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I asked this question here: http://math.stackexchange.com/questions/416885/checking-initial-condition-of-pde-is-satisfied-in-galerkin-methodhttps://math.stackexchange.com/questions/416885/checking-initial-condition-of-pde-is-satisfied-in-galerkin-method

But I did not receive the solution so I post it here.

The following is an argument I read, it is part of a proof of checking that the solution given by the Galerkin method satisfies the initial conditions.

From our weak formulation $$\langle u', v \rangle_{V',V} + a(u, v) = \langle f, v \rangle_{V',V}$$ which holds for all $v \in L^2(0,T;V)$, we can pick $v \in C^1(0,T;V)$ with $v(T) = 0$ and integrate this equation in time and use integration by parts to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u(0),v(0))_H.\tag{1}$$

Now we must use our Galerkin approximation $$\int_0^T\langle u_m', v_n \rangle_{V',V} + \int_0^Ta(u_m,v_n) = \int_0^T\langle f, v_n \rangle_{V',V}\tag{a}$$ and do some integration by parts and pass to the limit to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u_0,v(0))_H\tag{2}$$ where $v \in C^1(0,T;H)$ with $v(T) = 0$. This works because we know $u_m(0) \to u_0$, the initial condition.

Comparing (1) and (2), we can conclude $u(0) = u_0$.

Reasoning to obtain equation (2)

We have the convergences $u_m \rightharpoonup u$ in $L^2(0,T;V)$ and $u'_m \rightharpoonup u'$ in $L^2(0,T;V')$, where the $u_m \in V_m$. Pick $v_n \in V_n$ such that $v_n \to v$ in $L^2(0,T;V).$ Note that $v_n = \sum_{j=1}^n a_{j,n}w_j$ where $w_j$ is an o.n basis of $H$ and o.g basis of $V$.

We obtain from (a) the equation $$-\int_0^T\langle u_m, v_n' \rangle_{V',V} + \int_0^Ta(u_m, v_n) = \int_0^T\langle f, v_n \rangle + (u_m(0),v_n(0))_H$$ by IBP (and recall that $v \in C^1(0,T;V) \subset L^2(0,T;V)$ with $v(T) = 0$.)

My problem is, how to obtain (2) from here? Specificially, I don't understand why it is true that $$\int_0^T \langle u_m, v'_n \rangle \to \int_0^T \langle u, v' \rangle_{V'V}.$$

I am fine with the other terms in the equation. Why is true that $v'_n \to v$ in $L^2(0,T;V)$?

OK, so $v'$ exists and $v_n'$ probably exists assuming we can differentiate the coefficients. But why does it convergence to $v'$? No explanation is given anywhere.

I asked this question here: http://math.stackexchange.com/questions/416885/checking-initial-condition-of-pde-is-satisfied-in-galerkin-method

But I did not receive the solution so I post it here.

The following is an argument I read, it is part of a proof of checking that the solution given by the Galerkin method satisfies the initial conditions.

From our weak formulation $$\langle u', v \rangle_{V',V} + a(u, v) = \langle f, v \rangle_{V',V}$$ which holds for all $v \in L^2(0,T;V)$, we can pick $v \in C^1(0,T;V)$ with $v(T) = 0$ and integrate this equation in time and use integration by parts to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u(0),v(0))_H.\tag{1}$$

Now we must use our Galerkin approximation $$\int_0^T\langle u_m', v_n \rangle_{V',V} + \int_0^Ta(u_m,v_n) = \int_0^T\langle f, v_n \rangle_{V',V}\tag{a}$$ and do some integration by parts and pass to the limit to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u_0,v(0))_H\tag{2}$$ where $v \in C^1(0,T;H)$ with $v(T) = 0$. This works because we know $u_m(0) \to u_0$, the initial condition.

Comparing (1) and (2), we can conclude $u(0) = u_0$.

Reasoning to obtain equation (2)

We have the convergences $u_m \rightharpoonup u$ in $L^2(0,T;V)$ and $u'_m \rightharpoonup u'$ in $L^2(0,T;V')$, where the $u_m \in V_m$. Pick $v_n \in V_n$ such that $v_n \to v$ in $L^2(0,T;V).$ Note that $v_n = \sum_{j=1}^n a_{j,n}w_j$ where $w_j$ is an o.n basis of $H$ and o.g basis of $V$.

We obtain from (a) the equation $$-\int_0^T\langle u_m, v_n' \rangle_{V',V} + \int_0^Ta(u_m, v_n) = \int_0^T\langle f, v_n \rangle + (u_m(0),v_n(0))_H$$ by IBP (and recall that $v \in C^1(0,T;V) \subset L^2(0,T;V)$ with $v(T) = 0$.)

My problem is, how to obtain (2) from here? Specificially, I don't understand why it is true that $$\int_0^T \langle u_m, v'_n \rangle \to \int_0^T \langle u, v' \rangle_{V'V}.$$

I am fine with the other terms in the equation. Why is true that $v'_n \to v$ in $L^2(0,T;V)$?

OK, so $v'$ exists and $v_n'$ probably exists assuming we can differentiate the coefficients. But why does it convergence to $v'$? No explanation is given anywhere.

