
We explored the implementation of a Research-Based Learning (RBL) for the teaching and learning of probability in engineering education. The approach provided a student-centered learning environment, integrating practical activities within an active learning conception. A detailed RBL task in connection with weather forecasting was described, highlighting a methodical 10-day learning process that considers research, practical application, and reflection. The effectiveness of this RBL task was evaluated through thematic analysis of student reflective essays, focusing on their learning experiences and challenges. The analysis revealed enhanced understanding of probability, varied student engagement, and the value of practical learning. It also identified challenges in conceptualizing abstract probability concepts. The findings highlighted the RBL's potential in improving students' conceptual grasp and application of probability in real-world scenarios, providing key insights for future educational strategies in engineering. This study contributes to the broader discourse on innovative teaching methodologies in engineering education, demonstrating the impact of active, inquiry-based learning approaches.
Citation: José Luis Díaz Palencia, Yanko Ordónez Ontiveros. A classroom experience about the application of research-based learning for the teaching of probability in engineering[J]. STEM Education, 2024, 4(2): 127-141. doi: 10.3934/steme.2024008
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We explored the implementation of a Research-Based Learning (RBL) for the teaching and learning of probability in engineering education. The approach provided a student-centered learning environment, integrating practical activities within an active learning conception. A detailed RBL task in connection with weather forecasting was described, highlighting a methodical 10-day learning process that considers research, practical application, and reflection. The effectiveness of this RBL task was evaluated through thematic analysis of student reflective essays, focusing on their learning experiences and challenges. The analysis revealed enhanced understanding of probability, varied student engagement, and the value of practical learning. It also identified challenges in conceptualizing abstract probability concepts. The findings highlighted the RBL's potential in improving students' conceptual grasp and application of probability in real-world scenarios, providing key insights for future educational strategies in engineering. This study contributes to the broader discourse on innovative teaching methodologies in engineering education, demonstrating the impact of active, inquiry-based learning approaches.
In this paper, we are concerned with the motion of the supersonic potential flow in a two-dimensional expanding duct denoted by
f(0)>0,f′(x)>0,f′′(x)≥0,f′∞=limx→+∞f′(x)exists, | (1) |
and
φ(−y)=φ(y),φ(±f(0))=0,φ′(±f(0))=∓f′(0),φ′′≤0. | (2) |
At the inlet, the flow velocity is assumed to be along the normal direction of the inlet and its speed is given by
c∗<c1<q0(y)<c2<ˆq, | (3) |
where
vu=tanθ=±f′(x),ony=±f(x), | (4) |
where
The supersonic flow in the duct is described by the 2-D steady isentropic compressible Euler equations:
{∂∂x(ρu)+∂∂y(ρv)=0,∂∂x(ρu2+p)+∂∂y(ρuv)=0,∂∂x(ρuv)+∂∂y(ρv2+p)=0, | (5) |
where
∂u∂y=∂v∂x, | (6) |
then for polytropic gas, the following Bernoulli law holds
12q2+c2γ−1=12ˆq2, | (7) |
where
{(c2−u2)∂u∂x−uv(∂u∂y+∂v∂x)+(c2−v2)∂v∂y=0,∂u∂y−∂v∂x=0, | (8) |
Such problem (8) with initial data (3) and boundary condition (4) has already been studied by Wang and Xin in [25]. In their paper, by introducing the velocity potential
For better stating our results, we firstly give the description of domains
Our main results in the paper are:
Theorem 1.1. Assume that
|q′0(y)|≤−φ′′(y)1+(φ′(y))2q0c√q20−c2onΓin, | (9) |
then two alternative cases will happen in the duct, one contains vacuum, and the other does not. More concretely, the two cases are as follows:
(i) When vacuum actually appears in the duct, there exists a global solution
∂→nc2=0,onlup∪llow∖{M,N}, | (10) |
here
(ii) If vacuum is absent in the duct, the problem (8) with (3) and (4) has a global solution
Theorem 1.2. Assume that
f(0)>0,f′(0)>0,f′∞≥0,f′′(x)≤0. | (11) |
If
|q′0(y)|>−φ′′(y)1+(φ′(y))2q0c√q20−c2onΓin, | (12) |
then the
Remark 1. The condition (9) is equivalent to that the two Riemann invariants are both monotonically increasing along
{∂tu+u∂xu=0,u(0,x)=u0(x). | (13) |
As well known, if the initial data satisfies
minx∈Ru′0(x)<0, | (14) |
then the solution
{(∂x+λ−∂y)R+=0,(∂x+λ+∂y)R−=0,R±(0,y)=R0±(y). | (15) |
By the results of [16] and [30], there exists a global solution of the Cauchy problem (15) if and only if the two Riemann invariants
Remark 2. By the effect of expanding duct, as far as we know, it is hard to give a necessary and sufficient condition to ensure that vacuum must form at finite location in the duct. In Proposition 4.1, we will give a sufficient condition such that the vacuum will appear in finite place.
