Mixing
Using Euler's method, solve system of differential equations
$begingroup$
Using Euler's method, solve system of differential equations $Y'=AY$ if $$A= begin{bmatrix}
8 & -1 & 5 \
-2 & 3 & 1 \
4 & -1 & -1 \
end{bmatrix}$$
$$-$$ I've been reading about this method, and I saw that all examples with this metho have some time step $h$ or initial value $y(t_0)=y_0$, but here I don't have anything. Can I solve this like Nonhomogeneous constant-coefficient linear differential equations or to solve this with eigenvalues(I heard about this way, but I don't know how to do that)..
linear-algebra ordinary-differential-equations
asked Jan 31 at 20:25
FiggaroFiggaro
406
$endgroup$
add a comment |
$begingroup$
Using Euler's method, solve system of differential equations $Y'=AY$ if $$A= begin{bmatrix}
8 & -1 & 5 \
-2 & 3 & 1 \
4 & -1 & -1 \
end{bmatrix}$$
$$-$$ I've been reading about this method, and I saw that all examples with this metho have some time step $h$ or initial value $y(t_0)=y_0$, but here I don't have anything. Can I solve this like Nonhomogeneous constant-coefficient linear differential equations or to solve this with eigenvalues(I heard about this way, but I don't know how to do that)..
linear-algebra ordinary-differential-equations
asked Jan 31 at 20:25
FiggaroFiggaro
406
$endgroup$
2
$begingroup$
As you observed, the task description is incomplete without an initial value. One could take as canonical place-holder the identity matrix $Y_0=I$. This then would approximate the fundamental (matrix-valued) solution.
$endgroup$
– LutzL
Jan 31 at 21:35
add a comment |
$begingroup$
Using Euler's method, solve system of differential equations $Y'=AY$ if $$A= begin{bmatrix}
8 & -1 & 5 \
-2 & 3 & 1 \
4 & -1 & -1 \
end{bmatrix}$$
$$-$$ I've been reading about this method, and I saw that all examples with this metho have some time step $h$ or initial value $y(t_0)=y_0$, but here I don't have anything. Can I solve this like Nonhomogeneous constant-coefficient linear differential equations or to solve this with eigenvalues(I heard about this way, but I don't know how to do that)..
linear-algebra ordinary-differential-equations
asked Jan 31 at 20:25
FiggaroFiggaro
406
$endgroup$
Using Euler's method, solve system of differential equations $Y'=AY$ if $$A= begin{bmatrix}
8 & -1 & 5 \
-2 & 3 & 1 \
4 & -1 & -1 \
end{bmatrix}$$
$$-$$ I've been reading about this method, and I saw that all examples with this metho have some time step $h$ or initial value $y(t_0)=y_0$, but here I don't have anything. Can I solve this like Nonhomogeneous constant-coefficient linear differential equations or to solve this with eigenvalues(I heard about this way, but I don't know how to do that)..
linear-algebra ordinary-differential-equations
linear-algebra ordinary-differential-equations
asked Jan 31 at 20:25
FiggaroFiggaro
406
asked Jan 31 at 20:25
FiggaroFiggaro
406
asked Jan 31 at 20:25
FiggaroFiggaro
406
asked Jan 31 at 20:25
FiggaroFiggaro
406
asked Jan 31 at 20:25
FiggaroFiggaro
406
406
2
$begingroup$
As you observed, the task description is incomplete without an initial value. One could take as canonical place-holder the identity matrix $Y_0=I$. This then would approximate the fundamental (matrix-valued) solution.
$endgroup$
– LutzL
Jan 31 at 21:35
add a comment |
2
$begingroup$
As you observed, the task description is incomplete without an initial value. One could take as canonical place-holder the identity matrix $Y_0=I$. This then would approximate the fundamental (matrix-valued) solution.
$endgroup$
– LutzL
Jan 31 at 21:35
2
2
$begingroup$
As you observed, the task description is incomplete without an initial value. One could take as canonical place-holder the identity matrix $Y_0=I$. This then would approximate the fundamental (matrix-valued) solution.
$endgroup$
– LutzL
Jan 31 at 21:35
$begingroup$
As you observed, the task description is incomplete without an initial value. One could take as canonical place-holder the identity matrix $Y_0=I$. This then would approximate the fundamental (matrix-valued) solution.
$endgroup$
– LutzL
Jan 31 at 21:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
While you cannot solve it without an initial value, at least you can write down the recurrence.
Let $SJS^{-1}$ be a diagonalization of $A$.
Given an initial condition $y_{0}=Y(t_{0})$ and a step size $h$, Euler's method defines the sequence of iterates $y_{1},y_{2},ldots$ by
$$
y_{n+1}=left(I+hAright)y_{n}qquadtext{for }ngeq0.
$$
Equivalently,
$$
y_{n}=left(I+hAright)^{n}y_{0}qquadtext{for }ngeq0.
$$
Using the fact that
$$
I+hA=SS^{-1}+hSJS^{-1}=Sleft(I+hJright)S^{-1}
$$
we get
$$
y_{n}=Sleft(I+hJright)^{n}S^{-1}y_{0}qquadtext{for }ngeq0.
$$
As for the actual values of $S$ and $J$, according to Wolfram Alpha
they are
$$
Sapproxbegin{pmatrix}-0.502512 & -5.7963 & 2.63214\
-0.354525 & -26.7138 & -0.598336\
1 & 1 & 1
end{pmatrix}
$$
and
$$I+hJapproxbegin{pmatrix}1-2.65552h\
& 1+2.52861h\
& & 1+10.1269h
end{pmatrix}.
$$
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
While you cannot solve it without an initial value, at least you can write down the recurrence.
