Differential equations to eigenvalues, from Strang's textbook
$begingroup$
I am a little stumped about something in Strangs's (great) textbook "Linear algebra and it's Applications".
At the beginning of chapter 5 (Eigenvalues, page 233) he starts off with an example that connects an eigenvalue problem to systems of differential equations.
$$ begin{cases}frac{dv}{dt} = 4v - 5w\
frac{dw}{dt} = 2v - 3w end{cases}$$
He expresses the probelem as a matrix multiplication, clear.
$$begin{bmatrix}dv/dt \ dw/dtend{bmatrix} = begin{bmatrix}4 & -5\2 & -3end{bmatrix} begin{bmatrix}v \ wend{bmatrix}$$
Now this is where things are confusing to me. He expresses the matrix equation using only the symbols (A, u) and then shows us how we would handle the situation if instead of matrices and vectors we had simple scalars, which would be a basic differential equation that leads to a pure exponential solution.
$$frac{du}{dt} = mathbf{A} mathit{u}$$
$$frac{du}{dt} = cu to u(t) = e^{at}u_0$$
He then expresses the functions v and w as these pure exponential solutions, I don't understand why he can do this however because the original equations also contained terms with the other variable.
$$begin{cases}v(t) = e^{lambda t}y\
u(t) = e^{lambda t}zend{cases}$$
He then substitutes the pure exponentials into the original system of equations and from here on out it is pretty clear.
$$begin{cases}lambda e^{lambda t} y = 4e^{lambda t}y - 5e^{lambda t}z\
lambda e^{lambda t} z = 2e^{lambda t}y - 3e^{lambda t}zend{cases}$$
My question is basically, how do we get from the original equations to the pure exponentials? Why can the expressions for v(t) and w(t) be made with single terms, while the initial expressions each contain 2 terms?
ordinary-differential-equations eigenvalues-eigenvectors systems-of-equations
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
$endgroup$
|
show 2 more comments
$begingroup$
I am a little stumped about something in Strangs's (great) textbook "Linear algebra and it's Applications".
At the beginning of chapter 5 (Eigenvalues, page 233) he starts off with an example that connects an eigenvalue problem to systems of differential equations.
$$ begin{cases}frac{dv}{dt} = 4v - 5w\
frac{dw}{dt} = 2v - 3w end{cases}$$
He expresses the probelem as a matrix multiplication, clear.
$$begin{bmatrix}dv/dt \ dw/dtend{bmatrix} = begin{bmatrix}4 & -5\2 & -3end{bmatrix} begin{bmatrix}v \ wend{bmatrix}$$
Now this is where things are confusing to me. He expresses the matrix equation using only the symbols (A, u) and then shows us how we would handle the situation if instead of matrices and vectors we had simple scalars, which would be a basic differential equation that leads to a pure exponential solution.
$$frac{du}{dt} = mathbf{A} mathit{u}$$
$$frac{du}{dt} = cu to u(t) = e^{at}u_0$$
He then expresses the functions v and w as these pure exponential solutions, I don't understand why he can do this however because the original equations also contained terms with the other variable.
$$begin{cases}v(t) = e^{lambda t}y\
u(t) = e^{lambda t}zend{cases}$$
He then substitutes the pure exponentials into the original system of equations and from here on out it is pretty clear.
$$begin{cases}lambda e^{lambda t} y = 4e^{lambda t}y - 5e^{lambda t}z\
lambda e^{lambda t} z = 2e^{lambda t}y - 3e^{lambda t}zend{cases}$$
My question is basically, how do we get from the original equations to the pure exponentials? Why can the expressions for v(t) and w(t) be made with single terms, while the initial expressions each contain 2 terms?
ordinary-differential-equations eigenvalues-eigenvectors systems-of-equations
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
$endgroup$
$begingroup$
Using the left eigenvectors you get $(dot v-dot w)=2(v-w)$ and $(2dot v-5dot w)=-(2u-5w)$, which can be solved as independent scalar equations. Then recombine.