I asked this question here: https://math.stackexchange.com/questions/416885/checking-initial-condition-of-pde-is-satisfied-in-galerkin-method

But I did not receive the solution so I post it here.

The following is an argument I read, it is part of a proof of checking that the solution given by the Galerkin method satisfies the initial conditions.

From our weak formulation $$\langle u', v \rangle_{V',V} + a(u, v) = \langle f, v \rangle_{V',V}$$ which holds for all $v \in L^2(0,T;V)$, we can pick $v \in C^1(0,T;V)$ with $v(T) = 0$ and integrate this equation in time and use integration by parts to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u(0),v(0))_H.\tag{1}$$

Now we must use our Galerkin approximation $$\int_0^T\langle u_m', v_n \rangle_{V',V} + \int_0^Ta(u_m,v_n) = \int_0^T\langle f, v_n \rangle_{V',V}\tag{a}$$ and do some integration by parts and pass to the limit to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u_0,v(0))_H\tag{2}$$ where $v \in C^1(0,T;H)$ with $v(T) = 0$. This works because we know $u_m(0) \to u_0$, the initial condition.

Comparing (1) and (2), we can conclude $u(0) = u_0$.

Reasoning to obtain equation (2)

We have the convergences $u_m \rightharpoonup u$ in $L^2(0,T;V)$ and $u'_m \rightharpoonup u'$ in $L^2(0,T;V')$, where the $u_m \in V_m$. Pick $v_n \in V_n$ such that $v_n \to v$ in $L^2(0,T;V).$ Note that $v_n = \sum_{j=1}^n a_{j,n}w_j$ where $w_j$ is an o.n basis of $H$ and o.g basis of $V$.

We obtain from (a) the equation $$-\int_0^T\langle u_m, v_n' \rangle_{V',V} + \int_0^Ta(u_m, v_n) = \int_0^T\langle f, v_n \rangle + (u_m(0),v_n(0))_H$$ by IBP (and recall that $v \in C^1(0,T;V) \subset L^2(0,T;V)$ with $v(T) = 0$.)

My problem is, how to obtain (2) from here? Specificially, I don't understand why it is true that $$\int_0^T \langle u_m, v'_n \rangle \to \int_0^T \langle u, v' \rangle_{V'V}.$$

I am fine with the other terms in the equation. Why is true that $v'_n \to v$ in $L^2(0,T;V)$?

OK, so $v'$ exists and $v_n'$ probably exists assuming we can differentiate the coefficients. But why does it convergence to $v'$? No explanation is given anywhere.

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mattw
mattw
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Checking initial condition of PDE is satisfied in Galerkin method

I asked this question here: http://math.stackexchange.com/questions/416885/checking-initial-condition-of-pde-is-satisfied-in-galerkin-method

But I did not receive the solution so I post it here.

The following is an argument I read, it is part of a proof of checking that the solution given by the Galerkin method satisfies the initial conditions.

From our weak formulation $$\langle u', v \rangle_{V',V} + a(u, v) = \langle f, v \rangle_{V',V}$$ which holds for all $v \in L^2(0,T;V)$, we can pick $v \in C^1(0,T;V)$ with $v(T) = 0$ and integrate this equation in time and use integration by parts to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u(0),v(0))_H.\tag{1}$$

Now we must use our Galerkin approximation $$\int_0^T\langle u_m', v_n \rangle_{V',V} + \int_0^Ta(u_m,v_n) = \int_0^T\langle f, v_n \rangle_{V',V}\tag{a}$$ and do some integration by parts and pass to the limit to obtain $$\int_0^T -\langle u, v' \rangle_{V',V} + \int_0^T a(u,v) = \int_0^T\langle f, v \rangle_{V'V} + (u_0,v(0))_H\tag{2}$$ where $v \in C^1(0,T;H)$ with $v(T) = 0$. This works because we know $u_m(0) \to u_0$, the initial condition.

Comparing (1) and (2), we can conclude $u(0) = u_0$.

Reasoning to obtain equation (2)

We have the convergences $u_m \rightharpoonup u$ in $L^2(0,T;V)$ and $u'_m \rightharpoonup u'$ in $L^2(0,T;V')$, where the $u_m \in V_m$. Pick $v_n \in V_n$ such that $v_n \to v$ in $L^2(0,T;V).$ Note that $v_n = \sum_{j=1}^n a_{j,n}w_j$ where $w_j$ is an o.n basis of $H$ and o.g basis of $V$.

We obtain from (a) the equation $$-\int_0^T\langle u_m, v_n' \rangle_{V',V} + \int_0^Ta(u_m, v_n) = \int_0^T\langle f, v_n \rangle + (u_m(0),v_n(0))_H$$ by IBP (and recall that $v \in C^1(0,T;V) \subset L^2(0,T;V)$ with $v(T) = 0$.)

My problem is, how to obtain (2) from here? Specificially, I don't understand why it is true that $$\int_0^T \langle u_m, v'_n \rangle \to \int_0^T \langle u, v' \rangle_{V'V}.$$

I am fine with the other terms in the equation. Why is true that $v'_n \to v$ in $L^2(0,T;V)$?

OK, so $v'$ exists and $v_n'$ probably exists assuming we can differentiate the coefficients. But why does it convergence to $v'$? No explanation is given anywhere.