Let $SJS^{-1}$ be a diagonalization of $A$.
Given an initial condition $y_{0}=Y(t_{0})$ and a step size $h$, Euler's method defines the sequence of iterates $y_{1},y_{2},ldots$ by
$$
y_{n+1}=left(I+hAright)y_{n}qquadtext{for }ngeq0.
$$
Equivalently,
$$
y_{n}=left(I+hAright)^{n}y_{0}qquadtext{for }ngeq0.
$$
Using the fact that
$$
I+hA=SS^{-1}+hSJS^{-1}=Sleft(I+hJright)S^{-1}
$$
we get
$$
y_{n}=Sleft(I+hJright)^{n}S^{-1}y_{0}qquadtext{for }ngeq0.
$$
As for the actual values of $S$ and $J$, according to Wolfram Alpha
they are
$$
Sapproxbegin{pmatrix}-0.502512 & -5.7963 & 2.63214\
-0.354525 & -26.7138 & -0.598336\
1 & 1 & 1
end{pmatrix}
$$
and
$$I+hJapproxbegin{pmatrix}1-2.65552h\
& 1+2.52861h\
& & 1+10.1269h
end{pmatrix}.
$$
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
$endgroup$
add a comment |
$begingroup$
While you cannot solve it without an initial value, at least you can write down the recurrence.
Let $SJS^{-1}$ be a diagonalization of $A$.
Given an initial condition $y_{0}=Y(t_{0})$ and a step size $h$, Euler's method defines the sequence of iterates $y_{1},y_{2},ldots$ by
$$
y_{n+1}=left(I+hAright)y_{n}qquadtext{for }ngeq0.
$$
Equivalently,
$$
y_{n}=left(I+hAright)^{n}y_{0}qquadtext{for }ngeq0.
$$
Using the fact that
$$
I+hA=SS^{-1}+hSJS^{-1}=Sleft(I+hJright)S^{-1}
$$
we get
$$
y_{n}=Sleft(I+hJright)^{n}S^{-1}y_{0}qquadtext{for }ngeq0.
$$
As for the actual values of $S$ and $J$, according to Wolfram Alpha
they are
$$
Sapproxbegin{pmatrix}-0.502512 & -5.7963 & 2.63214\
-0.354525 & -26.7138 & -0.598336\
1 & 1 & 1
end{pmatrix}
$$
and
$$I+hJapproxbegin{pmatrix}1-2.65552h\
& 1+2.52861h\
& & 1+10.1269h
end{pmatrix}.
$$
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
$endgroup$
add a comment |
$begingroup$
While you cannot solve it without an initial value, at least you can write down the recurrence.
Let $SJS^{-1}$ be a diagonalization of $A$.
Given an initial condition $y_{0}=Y(t_{0})$ and a step size $h$, Euler's method defines the sequence of iterates $y_{1},y_{2},ldots$ by
$$
y_{n+1}=left(I+hAright)y_{n}qquadtext{for }ngeq0.
$$
Equivalently,
$$
y_{n}=left(I+hAright)^{n}y_{0}qquadtext{for }ngeq0.
$$
Using the fact that
$$
I+hA=SS^{-1}+hSJS^{-1}=Sleft(I+hJright)S^{-1}
$$
we get
$$
y_{n}=Sleft(I+hJright)^{n}S^{-1}y_{0}qquadtext{for }ngeq0.
$$
As for the actual values of $S$ and $J$, according to Wolfram Alpha
they are
$$
Sapproxbegin{pmatrix}-0.502512 & -5.7963 & 2.63214\
-0.354525 & -26.7138 & -0.598336\
1 & 1 & 1
end{pmatrix}
$$
and
$$I+hJapproxbegin{pmatrix}1-2.65552h\
& 1+2.52861h\
& & 1+10.1269h
end{pmatrix}.
$$
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
$endgroup$
While you cannot solve it without an initial value, at least you can write down the recurrence.
Let $SJS^{-1}$ be a diagonalization of $A$.
Given an initial condition $y_{0}=Y(t_{0})$ and a step size $h$, Euler's method defines the sequence of iterates $y_{1},y_{2},ldots$ by
$$
y_{n+1}=left(I+hAright)y_{n}qquadtext{for }ngeq0.
$$
Equivalently,
$$
y_{n}=left(I+hAright)^{n}y_{0}qquadtext{for }ngeq0.
$$
Using the fact that
$$
I+hA=SS^{-1}+hSJS^{-1}=Sleft(I+hJright)S^{-1}
$$
we get
$$
y_{n}=Sleft(I+hJright)^{n}S^{-1}y_{0}qquadtext{for }ngeq0.
$$
As for the actual values of $S$ and $J$, according to Wolfram Alpha
they are
$$
Sapproxbegin{pmatrix}-0.502512 & -5.7963 & 2.63214\
-0.354525 & -26.7138 & -0.598336\
1 & 1 & 1
end{pmatrix}
$$
and
$$I+hJapproxbegin{pmatrix}1-2.65552h\
& 1+2.52861h\
& & 1+10.1269h
end{pmatrix}.
$$
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
answered Feb 1 at 4:02
parsiadparsiad
18.7k32453
18.7k32453
add a comment |
add a comment |
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2
$begingroup$
As you observed, the task description is incomplete without an initial value. One could take as canonical place-holder the identity matrix $Y_0=I$. This then would approximate the fundamental (matrix-valued) solution.
$endgroup$
– LutzL
Jan 31 at 21:35