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$fracd^2ydx^2=f(x)$ with boundary conditions, how to find integration bounds
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Differential Equations - EigenfunctionsCan't match boundary conditions on a perturbation series solution to a non-linear ODE?Mysterious inconsistency in an inhomogeneous linear 2nd order ODE with specified boundary valuesLaplace Equation with non-const Dirichlet Boundary ConditionsPoisson partial differential equation under Neumann boundary conditionsquaestion to solution of the Laplace's Equation $u_tt+u_xx=0$ using method of separation of variablesWeak form of steady Navier-Stokes equations with special boundary conditionUsing Green's function to solve $y''=1$ with $y(0)=y(1)=0$Difficulty in working out trivial solutionMatrix representation of a finite difference with Neumann boundary conditions
$begingroup$
Given $$fracd^2ydx^2=f(x),quad y(-1)=y(1)=0,$$ I used $u=y'$ and $u(x_0)=u_0$ to get
$$
u(x)=u_0+int_x_0^xf(xi)dxi.
$$
Then we have $y'=u$, which we can integrate again using $y(x_0)=y_0$ to get
$$
y(x)=y_0+int_x_0^xBig(u_0+int_x_0^zeta f(xi)dxiBig)dzeta=(y_0-u_0x_0)+u_0x+int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta.
$$
Now my question is, what do I do with $y_0,x_0$ and $u_0$ to get it into the normal form using $C_1,C_2$? Wolfram Alpha gives the following expression:
$$
y(x)=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta,
$$
but how did they set $x_0=1$? Because we didn't use any boundary conditions yet. I understand $C_1$ and $C_2$ as just relabeling.
ordinary-differential-equations boundary-value-problem upper-lower-bounds
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
$endgroup$
add a comment |
$begingroup$
Given $$fracd^2ydx^2=f(x),quad y(-1)=y(1)=0,$$ I used $u=y'$ and $u(x_0)=u_0$ to get
$$
u(x)=u_0+int_x_0^xf(xi)dxi.
$$
Then we have $y'=u$, which we can integrate again using $y(x_0)=y_0$ to get
$$
y(x)=y_0+int_x_0^xBig(u_0+int_x_0^zeta f(xi)dxiBig)dzeta=(y_0-u_0x_0)+u_0x+int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta.
$$
Now my question is, what do I do with $y_0,x_0$ and $u_0$ to get it into the normal form using $C_1,C_2$? Wolfram Alpha gives the following expression:
$$
y(x)=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta,
$$
but how did they set $x_0=1$? Because we didn't use any boundary conditions yet. I understand $C_1$ and $C_2$ as just relabeling.
ordinary-differential-equations boundary-value-problem upper-lower-bounds
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
$endgroup$
add a comment |
$begingroup$
Given $$fracd^2ydx^2=f(x),quad y(-1)=y(1)=0,$$ I used $u=y'$ and $u(x_0)=u_0$ to get
$$
u(x)=u_0+int_x_0^xf(xi)dxi.
$$
Then we have $y'=u$, which we can integrate again using $y(x_0)=y_0$ to get
$$
y(x)=y_0+int_x_0^xBig(u_0+int_x_0^zeta f(xi)dxiBig)dzeta=(y_0-u_0x_0)+u_0x+int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta.
$$
Now my question is, what do I do with $y_0,x_0$ and $u_0$ to get it into the normal form using $C_1,C_2$? Wolfram Alpha gives the following expression:
$$
y(x)=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta,
$$
but how did they set $x_0=1$? Because we didn't use any boundary conditions yet. I understand $C_1$ and $C_2$ as just relabeling.
ordinary-differential-equations boundary-value-problem upper-lower-bounds
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
$endgroup$
Given $$fracd^2ydx^2=f(x),quad y(-1)=y(1)=0,$$ I used $u=y'$ and $u(x_0)=u_0$ to get
$$
u(x)=u_0+int_x_0^xf(xi)dxi.
$$
Then we have $y'=u$, which we can integrate again using $y(x_0)=y_0$ to get
$$
y(x)=y_0+int_x_0^xBig(u_0+int_x_0^zeta f(xi)dxiBig)dzeta=(y_0-u_0x_0)+u_0x+int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta.
