left(1-2x+3x^2right)
$$
and finally
$$
u(x,t) = frac 12left(1-2(x-t)+3(x-t)^2right) + frac 12left(1-2(x+t)+3(x+t)^2right)
$$
or
$$
u(x,t) = frac 12left(1+3t^2+x(3x-2)right)
$$
answered yesterday
CesareoCesareo
9,2563517
$endgroup$
$begingroup$
Why can we make such an assumption $f=g$?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
$f, g$ are general functions including the possibility $f = g$
$endgroup$
– Cesareo
yesterday
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
$$
u(x,t) = f(x-t)+g(x+t)
$$
$$
u(x,0) = f(x)+g(x) = 1-2x+3x^2
$$
so making $f = g$ we have
$$
f(x) = frac 12left(1-2x+3x^2right)
$$
and finally
$$
u(x,t) = frac 12left(1-2(x-t)+3(x-t)^2right) + frac 12left(1-2(x+t)+3(x+t)^2right)
$$
or
$$
u(x,t) = frac 12left(1+3t^2+x(3x-2)right)
$$
answered yesterday
CesareoCesareo
9,2563517
$endgroup$
$begingroup$
Why can we make such an assumption $f=g$?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
$f, g$ are general functions including the possibility $f = g$
$endgroup$
– Cesareo
yesterday
add a comment |
$begingroup$
$$
u(x,t) = f(x-t)+g(x+t)
$$
$$
u(x,0) = f(x)+g(x) = 1-2x+3x^2
$$
so making $f = g$ we have
$$
f(x) = frac 12left(1-2x+3x^2right)
$$
and finally
$$
u(x,t) = frac 12left(1-2(x-t)+3(x-t)^2right) + frac 12left(1-2(x+t)+3(x+t)^2right)
$$
or
$$
u(x,t) = frac 12left(1+3t^2+x(3x-2)right)
$$
answered yesterday
CesareoCesareo
9,2563517
$endgroup$
$begingroup$
Why can we make such an assumption $f=g$?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
$f, g$ are general functions including the possibility $f = g$
$endgroup$
– Cesareo
yesterday
add a comment |
$begingroup$
$$
u(x,t) = f(x-t)+g(x+t)
$$
$$
u(x,0) = f(x)+g(x) = 1-2x+3x^2
$$
so making $f = g$ we have
$$
f(x) = frac 12left(1-2x+3x^2right)
$$
and finally
$$
u(x,t) = frac 12left(1-2(x-t)+3(x-t)^2right) + frac 12left(1-2(x+t)+3(x+t)^2right)
$$
or
$$
u(x,t) = frac 12left(1+3t^2+x(3x-2)right)
$$
answered yesterday
CesareoCesareo
9,2563517
$endgroup$
$$
u(x,t) = f(x-t)+g(x+t)
$$
$$
u(x,0) = f(x)+g(x) = 1-2x+3x^2
$$
so making $f = g$ we have
$$
f(x) = frac 12left(1-2x+3x^2right)
$$
and finally
$$
u(x,t) = frac 12left(1-2(x-t)+3(x-t)^2right) + frac 12left(1-2(x+t)+3(x+t)^2right)
$$
or
$$
u(x,t) = frac 12left(1+3t^2+x(3x-2)right)
$$
answered yesterday
CesareoCesareo
9,2563517
answered yesterday
CesareoCesareo
9,2563517
answered yesterday
CesareoCesareo
9,2563517
answered yesterday
CesareoCesareo
9,2563517
9,2563517
$begingroup$
Why can we make such an assumption $f=g$?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
$f, g$ are general functions including the possibility $f = g$
$endgroup$
– Cesareo
yesterday
add a comment |
$begingroup$
Why can we make such an assumption $f=g$?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
$f, g$ are general functions including the possibility $f = g$
$endgroup$
– Cesareo
yesterday
$begingroup$
Why can we make such an assumption $f=g$?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
Why can we make such an assumption $f=g$?
$endgroup$
– Jaqen Chou
yesterday
$begingroup$
$f, g$ are general functions including the possibility $f = g$
$endgroup$
– Cesareo
yesterday
$begingroup$
$f, g$ are general functions including the possibility $f = g$
$endgroup$
– Cesareo
yesterday
add a comment |
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$begingroup$
$u_t(x,0)=phi(x)$ ???
$endgroup$
– Aleksas Domarkas
yesterday
$begingroup$
so we do need such a condition such that we could complete the left work?
$endgroup$
– Jaqen Chou
yesterday