Nth order linear ode, why do we have n general solutions?

[popopopd]

hi, I looked up the existence and uniqueness of nth order linear ode and I grasped the idea of them, but still kind of confused why we get n numbers of general solutions.

 

[SteamKing]

What does the characteristic equation for the ODE tell you?

http://en.wikipedia.org/wiki/Characteristic_equation_(calculus)

 

[HallsofIvy]

A basic property of linear homogeneous equations is that the set of solutions forms a vector space. That is, any linear combination of solutions, a_1y_1+ a_2y_2 is again a solution. One can show that, for an nth order homogeneous differential equation, this vector space has dimension n. That is, there exist n independent solutions such that any solution can be written in terms of those n solutions.

To prove that, consider the differential equation a_n(x)y^{(n)}(x)+ a_{n-1}y^{(n-1)}(x)+ \cdot\cdot\cdot+ a_2y''(x)+ a_1y'(x)+ a_0y(x)= 0.
with the following initial values:

I) y(0)= 1, y'(0)= y''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
By the fundamental "existence and uniqueness" theorem for initial value problems, there exist a unique function, y_0(x), satisfying the differential equation and those initial conditions.

II) y(0)= 0, y'(0)= 1, y''(0)= y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_1(x), satisfying the differential equation and those initial conditions.

III) y(0)= y'(0)= 0, y''(0)= 1, y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_2(x), satisfying the differential equation and those initial conditions.

Continue that, shifting the "= 1" through the derivatives until we get to
X) y(0)= y'(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0, y^{(n-1)}(0)= 1.

First, any solution to the differential equation can written as a linear combination of those n solutions:
Suppose y(x) is a solution to the differential equation. Let A_0= y(0), A_1= y'(0), etc. until A_{n-1}= y^{(n-1)}(0).
Then y(x)= A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x). That can be shown by evaluating both sides, and their derivatives, at x= 0.

Further, that set of n solutions are independent. To see that suppose that, for some numbers, A_0, A_1, \cdot\cdot\cdot, A_{n-1}, A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)= 0- that is, is equal to 0 for all x. Taking x= 0 we must have A_0(1)+ A_1(0)+ \cdot\cdot\cdot+ A_{n-1}(0)= A_0= 0. Since that linear combination is 0 for all x, it is a constant and its derivative is 0 for all x. That is A_0y_0'(x)+ A_1y_1'(0)+ \cdot\cdot\cdot+ A_{n-1}y'(0)= 0 for all x. Set x= 0 to see that A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each A_i, in turn, is equal to 0.

 

[popopopd]

A basic property of linear homogeneous equations is that the set of solutions forms a vector space. That is, any linear combination of solutions, a_1y_1+ a_2y_2 is again a solution. One can show that, for an nth order homogeneous differential equation, this vector space has dimension n. That is, there exist n independent solutions such that any solution can be written in terms of those n solutions.

To prove that, consider the differential equation a_n(x)y^{(n)}(x)+ a_{n-1}y^{(n-1)}(x)+ \cdot\cdot\cdot+ a_2y''(x)+ a_1y'(x)+ a_0y(x)= 0.
with the following initial values:

I) y(0)= 1, y'(0)= y''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
By the fundamental "existence and uniqueness" theorem for initial value problems, there exist a unique function, y_0(x), satisfying the differential equation and those initial conditions.

II) y(0)= 0, y'(0)= 1, y''(0)= y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_1(x), satisfying the differential equation and those initial conditions.

III) y(0)= y'(0)= 0, y''(0)= 1, y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_2(x), satisfying the differential equation and those initial conditions.

Continue that, shifting the "= 1" through the derivatives until we get to
X) y(0)= y'(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0, y^{(n-1)}(0)= 1.

First, any solution to the differential equation can written as a linear combination of those n solutions:
Suppose y(x) is a solution to the differential equation. Let A_0= y(0), A_1= y'(0), etc. until A_{n-1}= y^{(n-1)}(0).
Then y(x)= A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x). That can be shown by evaluating both sides, and their derivatives, at x= 0.