Remark 3. For the M-D compressible Euler equations, if the gases are assumed irrotational, by introducing the velocity potential
Remark 4. If the initial data contains a vacuum, especially for physical vacuum, the local well-posedness results of the compressible Euler equations have been studied in [8], [9], [12], [13], [18] and so on. But in general, such a local classical solution will blow up in finite time as shown in [3], [26] and the references therein.
Remark 5. For the case that Riemann invariants are both constants on
Remark 6. The symmetry of
The paper is organized as follows. In Section 2, we give some basic structures of the steady plane isentropic flow and discuss the monotonicity of two Riemann invariants
In this section, we start with some basic structures of the steady plane isentropic flow, which can be characterized by the Euler system (5).
Firstly, equation (8) can be written as the matrix form
(c2−u2−uv0−1)∂∂x(uv)+(−uvc2−v210)∂∂y(uv)=0. | (16) |
The characteristic equation of (16) is
(c2−u2)λ2+2uvλ+(c2−v2)=0. |
So for the supersonic flow, (8) is hyperbolic and we can get two eigenvalues
λ+=uv+c√u2+v2−c2u2−c2,λ−=uv−c√u2+v2−c2u2−c2. | (17) |
Correspondingly, the two families of characteristics in the
dy±dx=λ±. | (18) |
By standard computation, the Riemann invariants
R±=θ±F(q), | (19) |
where
{(∂x+λ−∂y)R+=0,(∂x+λ+∂y)R−=0, | (20) |
which can also be written as
∂±R∓=0 | (21) |
if we set
Let A be the Mach angle defined by
λ±=tan(θ±A). | (22) |
Under a proper coordinate rotation, the eigenvalues
From now on, we will discuss the monotonicity of the Riemann invariants
Lemma 2.1. For
|q′0(y)|≤−φ′′(y)1+(φ′(y))2q0c√q20−c2onΓin. | (23) |
Proof. Since
R′±(y)=−φ′′(y)1+(φ′(y))2±√q20−c2q0cq′0(y)onΓin. | (24) |
Thus
Due to Lemma 2.1, the following analysis will be focused on these two cases.
Case I.
Case II. At least one of
The next lemma shows the partial derivatives
Lemma 2.2. As the
∂λ+∂R−>0and∂λ−∂R+>0. | (25) |
Proof. By using (22) and the chain rule, it follows from direct computation that
∂λ+∂R−=(γ+1)q24(q2−c2)sec2(θ+A),∂λ−∂R+=(γ+1)q24(q2−c2)sec2(θ−A), | (26) |
thus (25) holds true for the supersonic flow.
In this section, the existence of the
Next we solve the solution in the border region
Let the
Lemma 3.1. If (9) holds true, then for the
Proof. We will prove the conclusion between these inner and border regions separately.
Step 1. Inner region
For any fixed two points
∂−R−=ddx(R−(x,y−(x)))≤0 | (27) |
along l in region
Step 2. Border region
Next we consider the solution in border region
R++R−=2arctanf′(x),f′′(x)≥0 | (28) |
on
Repeating this process in inner region
We can ensure that the hyperbolic direction is always in the x-direction by a proper rotation of coordinates if necessary. Without loss of generality, we may assume that
Lemma 3.2. If (9) holds true, then we have
Proof. In terms of the definition of
∂x=λ+∂−−λ−∂+λ+−λ−,∂y=∂+−∂−λ+−λ−. | (29) |
Thus we have
∂yR+=∂+R+λ+−λ−,∂yR−=−∂−R−λ+−λ−. | (30) |
Combining this with Lemma 3.1 yields the result.
