How can I evaluate P(x) here

Hello

How I can determine the function P(x) here if v= zero ?

v=(5*diff(P(x), x))/14 - (5*diff(P(x), x, x))/14 + (906154630400843*3^(1/2)*diff(P(x), x)^2)/(13194139533312*P(x)*(4645044914673847/(6871947673600*P(x)) + 6803141486033789/(42949672960000*P(x)^2) - 3171541396992589/1099511627776)) - (906154630400843*3^(1/2)*P(x)*diff(P(x), x)^2)/(13194139533312*(4645044914673847/(6871947673600*P(x)) + 6803141486033789/(42949672960000*P(x)^2) - 3171541396992589/1099511627776))

[Ahmad Almasri]

this is the full code

syms x y P(x)

l=20e-6;

e=0.06;

d=e*l/2;

k=0;

re=300;

rs=e;

rb=1/e;

ga=1.4;

pr=0.72;

ma=0.3;

kn0=0.01;

%ha=sqrt(re*rs*rb);

ha=sqrt(re*rs*rb);

sigmau=0.85;

thu=(2-sigmau)/sigmau;

P=P(x);

kn=kn0/P;

p=2;

%ma=sqrt((1/ga)*(1/(1-(1/p))*(e/((1/re)+(rb*rs)-e))))

ma=sqrt(e);

n1=((e*re)/(ga*ma^2*ha^2));

n2=(thu*ha*sinh(ha/2));

n3=(thu^2 * ha^2 * cosh(ha/2));

n4=-cosh(ha/2);

kn=kn0./P;

u=((k-n1*diff(P,x))/((n2*kn)+(n3*kn^2)+n4))*cosh(ha*y)+k-(n1*diff(P,x));

ux=diff(u,x);

n5=(ux+(n1*diff(P, x, x)))/cosh(ha*y);

n6=((k-n1*diff(P,x))/((n2*kn)+(n3*kn^2)+n4));

n7=k-(n1*diff(P,x));

y=0.5;

v=(n1*y*diff(P,x))-((n6*sinh(ha*y)*diff(P,x))/(ha*P))-((n7*y*diff(P,x))/P)-((n5*sinh(ha*y))/ha)+((n6*sinh(ha/2)*diff(P,x))/ha*P)+((n7*diff(P,x))/(2*P))+((n5*sinh(ha/2))/ha)-((n1*diff(P,x,x))/2);

Answer 1

[Star Strider]

When I attempted to integrate it analytically with my additions (including ‘dP’ and ‘d2P’), the result was an empty sym for ‘SPv’.

Not all differential equations have analytic integrations. If you want a numerical result, use the odeToVectorField and matlabFunction functions to create an anonymous function that the numerical ODE solvers can use ...

...such as:

syms x y P(x) P0 dP0 t Y 
dP = diff(P);
d2P = diff(dP);
l=20e-6;
e=0.06;
d=e*l/2;
k=0;
re=300;
rs=e;
rb=1/e;
ga=1.4;
pr=0.72;
ma=0.3;
kn0=0.01;
%ha=sqrt(re*rs*rb);
ha=sqrt(re*rs*rb);
sigmau=0.85;
thu=(2-sigmau)/sigmau;
% P=P(x);                                                   % <— This Eliminates The Function Definition Of ‘P(x)’!
kn=kn0/P;
p=2;
%ma=sqrt((1/ga)*(1/(1-(1/p))*(e/((1/re)+(rb*rs)-e))))
ma=sqrt(e);
n1=((e*re)/(ga*ma^2*ha^2));
n2=(thu*ha*sinh(ha/2));
n3=(thu^2 * ha^2 * cosh(ha/2));
n4=-cosh(ha/2);
kn=kn0./P;
u=((k-n1*diff(P,x))/((n2*kn)+(n3*kn^2)+n4))*cosh(ha*y)+k-(n1*diff(P,x));
ux=diff(u,x);
n5=(ux+(n1*diff(P, x, x)))/cosh(ha*y);
n6=((k-n1*diff(P,x))/((n2*kn)+(n3*kn^2)+n4));
n7=k-(n1*diff(P,x));
y=0.5;
% v=(n1*y*diff(P,x))-((n6*sinh(ha*y)*diff(P,x))/(ha*P))-((n7*y*diff(P,x))/P)-((n5*sinh(ha*y))/ha)+((n6*sinh(ha/2)*diff(P,x))/ha*P)+((n7*diff(P,x))/(2*P))+((n5*sinh(ha/2))/ha)-((n1*diff(P,x,x))/2);
v=(n1*y*dP)-((n6*sinh(ha*y)*dP)/(ha*P))-((n7*y*dP)/P)-((n5*sinh(ha*y))/ha)+((n6*sinh(ha/2)*dP)/ha*P)+((n7*dP)/(2*P))+((n5*sinh(ha/2))/ha)-((n1*d2P)/2);
% SPv = dsolve(v==0, P(0)==P0, dP(0)==dP0)
[VFv,Subsv] = odeToVectorField(v);
vfun = matlabFunction(VFv, 'Vars',{t,Y})