Further, that set of n solutions are independent. To see that suppose that, for some numbers, A_0, A_1, \cdot\cdot\cdot, A_{n-1}, A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)= 0- that is, is equal to 0 for all x. Taking x= 0 we must have A_0(1)+ A_1(0)+ \cdot\cdot\cdot+ A_{n-1}(0)= A_0= 0. Since that linear combination is 0 for all x, it is a constant and its derivative is 0 for all x. That is A_0y_0'(x)+ A_1y_1'(0)+ \cdot\cdot\cdot+ A_{n-1}y'(0)= 0 for all x. Set x= 0 to see that A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each A_i, in turn, is equal to 0.

<br /> Thanks! it helped alot!

 

[popopopd]

A basic property of linear homogeneous equations is that the set of solutions forms a vector space. That is, any linear combination of solutions, a_1y_1+ a_2y_2 is again a solution. One can show that, for an nth order homogeneous differential equation, this vector space has dimension n. That is, there exist n independent solutions such that any solution can be written in terms of those n solutions.

To prove that, consider the differential equation a_n(x)y^{(n)}(x)+ a_{n-1}y^{(n-1)}(x)+ \cdot\cdot\cdot+ a_2y&#039;&#039;(x)+ a_1y&#039;(x)+ a_0y(x)= 0.
with the following initial values:

I) y(0)= 1, y&#039;(0)= y&#039;&#039;(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
By the fundamental "existence and uniqueness" theorem for initial value problems, there exist a unique function, y_0(x), satisfying the differential equation and those initial conditions.

II) y(0)= 0, y&#039;(0)= 1, y&#039;&#039;(0)= y&#039;&#039;&#039;(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_1(x), satisfying the differential equation and those initial conditions.

III) y(0)= y&#039;(0)= 0, y&#039;&#039;(0)= 1, y&#039;&#039;&#039;(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_2(x), satisfying the differential equation and those initial conditions.

Continue that, shifting the "= 1" through the derivatives until we get to
X) y(0)= y&#039;(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0, y^{(n-1)}(0)= 1.

First, any solution to the differential equation can written as a linear combination of those n solutions:
Suppose y(x) is a solution to the differential equation. Let A_0= y(0), A_1= y&#039;(0), etc. until A_{n-1}= y^{(n-1)}(0).
Then y(x)= A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x). That can be shown by evaluating both sides, and their derivatives, at x= 0.

Further, that set of n solutions are independent. To see that suppose that, for some numbers, A_0, A_1, \cdot\cdot\cdot, A_{n-1}, A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)= 0- that is, is equal to 0 for all x. Taking x= 0 we must have A_0(1)+ A_1(0)+ \cdot\cdot\cdot+ A_{n-1}(0)= A_0= 0. Since that linear combination is 0 for all x, it is a constant and its derivative is 0 for all x. That is A_0y_0&#039;(x)+ A_1y_1&#039;(0)+ \cdot\cdot\cdot+ A_{n-1}y&#039;(0)= 0 for all x. Set x= 0 to see that A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each A_i, in turn, is equal to 0.