Corollary 1. Under the assumption that
|∂+R+|=λ+−λ−√λ2−+1|∇R+|and|∂−R−|=λ+−λ−√λ2++1|∇R−|. | (31) |
Lemma 3.3. Let
Proof. It follows from (29) that
∂l=v−uλ−λ+−λ−∂++uλ+−vλ+−λ−∂−. | (32) |
which implies that
∂lq=v−λ−uλ+−λ−∂+q+λ+u−vλ+−λ−∂−q=u(vu−λ−)λ+−λ−∂+q+u(λ+−vu)λ+−λ−∂−q. | (33) |
Since
Corollary 2. Combining Lemma 3.3 and the Bernoulli law yields that the sound speed is monotonically decreasing along the streamline, that is to say
We now focus on the solution in the inner regions
Set
{(∂x+λ−∂y)R+=0,(∂x+λ+∂y)R−=0,inD2i−1R+(x,y)=W+(x,y)onAiCi−1,R−(x,y)=W−(x,y)onBiCi−1. | (34) |
By standard iteration method, the local existence of the problem (34) can be achieved, one can also see [4] and [15] for more details. In order to get the global solution to the problem (34) in the region
Lemma 3.4. If (9) holds true, then
Proof. It comes from the definition of
∂−∂+R+=∂−∂+R+−∂+∂−R+=(∂−λ+−∂+λ−)∂yR+. | (35) |
Due to
∂−λ+−∂+λ−=∂λ+∂R−∂−R−−∂λ−∂R+∂+R+≤0 | (36) |
and
In terms of Lemma 3.1, one has
Lemma 3.5. Suppose that
Proof. Firstly, we give the estimates of
For any point
|R+(x,y)|≤‖W+‖C(AiCi−1),|R−(x,y)|≤‖W−‖C(BiCi−1) | (37) |
in the region
Fix
x0>x1>...>xN−1>xN,xN=φ(yN). | (38) |
If
W+(x0,y0)=2arctanf′(x1)−R−(x1,y1)=2arctanf′(x1)−R−(x2,y2)=2arctanf′(x1)+2arctanf′(x2)+R+(x2,y2)=N∑i=12arctanf′(xi)−R−(xN,yN). | (39) |
Similarly, if
W+(x0,y0)=N∑i=12arctanf′(xi)+R+(xN,yN). | (40) |
Combining (39) and (40) yields that
‖W+‖C(AiCi−1)≤2Narctanf′∞+max{‖R+‖C(Γin),‖R−‖C(Γin)}. | (41) |
Analogously, the estimate of
Next, we give the estimates of
It follows from Lemma 3.4 that
|∂+R+(x,y)|≤‖∂+W+‖C(AiCi−1),|∂−R−(x,y)|≤‖∂−W−‖C(BiCi−1). | (42) |
As the process shown in the estimates of
v−uλ−λ+−λ−∂+R++uλ+−vλ+−λ−∂−R−=±2uf′′(x)1+(f′(x))2,ony=±f(x). | (43) |
By the process of the reflection shown in the estimates of
∂+W+(x0,y0)≤∂+R+(x1,y1)=(2uλ+−λ−v−uλ−f′′1+(f′)2−uλ+−vv−uλ−∂−R−)(x1,y1)≤(2uλ+−λ−v−uλ−f′′1+(f′)2)(x1,y1)−(uλ+−vv−uλ−)(x1,y1)×∂−R−(x2,y2)=(2uλ+−λ−v−uλ−f′′1+(f′)2)(x1,y1)+(uλ+−vv−uλ−)(x1,y1)×(2uλ+−λ−uλ+−vf′′1+(f′)2)(x2,y2)+(uλ+−vv−uλ−)(x1,y1)×(v−uλ−uλ+−v)(x2,y2)×∂+R+(x2,y2) | (44) |
Since
2uf′′1+(f′)2,uλ+−vv−uλ−,λ+−λ−v−uλ−,λ+−λ−uλ+−v |
are uniformly bounded. Denote the maximum one of these bounds as the constant G, thus (44) can be expressed as
∂+W+(x0,y0)≤∂+R+(x1,y1)≤G2−G∂−R−(x1,y1)≤G2−G∂−R−(x2,y2)≤G2+G3+G2∂+R+(x2,y2)≤N∑i=1Gi+1+GN∂+R+(xN,yN). | (45) |
Similarly, if
∂+W+(x0,y0)≤N∑i=1Gi+1−GN∂−R−(xN,yN). | (46) |
Combining (45), (46) and (31) yields that
‖∂+W+‖C(AiCi−1)≤N∑i=1Gi+1+GNmax{‖∂+R+‖C(Γin),‖∂−R−‖C(Γin)}. | (47) |
Analogously, the same estimate of
Thus, we complete the proof of Lemma 3.5.
