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Find all second-order polynomial solutions to the Heat Equation
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The following problem is a homework and I am not asking for a solution but simply for a hint on how to approach it. The task is given as follows:
Find all second-order polynomial solutions to the heat equation:
$u_{t}-u_{xx}=0$
This is what I have done so far:
Given that $u(t, x)$ is a second-order polynomial, the equation will be the following:
$u(t, x) = ax^2+bxt+ct^2+dx+et+f$ (where $a, b, c, d, e, f$ are constants)
$u_{t}(t, x)=bx+2ct+e$
$u_{x}(t, x)=2ax+bt+d$
$u_{xx}(t, x)=2a$
From what follows that $bx+2ct+e-2a=0$
Is there anything else that I'm missing?
How should one approach this problem without any previous knowledge of PDEs?
Thanks.
polynomials pde heat-equation
asked Jan 31 at 18:34
AldersonAlderson
1338
$endgroup$
add a comment |
$begingroup$
The following problem is a homework and I am not asking for a solution but simply for a hint on how to approach it. The task is given as follows:
Find all second-order polynomial solutions to the heat equation:
$u_{t}-u_{xx}=0$
This is what I have done so far:
Given that $u(t, x)$ is a second-order polynomial, the equation will be the following:
$u(t, x) = ax^2+bxt+ct^2+dx+et+f$ (where $a, b, c, d, e, f$ are constants)
$u_{t}(t, x)=bx+2ct+e$
$u_{x}(t, x)=2ax+bt+d$
$u_{xx}(t, x)=2a$
From what follows that $bx+2ct+e-2a=0$
Is there anything else that I'm missing?
How should one approach this problem without any previous knowledge of PDEs?
Thanks.
polynomials pde heat-equation
asked Jan 31 at 18:34
AldersonAlderson
1338
$endgroup$
2
$begingroup$
As your last equality has to be satisfied for all $x$ and for all $t$, you can get more information about the coefficients.
$endgroup$
– Alex
Jan 31 at 18:52
$begingroup$
@Alex, after substituting I ended up with the following expression: $ax^2+2at-ct^2+dx+f=0$. Is this enough or do I need to perform more steps to consider the problem to be solved?
$endgroup$
– Alderson
Jan 31 at 19:07
1
$begingroup$
You had the equation $bx+2ct+e-2a=0$. As this equation has to hold forall x and t, you directly get $b=0=c$ and $e=2a$. From that you can write down you polynomial.
$endgroup$
– Alex
Jan 31 at 20:00
add a comment |
$begingroup$
The following problem is a homework and I am not asking for a solution but simply for a hint on how to approach it. The task is given as follows:
Find all second-order polynomial solutions to the heat equation:
$u_{t}-u_{xx}=0$
This is what I have done so far:
Given that $u(t, x)$ is a second-order polynomial, the equation will be the following:
$u(t, x) = ax^2+bxt+ct^2+dx+et+f$ (where $a, b, c, d, e, f$ are constants)
$u_{t}(t, x)=bx+2ct+e$
$u_{x}(t, x)=2ax+bt+d$
$u_{xx}(t, x)=2a$
From what follows that $bx+2ct+e-2a=0$
Is there anything else that I'm missing?
How should one approach this problem without any previous knowledge of PDEs?
Thanks.
polynomials pde heat-equation
asked Jan 31 at 18:34
AldersonAlderson
1338
$endgroup$
The following problem is a homework and I am not asking for a solution but simply for a hint on how to approach it. The task is given as follows:
Find all second-order polynomial solutions to the heat equation:
$u_{t}-u_{xx}=0$
This is what I have done so far:
Given that $u(t, x)$ is a second-order polynomial, the equation will be the following:
$u(t, x) = ax^2+bxt+ct^2+dx+et+f$ (where $a, b, c, d, e, f$ are constants)
$u_{t}(t, x)=bx+2ct+e$
$u_{x}(t, x)=2ax+bt+d$
$u_{xx}(t, x)=2a$
From what follows that $bx+2ct+e-2a=0$
Is there anything else that I'm missing?