$endgroup$
– LutzL
Jan 24 at 11:57
$begingroup$
I understand enough to carry out the method to solve the system (calculate eigenvectors etc). What I don't understand is how the system is initially converted to the exponential system.
$endgroup$
– BioHazZzZard
Jan 24 at 17:54
$begingroup$
I do not understand that last comment. From $dot u=Au$ you get $ddot u=Adot u=A^2u$ etc. to $u^{(k)}=A^ku$. Inserting into the Taylor series gives $u(t)=(I+sum_{k=1}^infty frac{t^k}{k!}A^k)u_0$, and that is the exponential series, $u(t)=exp(tA)u_0$.
$endgroup$
– LutzL
Jan 25 at 11:50
$begingroup$
Basically, how does one get from the initial system of equations to the system that contains the exponential terms?
$endgroup$
– BioHazZzZard
Jan 29 at 18:30
$begingroup$
You take $u(t)=exp(At)u_0$ and explore what happens when you express $u_0$ in an eigenbasis of $A$. Eigenspaces of $A$ stay invariant spaces for any analytical expression $f(A)$.
$endgroup$
– LutzL
Jan 29 at 18:50
|
show 2 more comments
$begingroup$
I am a little stumped about something in Strangs's (great) textbook "Linear algebra and it's Applications".
At the beginning of chapter 5 (Eigenvalues, page 233) he starts off with an example that connects an eigenvalue problem to systems of differential equations.
$$ begin{cases}frac{dv}{dt} = 4v - 5w\
frac{dw}{dt} = 2v - 3w end{cases}$$
He expresses the probelem as a matrix multiplication, clear.
$$begin{bmatrix}dv/dt \ dw/dtend{bmatrix} = begin{bmatrix}4 & -5\2 & -3end{bmatrix} begin{bmatrix}v \ wend{bmatrix}$$
Now this is where things are confusing to me. He expresses the matrix equation using only the symbols (A, u) and then shows us how we would handle the situation if instead of matrices and vectors we had simple scalars, which would be a basic differential equation that leads to a pure exponential solution.
$$frac{du}{dt} = mathbf{A} mathit{u}$$
$$frac{du}{dt} = cu to u(t) = e^{at}u_0$$
He then expresses the functions v and w as these pure exponential solutions, I don't understand why he can do this however because the original equations also contained terms with the other variable.
$$begin{cases}v(t) = e^{lambda t}y\
u(t) = e^{lambda t}zend{cases}$$
He then substitutes the pure exponentials into the original system of equations and from here on out it is pretty clear.
$$begin{cases}lambda e^{lambda t} y = 4e^{lambda t}y - 5e^{lambda t}z\
lambda e^{lambda t} z = 2e^{lambda t}y - 3e^{lambda t}zend{cases}$$
My question is basically, how do we get from the original equations to the pure exponentials? Why can the expressions for v(t) and w(t) be made with single terms, while the initial expressions each contain 2 terms?
ordinary-differential-equations eigenvalues-eigenvectors systems-of-equations
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
$endgroup$
I am a little stumped about something in Strangs's (great) textbook "Linear algebra and it's Applications".
At the beginning of chapter 5 (Eigenvalues, page 233) he starts off with an example that connects an eigenvalue problem to systems of differential equations.
$$ begin{cases}frac{dv}{dt} = 4v - 5w\
frac{dw}{dt} = 2v - 3w end{cases}$$
He expresses the probelem as a matrix multiplication, clear.
$$begin{bmatrix}dv/dt \ dw/dtend{bmatrix} = begin{bmatrix}4 & -5\2 & -3end{bmatrix} begin{bmatrix}v \ wend{bmatrix}$$
Now this is where things are confusing to me. He expresses the matrix equation using only the symbols (A, u) and then shows us how we would handle the situation if instead of matrices and vectors we had simple scalars, which would be a basic differential equation that leads to a pure exponential solution.
$$frac{du}{dt} = mathbf{A} mathit{u}$$
$$frac{du}{dt} = cu to u(t) = e^{at}u_0$$
He then expresses the functions v and w as these pure exponential solutions, I don't understand why he can do this however because the original equations also contained terms with the other variable.