Corollary 3. Based on Lemma 3.5, we have that
Proof. Substituting (31) into (47) gives that
|∇R±|=√λ2∓+1λ+−λ−|∂±R±|≲1c(N∑i=1Gi+1+GNmax{‖R+‖C1(Γin),‖R−‖C1(Γin)}). | (48) |
Combining the local existence and the Corollary 3 induces the global
We now turn to consider the border region
{(∂x+λ−∂y)R+=0,(∂x+λ+∂y)R−=0,inD2iR−(x,y)=Z−(x,y)onAiCi,R++R−=2arctanf′(x)onAiAi+1, | (49) |
where
Lemma 3.6. Suppose that
Proof. For any point
|R−(x,y)|≤‖Z−‖C(AiCi),|R+(x,y)|≤2arctanf′∞+‖Z−‖C(AiCi). | (50) |
in the region
|∂−R−(x,y)|≤‖∂−Z−‖C(AiCi),|∂+R+(x,y)|≤‖∂+R+‖C(AiAi+1). | (51) |
Moreover, by (43), we have that
‖∂+R+‖C(AiAi+1)≤G2+G‖∂−R−‖C(AiAi+1)≤G2+G‖∂−Z−‖C(AiCi). | (52) |
By the same process shown in Lemma 3.5, we know that
Corollary 4. Based on Lemma 3.6, we have that
Combining the local existence and the Corollary 4 induces the global
Theorem 3.7. For case I, there exists a global
∂+R+≥0,∂−R−≤0in(¯Ωnon∖{M,N}). | (53) |
Moreover, there exists a constant
|∂+R+|+|∂−R−|≤Cin(¯Ωnon∖{M,N}). | (54) |
So far we have established the global
Now we begin to consider the problem in case Ⅱ. We will prove Theorem 1.2.
Proof of Theorem 1.2. Firstly, we consider the special case
Suppose that
Let
{dy(0)−(x)dx=λ−,y(0)−(x0)=φ−1(x0). | (55) |
Then
∂∂x(∂R+∂y)+λ−∂∂y(∂R+∂y)+∂λ−∂R−∂R−∂y∂R+∂y=−∂λ−∂R+(∂R+∂y)2. |
Use the second equation in (20), we can get
∂R−∂y=∂xR−+λ−∂yR−λ−−λ+. |
Substitute it into the above identity to obtain that
∂∂x(∂R+∂y)+λ−∂∂y(∂R+∂y)+∂λ−∂R−∂xR−+λ−∂yR−λ−−λ+∂R+∂y=−∂λ−∂R+(∂R+∂y)2. | (56) |
Define
ddx(H0(x)∂yR+)−1=∂λ−∂R+H−10, | (57) |
where
If
∂R+∂y(x1,y1)=∂yR+(x0,y0)H0(x)(1+∂yR+(x0,y0)∫x1x0∂λ−∂R+H−10(s,y(0)−(s))ds). | (58) |
Since
H0(x)=exp{h0(R−(x,y(0)−(x))−h0(R−(x0,y0)}, | (59) |
and
Note that
∂R+∂y(x1,y1)≤∂yR+(x0,y0)H0(x)(1+∂yR+(x0,y0)K0(x1−x0)). | (60) |
By the boundary condition (43), one has that
∂yR−(x1,y1)=(v−uλ−uλ+−v∂yR+)(x1,y1)=(cos(θ+A)cos(θ−A)∂yR+)(x1,y1). | (61) |
The coefficient
∂yR−(x1,y1)<∂yR+(x1,y1). | (62) |
Analogously, let
{dy(1)+(x)dx=λ+,y(1)+(x1)=y1. | (63) |
one can obtain the similar equation of
ddx(H1(x)∂yR+)−1=∂λ+∂R−H−11, | (64) |
where
∂R−∂y(x2,y2)=∂yR−(x1,y1)H1(x)(1+∂yR−(x1,y1)∫x2x1∂λ+∂R−H−11(s,y(1)+(s))ds). | (65) |
Note that
∂R−∂y(x2,y2)≤∂yR−(x1,y1)H1(x)(1+∂yR−(x1,y1)K∗1(x2−x1)). | (66) |
By the same process and after series of reflection if necessary, the denominator of (66) will tend to
Next, we consider the case
The proof is very similar to the case
((v−uλ−)∂yR+)(x1,y1)−((uλ+−v)∂yR−)(x1,y1)=−2u(x1,y1)f′′(x1)1+(f′(x1))2. | (67) |
Thus we have
∂yR−(x1,y1)=(v−uλ−uλ+−v∂yR+)(x1,y1)+(2uuλ+−v)(x1,y1)×f″(x1)1+(f′(x1))2<∂yR+(x1,y1). | (68) |
This is same as (62). So we omit the details of proof for the case
Therefore, we complete the proof of Theorem 1.2.
Due to the conservation of mass, vacuum can not appear in the inner regions of the duct. So if vacuum exists, it must locate on the two walls. On the other hand, since the two walls are the streamlines, Corollary 2 shows that the sound speed is monotonically decreasing along the two walls. When the sound speed decreases to zero, it corresponds to the formation of vacuum.
