MHB How can we find the coefficients?

[mathmari]

Hey!

We have the initial value problem $$u'(t)=Au(t) \ \ , \ \ 0 \leq t \leq T \\ u(0)=u^0 \\ u \in \mathbb{R}^m$$ A is a $m \times m$ matrix

The eigenvalues of $A$ are $\lambda_j$ and the corresponding eigenvectors are $\phi^{(j)}$.

The general solution of initial value problem is $$u(t)=\sum_{j=1}^m c_j e^{\lambda_jt}\phi^{(j)}$$

right??

For $t=0$ we have $$u^0=\sum_{j=1}^m c_j \phi^{(j)}$$ How can we solve for $c_j$ ?? (Wondering)

Do we maybe have to use a dot product?? (Wondering)

 

[I like Serena]

Hey!

We have the initial value problem $$u'(t)=Au(t) \ \ , \ \ 0 \leq t \leq T \\ u(0)=u^0 \\ u \in \mathbb{R}^m$$ A is a $m \times m$ matrix

The eigenvalues of $A$ are $\lambda_j$ and the corresponding eigenvectors are $\phi^{(j)}$.

The general solution of initial value problem is $$u(t)=\sum_{j=1}^m c_j e^{\lambda_jt}\phi^{(j)}$$

right??

For $t=0$ we have $$u^0=\sum_{j=1}^m c_j \phi^{(j)}$$ How can we solve for $c_j$ ?? (Wondering)

Do we maybe have to use a dot product?? (Wondering)

Hi! (Wave)

Let's make that:
$$u^0=\sum_{j=1}^m c_j \phi^{(j)} = \Big(\phi^{(j)}\Big) \begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}$$
See how we can solve it for $c_j$? (Wondering)

 

[mathmari]

Let's make that:
$$u^0=\sum_{j=1}^m c_j \phi^{(j)}(0) = \Big(\phi^{(j)}(0)\Big) \begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}$$
See how we can solve it for $c_j$? (Wondering)

Are the eigenvectors $\phi^{(j)}$ a function of $t$?? (Wondering) Because you write $\phi^{(j)}(0)$.

$\Big (\phi^{(j)}(0)\Big )$ is a matrix, isn't it?? (Wondering) So, we have to find the inverse, or not??

 

[I like Serena]

Are the eigenvectors $\phi^{(j)}$ a function of $t$?? (Wondering) Because you write $\phi^{(j)}(0)$.

No I didn't! (Blush)

$\phi^{(j)}$ is a matrix, isn't it?? (Wondering) So, we have to find the inverse, or not??

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

So, $$u^0\Big (\phi^{(j)}\Big )^{-1}=\begin{bmatrix}
c_1\\
c_2\\
\cdot\\
\cdot\\
\cdot \\
c_m
\end{bmatrix}$$ Is this correct?? (Wondering)

Now we have the vector $\begin{bmatrix}
c_1\\
c_2\\
\cdot\\
\cdot\\
\cdot \\
c_m
\end{bmatrix}$. How can we write the formula for $c_j$ ?? (Wondering)

I found in my book the following solution $$u(t)=\sum_{j=1}^m e^{\lambda t}(u(0), \phi^{(j)})\phi^{(j)}$$ where $(\cdot , \cdot)$ is the euclidean dot product.
But how did we find that?? (Wondering)

 

[I like Serena]

So, $$u^0\Big (\phi^{(j)}\Big )^{-1}=\begin{bmatrix}
c_1\\
c_2\\
\cdot\\
\cdot\\
\cdot \\
c_m
\end{bmatrix}$$ Is this correct?? (Wondering)

The product is not commutative, so that should be
$$\Big (\phi^{(j)}\Big )^{-1} u^0=\begin{bmatrix}
c_1\\
c_2\\
\cdot\\
\cdot\\
\cdot \\
c_m
\end{bmatrix}$$

Now we have the vector $\begin{bmatrix}
c_1\\
c_2\\
\cdot\\
\cdot\\
\cdot \\
c_m
\end{bmatrix}$. How can we write the formula for $c_j$ ?? (Wondering)

That is a formula for $c_j$. To simplify it, we'd need more information, like $A$ being symmetric. (Wasntme)

I found in my book the following solution $$u(t)=\sum_{j=1}^m e^{\lambda t}(u(0), \phi^{(j)})\phi^{(j)}$$ where $(\cdot , \cdot)$ is the euclidean dot product.
But how did we find that?? (Wondering)

Looks there is an assumption in there that the eigenvectors are orthonormal.
I think that is only possible if the matrix $A$ is symmetric, but that does not seem to be given - or is it? (Wondering)