$$
Now my question is, what do I do with $y_0,x_0$ and $u_0$ to get it into the normal form using $C_1,C_2$? Wolfram Alpha gives the following expression:
$$
y(x)=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta,
$$
but how did they set $x_0=1$? Because we didn't use any boundary conditions yet. I understand $C_1$ and $C_2$ as just relabeling.
ordinary-differential-equations boundary-value-problem upper-lower-bounds
ordinary-differential-equations boundary-value-problem upper-lower-bounds
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
asked Apr 1 at 18:12
The Coding WombatThe Coding Wombat
342111
342111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is because
$$int_x_0^x f(t) dt=int_1^x f(t) dt - int_1^x_0 f(t) dt=int_1^x f(t) dt - k$$
For some $kinmathbbR$. Hence
$$int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta=int_x_0^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta$$
$$=int_1^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-int_1^xk_1dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-k_1(x-1)-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
So
$$y(x)=y_0-u_0x_0+u_0x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
$$=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta$$
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
$endgroup$
$begingroup$
But the expression I found has integration bounds from $x_0$ to $x$, so that isn't a constant right?
$endgroup$
– The Coding Wombat
Apr 1 at 18:25
$begingroup$
Yes the upper bound is variable, but as long as one of the bounds is constant, it can take any chosen value because the change in value only effects the result by a constant amount. This amount can be corrected by subtracting some constant which is written as part of $C_1$
$endgroup$
– Peter Foreman
Apr 1 at 18:27
$begingroup$
So we can set $x_0=1$ because we just subtract another integral with some other bound to correct for this? How did you get the integral in your answer?
$endgroup$
– The Coding Wombat
Apr 1 at 18:33
1
$begingroup$
Is that enough of an explanation for you now?
$endgroup$
– Peter Foreman
Apr 1 at 18:44
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is because
$$int_x_0^x f(t) dt=int_1^x f(t) dt - int_1^x_0 f(t) dt=int_1^x f(t) dt - k$$
For some $kinmathbbR$. Hence
$$int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta=int_x_0^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta$$
$$=int_1^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-int_1^xk_1dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-k_1(x-1)-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
So
$$y(x)=y_0-u_0x_0+u_0x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
$$=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta$$
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
$endgroup$
$begingroup$
But the expression I found has integration bounds from $x_0$ to $x$, so that isn't a constant right?
$endgroup$
– The Coding Wombat
Apr 1 at 18:25
$begingroup$
Yes the upper bound is variable, but as long as one of the bounds is constant, it can take any chosen value because the change in value only effects the result by a constant amount. This amount can be corrected by subtracting some constant which is written as part of $C_1$
$endgroup$
– Peter Foreman
Apr 1 at 18:27
$begingroup$
So we can set $x_0=1$ because we just subtract another integral with some other bound to correct for this? How did you get the integral in your answer?
$endgroup$
– The Coding Wombat
Apr 1 at 18:33
1
$begingroup$
Is that enough of an explanation for you now?
$endgroup$
– Peter Foreman
Apr 1 at 18:44
add a comment |
$begingroup$
This is because
$$int_x_0^x f(t) dt=int_1^x f(t) dt - int_1^x_0 f(t) dt=int_1^x f(t) dt - k$$
For some $kinmathbbR$. Hence
$$int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta=int_x_0^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta$$
$$=int_1^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-int_1^xk_1dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-k_1(x-1)-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
So
$$y(x)=y_0-u_0x_0+u_0x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
$$=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta$$
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
$endgroup$
$begingroup$
But the expression I found has integration bounds from $x_0$ to $x$, so that isn't a constant right?
$endgroup$
– The Coding Wombat
Apr 1 at 18:25
$begingroup$
Yes the upper bound is variable, but as long as one of the bounds is constant, it can take any chosen value because the change in value only effects the result by a constant amount. This amount can be corrected by subtracting some constant which is written as part of $C_1$
$endgroup$
– Peter Foreman
Apr 1 at 18:27
$begingroup$
So we can set $x_0=1$ because we just subtract another integral with some other bound to correct for this? How did you get the integral in your answer?
$endgroup$
– The Coding Wombat
Apr 1 at 18:33
1
$begingroup$
Is that enough of an explanation for you now?