First, we list some properties of the vacuum boundary, one can also see [4] and [25] for more details.
Lemma 4.1. When the vacuum area actually appears, the first vacuum point M and N must form at
Based on Theorem 3.7, we have obtained the
˜Rn±(y)=˜R±(y),y∈[−yn,yn],f(xM)−yn=1n,n=1,2,...|q′(y)|≤−ψ′′(y)1+(ψ′(y))2qc√q2−c2,y∈[−f(xM),−yn]∪[yn,f(xM)]. | (69) |
Consider the problem in the region
Let
R±(x,y)=limn→∞Rn±(x,y),(x,y)∈Ωvac. | (70) |
To sum up, we get that
Theorem 4.2. Assume that
∂+R+≥0,∂−R−≤0in(¯Ωvac∖{lup∪llow}). | (71) |
Next, we consider the regularity near vacuum boundary. In fact, there are many results about physical vacuum singularity. A vacuum boundary (or singularity) is called physical if
0<|∂c2∂→n|<+∞ | (72) |
in a small neighborhood of the boundary, where
Theorem 4.3. Let
Proof. For convenience, we only consider the case near the vacuum line
∂→nc2=−(γ−1)q∂→nq. | (73) |
By the relation
∂xq=qc2√q2−c2(∂xR+−∂xR−),∂yq=qc2√q2−c2(∂yR+−∂yR−). | (74) |
Substituting it into (73) gives that
∂→nc2=−(γ−1)q2c2√q2−c2[(n2−n1λ−)∂yR++(n1λ+−n2)∂yR−]. | (75) |
Combining this with (30) yields that
∂→nc2=−(γ−1)q2(u2−c2)4(q2−c2)[(n2−n1λ−)∂+R+−(n1λ+−n2)∂−R−]. | (76) |
Construct a line
∂−∂+R+=∂−λ+−∂+λ−λ+−λ−∂+R+. | (77) |
Integrating (77) from
∂+R+(Sh)=∂+R+(Ph)exp{∫ShPh∂−λ+−∂+λ−λ+−λ−ds}. | (78) |
This together with (26) yields that
∂+R+(Sh)=∂+R+(Ph)exp{∫ShPhγ+18c(−T1∂+R++T2∂−R−)ds}, | (79) |
where
T1=q2(u2−c2)sec2(θ−A)(q2−c2)32,T2=q2(u2−c2)sec2(θ+A)(q2−c2)32. |
Since
We claim that
We will use the proof by contradiction to obtain above assertions. Assume that
∂+R+(x,y)≥∂+R+(Sˉgn)≥ˉg2on(x,y)∈PˉgnSˉgn. | (80) |
By the fact that
∫SˉgnPˉgnγ+18c(−T1∂+R++T2∂−R−)ds≤∫SˉgnNˉgn−γ+18cT1∂+R+ds≤−γ+116c(Nˉgn)s0g0ˉg, | (81) |
where
∂+R+(Sˉgn)≤∂+R+(Pˉgn)exp{−γ+116c(Nˉgn)s0g0ˉg}. | (82) |
Note that
¯limn→∞∂+R+(Sˉgn)⩽ | (83) |
which is a contradiction. Thus, it follows
(84) |
This together with (71) gives that
(85) |
Combining it with (76) yields that
(86) |
Note that
(87) |
Similarly, one has that
(88) |
Thus, we prove the Theorem 4.2.
So far we have established the solution to the problem when vacuum actually appears. When it comes to the case that there is no vacuum in the duct, by the same prior estimates in Lemma 3.5 and Lemma 3.6, one only needs to check whether the duct can be covered by characteristics in finite steps.
Theorem 4.4. Assume that
Proof. It remains to check whether the duct can be covered by characteristics in finite steps. Since the duct is convex, then there exists
(89) |
Similarly, we can get that
(90) |
By the induction method, we have that
(91) |
Suppose that the image of the
(92) |
where
(93) |
where
Proof of Theorem 1.1. Under the assumptions of Theorem 1.1, Theorem 1.1 comes from Theorem 3.1 and Theorem 4.1-4.3.
Finally, we give a sufficient condition to ensure that vacuum must form at finite location in the duct.
Proposition 1. If
Proof. We will use the proof by contradiction. Assume that there is no vacuum on
(94) |
This together with (29) and (43) yields that
(95) |
By (95) and the Bernoulli law (7), we get that
(96) |
Since
then
(97) |
which is a contradiction. Thus, we complete the proof.
The authors would like to thank the referee for his (or her) very helpful suggestions and comments that lead to an improvement of the presentation.
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