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I have the following ODE

$$ u'' = -(1 + e^u), \quad u(0)=0,\quad u(1)=1$$

I want the divide the interval $[0,1]$ into $n-1$ equal subintervals each with length $h=1/(n-1)$ and we take approximate solution

$$ v(t,x) = \sum_{i=1}^{n} x_i t^{i-1} $$

polynomial of degree $n-1$ sothat $v(0,x)=0$ and $v(1,x)=1$, (boundary conditions are to be satisfied).

So, our goal is to determine $x_2,x_3,...,x_{n-1}$ interior points since we already know $x_1 = 0$. Once these are found, then we have an approximation to our true solution $u(t) \approx v(t,x)$.

My question is, how can we implement this in matlab? because solving such system is quite time consuming by hand.

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  • $\begingroup$ What are the $x_i$? The interior points or the polynomial coefficients? Do you want to find a polynomial $p$ of lowest degree so that $p''(x_i)=-(1+e^{p(x_i)})$ or do you include the higher order derivatives of $u$ so that simultaneously $u(x)-p(x)=O(x^s)$ for $x\approx 0$ and $u(x)-p(x)=O((1-x)^s)$ for $x\approx 1$? $\endgroup$ – LutzL Apr 22 at 11:51
  • $\begingroup$ Your question is not yet well-defined because you haven't specified a metric to use to measure how good a particular polynomial approximation is. One way you can do it is to require $v''=-(1+e^v)$ at your nodes $hi,i=1,2,\dots,n-1$ and require the boundary conditions to hold. (Here because you are using a polynomial, the derivatives can be done exactly). $\endgroup$ – Ian Apr 22 at 13:53
  • $\begingroup$ (Cont.) The resulting system of equations can be arranged into the form $F(\mathbf{v})=0$ where $\mathbf{v}$ is the vector $\{ v(hi) \}_{i=0}^n$ and $F$ is a vector-valued function. It is nonlinear, so it needs to be solved with a nonlinear function solver, such as Matlab's fsolve. $\endgroup$ – Ian Apr 22 at 13:54
  • $\begingroup$ Also, although I presume this doesn't help you with your actual task, I still feel compelled to point out that Matlab can do this entire job for you, most likely better than your implementation would be. Look into the functions bvp4c, bvp5c, and the helper function bvpinit. $\endgroup$ – Ian Apr 22 at 13:56
  • $\begingroup$ How do we solve the system of nonlinear equation in matlab? $\endgroup$ – Mikey Spivak Apr 22 at 15:19
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To perform collocation, we introduce the regular discretization $t_i = i/n$ with $0\leq i\leq n$ of the domain $[0,1]$ (i.e., a set of $n+1$ collocation points). Then, we approximate $u$ by the polynomial function $p(t) = \sum a_k t^k$ of degree $n$ which satisfies the above problem at the collocation points. Therefore, $-p''(t_i) = 1 + \exp\!\big(p(t_i)\big)$ in the interior domain $1\leq i \leq n-1$, and $p(t_0)=0$, $p(t_n) = 1$ at the boundaries. In terms of the coefficients $a_k$, this system writes as $$ -\sum_{k=0}^{n-2} (k+2) (k+1) a_{k+2} \frac{i^k}{n^k} = 1 + \exp\left(\sum_{k=0}^n a_k \frac{i^k}{n^k}\right) , \qquad 1\leq i \leq n-1 $$ and $a_0 = 0$, $\sum a_k = 1$. This nonlinear algebraic system can then be solved numerically in terms of the coefficients $a_k$, e.g. by using Matlab's fsolve.

