# I don't understand the case

yukcream
Given y' + ay = g(x) where y(0) =c,
also g(x) = g1, 0<=x<= xo
g(x) = g2, xo<x< infinity

for case x>xo
white
y = e^-ax Int(0->Xo){[e^aX1]g(X1)} dX1 + e^-ax Int(xo->x) [e^aX1] g(x1) dx1

I don't know why when calaulatung the case for x>Xo I have to consider the Intergration from 0 to Xo either? Can anyone tell me the answer?

Homework Helper
that should be
$$y=c+e^{-a x}\int_0^{x_0} e^{a t}g_1(t)dt+e^{-a x}\int_{x_0}^{x} e^{a t}g_2(t)dt$$
where t is used instead of x1 as dummy variable
the extra integral is used to insure that the initial condition is meet and the function is continuous.
in other words if y is defined
y=y1 0<x<x1
y=y2 x1<x
for continuity we require y1(x1)=y2(x1)
if we omit the first integral the function won't match up.
Past x0 we must start where we left off.

yukcream
lurflurf said:
that should be
$$y=c+e^{-a x}\int_0^{x_0} e^{a t}g_1(t)dt+e^{-a x}\int_{x_0}^{x} e^{a t}g_2(t)dt$$
where t is used instead of x1 as dummy variable
the extra integral is used to insure that the initial condition is meet and the function is continuous.
in other words if y is defined
y=y1 0<x<x1
y=y2 x1<x
for continuity we require y1(x1)=y2(x1)
if we omit the first integral the function won't match up.
Past x0 we must start where we left off.

Thanks for your help! But one more question is : what happens if the function is not continuous, also why we must make sure the function should be continuous??

Homework Helper
yukcream said:
Thanks for your help! But one more question is : what happens if the function is not continuous, also why we must make sure the function should be continuous??
The function is continuous where the differential equation holds, because the differential equation implies differentiability which implies continuity. Strictly speaking a differential equation should be given along with a domain of applicability. In this case it seams the implied domain is x<=0 where for x=0 it is true in a right hand limit sense. For example one could solve this differential equation
x*y'=y x!=0
y(-x)=y(x)
the solution of which being
y=|x|
or the equation
y'=y x!=0 y(0-)=-1 y(0+)=1
the solution being
y=-exp(x) x<0
y=exp(x) x>0
these are stange examples usually one expects the solution to be continuous and differentiable

Homework Helper
Well guys I've been reading right along and have taken the trouble to work through two examples, one continuous and the other discontinuous. This is what I found:

Case 1: g(x) is continuous at x=3:

$$y^{'}+y=g(x);\quad y(0)=1$$

$$g(x)=\left\{\begin{array}{cc}x,&\mbox{ for }0\leq x\leq 3 \\ \frac{x^2}{3},&\mbox{ for }x>3 \end{array}\right$$

Solving, we obtain:

$$y(x)=\left\{\begin{array}{cc}2e^{-x}+x-1 &\mbox{ for }0\leq x\leq 3 \\ e^{-x}(2e^3+2)+1/3(2-2x+x^2)-5/3 e^{3-x},&\mbox{ for }x>3 \end{array}\right$$

This solution is shown in the first plot. Note that it is continuous at x=3 but also differentiable at x=3 since the derivatives are equal there for both functions.

Case 2: g(x) is discontinuous at x=1:

$$y^{'}+y=g(x);\quad y(0)=1$$

$$g(x)=\left\{\begin{array}{cc}x,&\mbox{ for }0\leq x\leq 1 \\ x^2-2x+4,&\mbox{ for }x>1 \end{array}\right$$

Solving:

$$y(x)=\left\{\begin{array}{cc}2e^{-x}+x-1,&\mbox{ for }0\leq x\leq 1 \\ 2e^{-x}+8-4x+x^2-5e^{1-x},&\mbox{ for }x>1 \end{array}\right$$

This solution is shown in the second plot. It is continuous at x=1 but the derivative does not exist there since the derivatives of the two expression for y are not equal there.

So . . . is this always the case?

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Homework Helper
saltydog said:
This solution is shown in the second plot. It is continuous at x=1 but the derivative does not exist there since the derivatives of the two expression for y are not equal there.

So . . . is this always the case?
The solution can be written as an integral involving the forcing term g(x). Thus y will have 1 more level of smothness than g(x), since integration is a smothing operation. For example if g(x) is piecewise continuous y is continuous and y' is piecewise continuous. If g'(x) is piecwise continuous y' is continuous and g''(x) is piecwise continuous. And so on. A general solution involves and indefinite integration. If we chose to write the solution in a piecewise defined manner as the problem poser did we need to ensure that the piece that includes the initial condition has its constant chosen by the initial condition, and the other piece has its constant chosen by continuity of the solution at the border of the two regions.

Homework Helper
lurflurf said:
The solution can be written as an integral involving the forcing term g(x). Thus y will have 1 more level of smothness than g(x), since integration is a smothing operation. For example if g(x) is piecewise continuous y is continuous and y' is piecewise continuous. If g'(x) is piecwise continuous y' is continuous and g''(x) is piecwise continuous. And so on. A general solution involves and indefinite integration. If we chose to write the solution in a piecewise defined manner as the problem poser did we need to ensure that the piece that includes the initial condition has its constant chosen by the initial condition, and the other piece has its constant chosen by continuity of the solution at the border of the two regions.

Thanks Lurflurf. Don't quite understand your reply, particularly the constants you refer to. For the second case, I obtained expressions:

$$y(x)=e^{-x}+e^{-x}\int_0^x te^tdt;\quad 0\leq x\leq 1$$

$$y(x)=e^{-x}+e^{-x}\int_0^1 te^tdt+e^{-x}\int_1^x e^t(t^2-2t+4)dt;\quad x>1$$

Can we perhaps make some general statement about the behavior of the solution (maybe you did and I just don't understand it) depending on two cases at the border $x_m$:

$$g_1(x_m)=g_2(x_m)$$

and:

$$g_1(x_m)\ne g_2(x_m)$$

That is, at the point $x_m$, can differentiability be guaranteed for the first case but not the second?

When can continuity of the solution be guaranteed for either case?

Homework Helper
Thanks Lurflurf. Don't quite understand your reply, particularly the constants you refer to. For the second case, I obtained expressions:

What I mean about the constants is that if we write the solution
y=c1+y1 0<=x<x0
y=c2+y2 x1<=x
then c1 and c2 are not independent
we chose 1 (usually c1) according to the initial conditions
and the other according to the continuity of the solution
that is we require (initial condition y(0)=c)
c=y(0)=c1+y1(0)
y(x1-)=y(x0+) ->c1+y(x0)=c2+y2(x0)
Can we perhaps make some general statement about the behavior of the solution (maybe you did and I just don't understand it) depending on two cases at the border $x_m$:

yes we can and I did. The only tricky part is we are not explicitly told what properties g has and where the diffential equation is valid. I assume from context that the differential equation is valid for x<0 (x=0 in righthand sense).
And g1 and g2 are at least continuous and may be infinitely differentiable.

That is, at the point $x_m$, can differentiability be guaranteed for the first case but not the second?

When can continuity of the solution be guaranteed for either case?

So the function g is within an integral (similar aguments can be made in terms of the differential equation)
g continuous except at x0 ->y is every where continuous, continuously differentiable except at x0
g' continuous except at x0 ->y' is every where continuous, continuously differentiable except at x0
and so on
So in summary however smooth g is y will be one level more smooth
exceptions
if g is infinitely smoth so to will be y
if g is not integrable y may not exist
This all flows form g defining y by an integral transform, or by g being defined by a differentiable equation in y.