$endgroup$
– Peter Foreman
Apr 1 at 18:44
add a comment |
$begingroup$
This is because
$$int_x_0^x f(t) dt=int_1^x f(t) dt - int_1^x_0 f(t) dt=int_1^x f(t) dt - k$$
For some $kinmathbbR$. Hence
$$int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta=int_x_0^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta$$
$$=int_1^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-int_1^xk_1dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-k_1(x-1)-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
So
$$y(x)=y_0-u_0x_0+u_0x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
$$=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta$$
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
$endgroup$
This is because
$$int_x_0^x f(t) dt=int_1^x f(t) dt - int_1^x_0 f(t) dt=int_1^x f(t) dt - k$$
For some $kinmathbbR$. Hence
$$int_x_0^xBig(int_x_0^zeta f(xi)dxiBig)dzeta=int_x_0^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta$$
$$=int_1^xBig(int_1^zeta f(xi)dxi-k_1Big)dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-int_1^xk_1dzeta-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta-k_1(x-1)-k_2$$
$$=int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
So
$$y(x)=y_0-u_0x_0+u_0x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta+k_3x+k_4$$
$$=C_1+C_2x+int_1^xBig(int_1^zeta f(xi)dxiBig)dzeta$$
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
edited Apr 1 at 18:44
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
answered Apr 1 at 18:24
Peter ForemanPeter Foreman
8,1421321
8,1421321
$begingroup$
But the expression I found has integration bounds from $x_0$ to $x$, so that isn't a constant right?
$endgroup$
– The Coding Wombat
Apr 1 at 18:25
$begingroup$
Yes the upper bound is variable, but as long as one of the bounds is constant, it can take any chosen value because the change in value only effects the result by a constant amount. This amount can be corrected by subtracting some constant which is written as part of $C_1$
$endgroup$
– Peter Foreman
Apr 1 at 18:27
$begingroup$
So we can set $x_0=1$ because we just subtract another integral with some other bound to correct for this? How did you get the integral in your answer?
$endgroup$
– The Coding Wombat
Apr 1 at 18:33
1
$begingroup$
Is that enough of an explanation for you now?
$endgroup$
– Peter Foreman
Apr 1 at 18:44
add a comment |
$begingroup$
But the expression I found has integration bounds from $x_0$ to $x$, so that isn't a constant right?
$endgroup$
– The Coding Wombat
Apr 1 at 18:25
$begingroup$
Yes the upper bound is variable, but as long as one of the bounds is constant, it can take any chosen value because the change in value only effects the result by a constant amount. This amount can be corrected by subtracting some constant which is written as part of $C_1$
$endgroup$
– Peter Foreman
Apr 1 at 18:27
$begingroup$
So we can set $x_0=1$ because we just subtract another integral with some other bound to correct for this? How did you get the integral in your answer?
$endgroup$
– The Coding Wombat
Apr 1 at 18:33
1
$begingroup$
Is that enough of an explanation for you now?
$endgroup$
– Peter Foreman
Apr 1 at 18:44
$begingroup$
But the expression I found has integration bounds from $x_0$ to $x$, so that isn't a constant right?
$endgroup$
– The Coding Wombat
Apr 1 at 18:25
$begingroup$
But the expression I found has integration bounds from $x_0$ to $x$, so that isn't a constant right?
$endgroup$
– The Coding Wombat
Apr 1 at 18:25
$begingroup$
Yes the upper bound is variable, but as long as one of the bounds is constant, it can take any chosen value because the change in value only effects the result by a constant amount. This amount can be corrected by subtracting some constant which is written as part of $C_1$
$endgroup$
– Peter Foreman
Apr 1 at 18:27
$begingroup$
Yes the upper bound is variable, but as long as one of the bounds is constant, it can take any chosen value because the change in value only effects the result by a constant amount. This amount can be corrected by subtracting some constant which is written as part of $C_1$
$endgroup$
– Peter Foreman
Apr 1 at 18:27
$begingroup$
So we can set $x_0=1$ because we just subtract another integral with some other bound to correct for this? How did you get the integral in your answer?
$endgroup$
– The Coding Wombat
Apr 1 at 18:33
$begingroup$
So we can set $x_0=1$ because we just subtract another integral with some other bound to correct for this? How did you get the integral in your answer?
$endgroup$
– The Coding Wombat
Apr 1 at 18:33
1
1
$begingroup$
Is that enough of an explanation for you now?
$endgroup$
– Peter Foreman
Apr 1 at 18:44
$begingroup$
Is that enough of an explanation for you now?
$endgroup$
– Peter Foreman
Apr 1 at 18:44
add a comment |
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