For example, if $n=2$, we consider the collocation points $\lbrace 0, 0.5, 1\rbrace$, and we look for the coefficients $a_0$, $a_1$, $a_2$. These coefficients satisfy $a_0 = 0$ and the algebraic system $$ \left\lbrace \begin{aligned} &{-2} a_{2} = 1 + \exp\left(\tfrac{1}{2} a_1 + \tfrac{1}{4} a_2\right)\\ &a_1 + a_2 = 1 \end{aligned}\right. $$ which real solutions obtained numerically $$ (a_1,a_2) \in \lbrace (2.78945, -1.78945), (10.6101, -9.61014) \rbrace . $$ are represented below:

solutions

One observes that such solutions are not necessarily unique. Comparison with Matlab's bvp4c is given below, where the algorithm is initialized by the constant $0.5$:

matlab

sol = bvp4c(@(x,y) [y(2) -(1+exp(y(1)))], @(ya,yb) [ya(1) yb(1)-1], bvpinit(linspace(0,1,10), [0.5; 0.5]));
plot(sol.x,sol.y(1,:));
xlabel('t');
ylabel('u');

If initialization is performed with the constant $2.5$, then the higher solution is obtained.

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    $\begingroup$ If you take the two quadratic polynomials as initialization for the BVP solver, you get again two solutions. x=linspace(0,1,10); u = a(1)*x+a(2)*t^2; v=a(1)+2*a(2)*x; init=bvpinit(x,[u,v]); $\endgroup$ – LutzL Apr 26 at 15:46
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    $\begingroup$ Yes, that would be in the region of the "higher" solution. Plot from a python implementation. $\endgroup$ – LutzL Apr 26 at 15:54
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The polynomial can be determined for instance with the code (in octave, there might be slight differences for matlab).

  • First define a function that takes a coefficient array and returns an array containing the boundary conditions and the ODE residuals at the collocation points.

    function f = system(c)
      % degree of the interpolating polynomial
      % is also the number of subdivisions
      deg = length(c)-1;
      f = zeros(1,deg+1);
      % left boundary condition
      f(1)=polyval(c,0);
      % second derivative of polynomial
      c2 = polyder(polyder(c));
      % enforce differential equation at inner points
      for k = 1:deg-1
        xk = k*1.0/deg;
        f(k+1)= polyval(c2,xk) + (1+exp(polyval(c,xk)));
      end
      % right boundary condition
      f(deg+1)=polyval(c,1) - 1;
    end
    
  • This nonlinear system function can now be called in fsolve where the initialization array also determines the degree of the collocation polynomial

    clf;
    % define target accuracy
    options = optimset("TolX", 1e-9, "TolFun", 1e-6);
    % number of subdivisions = polynomial degree
    deg = 5;
    hold on;
    % loop over different initializations to find multiple solutions
    for k = 0:5
       c = zeros(1,deg+1);
       c(deg) = 1+3*k; c(deg-1) = -3*k;
       % log the initial coefficients
       c
       % solve system and log the result
       c = fsolve(@(c) system(c), c, options)
       % plot the solution
       x = linspace(0,1,101);
       plot(x, polyval(c,x))
    end
    hold off;
    

The execution of this code finds two solutions, which are repeated 3 times for the chosen initializations:

 0.24661  -0.38053    -0.31725   -1.02286    2.47402    0.00000
30.41419  -49.23356   11.70577   -1.37713    9.49074    0.00000

with the plot

plot from octave


Addendum: Taking the degree 2 solutions as in the answer of Harry49

 -1.78945    2.78945    0.00000
 -9.61015   10.61015    0.00000

and using these polynomials and their derivative to initialize the BVP solver (in python, as my version of octave does not have this functionality), gives the following plot, where the polynomials are drawn in gray below the more exact solutions:

enter image description here

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There is analytic solution of the ODE in quadratures.