$$begin{cases}v(t) = e^{lambda t}y\
u(t) = e^{lambda t}zend{cases}$$
He then substitutes the pure exponentials into the original system of equations and from here on out it is pretty clear.
$$begin{cases}lambda e^{lambda t} y = 4e^{lambda t}y - 5e^{lambda t}z\
lambda e^{lambda t} z = 2e^{lambda t}y - 3e^{lambda t}zend{cases}$$
My question is basically, how do we get from the original equations to the pure exponentials? Why can the expressions for v(t) and w(t) be made with single terms, while the initial expressions each contain 2 terms?
ordinary-differential-equations eigenvalues-eigenvectors systems-of-equations
ordinary-differential-equations eigenvalues-eigenvectors systems-of-equations
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
asked Jan 24 at 11:44
BioHazZzZardBioHazZzZard
1
1
$begingroup$
Using the left eigenvectors you get $(dot v-dot w)=2(v-w)$ and $(2dot v-5dot w)=-(2u-5w)$, which can be solved as independent scalar equations. Then recombine.
$endgroup$
– LutzL
Jan 24 at 11:57
$begingroup$
I understand enough to carry out the method to solve the system (calculate eigenvectors etc). What I don't understand is how the system is initially converted to the exponential system.
$endgroup$
– BioHazZzZard
Jan 24 at 17:54
$begingroup$
I do not understand that last comment. From $dot u=Au$ you get $ddot u=Adot u=A^2u$ etc. to $u^{(k)}=A^ku$. Inserting into the Taylor series gives $u(t)=(I+sum_{k=1}^infty frac{t^k}{k!}A^k)u_0$, and that is the exponential series, $u(t)=exp(tA)u_0$.
$endgroup$
– LutzL
Jan 25 at 11:50
$begingroup$
Basically, how does one get from the initial system of equations to the system that contains the exponential terms?
$endgroup$
– BioHazZzZard
Jan 29 at 18:30
$begingroup$
You take $u(t)=exp(At)u_0$ and explore what happens when you express $u_0$ in an eigenbasis of $A$. Eigenspaces of $A$ stay invariant spaces for any analytical expression $f(A)$.
$endgroup$
– LutzL
Jan 29 at 18:50
|
show 2 more comments
$begingroup$
Using the left eigenvectors you get $(dot v-dot w)=2(v-w)$ and $(2dot v-5dot w)=-(2u-5w)$, which can be solved as independent scalar equations. Then recombine.
$endgroup$
– LutzL
Jan 24 at 11:57
$begingroup$
I understand enough to carry out the method to solve the system (calculate eigenvectors etc). What I don't understand is how the system is initially converted to the exponential system.
$endgroup$
– BioHazZzZard
Jan 24 at 17:54
$begingroup$
I do not understand that last comment. From $dot u=Au$ you get $ddot u=Adot u=A^2u$ etc. to $u^{(k)}=A^ku$. Inserting into the Taylor series gives $u(t)=(I+sum_{k=1}^infty frac{t^k}{k!}A^k)u_0$, and that is the exponential series, $u(t)=exp(tA)u_0$.
$endgroup$
– LutzL
Jan 25 at 11:50
$begingroup$
Basically, how does one get from the initial system of equations to the system that contains the exponential terms?
$endgroup$
– BioHazZzZard
Jan 29 at 18:30
$begingroup$
You take $u(t)=exp(At)u_0$ and explore what happens when you express $u_0$ in an eigenbasis of $A$. Eigenspaces of $A$ stay invariant spaces for any analytical expression $f(A)$.
$endgroup$
– LutzL
Jan 29 at 18:50
$begingroup$
Using the left eigenvectors you get $(dot v-dot w)=2(v-w)$ and $(2dot v-5dot w)=-(2u-5w)$, which can be solved as independent scalar equations. Then recombine.