<script class="js-extraPhrases" type="application/json"> { "lightbox_close": "Close", "lightbox_next": "Next", "lightbox_previous": "Previous", "lightbox_error": "The requested content cannot be loaded. Please try again later.", "lightbox_start_slideshow": "Start slideshow", "lightbox_stop_slideshow": "Stop slideshow", "lightbox_full_screen": "Full screen", "lightbox_thumbnails": "Thumbnails", "lightbox_download": "Download", "lightbox_share": "Share", "lightbox_zoom": "Zoom", "lightbox_new_window": "New window", "lightbox_toggle_sidebar": "Toggle sidebar" } </script> <div class="bbImageWrapper js-lbImage" title="diff14.png" data-src="https://www.physicsforums.com/attachments/diff14-png.179218/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/diff14-png.179218/" data-url="" class="bbImage" data-zoom-target="1" style="" alt="diff14.png" title="diff14.png" width="414" height="54" loading="lazy" decoding="async" /> </div> (5)<br /> <div class="bbImageWrapper js-lbImage" title="diff27.png" data-src="https://www.physicsforums.com/attachments/diff27-png.179219/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/diff27-png.179219/" data-url="" class="bbImage" data-zoom-target="1" style="" alt="diff27.png" title="diff27.png" width="527" height="184" loading="lazy" decoding="async" /> </div> (17)<br /> <br /> - using picard&#039;s iteration in vector form, to prove nth order linear ODE&#039;s existence &amp; uniqueness.<br /> <br /> ex<br /> <div class="bbImageWrapper js-lbImage" title="diff31.png" data-src="https://www.physicsforums.com/attachments/diff31-png.179220/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/diff31-png.179220/" data-url="" class="bbImage" data-zoom-target="1" style="" alt="diff31.png" title="diff31.png" width="360" height="79" loading="lazy" decoding="async" /> </div> (21)<br /> <br /> <div class="bbImageWrapper js-lbImage" title="diff32.png" data-src="https://www.physicsforums.com/attachments/diff32-png.179221/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/diff32-png.179221/" data-url="" class="bbImage" data-zoom-target="1" style="" alt="diff32.png" title="diff32.png" width="425" height="443" loading="lazy" decoding="async" /> </div> (22)<br /> <br /> (<a href="http://ghebook.blogspot.ca/2011/10/differential-equation.html" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://ghebook.blogspot.ca/2011/10/differential-equation.html</a>)Hi, I actually did picard&#039;s iteration and found that without n initial conditions, nth order linear ODE will have n number of constants as we assume initial conditions are some arbitrary constants.<br /> <br /> since function spaces are vector spaces, solutions span n dimensional vector space (not very sure of this)<br /> <br /> If we do picard&#039;s iteration,<br /> <br /> y^(n-1)=y₀^(n-1)+∫y⁽ⁿ⁾dx<br /> y^(n-2)=y₀^(n-2)+∫y^(n-1)dx<br /> =y^(n-2)=y₀^(n-2)+∫y₀^(n-1)+∫y⁽ⁿ⁾dx<br /> .<br /> .<br /> .<br /> iteration goes on and on until the error is sufficiently decreased.if we assume each initial conditions are some constants, we will eventually sort out the solution function y w.r.t constants after sufficient number of iteration is done. Then it will look like,<br /> <br /> y (c₀ c₁ c₂ - - - - - - )[y1 y2 y3 y4 y5 - - - - - - ] &lt;-- (should be vertical)which is in the form of y = c1y1+c2y2+c3y3 . . .<br /> <br /> and so on<br /> is this correct?<br /> If not, how can we show that solution space has n number of basis?---------------------------------------------------------------------------------------------------------------------------<br /> <br /> Also, I have two questions about Strum Liouville 2nd order ODE.1. if we look at the Strum-Liouville 2nd order ODEs, there is an eigenvalue term within the equation. it seems like we are adding one more constant in the equation, which imposes a restriction to find a solution (n+1 constants with n conditions).<br /> <br /> [m(x)y&#039;]+[λr(x)-q(x)]y<br /> =m(x)[y&#039;&#039;+P(x)y&#039;+Q(x)y]<br /> =m(x)[y&#039;&#039;+P(x)y&#039;+(λr(x)-q(x))]<br /> =0<br /> <br /> ∴ [y&#039;&#039;+P(x)y&#039;+(λr(x)-q(x))]=0<br /> <br /> if we do Picard&#039;s iteration, then we have one more constant λ along with n constants..2. I don&#039;t understand how eigenvalue directly influence the solutions of 2nd order ODEThanks always, Your answers help me a lot!