How to input piecewise continuous differential function into chebfun?

I want to approximate the solution of a piecewise continuous differential equation with chebfun. But I did not figure out a way to pass the function into chebop.

Assume I have the following differential equation:

u''=x, 0<=x<=0.5

10*u''=x, 0.5<x<=1

With boundary condition u(0)=0, u(1)=1/80

How can I use chebop to solve u?

I tried to express the input with binary conditions but the approximate solution does not look right. Is that because I did not use chebop in the correct way, or is the problem with chebfun itself?

Below are the details:

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To verify, the analytic solution of the differential equation above is

u = 1/6*x^3-1/24*x, 0<=x<=0.5

u = 1/60*x^3-1/240*x, 0.5<x<=1

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I checked out the chebgui Demo, and found it has examples with discontinuity using either sign(x) or binary conditions, i.e. -(x>=5)*(x<=10).

Thus, I tried to input the differential equation with

L = chebop(0,1);
L.op = @(x,u)((x>=0.5)*10+(x<0.5)*1).*diff(u,2)-x;
L.lbc = @(u) u;
L.rbc = @(u) u-1/80;

But the result does not approximate the solution very well (the 'chebop' line in the figure).

On the other hand, if I plot the function with chebfun with the code

   L=chebfun({@(x)1/6*x^3-1/24*x, @(x)1/60*x^3-1/240*x}, [0 0.5 1]);

The plot approximates the true solution very well even with 'splitting' off (the 'chebfun' line in the figure, the line is identical with the true solution line 'ts').

Answer 1

[Torsten]

Without your boundary conditions, the solution of your differential equation is given by

u1(x)=1/6*x^3+a*x+b if 0<=x<=0.5
u2(x)=1/60*x^3+c*x+d if 0.5<=x<=1

So you have four degrees of freedom, but only two boundary conditions.

Thus the solution of your differential equation is not unique.

You may want to specify some transmission conditions at x=0.5, e.g. u1=u2 and du1/dx=du2/dx to make u1 and u2 unique.

Best wishes

Torsten.

[Torsten]

The above transmission conditions result in the solution from chebop:

u1(x)=1/6*x^3-19/240*x
u2(x)=1/60*x^3+1/30*x-3/80

Best wishes

Torsten.

[Eszter Feher]

Could you explain the method of specifying the transmission conditions? If I try to add an extra boundary condition at x=0.5, e.g.:

L.bc = @(u) diff(u(0.5),2);

I get: "Number of inputs to BCs do not match operator.".