Boundary Conditions Replace Initial Conditions

okay uh well my problem today is a little different because I don't have two initial conditions as we normally have for a second order differential equation instead I have two boundary conditions uh so let me show you the equation so I'm changing T to X because I'm thinking of this as a problem in space rather than in time so there's the second derivative the minus sign is for convenience this is the load but here's the new thing the the b i I'm on an interval 0 to one and at zero I let me take zero

for the two boundary conditions so my solution somehow does something like this maybe up and back down so it's zero there zero there and in between it solves the differential equation not a big difference but it it you'll see that it's entirely new type of problem okay as far as the solution to the equation goes there's nothing enormously new I still have a y particular the a particular solution who that solves the equation and then I still have the Y null the homogeneous solution any solution that solves the equation with zero on the right hand

side and in this example this is especially simple the null equation would be second derivative equals zero and those are the equ those are the functions linear functions that have second derivative equals z so there's the general solution and now I have to put in not the initial conditions but the boundary conditions okay so uh so I substitute xal 0 and I substitute x = 1 to into this I have to find y particular I'll do two examples I'll do two examples but but the general principle is to get these numbers these constants like C1

and C2 from the boundary conditions I'll put put in xal 0 and then I'll have y of 0 which is uh uh y particular at zero still Define plus C * 0 + D is that's that's the solution at the Left End which is supposed to be zero and then at the right hand end I have whatever this particular solution is at one plus plus now I'm putting in x = 1 I'm just plugging in X = 0 and then x = 1 at x = 1 I have C+ D C + D and that

gives me zero the two zeros come from there and there okay two equations they give me C and D so I'm all solved once I I know how to I know how to proceed once I find a particular solution so I'll just do two examples they'll have two particular Solutions and there are the most important examples in applications so let me uh start with the first example so my first example is going to be uh the equation minus DC y dx2 = 1 that will be my load f ofx is going to be one so

I'm looking for a particular solution to that equation and of course I can find a function whose second derivative is one or maybe minus one my function will be well if I want the second derivative to be one then probably a half X2 is the right thing and that would give me a minus so I think I have a half minus a half X2 that solves the equation and now I have the CX plus d the homogeneous the null solution and now I plug in and again I'm always taking y of Z to be zero

and also y one to be zero boundary conditions again boundary conditions not initial condition okay plug in xal 0 at xal 0 what do I learn xal Z that's zero that's zero so I learn that D is zero at x = 1 what do I learn this is minus a half D is zero now so and X is one so I think we learned C is plus a half okay with that at at xal 1 I'm supposed to get zero from the boundary condition so I have minus a half plus a half plus Z I

do get zero this is good so this answer is uh CX then is 12x - 12 x^2 that's it that's my solution that that function is zero at both ends and it's uh solves the differential equation so that's a simple example and maybe I can uh give you a application uh suppose I have a a rod here here's a bar and those lines that I put at the top and bottom are the ones that give me the boundary conditions and I have a weight a a weight just a weight of one maybe the bar itself

is it gives it will uh elastic for Gravity will pull displace the bar downwards because of its weight it's elastic and the this function gives me the this solution gives me the the the uh distribution if I go down a distance X then that tells me that this part of the bar originally at X will move down by an additional Y moves Mo so this is now at x + Y ofx and that's the Y and that is zero at the bottom zero at the top and positive in between okay that that was a pretty

quick description of an application and more to more important a pretty quick solution to the problem can I do a second example that won't be quite as easy okay so again my equation is going to be be minus the second derivative equals a load but now it will be a point load a point load that's a point load at x = a this is my friend the Delta function the Delta function you remember is zero except at that one point where this is zero this is zero at the point x equal a in my little

picture of a of a physical problem now I don't have any weight in the bar the the bar is uh weight thin weightless but I'm putting on at the point xal a right at this point I'm attaching a weight so this distance is xal a here's my weight my load hanging hanging at this point so I can see what will happen that load hanging down there will stretch the part above the bar above the load and compress the part below the load so it's a point load very important application okay now I have that equation

to solve okay I can solve it on the on one side of a x equal a and I can solve it on the other side of x equal a let me do that for X less than a I have minus the second derivative and what's the Delta function for X below a on the left side of the spike zero and X on the right side of the of the load again zero and what we what are the solutions to the null equation Y is CX + D on the left side of the load there and

now here it may have some different constants y equals uh what shall I say e x + f on the right side of the load and now I've got four numbers to find C D E and F and what do I know I know two boundary conditions always I know that y of 0 is zero from fixing the top of the bar so y of 0 equals 0 when I put in x equals 0 that will tell me D is zero and then also y of 1 is zero and that will be on this side

of the load so when I put in x = 1 that will tell me that e + f is0 at xal 1 so it tells me that f is minus E right so what do I know now D is gone zero f is minus E so can I just change this to like f is minus E so uh I had e x - e e * x -1 uh takes care of that boundary condition at x equal 1 it's gone okay but I still have two c and e to find so what what are my

two further uh conditions act the jump so so far I'm on the left of the jump the Spike the impulse the Delta function and on the right of it but now I've got to say what's happening at the impulse at the Delta function or at the point load okay well what's happening there I need two equations I've still got c and e to find so my first equation is that at that load the the bar is not going to break apart it's just going to be stretched above and compressed below but it's not going to

break apart so at the load at which is x equal a so now I'm ready for x equal a okay what happens at x equal a that's the same as that let me let me draw a picture of the solution here here is X this is X here's y here's xal 0 here's x = y I see uh a linear function CX up to the point xal a and here I have a a linear function coming back to zero you see that that's the picture of the solution the graph of the solution it has this

zero at the left boundary it has zero at the right boundary it has in between it is CX and e xus e and I have made it continuous at x equal a the the bar is not coming apart so that this solution runs into that Solution that's good that's one more condition but I need one further one final condition and somehow I have to use the Delta function and what does the Delta function tell me I'm just going to go go give you a the answer here rather than a theory of Delta functions that equation

so I you see what my solution is it it's a it's a broken line with a change of slope it's a ramp it it has a corner all those words describe functions like this so I have some slope going up here and some slope and let me tell you I I'll tell you what those slopes are I'll tell you what those slopes are in this so I'll tell you the answer and then uh would check so that c turns out to be 1 minus a so so in this region I have 1 - A *

X in that region and in this region below the so that's stretching the fact that it's positive displacement means stretching now this part is going to be in compression with that negative slope and I think in this region it's 1 - x * a which will be coming from there so there there is my solution because of the Delta function I need sort of like a two-part solution to the left of the Delta function the point load and to the right of the point load and then we could check that at the load x equal

a this is 1 - A * a this is 1 - A * a they do meet and now comes this mysterious fourth condition about the slopes the slope drops by one here the slope is 1 - A that's the 1 - A is the slope there and here the slope is min - A you see - x * a so the derivative is - A so it was 1 minus a the one dropped away and left me with minus a that's what the solution uh looks like and now I have to say one word

about why did the slope drop by one the slope dropped by one from 1 - A to minus a and that has to come from this Delta function and of course you remember about the Delta function the key point is when you integrate the Delta function you get one so when I integrate this equation I get a one on the right hand side from the Delta and on the left hand side I get I'm integrating the second derivative so I get the first derivative great the the first derivative at the end point minus the first

derivative at the start point should be the one and that's the drop of one I I'll uh do a fullscale job with Delta functions in another video uh I want to keep this one under control we're seeing the new idea is boundary conditions and here we're seeing a a a Delta function equation uh in this boundary value problem thank you