Differential equation unable to find a symbolic solution

Question

Rihards 2022-10-29

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https://ww2.mathworks.cn/matlabcentral/answers/1838948-differential-equation-unable-to-find-a-symbolic-solution

回答： Walter Roberson 2022-10-29

采纳的回答： Star Strider

在 MATLAB Online 中打开

Hello!

I'm seeking for some assitance in understanding the proper way of tackling this problem:

I am trying to find a solution for the following differential equation:

In my specific case, all of the variables are given, as depicted in the code example below, except eta - which in dependant on x.

I tried the dsolve approach, but while I'm expecting a numerical solution, it's warning be that no symbolic solution can be found instead.

The variable D has a given range, where at x=0, eta is assumed to be equal to 0, which I have listed as a condition for the differential equation, in order to remove the constant part.

I might have something fundumentally wrong with the approach here, but based on the examples, I am having a hard time figuring it out. The end goal here would be to plot out eta(x).

syms eta(x) D_off
X = 300000;
tau_0 = 0; 
tau_w = 17.0100;
rho_w = 1025;
g = 9.81;
s = 0.0050;
D_off(x)=X*s-x*s;    
ME = -g*diff(eta,x) +(tau_w-tau_0)/(rho_w*(D_off(x)+eta))==0
dsolve(ME,eta(0)==0)

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Answer 1

Star Strider 2022-10-29

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https://ww2.mathworks.cn/matlabcentral/answers/1838948-differential-equation-unable-to-find-a-symbolic-solution#answer_1086963

在 MATLAB Online 中打开

The equation is nonlinear in η, and dsolve will not solve nonlinear differential equations.

Integrating it numerically gives —

syms eta(x) D_off x Y

X = 300000;

tau_0 = 0;

tau_w = 17.0100;

rho_w = 1025;

g = 9.81;

s = 0.0050;

D_off(x)=X*s-x*s;

ME = -g*diff(eta,x) == -(tau_w-tau_0)/(rho_w*(D_off(x)+eta))

ME(x) =

[VF,Sbs] = odeToVectorField(ME)

VF =

Sbs =

etafcn = matlabFunction(VF, 'Vars',{x,Y})

etafcn = function_handle with value:

@(x,Y)[(1.89e+2./1.09e+2)./(x.*(-4.1e+1./8.0)+Y(1).*1.025e+3+1.5375e+6)]

xspan = [0 1E6];

[x,eta] = ode15s(etafcn, xspan, 0);

figure

semilogy(x, eta)

grid

xlabel('x')

ylabel('\eta(x)')

% dsolve(ME,eta(0)==0)

Make appropriate changes to get the desired result.

.

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Star Strider 2022-10-29

在 MATLAB Online 中打开

I took the easy way out with this, and defined two versions of ‘etafcn’, then used a loop for the two initial conditions and ‘x’ ranges —

syms eta(x) D_off1 D_off2 x Y

X = 300000;

tau_0 = 0;

tau_w = 17.0100;

rho_w = 1025;

g = 9.81;

s = 0.0050;

D_off1(x) = 40;

D_off2(x)=X*s-x*s;

ME1 = -g*diff(eta,x) == -(tau_w-tau_0)/(rho_w*(D_off1(x)+eta));

ME2 = -g*diff(eta,x) == -(tau_w-tau_0)/(rho_w*(D_off2(x)+eta));

[VF1,Sbs1] = odeToVectorField(ME1);

[VF2,Sbs2] = odeToVectorField(ME2);

etafcn{1} = matlabFunction(VF1, 'Vars',{x,Y});

etafcn{2} = matlabFunction(VF2, 'Vars',{x,Y});

xspan = [0 292E3 3E5];

eta0 = 0;

for k = 1:2

[x,eta] = ode15s(etafcn{k}, xspan((1:2)+(k-1)), eta0);

xc{k,:} = x;

etac{k,:} = eta;

eta0 = eta(end);

end

xv = cell2mat(xc);

etav = cell2mat(etac);

figure

plot(xv, etav)

grid

xlabel('x')

ylabel('\eta(x)')

Ax = gca;

Ax.XAxis.Exponent = 0;

xtickformat('%,7.0g');

xline(xspan(2), '--k')

This uses ‘etafcn1’ for the first interval and ‘etafcn2’ for the second interval. It uses the previous end value of ‘eta’ for the initial condition of the second iteration. The very end of this seems to approach the previous result, however it appears to be linear for the first range section. The vertyical dashed line demarcates the two ‘x’ ranges, for clarity.

It took a bit of experimenting to get this running.

.

Rihards 2022-10-29

Thanks! Will try to go trough it and see how all of that works.

Am I getting it right that I am unable to define the initial conditions anywhere else other than y(0)?

Star Strider 2022-10-29

As always, my pleasure!

I didn’t see your previous Comment until now. I used the previous end value of ‘eta’ as the initial conditon for the subsequent integration. It is of course acceptable to define the initial condition to be anything you want. It just has to be defined.

There is only one differential equation, and so only one initial condition, that being

.

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Answer 2

Walter Roberson 2022-10-29

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dsolve without the initial condition and then try to substitute in the location of the initial condition and try to solve for constants.

You will find that the solution without the initial condition is of the form A*x+b with A and B being specific numeric constants. When you substitute in x=0 you will get the constant b, which is non-zero and so cannot satisfy the required eta(0)==0