final value of x in PDE

if i have PDE as following:

u=M

t=t

x=r

and i want to calculate the value of x at t=0 u0, and final value of it t=infinity and u=final

i have the initial and final value of u

how can i write a command to solve this one?

[Torsten]

It's not clear what you want. Use the google translator to translate from your native language into English and copy the translation here.

[br]

i want to solve this equation for r at given time and concentration

[br]

where can write it?

this run command:

kp0=140270;
Eap=43900;
kd0=0.355;
Ead=17400;
T=333;
kp=kp0*exp(-Eap/(8.314*T));
kd=kd0*exp(-Ead/(8.314*T));
m=2;
t=linspace(0,3600,1000);
u=linspace(0.036405,111.3494528,10000);
x=pdepe(m,@(x,t,u,DuDx)eqn111(x,t,u,DuDx,T),@initial111,@(xl,ul,xr,ur,t)bc111(xl,ul,xr,ur,t,kp,kd),u,t);
t_indices = find(t >= 0 & t <= 3600);
figure
    hold on
    for j = t_indices
    plot(x(j,:),u);
    hold off
end
xlabel('dimensionless radius');
ylabel('concentration');

[Torsten]

Why don't you plot x against u for all given times t ? It's much more natural than plotting u against x.

But see my answer below.

[br]

becuse i want then to draw u and x but not at all time only at specific time i can chose

so i made it like this, but the remaning point is to calculate the end point of x, and how to make time in linspace from 0 to infinity

kp0=140270;

Eap=43900;

kd0=0.355;

Ead=17400;

T=333;

kp=kp0*exp(-Eap/(8.314*T));

kd=kd0*exp(-Ead/(8.314*T));

m=2;

t=linspace(0,999999,1000);

u=linspace(0.506405,110,1000);

x=linspace(0,0.0000000006,1000);

x1=pdepe(m,@(x,t,u,DuDx)eqn111(x,t,u,DuDx,T),@initial111,@(xl,ul,xr,ur,t)bc111(xl,ul,xr,ur,t,kp,kd),u,t);

u1=pdepe(m,@(x,t,u,DuDx)eqn11(x,t,u,DuDx,T),@initial11,@(xl,ul,xr,ur,t)bc11(xl,ul,xr,ur,t,kp,kd),x,t);

t_indices = find(t >= 0 & t <= 3600);

figure

hold on

for j = t_indices

plot(x1(j,:),u1(j,:));

hold off

end

xlabel('radius');

ylabel('concentration');

function [c,f,s]=eqn111(x,t,u,DuDx,T)

c=1;

f=(1.34*10^-6)*exp(-16000/(8.314.*T))*exp(-(0.57*1.225+0.43*1.005*0.524)/(0.57*2.91*(-8.675+T)+0.43*0.5*(-205+T)))*DuDx;

s=0;

end

Answer 1

[Vishisht]

To solve the given partial differential equation (PDE) using the method of characteristics, we can start by rewriting the PDE in characteristic variables.

Given:

- \( u = M \)

- \( t = t \)

- \( x = r \)

The PDE in characteristic variables becomes:

\[ \frac{\partial u}{\partial t} = 0 \]

\[ \frac{\partial x}{\partial t} = M \]

Now, we integrate these characteristic equations to obtain expressions for \( u \) and \( x \) in terms of \( t \) and initial conditions:

1. Integrating \( \frac{\partial u}{\partial t} = 0 \) gives \( u = u_0 \), where \( u_0 \) is the initial value of \( u \).

2. Integrating \( \frac{\partial x}{\partial t} = M \) gives \( x = Mr + x_0 \), where \( x_0 \) is the initial value of \( x \).

Given the initial and final values of \( u \) and the final value of \( x \), you can find the constants \( u_0 \) and \( x_0 \) using the provided information.

Once you have \( u_0 \) and \( x_0 \), you can use these expressions to find the values of \( x \) at \( t = 0 \) and the final value of \( x \) as \( t \rightarrow \infty \).

For example, if you know the initial value of \( u \) (\( u_0 \)), you can directly set \( x = x_0 \) to find the value of \( x \) at \( t = 0 \). And if you have the final value of \( u \), you can set \( u = \text{final} \) and solve for \( x \) as \( t \rightarrow \infty \).

Please provide the specific initial and final values of \( u \) and any other relevant information so that a more detailed solution can be provided.

[br]

Dov=(1.34*10^-7)*exp(-16000/(8.314.*T))*exp(-(0.57*1.225+0.43*1.005*0.524)/(0.57*2.91 (-8.675+T)+0.43*0.5*(-205+T)))*

u0=0.036405

u(final)=111.3494528

x0=0

t0=0

Answer 2

[Torsten]

i want to solve this equation for r at given time and concentration

I assume that the time for which you want to get r is in the output of "pdepe".

If c is the concentration vector at time t and r is the vector of radial discretization points, you can try

rq = interp1(c,r,cq)

where cq is the concentration for which you want to get the corresponding radius rq.

This might cause problems if c is not monotonic.

If so, try

c = [0 0 3 5 8 10 14];
r = [0 1/6 1/3 1/2 2/3 5/6 1];
cq = 6.25;
idx = find(diff(sign(c-cq)),1);
rq = ((cq-c(idx))*r(idx+1) + (c(idx+1)-cq)*r(idx))/(c(idx+1)-c(idx)) 

rq = 0.5694

[br]

sorry for that but can you explane that much easer, and how and were i put it, thanks for your time and help