Solving the heat equation gives constant solution
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2
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For the heat equation
$$u_t = 4 u_{xx}$$
$$u(x,0) = 1$$
$$u_x(0,t)=u_x(1,t)$$
$$0 leq x leq 1$$
we know that the general solution for Neumann boundary conditions is
$$u(x,t) = frac{A_0}{2} + sum_{n=1}^infty A_n e^{-(n pi)^2 4t} cos n pi x$$
but substituting the initial condition we get
$$u(x,0) = 1 = frac{A_0}{2} + sum_{n=1}^infty A_n cos n pi x$$
But the Fourier cosine series of $1$ is just $1$, so $A_0 = 2$ and $A_i = 0$ for $i > 0$
Then $$u(x,t) = 1$$
The fact that the solution is just $1$ makes me think that I did something wrong
proof-verification pde fourier-series heat-equation
asked Nov 16 at 17:32
The Bosco
495211
add a comment |
up vote
2
down vote
favorite
For the heat equation
$$u_t = 4 u_{xx}$$
$$u(x,0) = 1$$
$$u_x(0,t)=u_x(1,t)$$
$$0 leq x leq 1$$
we know that the general solution for Neumann boundary conditions is
$$u(x,t) = frac{A_0}{2} + sum_{n=1}^infty A_n e^{-(n pi)^2 4t} cos n pi x$$
but substituting the initial condition we get
$$u(x,0) = 1 = frac{A_0}{2} + sum_{n=1}^infty A_n cos n pi x$$
But the Fourier cosine series of $1$ is just $1$, so $A_0 = 2$ and $A_i = 0$ for $i > 0$
Then $$u(x,t) = 1$$
The fact that the solution is just $1$ makes me think that I did something wrong
proof-verification pde fourier-series heat-equation
asked Nov 16 at 17:32
The Bosco
495211
2
I mean, look at the heat equation. What does it tell you about how $u$ changes over time for the given initial conditions?
– epimorphic
Nov 16 at 17:43
1
Check that the constant function $1$ satisfies the required conditions. It does, doesn't it?
– DisintegratingByParts
Nov 16 at 19:08
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For the heat equation
$$u_t = 4 u_{xx}$$
$$u(x,0) = 1$$
$$u_x(0,t)=u_x(1,t)$$
$$0 leq x leq 1$$
we know that the general solution for Neumann boundary conditions is
$$u(x,t) = frac{A_0}{2} + sum_{n=1}^infty A_n e^{-(n pi)^2 4t} cos n pi x$$
but substituting the initial condition we get
$$u(x,0) = 1 = frac{A_0}{2} + sum_{n=1}^infty A_n cos n pi x$$
But the Fourier cosine series of $1$ is just $1$, so $A_0 = 2$ and $A_i = 0$ for $i > 0$
Then $$u(x,t) = 1$$
The fact that the solution is just $1$ makes me think that I did something wrong
proof-verification pde fourier-series heat-equation
asked Nov 16 at 17:32
The Bosco
495211
For the heat equation
$$u_t = 4 u_{xx}$$
$$u(x,0) = 1$$
$$u_x(0,t)=u_x(1,t)$$
$$0 leq x leq 1$$
we know that the general solution for Neumann boundary conditions is
$$u(x,t) = frac{A_0}{2} + sum_{n=1}^infty A_n e^{-(n pi)^2 4t} cos n pi x$$
but substituting the initial condition we get
$$u(x,0) = 1 = frac{A_0}{2} + sum_{n=1}^infty A_n cos n pi x$$
But the Fourier cosine series of $1$ is just $1$, so $A_0 = 2$ and $A_i = 0$ for $i > 0$
Then $$u(x,t) = 1$$
The fact that the solution is just $1$ makes me think that I did something wrong
proof-verification pde fourier-series heat-equation
proof-verification pde fourier-series heat-equation
asked Nov 16 at 17:32
The Bosco
495211
asked Nov 16 at 17:32
The Bosco
495211
asked Nov 16 at 17:32
The Bosco
495211
asked Nov 16 at 17:32
The Bosco
495211
asked Nov 16 at 17:32
The Bosco
495211
495211
2
I mean, look at the heat equation. What does it tell you about how $u$ changes over time for the given initial conditions?
– epimorphic
Nov 16 at 17:43
1
Check that the constant function $1$ satisfies the required conditions. It does, doesn't it?
– DisintegratingByParts
Nov 16 at 19:08
add a comment |
2
I mean, look at the heat equation. What does it tell you about how $u$ changes over time for the given initial conditions?
– epimorphic
Nov 16 at 17:43
1
Check that the constant function $1$ satisfies the required conditions. It does, doesn't it?
– DisintegratingByParts
Nov 16 at 19:08
2
2
I mean, look at the heat equation. What does it tell you about how $u$ changes over time for the given initial conditions?
– epimorphic
Nov 16 at 17:43
I mean, look at the heat equation. What does it tell you about how $u$ changes over time for the given initial conditions?
– epimorphic
Nov 16 at 17:43
1
1
Check that the constant function $1$ satisfies the required conditions. It does, doesn't it?
– DisintegratingByParts
Nov 16 at 19:08
Check that the constant function $1$ satisfies the required conditions. It does, doesn't it?
– DisintegratingByParts
Nov 16 at 19:08
add a comment |
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I mean, look at the heat equation. What does it tell you about how $u$ changes over time for the given initial conditions?
– epimorphic
Nov 16 at 17:43
1
Check that the constant function $1$ satisfies the required conditions. It does, doesn't it?
– DisintegratingByParts
Nov 16 at 19:08