Let $P(u)=\dfrac{du}{dx}.$

Then $$u''=\dfrac{dP}{dt} = \dfrac{dP}{du}\dfrac{du}{dx} = PP',$$ $$P^2 = -2(u+e^u+\mathrm{const}),$$ $$u'^2 = 2(C_1-u-e^u),$$ $$u' = \pm\sqrt2\sqrt{C_1-u-e^u},\tag1$$ $$x=C_2\pm\dfrac1{\sqrt2}\int \dfrac{du}{\sqrt{C_1 -u - e^u}}.\tag2$$ Assuming the derivative $P$ positive and taking in account the boundary conditions, the equation $(2)$ can be presented in the form of $$x=I(u,C_1)=\dfrac1{\sqrt2}\int\limits_0^u \dfrac{dv}{\sqrt{C_1 -v - e^v}},\tag3$$ where the constant $C_1$ can be defined from the equation $$I(1,C_1) = 1.\tag4$$ Taking in account that $C_1\ge e+1$ and using the substitution $w=1-v,$ one can get $$I(1,e) = \dfrac1{\sqrt2}\int\limits_0^1\dfrac{dv}{\sqrt{1-v+e(1-e^{v-1})}} =\dfrac1{\sqrt2}\int\limits_0^1\dfrac{dw}{\sqrt{w +e(1-e^{-w})}}$$ $$\le \dfrac1{\sqrt2}\int\limits_0^1\dfrac{dw}{\sqrt{w +e(w-\frac12w^2)}} = 2\int\limits_0^1\dfrac{d\sqrt w}{\sqrt{2+2e-e w}} = \dfrac2{\sqrt e}\int\limits_0^1\dfrac{dz}{\sqrt{2+2e^{-1}-z^2}}$$ $$= \dfrac2{\sqrt e}\arcsin\dfrac1{\sqrt{2+2e^{-1}}}]\approx 0.788<1$$ (Wolfram Alpha numeric calculations give $I(1,e)\approx 0.776236$)

This result contradicts with $(4),$ so the boundary conditions $u(0)=0$ and $u(1)=1\quad \color{green}{\mathbf{cannot\,be\,satisfied\,in\,the\,model\, with\,the\,constant\,sign\,of\, P}}$ (see also the comments of $\color{green}{\textbf{LutzL}}).$

This means that the function $u(x)$ has the global maximum, which accords to the branch point of $x(u).$

Let $(x_m,u_m)$ is the branch point, wherein $$u'=\begin{cases} \sqrt2\sqrt{C_1-u-e^u},\quad\text{if}\quad u < u_m\\[4pt] 0,\quad\text{if}\quad u = u_m\\[4pt] -\sqrt2\sqrt{C_1-u-e^u},\quad\text{if}\quad u > u_m. \end{cases}$$

$$u_m>1,\quad\dfrac{dx}{du}\bigg|_{\large u_m} =\infty,\quad C_1 = x_m = u_m+e^{u_m},\quad u_m = x_m-W(e^{x_m}),\tag6$$ Lambert W solution

where $W(x)$ is Lambert W function.

Then $$x_{1,2}(u) = x_m \pm \int\limits_u^{u_m}\dfrac{dv}{\sqrt{2(x_m - v-e^v)}},\tag7$$

$$\int\limits_0^{u_m} \dfrac{dv}{\sqrt{2(x_m - v-e^v)}} + \int\limits_1^{u_m} \dfrac{dv}{\sqrt{2(x_m - v-e^v)}} = 1,$$ $$\int\limits_0^{1} \dfrac{dv}{\sqrt{2(x_m - v-e^v)}} + 2\int\limits_1^{u_m} \dfrac{dv}{\sqrt{2(x_m - v-e^v)}} = 1,\tag8$$ with the numeric solutions $$\dbinom{u_m}{x_m}\in\left\{\approx\dbinom{1.08770}{4.05514},\dbinom{3.67011}{42.92633}\right\}.\tag9$$

Plot 1 Formula 1 Plot 2 Formula 2

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  • $\begingroup$ how can do this in matlab? $\endgroup$ – ILoveMath Apr 24 at 14:07
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    $\begingroup$ The constants for the existing solutions are $C=8.1102674$ for the lower one and $C=85.8540768$ for the upper one, so something in your argument is obviously wrong. One undiscussed assumption that you rely on is that the sign of $u'$ is always positive, which is not the case for the observed solutions, they have clear maxima inside the interval. So the claim that you have proven is that there are no monotonously increasing solutions. $\endgroup$ – LutzL Apr 26 at 22:51
  • $\begingroup$ @ILoveMath Any version of the integrator. $\endgroup$ – Yuri Negometyanov Apr 27 at 5:19
  • $\begingroup$ @LutzL thank you for the comment! The exact analytic solution is obtained. $\endgroup$ – Yuri Negometyanov Apr 27 at 5:21

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