final value of x in PDE

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br
br 2024-2-15
评论： br 2024-2-19
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?
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Torsten
Torsten 2024-2-18
编辑：Torsten 2024-2-18
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
br 2024-2-19
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

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回答（2 个）

Vishisht
Vishisht 2024-2-18
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.
  1 个评论
br
br 2024-2-18
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

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Torsten
Torsten 2024-2-18
编辑：Torsten 2024-2-18
Ran in:
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
  1 个评论
br
br 2024-2-19
sorry for that but can you explane that much easer, and how and were i put it, thanks for your time and help

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