How should one approach this problem without any previous knowledge of PDEs?
Thanks.
polynomials pde heat-equation
polynomials pde heat-equation
asked Jan 31 at 18:34
AldersonAlderson
1338
asked Jan 31 at 18:34
AldersonAlderson
1338
asked Jan 31 at 18:34
AldersonAlderson
1338
asked Jan 31 at 18:34
AldersonAlderson
1338
asked Jan 31 at 18:34
AldersonAlderson
1338
1338
2
$begingroup$
As your last equality has to be satisfied for all $x$ and for all $t$, you can get more information about the coefficients.
$endgroup$
– Alex
Jan 31 at 18:52
$begingroup$
@Alex, after substituting I ended up with the following expression: $ax^2+2at-ct^2+dx+f=0$. Is this enough or do I need to perform more steps to consider the problem to be solved?
$endgroup$
– Alderson
Jan 31 at 19:07
1
$begingroup$
You had the equation $bx+2ct+e-2a=0$. As this equation has to hold forall x and t, you directly get $b=0=c$ and $e=2a$. From that you can write down you polynomial.
$endgroup$
– Alex
Jan 31 at 20:00
add a comment |
2
$begingroup$
As your last equality has to be satisfied for all $x$ and for all $t$, you can get more information about the coefficients.
$endgroup$
– Alex
Jan 31 at 18:52
$begingroup$
@Alex, after substituting I ended up with the following expression: $ax^2+2at-ct^2+dx+f=0$. Is this enough or do I need to perform more steps to consider the problem to be solved?
$endgroup$
– Alderson
Jan 31 at 19:07
1
$begingroup$
You had the equation $bx+2ct+e-2a=0$. As this equation has to hold forall x and t, you directly get $b=0=c$ and $e=2a$. From that you can write down you polynomial.
$endgroup$
– Alex
Jan 31 at 20:00
2
2
$begingroup$
As your last equality has to be satisfied for all $x$ and for all $t$, you can get more information about the coefficients.
$endgroup$
– Alex
Jan 31 at 18:52
$begingroup$
As your last equality has to be satisfied for all $x$ and for all $t$, you can get more information about the coefficients.
$endgroup$
– Alex
Jan 31 at 18:52
$begingroup$
@Alex, after substituting I ended up with the following expression: $ax^2+2at-ct^2+dx+f=0$. Is this enough or do I need to perform more steps to consider the problem to be solved?
$endgroup$
– Alderson
Jan 31 at 19:07
$begingroup$
@Alex, after substituting I ended up with the following expression: $ax^2+2at-ct^2+dx+f=0$. Is this enough or do I need to perform more steps to consider the problem to be solved?
$endgroup$
– Alderson
Jan 31 at 19:07
1
1
$begingroup$
You had the equation $bx+2ct+e-2a=0$. As this equation has to hold forall x and t, you directly get $b=0=c$ and $e=2a$. From that you can write down you polynomial.
$endgroup$
– Alex
Jan 31 at 20:00
$begingroup$
You had the equation $bx+2ct+e-2a=0$. As this equation has to hold forall x and t, you directly get $b=0=c$ and $e=2a$. From that you can write down you polynomial.
$endgroup$
– Alex
Jan 31 at 20:00
add a comment |
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$begingroup$
As your last equality has to be satisfied for all $x$ and for all $t$, you can get more information about the coefficients.
$endgroup$
– Alex
Jan 31 at 18:52
$begingroup$
@Alex, after substituting I ended up with the following expression: $ax^2+2at-ct^2+dx+f=0$. Is this enough or do I need to perform more steps to consider the problem to be solved?
$endgroup$
– Alderson
Jan 31 at 19:07
1
$begingroup$
You had the equation $bx+2ct+e-2a=0$. As this equation has to hold forall x and t, you directly get $b=0=c$ and $e=2a$. From that you can write down you polynomial.
$endgroup$
– Alex
Jan 31 at 20:00