producing:

vfun =

  function_handle with value:

    @(t,Y)[Y(2);Y(2)+(sqrt(3.0).*Y(2).^2.*1.92299994911867e2)./(Y(1).*(6.759429983029036e2./Y(1)+1.0./Y(1).^2.*1.58397981106159e2-2.884500097018271e3))-(sqrt(3.0).*Y(1).*Y(2).^2.*1.92299994911867e2)./(6.759429983029036e2./Y(1)+1.0./Y(1).^2.*1.58397981106159e2-2.884500097018271e3)]

or, more conveniently (with a bit of manual editing):

    vfun =  @(t,Y) [Y(2);Y(2)+(sqrt(3.0).*Y(2).^2.*1.92299994911867e2)./(Y(1).*(6.759429983029036e2./Y(1)+1.0./Y(1).^2.*1.58397981106159e2-2.884500097018271e3))-(sqrt(3.0).*Y(1).*Y(2).^2.*1.92299994911867e2)./(6.759429983029036e2./Y(1)+1.0./Y(1).^2.*1.58397981106159e2-2.884500097018271e3)];

[Ahmad Almasri]

Thank you very much

But I need the function P(x) from equation (v) when v equal to zero

I don't see it in this solution

[Star Strider]

My pleasure.

It is automatically in the solution. Set your initial conditions to: [0; 0].

[Ahmad Almasri]

I'm sorry

but please can you give me the final answer of P(x) if P(0)=2 and P(1)=1

[Star Strider]

No apologies necessary!

Your differential equaiton is nonlinear, and only a few nonlinear differential equations have analytic solutions. (Yours is not one of them.)

You did not say that this was a boundary-value problem before.

Try this:

vfun = @(t,Y) [Y(2);Y(2)+sqrt(3.0).*Y(2).^2.*1.562241717104278e-2-((sqrt(3.0).*2.022457456864508e33-sqrt(3.0).*Y(1).*1.399056394183326e35).*Y(2).^2.*8.408108779363497e-19)./(Y(1).^2.*-1.982213373120368e18+Y(1).*4.645044914673847e17+1.088502637765406e17)+sqrt(3.0).*Y(1).*Y(2).^2.*6.666666266042039e-2];
twobc = @(ya,yb) [ ya(1) - 2; yb(1) - 1 ];
solinit = bvpinit(linspace(0,1,9),[-1 0]);
sol = bvp4c(vfun,twobc,solinit);

figure
plot(sol.x, sol.y)
grid
legend('$$P(x)$$', '$\frac{dP(x)}{dx}$', 'Location','W', 'Interpreter','latex')

This creates a numerical solution that the plot call then plots.

[Ahmad Almasri]

Thank you very much

[Star Strider]

As always, my pleasure.