$endgroup$
– LutzL
Jan 24 at 11:57
$begingroup$
Using the left eigenvectors you get $(dot v-dot w)=2(v-w)$ and $(2dot v-5dot w)=-(2u-5w)$, which can be solved as independent scalar equations. Then recombine.
$endgroup$
– LutzL
Jan 24 at 11:57
$begingroup$
I understand enough to carry out the method to solve the system (calculate eigenvectors etc). What I don't understand is how the system is initially converted to the exponential system.
$endgroup$
– BioHazZzZard
Jan 24 at 17:54
$begingroup$
I understand enough to carry out the method to solve the system (calculate eigenvectors etc). What I don't understand is how the system is initially converted to the exponential system.
$endgroup$
– BioHazZzZard
Jan 24 at 17:54
$begingroup$
I do not understand that last comment. From $dot u=Au$ you get $ddot u=Adot u=A^2u$ etc. to $u^{(k)}=A^ku$. Inserting into the Taylor series gives $u(t)=(I+sum_{k=1}^infty frac{t^k}{k!}A^k)u_0$, and that is the exponential series, $u(t)=exp(tA)u_0$.
$endgroup$
– LutzL
Jan 25 at 11:50
$begingroup$
I do not understand that last comment. From $dot u=Au$ you get $ddot u=Adot u=A^2u$ etc. to $u^{(k)}=A^ku$. Inserting into the Taylor series gives $u(t)=(I+sum_{k=1}^infty frac{t^k}{k!}A^k)u_0$, and that is the exponential series, $u(t)=exp(tA)u_0$.
$endgroup$
– LutzL
Jan 25 at 11:50
$begingroup$
Basically, how does one get from the initial system of equations to the system that contains the exponential terms?
$endgroup$
– BioHazZzZard
Jan 29 at 18:30
$begingroup$
Basically, how does one get from the initial system of equations to the system that contains the exponential terms?
$endgroup$
– BioHazZzZard
Jan 29 at 18:30
$begingroup$
You take $u(t)=exp(At)u_0$ and explore what happens when you express $u_0$ in an eigenbasis of $A$. Eigenspaces of $A$ stay invariant spaces for any analytical expression $f(A)$.
$endgroup$
– LutzL
Jan 29 at 18:50
$begingroup$
You take $u(t)=exp(At)u_0$ and explore what happens when you express $u_0$ in an eigenbasis of $A$. Eigenspaces of $A$ stay invariant spaces for any analytical expression $f(A)$.
$endgroup$
– LutzL
Jan 29 at 18:50
|
show 2 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085776%2fdifferential-equations-to-eigenvalues-from-strangs-textbook%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085776%2fdifferential-equations-to-eigenvalues-from-strangs-textbook%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Using the left eigenvectors you get $(dot v-dot w)=2(v-w)$ and $(2dot v-5dot w)=-(2u-5w)$, which can be solved as independent scalar equations. Then recombine.
$endgroup$
– LutzL
Jan 24 at 11:57
$begingroup$
I understand enough to carry out the method to solve the system (calculate eigenvectors etc). What I don't understand is how the system is initially converted to the exponential system.
$endgroup$
– BioHazZzZard
Jan 24 at 17:54
$begingroup$
I do not understand that last comment. From $dot u=Au$ you get $ddot u=Adot u=A^2u$ etc. to $u^{(k)}=A^ku$. Inserting into the Taylor series gives $u(t)=(I+sum_{k=1}^infty frac{t^k}{k!}A^k)u_0$, and that is the exponential series, $u(t)=exp(tA)u_0$.
$endgroup$
– LutzL
Jan 25 at 11:50
$begingroup$
Basically, how does one get from the initial system of equations to the system that contains the exponential terms?
$endgroup$
– BioHazZzZard
Jan 29 at 18:30
$begingroup$
You take $u(t)=exp(At)u_0$ and explore what happens when you express $u_0$ in an eigenbasis of $A$. Eigenspaces of $A$ stay invariant spaces for any analytical expression $f(A)$.
$endgroup$
– LutzL
Jan 29 at 18:50